Chemistry
Unit 6: Moles and Stoichiometry
Chemistry
Learning Objectives Moles and Stoichiometry
Essential knowledge and skills:
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Perform conversions between mass, volume, particles, and moles of a substance.
Perform stoichiometric calculations involving the following relationships:
mole-mole;
mass-mass;
mole-mass;
mass-volume;
mole-volume;
volume-volume;
mole-particle;
mass-particle; and
volume-particle.
Identify the limiting reactant (reagent) in a reaction.
Calculate percent yield of a reaction.
Calculate the % Composition
Calculate the Empirical & Molecular Formula
Calculation of Formula (Molar) Mass
Essential understandings:
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Atoms and molecules are too small to count by usual means. A mole is a way of counting any type of particle (atoms,
molecules, and formula units).
Avogadro’s number = 6.023 × 1023 particles per mole.
Molar mass of a substance is its average atomic mass in grams from the Periodic Table.
Molar volume = 22.41 L/mol for any gas at standard temperature and pressure (STP).
Stoichiometry involves quantitative relationships. Stoichiometric relationships are based on mole quantities in a
balanced equation.
Total grams of reactant(s) = total grams of product(s).
The empirical formula shows the simplest whole-number ratio in which the atoms of the elements are present in the
compound. The molecular formula shows the actual number of atoms of each element in one molecule of the substance.
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Cheaper by the Mole?
Moles Explained
Think of moles as a "chemist's dozen". Just as 12 eggs is a dozen eggs, 6.023 × 1023 eggs is a mole of
eggs. 6.023 × 1023 molecules of oxygen is a mole of oxygen. The number of grams in a mole is different
from substance to substance. If
you're like most students, it's this that's
confusing you. You can find the
number of grams by looking on the periodic
table. Just look at the atomic mass
of the element.
Atomic Mass tells you that a carbon
atom has a mass of 12.011 atomic
mass units (amu). One mole of
carbon has 12.011 grams. You will
use this number as the molar mass
for carbon
Picture it this way: a dozen elephants have a different mass than a dozen rabbits- but in each case, you
have a dozen animals. Similarly, a mole of oxygen gas has a different mass than a mole of water- but in
each case, you have 6.023×1023molecules.
A mole is a standard scientific unit for measuring large quantities of very small entities such
as atoms, molecules, or other specified particles. The mole designates an extremely large number of
units, 6.02214179 × 1023, which is the number of atoms determined experimentally to be found in 12
grams of carbon-12. Carbon-12 was chosen to serve as the reference standard of the mole unit for the
International System of Units (SI).
These entities could be atoms, molecules, formula units, ions
1 mole 12C = 6.023 x 1023 carbon atoms
1 mole H2O = 6.023 x 1023 H2O molecules
1 mole NaCl = 6.023 x 1023 formula units
1 mole of electrons = 6.023 x 1023 electrons
1 mole of Na+ ions = 6.023 x 1023 Na+ ions
1 mole of electrons = 6.023 x 1023 electrons
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Molar Mass of Compounds
Molar Mass Tutorial
Molar mass is the mass in grams per mol. of a substance. Molar mass is determined by using the average
atomic masses of each element. Also called molecular weight or gram formula mass.
Determine the molar mass of nitrogen (III) oxide
N2O3
N
2 x 14.01
O
3 x 16.00
76.01 g/mol
Determine the molar mass of copper (II) sulfate pentahydrate
CuSO4 ●5H2O
Cu
1 x 63.55
S
1 x 32.08
O
4 x 16.00
H2O
5 x 18.02
249.73 g/mol
Calculation of Molar Mass worksheet
Part 1.
Formula
Molar mass (g/mol)
Name of compound
1. CH4
_________________ g/mol
_________________________
2. CaCO3
_________________ g/mol
_________________________
3. SO2
_________________ g/mol
_________________________
4. NaClO4
_________________ g/mol
_________________________
5. NaMnO4
_________________ g/mol
_________________________
4
6. LiCl
_________________ g/mol
_________________________
7. H2SO3
_________________ g/mol
_________________________
8. FeSO4
_________________ g/mol
_________________________
9. CS2
_________________ g/mol
_________________________
10. K3PO4
_________________ g/mol
_________________________
Part 2.
