MCAT Review

The MCAT-Review.org Project

Tricks and Shortcuts Use d = vt in acceleration problems. Where v is the average of the initial and final velocity. Works only for uniform acceleration (almost all acceleration problems in the MCAT is uniform). Q: a car uniformly accelerates from rest to 60 mph in 36 seconds (36/3600 hour). How far did the car travel? A: d = 30 x (36 / 3600) = 0.3 mile. (the average of 0 and 60 mph is 30 mph). Without this formula, youd have to first calculate the acceleration, then use another formula to calculate the distance. Q: a ball is thrown downward at 10 m/s, 2 seconds later, it is falling at 30 m/s. How far did the ball travel? A: d = 20 x 2 = 40 m (you get v = 20 by averaging 10 and 30). Look how much time this saves you compared to s = v0t + at2 = 10x2 + x10x22 = 20 + 20 = 40. For a question regarding the incline block with an object accelerating downward on it. I have had two practice questions ask for the acceleration... Without any calculations I eliminated two choices on one of the questions and three on the other question. The absolute minimum and maximum are 0 m/s^2 (perpendicular - no force acting in the parallel direction) and 10 m/s^2 (gravity approx. for MCAT). On this specific question two of the choices were above 10 m/s/s and one was zero. I knew the value to be somewhere in between and was able to answer and continue with no calculations. In any section of the MCAT (usually verbal) with a question that lists three statements and asks which ones are true: 1. Look at all of the answer choices first and see which one has the same Roman numeral in two of the answer choices. 2. Read the Roman numeral you have chosen and decide if it is true/false. 3. If true, cross out the two answer choices that don't contain that Roman numeral. If false cross out the two answer choices that do contain the roman numeral.

Copyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

4. Compare the two answer choices left and read the Roman numeral that is only in one answer choice. 5. If it's true, that's the answer. If false, it's the other answer. This trick is used to answer these types of questions faster by only having to decide if two Roman numerals are true instead of three. Example: A change in which of the following will affect the buoyant force experienced by an object that is totally submerged in a liquid? I. Density of the liquid II. Density of the object III. Depth of the object A. I only B. II only C. I and III only D. I, II, and III only Since ( II ) is in two of the answer choices, we look at that one first. Since Roman numeral II is false, we can cross out B and D. Next, we look at A and C. Since III is only in C, we see if that one is true. It's false, so we can cross out C and choose A as the correct answer without ever looking at Roman numeral ( I ) Its safe to use symmetry in kinematics problems. You can use symmetry to break down complex problems into simple parts. Q: you throw a ball up at 10 m/s, whats its speed when it falls back into your hands? A: 10 m/s downward. Q: if the entire trip of the ball takes 2 seconds, at what time did the ball reach the maximum height? A: 1 second. For questions with a very long stem, read the last sentence of the question first! Sometimes a question is just worded really long and confusing and then after you spend much time reading it, you realize its asking for something real simple. So, to summarize: if the question is a few lines long, read the last sentence first to get what its asking (you can even glance at the answer choices at this point).

Balancing electrochemical equations: dont balance from scratch. Look at the answer choices, and choose the one that has balanced charge, atoms, and shows the correct reaction.

Q: What is the correctly balanced equation for the following reaction? K2Cr2O7 (aq) + HCl (aq) KCl (aq) + CrCl3 (aq) + H2O (l) + Cl2 (g)


a. Cr2O72- (aq) + 14 HCl (aq) 2CrCl3 (aq)+ 7H2O (l) + 3Cl2 (g) + 2Cl (aq) b. 2K2Cr2O7 (aq) + 14 HCl (aq) 2CrCl3 (aq)+ 7H2O (l) + 3Cl2 (g) + 2KCl (aq) c. Na2Cr2O7 (aq) + 14 HCl (aq) 2CrCl3 (aq)+ 7H2O (l) + 3Cl2 (g) + 2NaCl (aq) d. K2Cr2O7 (aq) + 14 HCl (aq) 2CrCl3 (aq)+ 7H2O (l) + 3Cl2 (g) + 2KCl (aq) A: Choice a doesnt have charge balanced. Choice b doesnt have atoms balanced. Choice c isnt the right reaction. Choice d has everything balanced and is the right reaction. Look at any figures and tables before reading the passage. This is a simple rule, but very effective. Youll understand better and be able to read faster this way. Axial/equatorial without drawing out the chair form: When you are learning to draw out the chair form in organic chemistry, some of you may have noticed a pattern. Axials point up and down alternatingly. Similarly, equitorials point up and down in an alternating fashion. Thus if two substituents are 1 or 3 positions away on the cyclohexane, then their up and down orientation doesnt correspond to their axial/equatorial orientation (odd positions dont correspond). For example, if theyre both up, then one of them is axial, the other is equatorial. However, if two substituents are 2 positions apart on the cyclohexane, then their up and down orientation corresponds to their axial and equatorial (even positions correspond). For example, if both point up, then both will be axial, or both will be equatorial. This is better shown by example, and if you ever encounter this question type, it will save you time drawing out the chair. Q: what is the orientation of the substituents below?

A: t-Bu is equatorial bcz its large. Cl is 1 position away (doesnt correspond), so pointing up just like t-Bu means its the opposite orientation: axial.

A: t-Bu is equatorial bcz its large. Cl is 2 positions away (correspond), so pointing the same direction as t-Bu means that its also the same equatorial orientation.


A: t-Bu is equatorial bcz its large. Cl is 3 positions away (doesnt correspond), so pointing up like t-Bu means its the opposite orientation: axial.

A: t-Bu is equatorial. Cl is 2 positions away (correspond). So by pointing in the opposite direction, Cl is the opposite orientation- axial. For mirrors and lens questions, its faster if you just memorize the following instead of using the 1/f = 1/p + 1/q equation every time: All real images are inverted. All virtual images are erect. For diverging mirrors and lenses, the image is always smaller than the object and virtual. For mirrors, real images are in front of the mirror (light reflect off mirror and converge). For lenses, real images are behind the lens (light shines through the lens and converge). Virtual images are on the opposite side of the mirror/lens compared to the real images. Now that the easy stuff is out of the way, here comes the hard stuff - for converging mirrors and lenses: Object far away (p > R), image is smaller than object and real. Think of focusing the suns light with a magnifying glass. The image is small and real - real enough to burn ants. Object at R (p = R), image is same size as object, and real. R stands for real. Object between R and f (R > p > f), image is larger than object, and real. Just continuing the trend: smaller far away, same size at R, larger past R. Object at f (p = f), image is undefined / at infinity. f stands for undefined or infinity. Object closer than f (p < f), image larger than object, and virtual. When the real crosses the undefined at f, it becomes virtual. To help you memorize all this, refer to the figures in the light and geometrical optics notes at mcat-review.org. Q: what kind of the image is formed for an object 5 cm in front of a convex mirror? (the radius of curvature for the mirror is 10 cm) A: virtual, erect, smaller. You should arrive at this answer in an instant. You dont even need to look at the radius of curvature. All you need to know is that convex mirror is diverging, and all diverging mirrors/lenses make virtual images that are smaller than the object. Q: what is the height of the image formed by a 2 cm tall object standing 20 cm in front of a convex lens with radius of curvature of 4 cm. a. 0.2 cm, erect. b. 4 cm, erect.


c. 0.2 cm, inverted. d. 4 cm, inverted. It should come to you right away that this is a converging lens, and the object is very far (p>R). Image should be smaller than object and real. All real images are inverted. So the choice is c. For ray tracing problems, use the above method to first determine what the image should look like. If you really mastered the above techniques, you should be able to tell what the image should look like in an instant. This should eliminate half the answer choices, which has the image totally wrong. When youre left with just two choices, eye ball the diagrams and you should spot the rightly drawn ray diagram. For a really complex pulleys problem, there is a way to arrive at the answer without analyzing the pulley if the question gives you the distances that you pulled and the distance that the pulley weight moved. For example, if the question said you pulled on the pulley for 20 cm, and the 50 N weight moved 10 cm (you had to pull 2x the distance), then the force you exerted must be the weight of the pulley. In this case, without analyzing the pulley system, you know that a force of 25 N is needed to move the 50N weight on this pulley. Not sure if you can estimate this problem by turning that 5.5 into a 6, or that 343 into a 300? Glance at the answer choices. If they are an order of magnitude apart, then you can go wild with your estimations. If they are really close (they rarely are), then estimate conservatively. If you are stuck on a physics question: If the question stem only gives you 1-3 variables, chances are they are all important. Write out all these variables, then try to remember an equation that utilizes these variables. If you still cant figure it out, then use the units of the given variables and do dimensional analysis to arrive at an answer with the correct unit. If the question gives you many variables like 5 or more, dont panic, get confused, then read and re-read the question to try and understand it. Jump to what the question is asking for (you can do this by looking at the answer choices given). For example, if the question asks you to solve for density, then you know you only need mass and volume. Once you know that, you can focus on the variables that actually let you arrive at mass and volume. If you are stuck because you cant think of any formula that can solve the question, and you know youve memorized the most common one, then relax, because the formula is most probably given to you in the passage. The MCAT dont test you on memorizing abstract formula, which means that it will give you any abstract formulas you need to solve the problem. So, if you feel like you dont know a formula to solve something, chances are, the formula is in the passage. Memorize this formula instead of deriving it on test day: Fraction immersed in water = density of object/density of water This equation takes ages to derive, but if you just memorize the above, youll easily be able to solve those difficult density/buoyancy problems.


