AP Chemistry A Review of Analytical Chemistry
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AP Chemistry
A Review of Analytical
Chemistry
AP Chemistry
Ch 1
(Prentice Hall)
What Temperature Do You Read? • A measurement always has
some amount of
uncertainty
• To indicate the uncertainty
of a single measurement
scientists use a system
called significant figures
• What temperature do you
read?
• Are you certain there
is a 30?
• Are you certain about
32?
• Are you UNcertain
about the decimal
point?
Uncertainty in Measurement
Consider:
• 32 degrees
• 32.5 degrees
• 32.3 degrees
• 32. 7 degrees
• It is important to know which digits of the reported number are uncertain
• The last number may be different from person to person and is a visual estimate
• Therefore, the third digit is an uncertain number
Reporting Measurements
• Unless stated otherwise, uncertainty is in the last
digit 1
• For example, 32.5 0.1, means the temperature
ranges from 32.4 to 32.6 degrees
• The numbers recorded in a measurement (all the
certain numbers plus the first uncertain number)
are called significant figures
• Uncertainty comes from limitations of the techniques used for comparison (the limitation is due to the instrument)
• Let’s weigh the same item using three different scales…
• Let’s write down these number & we sill use them for our practice calculations
Cu = Cu = Cu =
Rules For Counting Significant
Figures
1. Nonzero integers are always significant:
– For example:
1234 has four sig figs
2. Dealing With Zeros. There are three classes:
– Leading zeros never significant, for example:
0.0025 has 2 sig figs
– Captive zeros are significant, for example:
1.008 has 4 sig figs
– Trailing zeros are significant if the number has a decimal point, for example:
100 has 1 sig fig
100. Has 3 sig figs
3. Exact Numbers are numbers known with
certainty and were not obtained by measuring
devices: (They have an unlimited number of
significant figures)
• They arise from definitions (the periodic table)
– Amu (___g/1 mole)
– 22.4 L/1 mole
• They arise from counting numbers
– Number of moles
– Number of coefficients
4. Exponents do not count as sig. figs.
For example:
6.02 x 1023 = 3 sig. figs
6.02214199 x 1023 = 9 sig. figs
Rules for Rounding Off
If the number is less than 5, the preceding digit stays the same, for example:
1.33 rounds to 1.3
If the number is equal to or greater than 5, the preceding digit is increased by 1, for example:
1.36 rounds to 1.4
(worked problems in the text show the correct number of sig figs in each step)
Calculations Using Significant Figures
• Calculators do not know about sig. Figs
– For example, try this on your calculator:
8.315 = 0.027902685
298
• Answers to calculations must be rounded to the
proper number of significant figures (answers
must reflect the measuring device)
Multiplication and Division Using Sig Figs
• The answer has the same number of sig figs as
the number in the problem with the least number
of sig figs
• For example: 4.56 x 1.4 = 6.384 Final answer 6.4
• Now try this rule using our three measuring
devices, convert to moles:
Adding and Subtracting Using Sig Figs
• Go with the least number of decimals
• For example:
12.11
18.0 least number of decimals
+ 1.013
31.123 Final answer 31.1
Now let’s add our measured masses together…
Accuracy vs. Precision
• Accuracy refers to how close a measurement is
to some accepted value, we’ll use % error
% error = exp value – accepted value x 100%
accepted value
• Precision refers to the repeatability of a
measurement
– Using sig figs
– Using Absolute Deviation
Remember Scientific Notation:
• …is a technique used to express very large
or very small numbers
• …is based on the power of 10
• To use numbers written in scientific
notation (calculator):
– Use the EE or Exp button, this means x 10
– Never hit x 10 or you’ll be off by 10 times!
Let's practice scientific notation...
You will need to memorize the following common SI Prefixes
& conversions: nm m
1,000,000,000 nm = 1 m
or
1.0 x 109 nm = 1 m
Let's do a few...
1. 500 nm to m (color of blue)
2. 3.4 m to nm (FM radiowave)
Required way to show your work
• You have two jobs in this class:
1. To be able to perform the conversions
2. To be able to prove that you know why
the answer is correct
3. In short, SHOW YOUR WORK!!!
4. Note the 2006 AP Exam Scoring
Guidelines…
• Note the grading rubric & how units
are given, even for multiplying…
• Note how this answer
earned all points…
• Note how this answer lost
points due to significant
figures
AP Chem
Ch 4-1
(Prentice Hall)
1,000,000,000 nm = 1 m
• This conversion comes in handy when dealing with wavelengths
with atomic spectra chemistry:
Frequency (v) = c = speed of light (3 x 108m/s)
wavelength
What’s the frequency of a wavelength of light of 500 nm?
