Chapter 2 Chemical formulas and composition stoichiometry

Chapter 2 Chemical formulas and

composition stoichiometryAP Chemistry

Milam

2-1 chemical formulas

• Allotropes are elements with two different forms like graphite and diamond are both carbon

• Organic compounds all contain carbon and usually contain hydrogen, oxygen, sulfur and nitrogen

2-2 ions and ionic compounds

• Cations are ions with a positive charge• Anions are ions with a negative charge• Cations and ions form ionic compounds,

they form crystalline structures and the ratio of cation to anion is how the chemical formula is written

• The formula can be deduced using the charges of the ions

2-2

• The smallest possible whole number ratio is called the formula unit

• Ex. NaCl, CaBr2, Al2O3

• Polyatomic ions are groups of atoms bonded together that have a net charge yet are stable enough to act as a single unit

• It is to your advantage to memorize the common polyatomic ions within a week

2-2

• Polyatomic ions in compounds have parentheses when more than one is necessary to balance charge

• Na2CO3, (NH4)2CO3, Al2(SO4)3

2-3 Names and formulas of some ionic compounds

• You need to be able to write names and formulas for ionic compounds without thinking!

• The name of a compound always gives enough information to figure out the formula, and vice versa

• To write the formula, balance the + and - charges

2-3• If you have Na and Cl, Na has a +1 charge

and Cl has a –1 charge, therefore you need equal numbers of Na and Cl – formula unit NaCl

• If you have Ca and Cl, Ca has a +2 charge so you need double the Cl ions – formula unit CaCl2

• To name, you just give the names of the ions, cation first and change the ending to ide

2-3

• Polyatomic ions do not change their ending to ide

• Some elements have multiple charges, mostly transition and inner transition metals, we indicate the charge in the name with Roman numerals

• There is also a classic naming system that uses an ic/ous ending to indicate the larger charge

2-3

• Ex. Fe+2 and Fe+3

• Ferrous and Ferric ions – classic naming• Iron (II) and Iron (III) – stock naming

system• FeCl2 – because there are two chlorides

with a –1 charge, we can infer that iron has a +2 charge, therefore this is Iron (II) chloride or Ferrous chloride

2-4 Atomic Weights

• Atomic mass units are approximately the mass of 1 proton or neutron

• The definition is 1/12th the mass of a C-12 atom

• The atomic weights are listed on the periodic table

2-5 The Mole!

• We use moles all of the time in chemistry, because we use them in complicated situations, they seem complicated. Always remember that moles are just like dozens! They are a number, 6.02 x 1023 of anything is a mol!

• A mol is a huge number, 602,000,000,000,000,000,000,000 or 602 sectillion

2-5• Like amu’s, the mole is defined by the # of

particles in 12 grams of C-12, so the actual number of 6.0221415 x 1023 but we will stick with 6.02 x 1023 in here.

• You must be able to convert between grams, moles and # of particles and later we will use moles in concentration, gas laws, heat transfer, stoichiometry, electrochemistry, titrations and other subjects

2-6 Formula weights, molecular weights and moles

• Formula weight, molecular weight, atomic weight are all terms that specify that you are talking about molar mass and what type of substance you have (ionic, molecular or elemental)

• In here I will always say molar mass and that’s what you will see for the most part everywhere else as well

2-6• Molar mass is the mass of one mole of a

substance• Molar masses are obtained from the

periodic table and are equivalent numerically to the amu’s of a molecule or formula unit or atom

• To find the molar mass of something, take the atomic mass of each element and multiply by the subscripts and then add them all

2-6 Molar mass of H2SO4

• Molar mass = 2*1.0 grams/mol + 1*32.1 g/mol + 4*16.0 g/mol = 98.1 g/mol

• This means that if I had 1 mol of sulfuric acid I would have 98.1 grams of it as well

• Follow up, find the molar mass of Ca(OH)2.• Ans. 74.1 g/mol• In this class we will always round atomic

masses to the tenths place before calculating

2-7 Percent composition and formulas of compounds

• To find the % composition, you take the formula, find the molar mass, and the masses of the individual elements taking into account the subscripts then divide elemental mass by total and convert to a percentage.

• Ex. H2SO4 has a molar mass of 98.1 g/mol

2-7

• There are 2 hydrogens, so 2 g/mol divided by 98.1 g/mol gives 0.0203 or 2.03%

• 1 Sulfur is 32.1 g/mol divided by 98.1 g/mol gives .3272 or 32.72 %

• 4 oxygens are 64 g/mol divided by 98.1 g/mol gives .6524 or 65.24%

• The total is 2.03 + 32.72 + 65.24 = 99.99 which is close enough to 100%

2-8 Derivation of formulas from elemental composition

• Empirical formulae are formulae with the subscript ratios reduced to the lowest possible whole number ratios, hydrogen peroxide has a formula H2O2, and its empirical formula is HO

• If you know the % composition of a substance, you can calculate the empirical formula and this can be slightly difficult

2-8

• Determine the empirical formula for a compound with a % composition of 87.5 % nitrogen and 12.5% hydrogen

• Step 1 – assume you have 100 grams of total stuff, so now you’ll have 87.5 grams of N and 12.5 g of H

• Step 2 – convert each into moles• 87.5 grams N = 6.25 moles N• 12.5 g H = 12.5 mol H

2-8• Step 3 – take the smallest # of moles and

divide all of the moles by that number• 6.25 mol N/ 6.25 mol = 1 mol N• 12.5 mol H/6.25 mol = 2 mol H• These amounts become your subscripts• Ans. NH2

• Most times the final answers will be decimals such as 1.98 or 3.01, just round them

2-8• Also if you end up with numbers like 1 and

2.5 and 2, you need to double all of them to get 2, 5 and 4 so that you only have whole numbers

• Most people get stuck at step 3 and don’t know what to do when they get decimal results, remember that you are using an arbitrary starting point (100 g) and you need to adjust by dividing by the smallest number of moles to get whole numbers

2-9 Determination of molecular formulas

• Determining empirical formulas is difficult and this is an extra step added on, but empirical formulas are not what we always want. Most of the time we want more information so we want to find the molecular formula. For example, if you tell me something has an empirical formula of CH, I don’t know whether it is benzene (C6H6) or ethyne (C2H2) so we need more information

2-9 • Once you find an empirical formula, the

extra information to find the molecular formula is the molar mass

• For example, you determine CH to be the empirical formula, which has a molar mass of 13 g/mol

• If I tell you the molar mass if 78 g/mol, you know there must be 6 times the mass, so your molecular formula is C6H6

2-9

• Going from empirical formula to molecular is a simple step, but when added into the whole problem it makes steps 3 and 4 a bit more confusing

2-10 Some other interpretations of Chemical Formulas

• At this point there are now too many problems to remember all of the steps to follow for each type of problem, you need to be able to solve what the steps are by understanding the problem.

• In these situations it is useful to try and find the first and last steps so you know what you have and what you need to end up with

2-10

• Problem types include:– Finding the mass of an element with the mass

of the compound and vice-versa– Finding the formula of a hydrate*** this will be

done in lab and is likely to show up on an AP exam

2-11 Purity of samples

• Not likely to end up on an AP exam, if it does it should be solveable or qualitative in nature