From the above equation: Fraction immersed = Volume in water / Total volume of floating object = object/ water Fraction above water = Volume outside water / Total volume of floating object = 1 Fraction immersed. For verbal, this little technique will help you read: nod when you agree with something the author says, frown if you dont like something the author says, and make whatever subtle gestures that come natural to you as you are absorbing what the author is saying. This trick will prevent you from zoning out because you are actively participating in your reading. Also, this trick engages you, and you retain the material much better. Where pertinent, form a mental image when youre reading. Be it action, a picture, or a timeline. Again, doing this actively engages you in reading, and you retain stuff better. Dont read from sentence to sentence. Read from idea to idea. Sometimes an idea is expressed in one sentence; sometimes it takes a few sentences to express an idea. If you follow the passage from one idea to the next, youll read faster and comprehend more. At the end of every passage, in small letters, is the title, year, and author. Look at the title before you read the passage. It doesnt take but a second, and it can give you the gist of the passage and guide your reading. Titles for sciences passages are especially insightful because many scientists will report their conclusion in their title. Even for abstract passages, their title can still offer valuable insights. For example, if a title says, On Thought and Memories, youll know that the incoming passage is going to be an abstract philosophical dissertation, so youll be mentally prepared for it. If you are stuck between two answer choices for an MCAT question, pick the choice that you feel most directly answers the question asked. In fact, the right answer choice should always answer the question. A correct statement that is not related to the question at hand is wrong. On a complex physics question, using the answer choices and work backwards may be easier and faster than solving the question the regular way. I found that if I first write my synthesis (last paragraph) first, the antithesis and intro paragraphs are a lot easier to come up with afterward. In addition, this helps in choosing examples and constructing statements that flow into the synthesis. Even if I wrote a quick outline before I started writing, I sometimes found that by the time I got to the synthesis, I would change my mind about what I wanted to say, and then there wasn't enough time to change everything else so that is was cohesive. By writing the synthesis first, I made sure this didn't happen. I also stuck to the three paragraph format of intro, antithesis and synthesis, and allotted 8-9 minutes for each, so it was easier to pace myself.


The rule of 9's: When asked to predict the relative percentage of protonated or deprotonated for weak acids and weak bases, normally you'd have to use the Henderson Hasselbach Eqn. of pH= pKa + Log ([WB]/[WA]) but this is a lot easier and faster. So if the solution pH is 4.5 and the pKa of the compound is 4.5 then there is a 1:1 ration of (WB/WA), that would make it 50 % WA/protonated, 50% WB/deprotonated. Now for the trick, for every pH unit difference in the solution you add one nine in the percentage. For Example: 1. pH of solution is 3.5, where the pKa of the compound is 4.5, that's -1 difference, so you add one nine to the percentage, so it would be 90% WA/protonated, and 10% WB/deprotonated. 2. pH of solution is 2.5, pKa of compound is 4.5, that's a -2 difference, so you add TWO nine's. So it would be 99.0% WA/protonated, and 1% WB/deprotonated. 3. pH of solution is 7.5, pKa of compound is 4.5, that's a difference of +3, so you add THREE nine's. So it would be a 99.9% WB/deprotonated, and 0.1% WA/protonated. So the general rule: For every one pH units, you add one nine in the percentage. A fast and easy way to calculate pH when the [H+] isnt a simple number: Instead of messing with logs and trying to calculate on test day, try this: H+= 3x10^-4, instead of trying to figure out -log(3x10^-4), simply subtract 1 from the exponent (4-1=3) this will give you 3.000 Then subtract the base number from 9 (9-3=6), put this after the decimal. gives you a pH of 3.6. not 100 % accurate, but will get you damn close and save time. actual answer is 3.522 If given the pH and asked to find the H+ concentration, then do it backwards. Reagents in OChem: What if you encounter a reagent youve never seen before? Remember these generalizations: anything with a lot of oxygens will oxidize stuff (eg. KMnO4, CrO3). Anything with a lot of hydrogens will reduce stuff (eg. LiAlH4, NaBH4). Also, try to guess based on similarities to what you already know. For example: you know that PCC (pyridine chlorochromate) will oxidize alcohols to aldehydes/ketones. So, if the question gives you pyridine dichromate, then its safe to assume it the same as PCC! When asked what type of solvent you would use with a reaction, and you dont know what solvent is used with a particular reaction type, then simply look for the chemical that will not react. Simple concept but often over looked by many people and often brings


you to the right answer. For example, a Grignard reagent reacts with alcohols and carbonyl compounds, so a solvent is something that will NOT react with these.

Careless Mistakes Axial / equatorial of two or more substituents: two substituents both pointing up does not mean that they are either both axial or both equatorial. When in doubt, always draw out the chair diagram. What is the orientation of the substituents in this molecule?

Ans: t-Bu is equatorial, Cl is axial. A careless mistake is to think oh, both are pointing up, so both must be equatorial. The Multiple Trap: Answer choices that are exact multiples of one another means it is a trap for carelessness, especially when a lot of math is involved. Watch out for your units. Make sure you either convert everything to SI, or the non-SI end up cancelling out. Be especially wary of questions that combine chemistry and physics, because molecular weight in chemistry is in g/mol, but the SI unit for mass in physics is kg. Q: how much would 22.4 L of oxygen weigh at standard temperature and pressure? a. 320 N b. 0.32 N c. 160 N d. 0.16 N A: therere two traps here. First, molecular oxygen is a diatomic, so its 32 g/mol, not 16 g/mol. Second, the m in F = ma is in kg, but youre using 32 g/mol. Only if youre careful, youll end up with correct answer choice b (22.4 L / 22.4 L/mol = 1 mol of oxygen. 1 mol x 32 g/mol = 32 g. F = ma = 0.032 kg x 10 = 0.32 N). Questions like this combines chemistry with physics, so watch out. Strong acid = high Ka = low pKa = high [H+] = low pH. Strong base = high Kb = low pKb = high [OH-] = low pOH. Q: What happens if I increase the Ka? Acidity increases. Q: What if I increase the proton concentration? Acidity increases. Q: What if I increase the pH? Acidity decreases. Dont make a careless mistake here. Same goes for pKa: increasing pKa decreases acidity. Stoichiometry: be especially careful when working this type of problem.


Pay attention to subscripts for questions like: how many moles of potassium ions are present in 15 mL of 2 M K2SO4? Ans 0.015 L x 2 M x 2 = 0.06 mols. Subscripts are easy to ignore if youre not careful. The molecular weight of oxygen in the air is 16x2=32 because oxygen exists as a diatomic. Same thing goes for other diatomics. DNA vs RNA: DNA uses dATGC, RNA uses AUGC. This subtle difference presents a dangerous opportunity for careless errors. For example, if the right answer choice is AUGC for an RNA, then one of the wrong answer choices might be ATGC, just to trick you. Dont fall for it! Be aware that on the real mcat you may not have the scroll feature on your mouse. Therefore, when practicing do not use the scroll button on your mouse but rather, manually slide the page down with the pointer arrow. This is not only important in that it allows you to simulate the real MCAT. It's also important because if you forget that you do not have the scroll feature on the real mcat, you may try to scroll and when the screen doesn't move you may mistakenly think that you are on the last discrete of that page for example and actually skip questions.(i know someone who this happened to) Confusing Content and Misconceptions Inclined planes: Theres an easy way to remember which component is sin and which is cos. sin is zero when the angle is zero, so sin is the component parallel to the inclined plane (when the angle is zero, the parallel force is zero, the block wont slide). Or, if you are not a math person, just remember, sin stands for slide; its the component that causes the block to slide. Thermodynamics, deltaG, Equilibrium constant, tell you what direction a reaction will occur. Kinetics, activation energy, rate constant, tell you how fast a reaction will occur. Q: Reaction a has Keq of 1x103, reaction b has Keq of 2x103, which reaction occurs faster? A: Keq doesnt say anything about rate. Q: Reaction a has a rate constant of 1x103 at sea level, but the exact same reaction has a rate constant of 2x103 at 1000 ft above sea level. Which reaction occurs faster? A: the one with the higher rate constant. 1/f = 1/p + 1/q is not the only equation you need to memorize for mirrors and lenses. Dont forget that height of the object/image directly correlates with the distance of object/image: h / h = q / p = magnification. Where h is height of image, h is height of object, q is distance of image, p is distance of object. When calculating cell potential in electrochemistry, use the values in the table of reduction potentials that is given to you. Unlike stoichiometry and enthalpy calculations, you should NOT multiply them by the coefficients of the balanced equation.