The frequency of a red light is 4.74 x 1014 calculate its wavelength.
sec
AP Chemistry
Solution Chemistry
Ch 15.1-15.4
(Prentice Hall)
• Solution: is a homogeneous mixture, where the components are uniformly intermingled.
– Solvent: the substance present in the largest amount (usually the component doing the dissolving).
– Solute: the substance present in the smaller amount
(usually the component being dissolved).
• Solutions can be categorized by their physical state:
– Solid solutions
– Gas solutions
– Liquid solutions
• Solid solutions: contains two or more metals and are called alloys
• The more abundant element = solvent
• The less abundant element = solute
• For example steel = carbon + iron
What is the solvent?
What is the solute?
– Gaseous solutions: contain two or more gases that do not react with each other
• For example: Air:
• 78% N2
• 21% O2
• 1% other gases
– Liquid solutions: have a liquid solvent and the solute can consist of either a gas, liquid, or solid
• Aqueous solutions are special solutions that have water as the liquid solvent
You will need to memorize the following common SI
Prefixes & conversions:
mL L 1000 mL = 1 L
Let's practice a few, use the "picket fence method"
1. 459 L to mL
2. 0.0032 mL to L
You will need to memorize the following common
SI Prefixes & conversions:
mL L
1000 mL = 1 L
• For example, this conversion comes in handy when dealing with
solution chemistry:
Molarity (M) = moles of solute = mol
Liters of solution L
8.320g NiCl2 added to 250 mL of water, what’s the concentration?
How many grams of KCl are needed to prepare 0.750 L of a 1.50 M
1000 g = 1 kg
Molality (m) = moles of solute = mol
Kg of solvent kg
What is the molality of a solution containing 75.2 g of
AgClO4 dissolved in 885 g of benzene?
What is the molality of a solid solution of 0.125 g Cr &
81.3 g of Fe?
• Mass Percent: is one way of describing a
solution’s composition, sometimes called
weight percent, (%w/w) which is the mass of a
solute in a given mass of solution.
• Mass Percent = mass of solute x 100%
mass of solution
or Mass Percent = g of solute x 100%
g of solute + g of solvent
Does this look familiar? part/whole x 100%
• Milk is a solution it is composed of a dissolved
sugar called lactose.
• Cow’s milk typically contains 4.5% by mass of
lactose.
• Calculate the mass of lactose present in 175 g of
milk. Use the formula below…
Mass Percent = mass of solute x 100%
mass of solution
• Dissolved sugar called lactose = solute
• Mass % = 4.5%
• Solution mass = 175 g of milk
4.5% = mass of solute (sugar) x 100%
175 g
7.9 g = mass of solute (sugar lactose)
Switch To Clickers
• A sample of brass contains 68 g copper and 7 g zinc.
• What is the mass percent of this solid solution, (alloy)?
(remember to first identify the solute and solvent)
• A sample of brass contains 68 g copper and 7 g zinc.
% Cu = 68 g x 100%
68 g + 7 g
= 68 g x 100%
75 g
% Cu = 90.67%
10-1
Calculating AMU
• Next, let’s determine the mass of each atom.
• As we learned, there are different types of
atoms known as elements.
• Each element has a different mass and it is
called it’s amu
• atomic mass units
• Another name for amu is molar mass
• Molar mass is obtained by summing the masses of the component atoms.
For example: NH3 has the following amu:
N = 1 atom x 14.01 amu = 14.01 amu
H = 3 atoms x 1.01 amu = 3.03 amu
total amu = 17.04 amu
or 17.04 molar mass
Percent Composition
Ch 10-3
• If we wanted to know the percent boys in the class currently, how could we do this?…
• How about the percent girls in the class?
• Correct, count the number of boys/girls (part)
• Count total number in the class (whole)
• Then divide the part over the whole and multiply by 100 %
Percent composition = Part x 100%
whole
Does this formula make sense?