People are used to associate osmolarity with salt content. However, anything dissolved contribute to osmolarity. For example, albumin in blood contributes to its osmolarity. Impurities dont just broaden the melting range of a compound, but also decreases it. This is because impurities interfere with intermolecular attractions between the molecules of your compound. Apart from water absorption, a less well-known but tested concept is that the large intestine also makes vitamin K and B12 (bacteria in large intestine makes them). Amoeboid movement is not rocket propulsion: cells that move by amoeboid mechanism shoots out a filament. The cell will move in the direction of the filament growth (+ end). Do not make the mistake of thinking about Newtons third law and guess that the cell will move in the opposite direction of filament growth. Arrow is filament. Filament is growing to the left (+ end). This cell will move to the left. Acids are soluble in bases, bases are soluble in acids. This goes against like dissolves like, so some people get it wrong. Light passing through a different media changes in velocity and wavelength but the frequency stays the same. The reason we observe same color is because we observe frequency, not wavelength. The following is a trick question that can pop up on the MCAT: Which object has the highest inertia mass? A) a 12kg object moving at a speed of 12m/sec B) a 4kg object moving at a speed of 17m/sec C) a 15kg object moving at 3 m/sec D) a 2kg object moving at 20m/sec Answer: C is correct because it has the highest mass of all the answer choices. Inertia is defined as an objects resistance to change in motion. But practically speaking, its the same as mass. Dont be fooled by quantities like momentum or speed. For questions asking asking about inertia: simply replace the word inertia in the question with the word mass. This is 100% fool proof! You will never misinterpret inertia problems again.

Submit Yours Good luck on the MCAT. Dont forget to submit your ideas, tricks, shortcuts to [email protected] If we choose to incorporate anything you submit into our project,


we will give back to you the full amount of your original donation (minus any paypal transaction fees, if any). Submit up to 3 of your best stuff. It must be the kind of stuff that make people go wow, this is neat, why havent I thought of this before? Dont rush your submission. Submit only after you have taken the real MCAT. This way, youll be able to offer better ideas because youll have more time to prep, and a real MCAT experience under your belt. We pick the submission with the higher quality for identical ideas that were submitted at similar times.


MCAT Review

Note: Content online at http://mcat-review.org is actively maintained and updated. When you have internet access, use the online version for the most up to date content.

Physical Sciences SectionGeneral Chemistry Reviewq

Electronic Structure and Periodic Table Bonding Phases and Phase Equilibria Stoichiometry Thermodynamics and Thermochemistry

q

q

q

q


q

Rate Processes in Chemical Reactions - Kinetics and Equlibrium Solution Chemistry Acids/Bases Electrochemistry

q

q

q

Physics Reviewq

Translational Motion Force and Motion, Gravitation Equilibrium and Momentum Work and Energy Waves and Periodic Motion Sound Fluids and Solids Electrostatics and Electromagnetism Electronic Circuit Elements Light and Geometrical Optics Atomic and Nuclear Structure

q

q

q

q

q

q

q

q

q

q

Biological Sciences SectionBiology ReviewCopyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

q

Molecular Biology: Enzymes and Metabolism Molecular Biology: DNA and Protein Synthesis r DNA r Protein Synthesis Molecular Biology: Eukaryotes Genetics Microbiology Generalized Eukaryotic Cell Specialized Eukaryotic Cells and Tissues Nervous and Endocrine Systems Circulatory, Lymphatic, and Immune Systems Respiratory System Skin System Digestive and Excretory Systems r Digestive System r Excretory System Muscle and Skeletal Systems Reproductive System and Development Evolution

q

q

q

q

q

q

q

q

q

q

q

q

q

q


Organic Chemistry Reviewq

The Covalent Bond Molecular Strucure and Spectra Separations and Purifications Hydrocarbons Oxygen Containing Compounds r Alcohols r Aldehydes and Ketones r Carboxylic Acids r Acid Derivatives r Keto Acids and Esters Amines Biological Molecules r Carbohydrate r Amino Acids and Proteins r Lipids r Phosphorus Compounds

q

q

q

q

q

q

Verbal Sectionq

Comprehension Evaluation Application Incorporation of new information

q

q

q


Writing Sectionq

Developing a central idea Synthesizing concepts and ideas Presenting ideas cohesively and logically Writing clearly, following accepted practices of grammar, syntax, and punctuation, consistent with timed, first-draft composition

q

q

q

Source: Official testing topics were obtained directly from the AAMC.

How to Review for the MCATYou want to review for the MCAT in the most efficient way. First, review only the topics that will be on the exam. Second, review these exam topics with the test in mind. What this means is that you must do content review in conjunction with working real exam questions. If you just do content review alone, then you'll end up forgetting most of what you went over. However, if you work MCAT problems while doing content review, you'll retain most of what you learned. Incidentally, the stuff that you retain will be the stuff that will be tested on the MCAT. There's only so many concepts an introductory course can cover, and there's only so many ways one can test these concepts. So, if you know the aamc topics, and you do enough past MCAT questions, then you'll figure things out. What this means is that a question may at first appear to be very difficult because of the way it's worded, or because it is given in the context of a strange experiment that you've never heard before. But then you'll realize that the MCAT doesn't test you on strange experiments that are not covered in a basic introductory course. After realizing this, you'll be able to see right through the strange experiment and realize that the question is simply testing you on some very basic concept, like Le Chatelier's principle, just disguised in the context of some unfamiliar situation. To sum things up, review for the MCAT effectly by going over the official test topicsCopyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

published by the AAMC. Use the online notes here to guide your studies. And remember, do not wait until the last minute to take past MCAT exams. Start doing MCAT questions as early as possible, in conjunction with content review. Have fun reviewing for the MCAT!


MCAT General Chemistry Review

Electronic Structure and Periodic TableElectronic structure Orbital structure of hydrogen atom, principal quantum number n, number of electrons per orbital Ground state, excited states Absorption and emission spectra Quantum numbers l, m, s, and number of quantum states (electrons) per orbital Common names and geometric shapes for orbitals s, p, d Conventional notation for electronic structure Bohr atom Effective nuclear charge The periodic table: classification of elements into groups by electronic structure; physical and chemical properties of elements Alkali metals Alkaline earth metals Halogens Noble gases Transition metals Representative elements Metals and nonmetals Oxygen group The periodic table: variations of chemical properties with group and row Electronic structure the representative elements the noble gases transition metals


Valence electrons First and second ionization energies definition prediction from electronic structure for elements in different groups or rows Electron affinity definition variation with group and row Electronegativity definition comparative values for some representative elements and important groups Electron shells and the sizes of atoms

BondingThe ionic bond (electrostatic forces between ions) Electrostatic energy q1 q2 /r Electrostatic energy lattice energy Electrostatic force q1 q2 /r2 The covalent bond sigma and pi bonds hybrid orbitals (sp3 , sp 2 , sp and respective geometries) valence shell electron pair repulsion (VSEPR) theory, predictions of shapes of molecules (e.g., NH 3 , H2 O, CO 2 ) Lewis electron dot formulas resonance structures formal charge Lewis acids and bases Partial ionic character role of electronegativity in determining charge distribution dipole moment


Phases and Phase EquilibriaGas phase Absolute temperature, K scale Pressure, simple mercury barometer Molar volume at 0 degrees Celcius and 1 atm = 22.4 L/mol Ideal gas definition ideal gas law PV=nRT Boyle's law Charles' law Avogadro's law Kinetic molecular theory of gases Deviation of real-gas behavior from ideal gas law qualitative quantitative (Van der Waals' equation) Partial pressure, mole fraction Dalton's law relating partial pressure to composition Liquid phase: intermolecular forces Hydrogen bonding Dipole interactions Van der Waals' forces (London dispersion forces) Phase equilibria Phase changes and phase diagrams Freezing point, melting point, boiling point, condensation point Molality Colligative properties vapor pressure lowering (Raoult's law) boiling point elevation (T b = Kb *m)


freezing point depression (T f = -Kf *m) osmotic pressure Colloids Henry's Law

StoichiometryMolecular weight Empirical formula versus molecular formula Metric units commonly used in the context of chemistry Description of composition by % mass Mole concept; Avagadro's number Definition of density Oxidation number common oxidizing and reducing agents disproportionation reactions redox titration Description of reactions by chemical equations conventions for writing chemical equations balancing equations, including oxidation-reduction equations limiting reactants theoretical yields

Thermodynamics and ThermochemistryEnergy changes in chemical reactions- thermochemistry Thermodynamic system, state function Conservation of energy Endothermic/exothermic reactions enthalpy H and standard heats of reaction and formation Hess' law of heat summation Bond dissociation energy as related to heats of formation


Measurement of heat changes (calorimetry), heat capacity, specific heat capacity (specific heat of water = 4.184 J/gk) Entropy as a measure of "disorder"; relative entropy for gas, liquid, and crystal states Free energy G Spontaneous reactions and G Thermodynamics Zeroth law (concept of temperature) First law (E = q + w, conservation of energy) Equivalence of mechanical, chemical, electrical and thermal energy units Second law (concept of entropy) Temperature scales, conversion Heat transfer (conduction, convection, radiation) Heat of fusion, heat of vaporization PV diagram (work done = area under or enclosed by curve) Calorimetry

Rate Processes in Chemical Reactions - Kinetics and EqulibriumReaction rates Dependence of reaction rate upon concentration of reactants; rate law rate constant reaction order Rate determining step Dependence of reaction rate on temperature activation energy activated complex or transition state interpretation of energy profiles showing energies of reactants and products, activation energy, H for the reaction Arrhenius equation Kinetic control versus thermodynamic control of a reaction Catalysts; the special case of enzyme catalysis


Equilibrium in reversible chemical reactions Law of Mass Action the equilibrium constant application of LeChatelier's principle Relationship of the equilibrium constant and standard free energy change

Solution ChemistryIons in solution Anion, cation (common names, formulas and charges for familiar ions; e.g., NH 4 + , ammonium; PO 4 3- , phosphate; SO 4 2- , sulfate) Hydration, the hydronium ion Solubility Units of concentration (e.g., molarity) Solubility product constant, the equilibrium expression Common-ion effect, its use in laboratory separations Complex ion formation Complex ions and solubility Solubility and pH

Acids/BasesAcid / base equilibria Bronsted definition of acid, base Ionization of water Kw, its approximate value (Kw = [H+][OH-] = 1*10^-14 at 25C) definition of pH; pH of pure water Conjugate acids and bases (e.g., amino acids) Strong acids and bases (common examples, e.g., nitric, sulfuric) Weak acids and bases (common examples, e.g. acetic, benzoic) dissociation of weak acids and bases with or without added salt hydrolysis of salts of weak acids or bases


calculation of pH of solutions of salts of weak acids or bases Equilibrium constants Ka and Kb: pKa, pKb Buffers definition and concepts (common buffer systems) influence on titration curves Titration Indicators Neutralization Interpretation of titration curves Redox titration

ElectrochemistryElectrolytic cell electrolysis anode, cathode electrolyte Faraday's law relating amount of elements deposited (or gas liberated) at an electrode to current electron flow; oxidation, and reduction at the electrodes Galvanic or voltaic cell half reactions reduction potentials; cell potential direction of electron flow

Old topicsThe topics below are outdated. They have been either modified or replaced by the most recent aamc publication. E = kQ1Q2/d E = lattice energy Force attraction = R(n+e)(n-e)/d^2Copyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

Measurement of heat changes (calorimetry); heat capacity; specific heat (specific heat of water = 1 cal per degrees Celcius) First law: E = Q - W (conservation of energy)

Source: Official MCAT general chemistry topics were obtained directly from the AAMC.


Electronic Structure and Periodic TableElectronic structureq

Orbital structure of hydrogen atom, principal quantum number n, number of electrons per orbital r In the Bohr model, the hydrogen electron orbits the nucleus. r In quantum mechanics, hydrogen electron exists in a spherical probability cloud around the nucleus. r The principle quantum number, n, defines what shell the electron is in. r n values start from one: 1,2,3 ...etc. r Higher n shells are higher in energy (if subshells are the same). r There are n squared orbitals per shell. r There are 2 electrons per orbital. r Thus, there are 2n^2 electrons per shell. Ground state, excited states r Electrons are normally in their ground state. r When they absorb energy, they get promoted to excited states. r Excited states are higher in energy than ground states. r Excited states come back down to the ground state via release of energy. Absorption and emission spectra r The absorption spectrum shows what wavelengths of light are absorbed. r The absorption spectrum looks like black lines on a rainbow background. r The emission spectrum shows what wavelengths of light are emitted. r The emission spectrum looks like colored lines on a black background. r The absorption spectrum corresponds to the emission spectrum in pattern. r The emission spectrum shifts to a slightly longer wavelength. Quantum numbers l, m, s, and number of quantum states (electrons) per orbital r l is the angular momentum quantum number: l are integers that range from 0 to n1. s spdf: l=0,1,2,3 for s,p,d,f respectively. s spdf designates subshells. s s subshells hold 1 orbital. p holds 3, d holds 5, f holds 7. s each orbital holds a maximum of 2 electrons. s s subshells hold a maximum of 1x2=2 electrons, p: 3x2=6,Copyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

q

q

q

r

r

d: 5x2=10, f: 7x2=14. s A generalized formula for the above pattern: for any subshell, 4l+2 electrons can be held. s for a given shell, higher subshells have higher energy. s a low shell with a high subshell may be higher in energy than a higher shell with a low subshell. m is the magnetic quantum number: m are integers that range from -l to l, including zero. s is the spin quantum number: s is either +1/2 or -1/2.

q

Common names and geometric shapes for orbitals s, p, d

r r r

Electrons are filled by occupying the lowest energy subshells first. Subshell arranged in increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d s The best way to memorize the above is by interpreting the periodic table:


s s

s

s

Starting from the first row, going across, both hydrogen and helium is 1s. Next row: 2s then 2p. Third row: 3s then 3p. Fourth row: 4s, 3d, then 4p. Fifth row: 5s, 4d, then 5p. Sixth row: 6s, 4f, 5d, then 6p. Last row: 7s, 5f, then 6d. The pattern we get from looking at the periodic table is exactly in the order of increasing energy. For a given subshell, the columns represent how many electrons are in that subshell. For example, the fifth column of the d subshells contain elements that have 5 electrons in that subshell. The number of columns for each subshell indicate the maximum number of electrons that subshell can hold. For example, the d subshells have 10 columns showing that d orbitals can hold 10 electrons total.

q

Conventional notation for electronic structure

r

Conventional notation:


r r

r

r

r

Orbital diagrams: Aufbau principle: shells / subshells of lower energy gets filled first (This is the most obvious rule. For example, 1s fills first, then 2s, then 2p ...etc. Review the exact order of energies because later on, the d subshells get filled after the s. Hund's rule: when you fill a subshell with more than 1 orbital (p, d, f), you first fill each orbital with a single electron and with the same spin (check out electrons 5, 6, and 7 in the orbital diagram, which fills according to Hund's rule). The reason for Hund's rule is that electron-electron repulsion in doubly occupied orbitals make them higher in energy than singly occupied orbitals. Pauli exclusion principle: 2 electrons in the same orbital must be of different spins (for example, check out electrons 5 and 8 in the orbital diagram). Watch out for d4 and d9 elements. Instead of s2d4, it's s1d5 and s1d10 because they want to achieve a half-full or full d subshell.

q

Bohr atom r Electron orbiting the nucleus in a circular orbit. r Larger n values have larger orbiting radii. r ...more on Bohr in chemistry Effective nuclear charge r Effective nuclear charge = nuclear charge - shielding electrons. r Shielding electrons are those that stand between the nucleus and the electron weCopyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

q

r

r

r

r

are interested in. Shielding electrons are those that are in subshells closer to the nucleus (lower in energy) than the electron we are interested in. MCAT questions usually give you a diagram of the Bohr model, in which case, shielding electrons are those that orbits at a smaller radius. The higher the effective nuclear charge for an electron, the more stable it is (higher ionization energy, not easily knocked off). Effective nuclear charge increases for outer electrons as you go across (left to right) the periodic table.

The periodic table: classification of elements into groups by electronic structure; physical and chemical properties of elements

q

q

Alkali metals r Single valence electron - low ionization energy, very reactive. r Wants to lose that electron to achieve empty valence shell. r More reactive as you go down because of increasing radii. r Reacts with oxygen to form oxides. r Reacts with water to form hydroxides and releases hydrogen. r Reacts with acids to form salts and releases hydrogen. r Most commonly found in the +1 oxidation state. Alkaline earth metals r 2 valence electrons - relatively low in ionization energy, quite reactive. r Wants to lose both electrons to achieve empty valence shell. r More reactive as you go down because of increasing radii. r Reacts with oxygen to form oxides. r Reacts with water to form hydroxides and releases hydrogen.Copyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

q

r r

Reacts with acids to form salts and releases hydrogen. Most commonly found in the +2 oxidation state.

q

Halogens r 7 valence electrons (2 from s subshell and 5 from p subshell) - high electron affinity, very reactive. r Wants to gain one electron to achieve full valence shell. r More reactive as you go up because of decreasing radii. r Reacts with alkali metals and alkaline earth metals to form salts. r Most commonly found in the -1 oxidation state. Noble gases r Full valence shell of 8 - high ionization energy couple with low electron affinity. r Don't react. r Found in the oxidation state of 0. Transition metals r High conductivity due to free flowing (loosely bound) outer d electrons. r In the presence of ligands (when in a chemical complex), the d orbitals become nondegenerate (different in energy). r Electron transitions between nondegenerate d orbitals gives transition metal complexes vivid colors. r Varied oxidation states - but always +. Representative elements r Representative elements include the s block and the p block of the periodic table. r No free flowing (loosely bound) outer d electrons. r Valence shell fills from left (1 electron) to right (8 electrons). r Standard nomenclature from left to right: I A, II A, III A, IV A, V A, VI A, VII A, VIII A. Metals and non-metals r Metals are to the left of metalloids. r Non-metals are to the right of metalloids. r Metalloids: diagonal line from Boron to Polonium: B, Si, As, Te, Ge, Sb, (Po).r

q

q

q

q

Chemical properties Metals

Non-metals


Likes to lose electrons to gain a + oxidation state (good reducing agent). Lower electronegativity - partially positive in a covalent bond with nonmetal. Forms basic oxides. Physical properties Good conductor of heat and electricity Malleable, ductile, luster, solid at room temp(except Hg)q

Likes to gain electrons to form a oxidation state (good oxidizing agent). Higher electronegativity - partially negative in a covalent bond with metal. Forms acidic oxides. Poor conductor of heat and electricity Solid, liquid, or gas at room temp. Brittle if solid and without luster.

Oxygen group r The group (column) that contains oxygen. r Oxygen and sulfur chemically similar (if a question asks you what element you can substitute for oxygen and still keep the same chemical reactivity, then choose sulfur). r Se - Te - Po = non-metal - metalloid - metal (or metalloid).