• Let’s apply this formula
% Composition = part x 100%
whole
by using the formula of the compound…
What is the % N to % H of
Windex, which is ammonia,
(NH3)
• First let’s determine the number of elements
in the formula NH3:
• N:1 atom
• H:3 atoms
Now do part x 100%
whole
N = 1 atom x 14.01 amu = 14.01 x 100%
17.03
= 82.21 % N
H = 3 atoms x 1.01 amu = 3.03 x 100%
17.03
= 17.79 % H
• The percentages may not always total to 100% due to rounding,
for example if you go to “1 decimal spot”
82.2% N
+ 17.7% H
99.9 % total
• But if you go to “2 decimal spots” you get closer to 100%
82.21% N
+ 17.79% H
100.00 % total
• Overall your numbers should
add up close to 100%
Switch To Clickers
You and your lab partner find the %Cl in
this formula (CCl2F2).
• You should have calculated
• 58.64 % Cl
Calculating % Composition By
Given Masses
• http://videos.howstuffworks.com/science-
channel/29291-100-greatest-discoveries-
atomic-weight-video.htm
• Dalton’s discovery of relative weights…
• According to the Law of Constant
Composition, any sample of a pure
compound always consists of the same
elements combined in the same proportion
by mass.
Percent composition can be determined of each
element in a compound by its mass:
% Composition = part mass x 100%
whole mass
This formula can be applied by using the
formula of the compound or by experimental
mass analysis of the compound
A sample of butane (C4H10)--lighter fluid--
contains 288 g carbon and 60 g hydrogen.
Find %C and %H in butane
First find total mass of sample (“whole”)
288 g C + 60 g O = 348 g
Next find the percent of each element in the formula… Use the following equation:
Percent composition = Part x 100%
whole
288 g C + 60 g O = 348 g
Part Part whole
Can you figure it out?…
288 g C (C part) x 100 % = 82.8 % C
348 g (whole)
60 g O (O part) x 100% = 17.2 % O
348 g (whole) + ---------------
100.0 %
Percents should add up close to 100%
(It may be a little over or under, that’s ok…)
Switch To Clickers
• Now you and your lab partner try this one…
• What is the percentage composition of a
carbon and oxygen compound, that contains
40.8 g of carbon and 54.4 g of oxygen.
• The total mass of the compound is 95.2 g.
% C = ? And % O = ?
• You should have calculated…
42.86 % C
57.14 % O
Ch 10-1
The Mole
Molar Mass
• The mass of one mole of atoms of any element is the molar mass which is numerically equal to the atomic mass unit (amu), but in grams…
• Molar mass = __g = 1 amu
1 mole
• Therefore CO2 has an amu = 44.01
or 44.01 g
1 mol
Using The Mole Map
Converting
from grams to moles
EXAMPLE: A student weighs out 88 grams
of solid CO2 (dry ice), how many moles
does the student have?
First, find the amu of CO2 :
C = 1 atom x 12.01 amu = 12.01 amu
O = 2 atoms x 16.00 amu = 32.00 amu
CO2 total amu = 44.01 amu
CO2 = 44g
1 mol
• Then convert from grams to moles
• Use your mole map…
Using The Mole Map
(What’s Given) x 1 mole
(molar mass)
88 g CO2 x 1 mole = 2 moles CO2
44 g
• Now you try to find the moles of various
compounds…
ALWAYS
SHOW YOUR WORK!!!
Converting To Particles
• A particle can
be defined as:
– an atom,
– a molecule,
– or formula unit
(ionic
compounds)
• Next, convert from mol to molecules:
• Use your mole map…
(moles calculated) x (Avogadro’s #)
1 mole
2 mole CO2 x 6.02 x 1023 molecules
1 mole
= 1.204 x 1024 molecules CO2
Empirical Formulas
and
Molecular Formulas
Ch 10-3
• Empirical Formulas are the simplest
(lowest) whole number ratio of atoms in a
molecule or ionic compound
• For example:
• C6H6 = CH
• H2O2 = HO
• C6H12O6 = CH2O
Empirical Formulas
…can be determined from % composition,
here is the “process:”
1. % is the same as grams
2. Convert from grams to moles
3. Next divide by the smallest # of moles
4. …this gives the empirical formula
Empirical Formula of Eugenol a
Component of Clove Oil?