The periodic table: variations of chemical properties with group and rowq

Electronic structure ...a repeat of electronic structure section above r representative elements s Representative elements include the s block and the p block of the periodic table. s No free flowing (loosely bound) outer d electrons. s Valence shell fills from left (1 electron) to right (8 electrons). s Standard nomenclature from left to right: I A, II A, III A, IV A, V A, VI A, VII A, VIII A. r noble gases s Full valence shell of 8 - high ionization energy couple with low electron affinity. s Don't react. s Found in the oxidation state of 0. r transition metals s High conductivity due to free flowing (loosely bound) outer d electrons. s In the presence of ligands (when in a chemical complex), the d orbitals become nondegenerate (different in energy).Copyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

s

s

Electron transitions between nondegenerate d orbitals gives transition metal complexes vivid colors. Varied oxidation states - but always +.

q

Valence electrons r Electrons in the outer shell. r Ranges from 1 to 8 from left to right of the representative elements. r The valence electron rule does not apply to transition metals. First and second ionization energies r definition of first ionization energy: The energy needed to knock off the first valence electron. r definition of second ionization energy: The energy needed to knock off the second valence electron. r prediction from electronic structure for elements in different groups or rows

q

s s s s s s s

s

s

Ionization energy decreases as you go down because of increasing radii. Ionization energy increases as you go right because of decreasing radii. Highest peaks are noble gases. Lowest troughs are alkali metals. Local maxima occurs for filled subshells and half-filled p subshells. Second ionization energy is always higher than the first ionization energy (usually a lot higher). Alkali metals and hydrogen: first ionization energy very low. Second ionization much higher. Alkaline earth metals: first ionization energy low. Second ionization energy also low.


q

Electron affinity r definition - electron affinity is the amount of energy released when something gains an electron (how easily it can gain an electron). r variation with group and row

s s s s s s

As you go down a group, electron affinity decreases because of larger radii. As you go across (left to right) a row, electron affinity increases. Highest peaks are for the halogens. Lowest for noble gases. Local minima occurs for filled subshells and half-filled p subshells.

q

Electronegativity r definition - electronegativity is how much something hordes electrons in a covalent bond. r comparative values for some representative elements and important groups s Electronegativity increases toward the top right. s Fluorine is the most electronegative element. s Things around fluorine are highly electronegative: N, O, F, Cl, Br. s Halogens are electronegative, especially toward the top of the group. s Noble gases can be very electronegative if they participate in bond formation (Kr and Xe). s Non-metals are more electronegative than metals. s Covalent bond is a sharing of electrons between elements. s The more electronegative element in a covalent bond gets a larger share of the electrons and has a partial negative charge s The less electronegative (more electropositive) element in a covalent bondCopyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

s

s

gets a smaller share of the electrons and has a partial positive charge. If the electronegativity difference is too great, an ionic bond occurs instead of a covalent one. Ionic bonds result from a complete transfer of electrons from the electropositive element to the electronegative element.

q

Electron shells and the sizes of atoms r Electron shells s Electron shells are defined by the principle quantum number - the n value. s Going down the periodic table means jumping to the next shell. s As you fill to the next shell (Ne to Na), the effective nuclear charge decreases because the old shell stands in between the nucleus and the new shell. s Filling to the next shell causes a jump in atom size because of decreased effective nuclear charge. s As you go down a group (Na to K), the atomic size increases even though the effective nuclear charge stays the same, because higher shells have a larger radius than lower shells. s Going across the periodic table means filling up the same shell (by going through subshells). s As you fill up a shell, the effective nuclear charge increases because the atomic number (protons) is increasing while the same-shell electrons you add do not shield one another. s With increasing effective nuclear charge, the electrostatic attraction (F=kQq/ r^2) between the nucleus and the electrons increases, so the atom becomes more compact. s The increasing effective nuclear charge and electrostatic attraction is why going across a periodic table means decreasing atomic size. r Sizes of atoms


s s s s

Size increases as you go down a column. Size decreases as you go across (to the right of) a row. Atomic sizes may overlap if you zigzag on the periodic table.


Bonding

The ionic bond (electrostatic forces between ions)The ionic bond forms when electrons transfer completely from one atom to another, resulting in oppositely charge species that attract each other via electrostatic interaction. Electrostatic energy q1 q2 /r Electrostatic Energy = Electrostatic potential x charge = kq1 /r x q2 = kq1 q2 /r Electrostatic energy is negative because q1 and q2 are opposite in charge (If q1 and q2 are not opposite in charge, then they would repel each other, and no ionic bond would form). Frequently, the negative sign is dropped and only the magnitude of the electrostatic energy is used. The greater the magnitude of electrostatic potential, the stronger the ionic bond. Strong ionic bonds are promoted by high charge magnitudes (q values) that are close together (small r value). Ions that form strong ionic bonds have high charge density, that is, the charge to size ratio is high.

Electrostatic energy lattice energy Lattice energy measures the ionic bond strength. Lattice energy is the energy required to break the ionic bond. The larger magnitude of the lattice energy, the stronger the ionic bond and the harder it is to break. The lattice energy is proportional to the electrostatic attraction between the ions.

Electrostatic force q1 q2 /r2 Coulomb's law: F = kq1 q2 /r2 Larger charge magnitudes + charges being closer together greater electrostatic force. The Coulomb's constant, k, is 9E9. Opposite charges attract (negative F), same charges repel (positive F). If q1 doubles, the electrostatic force doubles.Copyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

If r halves, the electrostatic force increase by a factor of 22 = 4. Coulomb's law is analogous to the universal law of gravitation: F = Gm1 m 2 /r2 G is analogous to k and m is analogous to q. The big difference is that G is tiny compared to k, because gravitational force is weaker compared to the much stronger electrostatic force.

The covalent bondThe covalent bond results when there is a sharing of electrons between two atoms, resulting in the overlap of their electron orbitals.

sigma and pi bonds bonds are single bonds. They also make up the first bond of double and triple bonds. bonds are double and triple bonds. They make up the second bond in a double bond, and both the second and the third bond in a triple bond.

hybrid orbitals: sp3, sp2, sp and respective geometries

Hybrid orbitals are produced by hybridizing (mixing) electron orbitals to produce geometries that facilitate bonding. Sp3: a hybrid between one s with 3 p orbitals. Tetrahedral in geometry. Contains single bonds only. Sp2: a hybrid between one s with 2 p orbitals. Trigonal planar in geometry. Contains a double bond. Sp: a hybrid between one s with one p orbital. Linear in geometry. Contains a triple bond.


Hybrid orbitals are most commonly used with carbon as the center atom.

valence shell electron pair repulsion and the prediction of shapes of molecules (e.g., NH3, H2O, CO2)

In short, it is the VSEPR theory. The VSEPR theory is used to predict the geometry of molecules. The shapes of molecules are determined by the molecular geometry. Radicals also count as an electron pair.Copyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

The VSEPR number is the total number of bonds + unbonded electron pairs. When calculating the VSEPR number, always use the electron/bond configuration about the central atom. NH 3 has a vsepr number of 4 (3 bonds to H and 1 unbonded pair). If you look up the table for VSEPR # = 4 and # unbonded electron pairs = 1, then you'll find that NH 3 is trigonal pyramidal. H2 O has 2 bonds, 2 unbonded electron pairs - it is bent. CO 2 has 2 double bonds and 0 unbonded electron pairs - it is linear.

Lewis electron dot formulasEvery dot represents 1 electron. Every line represents 1 bond (2 electrons). A "lone pair" is represented by two dots. Formulas are drawn in such a way that an octet is achieved on each atom. Exceptions include the boron column (they form 3 bonds and have a six-tet), large elements (3rd row and below such as the 10-tet P in PO 4 3- and the

12-tet S in SO 4 2- ), and radicals (compounds with an odd # total electrons that result in a single, unpaired electron).

All electrons in a bond are shared and can be used to satisfy the octet for both atoms on either side of the bond. Rule of thumbs for Lewis structures Carbon: 4 bonds total (meaning 4 total bonds. It can either be 4 single bonds or two double bonds ...etc) and no lone pairs. eg. CH 4 , CO 2 Oxygen: O can be O: 2 bonds total, 2 lone pairs. eg. H2 O, O2 O1- : 1 bond, 3 lone pairs, formal charge of -1. O1+: 3 bonds, 1 lone pair, formal charge of +1. Nitrogen: N can be N: 3 bonds total, 1 lone pair. eg. Amine or ammonia NH 3 N+ : 4 bonds, 0 lone pair, formal charge of +1. eg. Ammonium NH 4 + Halogens: 1 bond, 3 lone pairs. eg. CCl4 Hydrogen: 1 bond, 0 lone pair (exception to octet rule). Carbocation: C+ has 3 bonds, no lone pairs, formal charge +1. Carbanion: C- has 3 bonds, 1 lone pair, formal charge -1.


Boron: 3 bonds, 0 lone pairs (exception to the octet rule). eg. BH 3 Common Lewis structures Hydrogen Lewis structures Hydrogen Proton: Hydride ion: Boron Lewis structures

Borane:

Borohydride ion: Carbon Lewis structures

Methane:

Carbocation:

Carbanion: Nitrogen Lewis structures


Amine / Ammonia:

Ammonium:

Imine: Oxygen Lewis structures Molecular oxygen:

Water, alcohol, and ethers:

Ozone: Halogen Lewis structures Hydrogen fluoride: Chloromethane: Bromide ion: R in the figures are either carbon or hydrogen. Lewis structures for elements in the same column (group) of the periodic table are similar to one another. For example, sulfur can be substituted for oxygen in lewis structures of oxygen.

resonance structures


When there are more than 1 satisfactory Lewis structures for a molecule, they are called resonance structures. You can visualize the molecule "shifts" between each of its resonance structure really fast, spending more time in the more stable resonance structures. Or more accurately, the structure of the molecule is a "combination" of its resonance structures, taking on more character from the most stable resonance structures. Eg. The bond length of a molecule that has both a single and a double bond resonance structure is intermediate between a single bond and a double bond. The molecule spends most of its time in the most stable resonance structure. Stable properties: Octet rule is satisfied in every atom (except for boron group and hydrogen). No formal charges. If there must be formal charges, like charges are apart and unlike charges are close together.

formal chargeFormal charge = valence electron # in the unbonded atom - electron # in the bonded atom. Electron # in the bonded atom = dots around the atom + lines connected to the atom. The dots around the atom represent electrons that are held entirely by the atom. The lines connected to the atom represent bonding electron pairs, in which the atom only gets one of the two electrons. Formal charges (other than 0) must be labeled next to the atom with the formal charge. Common formal charges: Oxygen with only a single bond: -1. Oxygen with no bond but have an octet: -2. (Oxygen usually exists as the diatomic O 2 and have a double bond to themselves) Carbon with only 3 bonds: either +1 if carbocation or -1 if carbanion. Nitrogen with 4 bonds: +1.Copyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

Halogen with no bonds, but have an octet: -1. (Halogens usually exist as a diatomic and have a single bond to themselves such as Cl 2 ) Boron with 4 bonds: -1. eg. BH 4 -

Lewis acids and basesLewis acid accept electron pairs. They don't have lone pairs on the central atom. eg. BF3 Lewis bases donate electron pairs. They have lone pairs on their central atom. eg. NH 3

Partial ionic characterCovalent bonds between atoms with dissimilar electronegativities have a partial ionic character. role of electronegativity in determining charge distribution The more electronegative atom receives a partial negative charge. The less electronegative atom receives a partial positive charge.

dipole moment Molecules with asymmetrical partial charge distribution have a dipole moment. eg. H 2 O has a dipole moment because the molecule is bent and the oxygen-side of the molecule is partially negative. Dipole moment depends on charge and distance. The greater electronegativity difference, the greater the charge and hence the dipole moment. The greater the distance separating the charges, the greater the dipole moment. Molecules with symmetrical partial charge distribution do not have dipole moments. eg. CCl4 do not have a dipole moment because the partially negative chlorine atoms are arranged symmetrically in a tetrahedron. The symmetry cancels out their individual dipole moments. Things with a dipole moment are said to be polar. Are the individual bonds in CCl4 polar? Ans: yes. Is the entire molecule CCl4 polar? Ans: no.


Old topicsThe topics below are outdated. They have been either modified or replaced by the most recent aamc publication.

E = kQ1Q2/d Energy = Electrostatic potential x charge = kQ1 /d x Q2 = kQ1 Q2 /d E is negative because Q1 and Q2 are opposite in charge. The more negative E is, the stronger the ionic bond. Strong ionic bonds are promoted by high charge magnitudes (Q values) that are close together (small d value).

E = lattice energy The name used for E is the lattice energy, and it measures the ionic bond strength. Lattice energy is the energy required to break the ionic bond. The larger magnitude of the lattice energy, the stronger the ionic bond and the harder it is to break.

Force attraction = R(n+e)(n-e)/d^2 The above equation describes the force of attraction between the cation n+ and the anion n- at a distance d apart. R is Coulomb's constant (usually written as k). n+e = charge of cation in coulombs = positive charge (n+) times coulombs per electron (e). n-e = charge of anion in coulombs = negative charge (n-) times coulombs per electron (e). The elementary charge or coulombs per electron (e) is 1.6E-19, but you don't have to memorize it. The MCAT will give it to you. The Coulomb's constant is 9E9. The official Coulomb's law states: F = kQ1 Q2 /r2


Phases and Phase Equilibria

Gas phase

Absolute temperature, K scaleK Absolute zero Freezing point of water / melting point of ice Room temperature Body temperature 0 C F 32 77 99

-273 -460

273 0 298 25 310 37

Boiling point of water / condensation of steam 373 100 212 K = C + 273 F = C x 1.8 + 32

Pressure, simple mercury barometerPressure is the force exerted over an area: P = F/A Due to gravity, the atmosphere exerts a pressure of 101 kPa at sea level. For convenience, 101 kPa = 1 atm. Pressure decreases at higher elevations. The mercury barometer measures atmospheric pressure by allowing the atmospheric pressure to "push" on a column of mercury. The barometer is open at one end and closed off (vacuum) at the other. The atmosphere "pushes" at the open end, which results in the mercury rising up in the closed end. The measured atmospheric pressure P = F/A. F is the weight of the mercury that got pushed up and A is the cross-section area of the column that the mercury got pushed through. Standard mercury barometers are calibrated such that 1 atm of pressure will push theCopyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

mercury up by 760 mm. For convenience, mm Hg is also called the Torr. So, you don't have to do the P=F/A calculation to find out the pressure reading from a barometer. Just know that 1 atm = 760 mm Hg = 760 torr. 1 atm = 101 kPa = 101,000 Pa = 760 mm Hg = 760 Torr. When performing P = F/A calculations, make sure that F is in Newtons, A is in meter squared and the resulting P will be in Pascals. You can then convert the Pascals to whatever units the answer choices are in.

Molar volume at 0 degrees Celsius and 1 atm = 22.4 L/molYou must memorize this: ideal gases occupy 22.4 L per mol of molecules. Do not get this mixed up - it is 22.4 liters per mole, not the other way around. The way to remember this is that the mol is a huge number - 6.02E23 molecules. These gazillions of molecules occupy a lot of space - 22.4 L to be exact. Another way you can remember this is to look at the periodic table: Air is made up mostly of nitrogen, which has an atomic mass of 14. In the diatomic form, N2 weighs 14x2 = 28 grams per mol. Now, air is really light. In order for you to grab 28 grams of air, you need more than just a bottle of air, you need a huge tank totaling 22.4 L.

Ideal gas

definition An ideal gas consists of pointy dots moving about randomly and colliding with one another and with the container wall. The ideal gas obeys the kinetic molecular theory of gases and has the following properties. Random molecular motion. No intermolecular forces. No (negligible) molecular volume. Perfectly elastic collisions (conservation of total kinetic energy). You can treat gases as ideal gases at: Low pressures High temperatures Deviation from the ideal occurs at high pressure and low temperature. At theseCopyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

conditions, the gas molecules are "squished" together. When the gas molecules are so close together, they experience intermolecular interactions. Also, the molecular volume becomes significant when the total volume is squished down so much. The intermolecular attractions will cause collisions to be sticky and inelastic. At the extremely high pressures and low temperatures, gases cease to be gases at all they condense into liquids. Ideal gases behave according to the ideal gas law.

ideal gas law PV=nRT, where P is pressure, V is volume, n is # mols of gas, R is the gas constant, and T is temperature.

Combined gas law: Because nR is constant (n is the # mols and R is the gas constant), PV/T must also be constant. Boyle's law and Charles' law can all be derived from the combined gas law.

Boyle's law: at constant temperature, P1 V1 = P2 V2


Charles' law: at constant pressure, Charle's law extrapolates to absolute zero, where volume also goes to zero (this is only an extrapolation). Avogadro's law: Equal volumes of two gases will also contain equal number of mols of each gas (given ideal conditions: ideal gas at STP). PV = nRT R is constant, and at STP, pressure and temperature is also constant. V/n = RT/P If you plug in STP values, you'll end up with V/n = 22.4 L/mol. All ideal gases at STP will occupy 22.4 L per mol of gas molecules.

Kinetic molecular theory of gasesThe ideal gas laws can be derived from the kinetic theory of gases. The kinetic theory holds the following assumptions Random molecular motion. No intermolecular forces. No (negligible) molecular volume. Perfectly elastic collisions (conservation of total kinetic energy). The kinetic theory holds the following concepts: Pressure of a gas is due to its molecules constantly colliding with the walls of its container. Pressure is equally distributed over the walls of the container because molecular motion is random. Temperature is a measure of the average kinetic energy of the gas molecules. Higher temperature means the molecules are traveling faster, lower temperatures means slower molecules. Diffusion and Effusion Diffusion: random molecular motion, causing a substance to move from an area of higher concentration to an area of lower concentration (diffusion down its concentration gradient).