…is 73.14% C, 7.37% H and 19.49 g O
Remember, % is the same as grams (g)
Empirical Formula of Eugenol,
continued…
Next, convert to the central unit, the mole
73.14 g C x 1 mole = 6.09 mol C
12.01 g
7.37 g H x 1 mole H = 7.31 moles H
1.0079 g
19.49 g O x 1 mole O = 1.22 moles O
15.9994 g
Empirical Formula of Eugenol,
continued…
Finally divide by the smallest # of moles
6.09 mol C 7.31 moles H 1.22 moles O
1.22 mol 1.22 mol 1.22 moles
C: 4.99 H: 5.99 O: 1.00
Or C: 5 atoms H: 6 atoms O: 1.00 atoms
Therefore C5H6O
is Eugenol’s empirical formula
Eugenol
Count the number of
carbon atoms,
hydrogen atoms and
oxygen atoms, does
this fit the empirical
formula that we just
derived? No, because
we did not find the
molecular formula…
Molecular Formulas
Molecular formulas are also known as the
“true formula” of a molecule.
To derive this use amu:
Molecular Formula = True amu
empirical amu
True Formulas
• The molar mass of Eugenol is 164.2 g/mol,
what’s the molecular formula of Eugenol?
• Use: True amu
empirical amu
164.2 g/mol = 164.2 g/mol = 2
C5H6O 82 g/mol
Therefore 2(C5H6O) = C10H12O2
Hydrated Compounds
Ch 7-3
&
Ch 14-3
Hydrated Compounds
• If ionic compounds
are prepared water
solutions and then
isolated as solids, the
crystals often have
molecules of water
trapped in the lattice,
for example CuSO4
CuSO4 • 5 H2O
Should you add the water when figuring the molar mass?
Yes, therefore
CuSO4 • 5 H2O has an amu of 249.7 g/mol
Notice the color difference of the dehydrated crystals
Hydrates • A compound that is hydrated is called a hydrate,
they form solids that includes water in their crystal structure
• Water can be driven from a hydrate to leave an anhydrous compound
•
Naming Hydrates
• To name hydrates:
1. Name the compound
2. Plus the word hydrate—use prefixes to indicate how
many waters are associated with the compound
3. Example: Copper (II) Sulfate pentahydrate
4. To write their formulas
Write: the name of the compound • number of H2O
CuSO4 • 5 H2O
Units of Hydration • A student heats hydrated crystals of CuSO4, how
many moles of water are associated with the
crystals?
• Step 1: Find the mass of the crystals:
1.023 g of CuSO4 • x H2O
• Step 2: Subtract the dehydrated crystal mass from
the initial crystal mass = mass of water
1.023 g of CuSO4 • x H2O – 0.654 g of CuSO4
= 0.369 g water
Units of Hydration Continued…
• Step 3: Determine the number of moles
0.369 g H2O x 1 mol
18.02 g = 0.0205 mol H2O
0.654 g CuSO4 x 1 mol/159.6 g = 0.00410 mol CuSO4
• Step 4: Determine the molar ratio (see above)
0.0205 mol H2O 0.00410 mol CuSO4
0.00410 mol CuSO4 0.00410 mol CuSO4
1 CuSO4 • 5 H2O
Ch 11-1
Mole-Mole Relationships
Using Mole Ratios
Ammonia (NH3) is used in huge quantities as a fertilizer. It is manufactured by combining nitrogen and hydrogen according to the following equation:
N2 + 3 H2 2 NH3
How many moles of NH3 can be made from 1.30 mol H2?
N2 + 3 H2 2 NH3
1.30mol ? mol
• Write what’s given first:
1.30 mol H2
• Next look at the balanced equation to
convert from moles of one thing to moles of
another:
1.30 mol H2 x 2 mol NH3 = 0.867 mol NH3
3 mol H2
http://spaceflight.nasa.gov/gallery/video/shuttle/sts-107/html/fd11.html
What if you had the
task of figuring out
how much LiOH was
needed on board a
space shuttle flight?
Solid LiOH is used to
take out CO2 from the
shuttle’s environment.
SAMPLE PROBLEM
• Solid lithium hydroxide is used in space
vehicles to remove exhaled carbon dioxide
from the living environment.
STEP 1 Balance the equation for the reaction:
LiOH + CO2 Li2CO3 + H2O
• STEP 1 Balance the equation for the
reaction:
2 LiOH + CO2 Li2CO3 + H2O
STEP 1 Balance the equation for the reaction:
2 LiOH + CO2 Li2CO3 + H2O
You need 41.8 mol LiOH in order to maintain the
atmosphere of the space shuttle for the
astronauts
How many moles of CO2 will be taken out of the
space shuttle?