Effusion: random molecular motion, causing a substance to escape a container through a very small openning. Graham's Law (applies both to diffusion and effusion for the purposes of the MCAT).Rate1/ Rate2

= M 2/ M 1

Rate = rate of diffusion or effusion. M = molecular weight of gas molecule. A possible question on the MCAT is two gasses diffuse down a tube from opposite ends. Where will the gases meet? The gist of this is that the lighter gas will travel faster, and the gases will meet at a point that is farther from the end of the lighter gas. Graham's Law is derived from the Kinetic theory Temperature = average kinetic energy At a given temperature all gases have the same kinetic energy. m1 v 1 2 = m2 v 2 2 m1 v 1 2 = m2 v 2 2v12/ v 1/ v22

= m 2/ m 1

v2

= m 2/ m 1

Deviation of real-gas behavior from ideal gas law

qualitative When molecules are far apart (under conditions of low P, high T), they are ideal. When molecules are brought close together (higher P, lower T), they experience intermolecular attraction. When molecules are brought so close together that they clash into one another, they experience steric repulsion.

quantitative (Van der Waals' equation)


b for bounce. The term with the constant b is the repulsion term. The greater b is, the more repulsion, which leads to greater pressure. a for attraction. The term with the constant a is the attraction term. The greater a is, the more attraction, which leads to less pressure.

Partial pressure, mole fractionPartial pressure = a component of the total pressure exerted by a species in a gas mixture. The total pressure of a mixture of gas = The sum of all the partial pressures. Mole fraction = a component (fraction) of the total # mols that belongs to a species in a gas mixture. Mole fraction for species A = # mols of A / # mols of the entire gas mixture. = # mols of A / # mols of A, B, C ... Dalton's law relates partial pressure to mole fraction.

Dalton's law relating partial pressure to compositionPi = iP total Ptotal = P i = iP total Ptotal is total pressure. Pi is partial pressure of species i. i is the mole fraction of species i.

Liquid phase: intermolecular forces

Hydrogen bondingCopyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

Hydrogen bonding is a weak interaction between a partially positive H and a partially negative atom. Technically, hydrogen bonds are a special type of dipole-dipole interaction. Hydrogen bonding increases the boiling point. Partially positive H are also called hydrogen bond donors. They are hydrogens that are bonded to either F, O, or N. Partially negative atoms are also called hydrogen bond acceptors. They are most commonly F, O, or N. Do ethers form hydrogen bonds with other ethers? Ans: no, because ethers do not have a partially positive H (donor). The more polar a bond is, the stronger the hydrogen bond. The H-F bond is the most


polar, followed by the H-O bond, and lastly the H-N bond.

Dipole interactions

All polar molecules exhibit dipole-dipole interactions. This is where the polar molecules align such that opposites attract. Dipole-dipole interactions increase the boiling point, though not as significantly as hydrogen bonding. Dipole interactions are stronger the more polar the molecule is. Ion-dipole interactions are similar to dipole-dipole interactions, but it's stronger because it is no longer an interaction involving just partial charges. Instead, it is an interaction between a full charge (ion) and a partial charge (dipole). Ion-dipole interactions get stronger when you have larger charge magnitude of the ion, and large polarity of the dipole molecule.

Van der Waals' forces (London dispersion forces)Also called dispersion forces. Dispersion forces exists for all molecules, but are only significant for non-polar molecules. For polar molecules, dipole forces are predominant. Dispersion forces result from induced and instantaneous dipoles. Induced dipoles: when a polar molecule interacts with a non-polar molecule, then polar molecule induces a dipole in the non-polar molecule. Instantaneous dipoles: Non-polar molecules have randomly fluctuating dipoles that tend to align with one another from one instant to the next. Dispersion forces get stronger for larger molecules. For example, decane (C 10H 22)has a stronger dispersion force than ethane (C 2 H 6 ).

Phase equilibriaCopyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

Phase changes and phase diagrams

Solid: atoms/molecules vibrate about a fixed position. Hard to compress. Does not flow to fill a container. Liquid: atoms/molecules move about, but are close together and bound by intermolecular forces. Hard to compress. Flows to fill a container. Gas: atoms/molecules fly about far apart from one another and do not experience intermolecular forces. Easy to compress. Flows to fill a container. Solid-liquid boundary: solid and liquid exist in equilibrium. Solid-gas boundary: solid and gas exist in equilibrium. Liquid-gas boundary: liquid and gas exist in equilibrium. Triple point: the temperature and pressure at which all three phases of matter coexist in an equilibrium. Critical point: the temperature and pressure at which liquids and gases become indistinguishable. Critical temperature: the temperature above which you can no longer get a liquid no matter how much pressure you press on it. Water phase diagram is different from others because the solid-liquid boundary is slanted to the left. This is because water (liquid) is more dense than ice (solid), and if you increase the pressure at a given temperature, then you turn ice into water. Mnemonic for remembering which section of the phase diagram is for gases: "gas comes out this way."Copyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

Freezing point, melting point, boiling point, condensation pointFreezing point: temperature (at a given pressure) that liquids begins to freeze into a solid. Melting point: temperature (at a given pressure) that a solid begins to melt into a liquid. Boiling point: temperature (at a given pressure) that a liquid begins to turn into a gas. Condensation point: temperature (at a given pressure) that a gas begins to condense into a liquid. Freezing point and melting point are the same, they can both be found along the solidliquid phase boundary. Boiling point and condensation point are the same, they can be found along the liquid-gas boundary. Sublimation: conversion of a solid directly into a gas. Conditions for sublimation can be found along the solid-gas boundary.

MolalityMolality is a measure of the concentration of solutes in a solution. Molality is given the symbol m (don't confuse the small case m with the large case M that is molarity) Molality = mols of solute / mass (in kg) of solvent. Compare molality (mol solute/kg solvent) to molarity (mol solute/L solution).

Colligative propertiesColligative properties = properties that depend on the # of solute particles, but not on the type. Solute particles in solution likes to keep the solution in liquid phase. This is why it makes it harder to boil (raises its boiling point) and also makes it harder to freeze (lowers the freezing point). Lowering the vapor pressure is just another fancy name for raising the boiling point. Van't Hoff Factor (i): all colligative properties take into consideration of the Van't Hoff factor. Basically, it means convert concentration to reflect the total number of particles in solution. For example, glucose has i of 1 because it doesn't break up in solution. NaCl has i of 2, because in solution, it breaks up into 2 particles Na+ and Cl-.Copyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

vapor pressure lowering (Raoult's law) P = solvent P solvent P = solute P solvent P is the vapor pressure. P is the decrease in vapor pressure. solute = mol fraction of the solute = # mols of solute / # total mols of both solute and solvent solvent = mol fraction of the solvent = # mols of solvent / # total mols of both solute and solvent P solvent is the vapor pressure of the pure solvent alone. When you are calculating solute , make sure you take into account of van't Hoff. ie. 1 mols of NaCl in solution is actually 2 mols of particles.

boiling point elevation (deltaTb = kb*m *i) T b = kb mi T b is the increase in boiling point. k b is the molal boiling point constant (like almost every other constants, the MCAT will give it to you). m is the molality (mol solute/kg solvent). i is van't Hoff factor.

freezing point depression (deltaTf = -kf*m *i) T f = -k fmi T f is the decrease in freezing point (the negative sign shows that the change is a decrease). k f is the molal freezing point constant. m is the molality (mol solute/kg solvent). i is van't Hoff factor.


osmotic pressure = MRT *i is the osmotic pressure. M is the molarity in mol/L. R is ideal gas constant. T is the temperature in K. Osmotic pressure determines whether and in what direction osmosis will occur. Osmosis is the movement of solvent across a semi-permeable membrane from an area of low solute concentration (high solvent concentration) to an area of high solute concentration (low solvent concentration). Solvent will move from an area with low value to an area with high value.

ColloidsSolution: things are mixed at the molecular level and will always stay mixed. When you use the term dissolve, you are making a solution. Colloids: things are mixed at a "semi-molecular level" with solute aggregates that are really really tiny. Colloids will stay mixed until you centrifuge it. Suspension: things are mixed at a particle level and will NOT stay mixed. The famous colloid example is milk. Also, when you shake water and oil vigorously, you can get an emulsion, which is a colloid.

Henry's LawPsolute = k [solute] Psolute is the partial pressure of the solute at the solution's surface. k is a constant. [solute] is the solute concentration in solution. The partial pressure of a solute just above the solution's surface is directly proportional to its concentration.


Stoichiometry

Molecular weightMolecular weight is numerically equal to molecular mass (amu) 1 amu = 1 g/mol12Carbon

has 12 amu and weighs 12 g/mol

Empirical formula versus molecular formulamolecular structure molecular formula empirical formula

C6 H12O6

CH 2 O

empirical formula is what you get after dividing everything in the molecular formula by the highest common factor.

Metric units commonly used in the context of chemistryMolarity = M = mol/L molality = m = mol/kg mass = kg. molar mass = g/mol.