Next Step Using this mole ratio, we can calculate
the moles of CO2 needed to react with the given
moles of LiOH:
2 LiOH + CO2 Li2CO3 + H2O
41.8 mol LiOH X 1 MOL CO2 = 20.9 mol CO2
2 mol LiOH
What if we start off with grams?
11-2
Mass-Mass Calculations
Steps for Calculating the Masses of Reactants &
Products in Chemical Reactions
STEP 1 Balance the equation for the reaction.
STEP 2 Convert the masses of reactants or products to
moles
STEP 3 Use the balanced equation to set up the
appropriate
mole ratio(s).
STEP 4 Use the mole ratio(s) to calculate the number
of
moles of the desired reactant or product.
STEP 5 Convert from moles back to mass.
SAMPLE PROBLEM
Solid lithium hydroxide is used in space
vehicles to remove exhaled carbon dioxide
from the living environment.
The products are solid lithium carbonate and
liquid water.
What mass of gaseous carbon dioxide can
1000 g of lithium hydroxide absorb?
http://spaceflight.nasa.gov/gallery/video/shuttle/sts-107/html/fd11.html
What if you had the
task of figuring out
how much LiOH was
needed on board a
space shuttle flight?
Solid LiOH is used to
take out CO2 from the
shuttle’s environment.
What mass of gaseous carbon dioxide
can
1000 g of lithium hydroxide absorb?
2 LiOH + CO2 Li2CO3 + H2O
1000 g ? g
STEP 2 Convert the masses of reactants or products to moles:
1.00 x 10 3 g LiOH x 1 mol LiOH = 41.8 mol LiOH
23.95 g LiOH
STEP 3 The appropriate mole ratio is:
2 LiOH + CO2 Li2CO3 + H2O
1 mol CO2
2 mol LiOH
Step 4 Using this mole ratio, we calculate
the moles of CO2 needed to react with
the given mass of LiOH:
41.8 mol LiOH X 1 MOL CO2 = 20.9 mol CO2
2 mol LiOH
Step 5 We calculate the mass of CO2 by using
its molar mass (44.01 g):
20.9 mol CO2 x 44.01 g = 920. g CO2
1 mol CO2
Thus 1.00x103 g of LiOH(s) can absorb 920. g of CO2
(g).
Now You Try! Phosphorus is placed in a flask of chlorine gas, heat and
light is given off forming phosphorus trichloride:
Step 1: Write and balance the Equation:
P4 + Cl2 PCl3
QuickTime™ and aCinepak decompressor
are needed to see this picture.
Step 1: Write and balance the Equation:
The Balanced Equation
P4 + 6 Cl2 4 PCl3
Now You Try!
How many grams of Cl2 will react with 1.24 g
of P4?
P4 + 6 Cl2 4 PCl3
1.24 g ? g
P4 + 6 Cl2 4 PCl3
1.24 g ? g
g P4 mol P4 mol: mol ratio g Cl2
1.24 g P4 x 1 mol P4 = 0.0100 mol P4
123.88 g
0.0100 mol P4 x 6 mol Cl2 = 0.0601 mol Cl2
1 mol P4
0.0601 mol Cl2 x 70.90 g Cl2 = 4.26 g Cl2
1 mol Cl2
Limiting Reactants
Ch 11-3
• Stoichiometry:
is the process of using a chemical equation
(mole ratio) to calculate the masses of
reactants and products in a reaction
• There are three parts to a reaction when
using stoichiometry…
The reactant that runs out first limits the amount
of product that can be formed and is called the
limiting reactant (or limiting reagent).
The reactant that is left over when the
reactions stops is called the excess of
reactant (or reagent).
A stoichiometric quantity is when reactants are
mixed in exactly correct amounts so that all
reactants “run out” at the same time.
The balanced equation is the correct
“stoichiometric quantity
– nothing is “left over”
– nothing “runs out first”
Let’s Do A Little Activity…
• You will receive a small kit of molecular
models to build.
• We will investigate a simple reaction.
• But we will build by using different mole
relationships to determine the Limiting
Reactant and Excess Reactant.
• Let’s begin…
1. 3 H2 + 1 O2 --> ? H2O
* What is the limiting and excess reactant?
?
1. 3 H2 + 2 O2 --> ? H2O
* What is the limiting and excess reactant?
?
1. 2 H2 + 1 O2 --> ? H2O
* What is the limiting and excess reactant?
?
Now Balance The Equation (This is the correct Stoichiometric Quantity)
H2 + O2 --> H2O
Are There Any Limiting or
Excess Reactants?
2 H2 + O2 --> 2 H2O
• Using this concept we can complete a
“special mass to mass calculation” to
figure out how much of our reactants
and products we need to get the
“greatest” amount of our product.
What is the correct ratio of Zn to HCl to produce
the maximum amount of H2?
(moles of HCl will be held constant…)
7.00 g 3.27 g 1.31 g 7.00 g 3.27 g 1.31 g
0.107 mol Zn
0.100 mol HCl
0.050 mol Zn
0.100 mol HCl
0.020 mol Zn
0.100 mol HCl
Excess Zinc Reactants match Excess HCl
perfectly; all is
consumed
Steps To Determine The Limiting
1. Compare the moles of reactant 1 to reactant 2
2. This will determine how many moles needed for reactant 2
3. Then determine how many moles you actually have for reactant 2
4. Compare the moles of needed to actual
5. This will help you to determine what is limiting and what is excess
Steps for Solving Stoichiometry Problems
Involving Limiting Reactants
STEP 1 Write and balance the equation for the reaction.
STEP 2 Convert known masses of reactants to moles.
STEP 3 Using the numbers of moles of reactants and the
appropriate
mole ratios, determine which reactant is limiting.
STEP 4 Using the amount of the limiting reactant and the
appropriate mole ratios, compute the number of moles of the
desired product.
STEP 5 Convert from moles of product to grams of product,
using the molar mass (if this is required by the problem).
Calculate the mass of ammonia produced…
25,000 g of nitrogen gas and 5000 g
of hydrogen gas are mixed and reacted to form
ammonia.
Remember, the limiting reactant determines
the amount of product formed…
N2 (g) + 3 H2(g) 2 NH3 (g)
25,000 g 5000 g ?
N2 (g) + 3 H2 (g) 2 NH3 (g) 25,000 g 5000 g
• What is the Limiting & Excess Reactant?
N2 (g) + 3 H2 (g) 2 NH3 (g)
N2 25,000 g N2 x 1 mol N2
28.02 g N2
= 892 mol N2
Mol of N2 available
892 mol N2 x 3 mol H2
1 mol N2
= 2680 mol H2 needed to react
H2
5000 g H2 x 1 mol H2
2.016 g H2
= 2480 mol H2
Mol of H2 available
We have 2480 mol H2 but NEED 2680 mol H2. Therefore H2 will run out first. N2 is excess
25,000 g 5000 g
• Now determine the maximum amount of
product produced if hydrogen is the limiting
reactant…
• Remember 2480 mol H2 Mol of H2 available
N2 (g) + 3 H2 (g) 2 NH3 (g)
Calculate the mass of ammonia produced if
hydrogen is the limiting reactant…
25 g of nitrogen gas and 5 g of hydrogen gas
are mixed and reacted to form ammonia.
Remember, the limiting reactant determines
the amount of product formed…
N2 (g) + 3 H2(g) 2 NH3 (g)
25 g 5 g ?
N2 (g) + 3 H2(g) 2 NH3 (g)
5 g H2 x x 1 mol H2 = 2.48 mol H2
2.016 g H2
2.48 mol H2 x 2 mol NH3 = 1.65 mol NH3
3 mol H2
1.65 mol NH3 x 17.03 g NH3 = 28.1 g NH3
1 mol NH3
Therefore: N2 (g) + 3 H2(g) 2 NH3
25 g 5 g (limiting) ?
25 g 5 g 28.1 g NH3
Percent Yield
Chapter 11-3
Theoretical Yield
• The amount of product formed is controlled
by the limiting reactant—products stop
forming when on reactant runs out.
• The amount of product calculated in this
way is called the theoretical yield.
• This is the amount of product predicted
from the amount of reactants used.
Actual Yield
• However, the amount of product predicted
(the theoretical yield) is seldom obtained.
• One reason for this is the presence of side
reactions (other reactions that consume one
or more of the reactants or products).
• The actual yield of product, is the amount
of product actually obtained.
Percent Yield
• The comparison of the product actually
obtained and theoretically obtained is called
the percent yield:
• Percent Yield = Actual Yield x 100%
Theoretical Yield
Chapter 1 Summary
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