Description of composition by % mass


%mass = mass of species of interest / total mass * 100

Mole concept; Avogadro's number1 mole = 1 mol = 1 Avogadro's number = 6.02E23 molecules

Definition of densitydensity = mass / volume = kg/m3 often in chemistry, specific gravity is used. specific gravity = number of times the density of water = density of substance / density of water density of water = 1 g/mL = 1 g/cm3 specific gravity of water = 1 g/cm3 / 1 g/cm3 = 1 density of lead = 11 g/cm3 specific gravity of lead = 11 g/cm3 / 1 g/cm3 = 11 specific gravity is unitless

Oxidation numbercommon oxidizing and reducing agents oxidizing agents Oxygen O2 , Ozone O3 , Permanganates MnO4 - ,

reducing agents

Hydrogen H2 , metals (such as K), Zn/HCl, Sn/HCl, LAH Chromates CrO4 2- , Dichromates Cr2 O7 2- , peroxides (Lithium Aluminium Hydride), NaBH4 (Sodium Borohydride), lewis bases, stuff with a lot of hydrogens H2 O2 , lewis acids, stuff with a lot of oxygens disproportionation reactions An element in a single oxidation state reacts to form 2 different oxidation states. Disproportionation can occur when a species undergo both oxidation and reduction. For example: 2Cu+ Cu + Cu2+ Here, the Cu+ acts as both oxidizing and reducing agent and simultaneously reduce and oxidize itself.+ 2+


The oxidized Cu becomes Cu The reduced Cu+ becomes Cu

redox titration Some terms and concepts A = analyte = stuff with the unknown concentration that you want to find out by titration. Aox = analyte that is an oxidizing agent = analyte in its oxidized state. Ared = analyte that is a reducing agent = analyte in its reduced state. T = titrant = stuff that you add drip by drip to determine how much of it is needed to complete the titration. Tox = titrant that is an oxidizing agent = titrant in its oxidized state. Tred = titrant that is a reducing agent = titrant in its reduced state. S = standard = something with an accurately known amount or concentration. You use it in a reaction that accurately (stoichiometrically) produces a known amount or concentration of I 2 . Sox = standard that is an oxidizing agent = standard in its oxidized state. Sred = standard that is a reducing agent = standard in its reduced state. X = reactions intermediate = a species that is not present in the net equation of the overall reaction. Xox = intermediate that is an oxidizing agent = intermediate in its oxidized state. Xred = intermediate that is a reducing agent = intermediate in its reduced state. Iodimetric titration: Ared + I2 Aox + 2I Iodometric titration: 1) Aox + 2I - Ared + I2 2) Tred + I2 Tox + 2I Using a standard Iodimetric titration with standard: 1) Sox + 2I - Sred + I2Copyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

2) Ared + I2 Aox + 2I notes: step 1 makes sure that the I 2 produced is of accurate amount/concentration by the use of the standard. Iodometric titration with standard: 1) Sox + Xred Sred + Xox 2) Xox + Ared(limiting reagent) Xred + Aox 3) Xox(left over) + 2I - Xred + I2 4) I 2 + Tred 2I - + Tox notes: step 1 makes an intermediate of accurately known amount. step 2: the analyte eating up an unknown, but calculatable, amount of the intermediate. step 3: the remaining intermediate going on to make I 2 step 4: Here, you will find out how much T is needed to eat up all the I 2 produced from step 3. From this, you'll know the amount of Xox(left over). You also can calculate the amount of Xox originally produced by the standard. Thus Xox - Xox(left over) = the amount of analyte. Important note: this is usually not a simple subtraction because you need to take stochiometric ratios into consideration. Iodine is used in redox titrations because in the presence of starch, I 2 is dark blue while I - is colorless. You can only accurately titrate something going from dark to colorless ( I 2 2I - ), but not the otherway round. A redox titration does not necessarily need the presence of Iodine. As long as some type of color change can be seen at the equivalence point of the redox reaction, then it will work. For example: 5 H2 O2 + 6 H+ + 2 MnO4 - 5 O2 + 2 Mn 2+ + 8 H2 O Goes from purple to colorless because of MnO4 - Mn 2+ transition. Redox titrations are similar to acid-base titrations, except instead of measuring pH, you look for a color change. Practice question: 1) Sox + 5Xred 3Sred + 3Xox 2) 3Xox + Ared(limiting reagent) 3Xred + Aox 3) Xox(left over) + 2I - 2Xred + I2 4) I 2 + 2Tred 2I - + Tox


after a long time doing drip by drip titration, you finally saw the dark color change to colorless. You noted down the initial and final volume reading of your pippette to be 300 mL and 200 mL, respectively. The concentration of the titrant you used was 10 M. You dissolved 1/2 mols of the standard to begin with. How much analyte was there? First, convert everything to mols (amount). n = MV. For the titrant (T red) it is 10 M x (0.3 L - 0.2 L) = 1 mol For the standard (S ox), it is already given to you in mols. However, if it's not, you have to convert it to mols. We know from the notes above that Xox - Xox(left over) = the amount of analyte, after taking into account of stochiometric ratios. Here are the stochiometric ratios: From step 4 I 2 : 2Tred From step 3 Xox(left over) : I 2 From step 2 3Xox : Ared(limiting reagent) From step 1 Sox : 3Xox Xox = 0.5 mol Sox * 3Xox / Sox = 1.5 mol Xox Xox(left over) = 1 mol Tred * I 2 / 2Tred * Xox(left over) / I 2 = 0.5 mol Xox(left over) For every Ared(limiting reagent), you eat up 3 Xox, thus: Xox - 3Ared(limiting reagent) = Xox(left over) 1.5 - 3 * Ared(limiting reagent) = 0.5 Ared(limiting reagent) = 1/3 mol This is why you always look at the stoichiometry of the reaction in calculations. It's almost never a simple addition or subtraction. The reaction in the question is actually a real redox titration taken from wikipedia.

Description of reactions by chemical equationsconventions for writing chemical equations


Phases (s) = solid (l) = liquid (g) = gas (aq) = aqueous (dissolved in water) Coefficient an equation with coefficients is a balanced equation. Direction A single head arrow denotes the reaction goes to completion in the direction of the arrow. A double-sided arrow denotes a reaction in equilibrium. A double-sided arrow with one side larger than the other denotes an equilibrium in favor of the side of the larger arrow. Charge Denotes charge and magnitude, for example +, -, 2+, 5- ...etc. Neutral charges are not denoted.

balancing equations, including oxidation-reduction equations balance the combustion of propanol: C3 H8 O + O2 CO 2 + H2 O pick out the atom (or group) that is the easiest to balance (usually represented in only 1 term on both side of the equation. In this case it is carbon. C3 H 8 O + O2 3CO2 + H2 O The next easiest to balance is hydrogen C3 H 8 O + O2 3CO2 + 4H 2 O Leave the hardest to last, oxygen. O is present in every term of the equation, so if we tried to balance O first, we'd be having a hard time. However, now that we balanced every other term, this leaves only one term left that contains O and that we haven't balanced yet. Do a quick count of oxygen atoms: there's 1 from C3 H8 O, 3x2 from 3CO 2 , and 4x1 from 4H2 O. Set up this equation: 1 + 2x = 3x2 + 4x1, where x would be the coefficient of our last term, O2 . Solve for x C3 H 8 O + 9 / 2 O 2 3CO2 + 4H 2 OCopyright (c) 2010-2012 - MCAT-Review.org - All Rights Reserved - Edition 4.7

Even though we balanced out every term, we're not done yet. We need to get rid of any fractions, so multiply every term by 2. 2C3 H 8 O + 9O 2 6CO2 + 8H 2 O Balancing oxidation-reduction (redox) equations 1. Separate into half reactions. There will be 2 half equations: one will be oxidation, the other reduction. Half equations contain only species of interest - those containing the atom that undergoes a change in oxidation state. Anything that is not covalently attached to the atom is not part of the species of interest. Anything that does not undergo a change in oxidation state is a spectator ion/species. 2. Balance each of the half reactions. Balance both charge and atoms. To balance one oxygen atom: Under acidic conditions: add H2 O to the side that needs the oxygen atom, then add H+ to the other side. Under basic conditions: add 2OH - to the side that needs the oxygen atom, then add H2 O to the other side. The Ion-Electron Method: you balance out the atoms first, then charge. The Oxidation-State Method: treat the species of interest as a single atom (those that undergo a change in oxidation number) and then balance it. 3. Recombine the half reactions. Multiply each half reaction by a factor, such that when you add them together, the electrons cancel out. It's like you're trying to solve a simultaneous equation and you want to eliminate the electron term. 4. Finishing touches Combine any idendical species on the same side of the equation. Cancel out any identical species on opposite sides of the equation. Add back in the spectator ions. For the oxidation-state method, now is also the time to balance out the oxygens and hydrogens. Check to make sure that both sides of the equation have equal number of atoms and neutral net charge.


Example using ion-electron method: K2 Cr2 O7 (aq) + HCl (aq) KCl (aq) + CrCl 3 (aq) + H2 O (l) + Cl 2 (g) 1. Separate into half reactions. Reduction: Cr2 O7 2- Cr3+ Oxidation: Cl - Cl 2 Species of interest for the oxidation reaction is Cl - , not HCl, because the H+ is not covalently attached to our atom of interest, and the hydrogen proton breaks off in aqueous solution. Similarly, we use Cr2 O7 2- and not K2 Cr2 O7 K+ is the spectator ion. 2. Balance each of the half reactions. The Ion-Electron Method: you balance out the atoms first, then charge. Balancing atoms for the reduction half reaction (Ion-electron method): 1. Cr2 O7 2- Cr3+ 2. Cr2 O7 2- 2Cr3+ 3. Cr2 O7 2- + 14 H+ 2Cr3+ +

MCAT Review

Documents

2cr6 6e

microtubule

recent aamc

protein kinase

6e

programmed

2ag 2e

van der waals