Formula
Molar mass (g/mol)
Name of compound
11. C2H5OH
_________________ g/mol
_________________________
12. Cu(OH)2
_________________ g/mol
_________________________
13. Ca3(PO4)2
_________________ g/mol
_________________________
14. Ba(NO3)2
_________________ g/mol
_________________________
15. Al2(SO4)3
_________________ g/mol
_________________________
Formula
Molar mass (g/mol)
Name of compound
16. Zn(HCO3)2
_________________ g/mol
_________________________
17. NaHCO3
_________________g/mol
_________________________
18. C11H22O12
_________________g/mol
_________________________
19. Mg(NO2)2
_________________g/mol
_________________________
20. Cu(NO3)2
_________________ g/mol
_________________________
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Part 3. Write the formula of each compound. Then calculate its molar mass.
Name
Formula
Molar Mass (g/mol)
Potassium carbonate
Copper (II) sulfate
Carbon tetrachloride
Sodium phosphate
Iron (III) hydroxide
Challenge Problems
Formula
Molar mass (g/mol)
Name of compound
21. Pb(C2H3O2)2
______________ g/mol
_________________________
22. (NH4)2Cr2O7
______________ g/mol
_________________________
23. Cr2(SO4)3
______________ g/mol
_________________________
Percent Composition
Percent Mass
The percent composition is the percentage of each element present in a compound by mass. To
find the percent composition, you calculate how much of the molar mass is supplied by a
particular element, divided by the molar mass, then multiplied by 100 to convert to a percentage.
Here are two examples of this type of calculation using the examples above for calculating molar
masses.
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The percent composition of NaCl would be as follows:
% Na =
%Cl =
22.99 g/mol Na
×100 = 39.33% Na
58.44 g/mol NaCl
35.45 g/mol Cl
×100 = 60.66% Cl
58.44 g/mol NaCl
Notice that although the percentages should add up to 100%, these only add up to 99.99%,
probably because of rounding. The percent composition of NH3 would be as follows:
%N =
14.01 g/mol N
×100 = 82.22% N
17.04 g/mol NH3
%H =
3 mol H 1.08 g/mol H
×100 = 17.8% H
17.04 g/mol NH3
Percent Composition Worksheet
Determine the percentage composition of each of the compounds below
1. CuBr2
Cu: _________
Br: __________
2. NaOH
Na: ___________
O: ___________
H: ___________
3. (NH4)2S
N: ___________
H: ___________
S: ___________
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4. N2H2
N: ___________
S: ___________
5. KMnO4
K: ___________
Mn: ___________
O: ___________
6. (NH4)3PO4
N: ___________
H: ___________
O: ___________
P: ___________
Composition of Hydrates
Hydrate Composition
Empricial Formula of Hydrates
A hydrate is an ionic compound with water molecules loosely bonded to its crystal structure. The water is
in a specific ratio to each formula unit of the salt. For example the formula CuSO45H2O indicates that
there are five water molecules for every one formula unit of CuSO4.
Hydrate – contains water molecules
Anhydrous – water has been removed
1. What percentage of water is found in CuSO45H2O
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2. What percentage of water is found in Na2S9H2O
3. A 5.0 g sample of a hydrate of BaCl2 was heated and only 4.3 g of the anhydrous salt remained.
What percentage of water was in the hydrate?
4. A 2.5 g sample of a hydrate of Ca(NO3)2 was heated and only 1.7 g of the anhydrous salt
remained. What percentage of water was in the hydrate?
5. A 3.0 g sample of Na2CO3H2O was heated to a constant mass. How much anhydrous salt
remains?
6. A 5.0 g sample of Cu(NO3)2xH2O is heated and 3.9 g of the anhydrous salt remains. What is the
value of x?
Empirical and Molecular formulae
Empirical formula (also called the simplest formula) is the smallest whole number ratio of atoms in
a compound.
Example:
C6H12O6 is the molecular formula
CH2O is the empirical formula for this compound
Steps in determining the empirical formula:
Problem: Find empirical formula for a compound that contains:
18.8 % Na
29% Cl
52.2 % O
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Step1: Change % to grams using the assumption that there is 100 g of compound
18.8 g Na
29 g Cl
52.2 g O
Step 2: Convert grams to moles by dividing by molar mass
18.8 𝑔
𝑔
𝑚𝑜𝑙
22.99
29 𝑔
35.45
𝑔
𝑚𝑜𝑙
= 0.818 𝑚𝑜𝑙 𝑁𝑎
= 0.818 𝑚𝑜𝑙 𝐶𝑙
52.2 𝑔
16.00 𝑔/𝑚𝑜𝑙
= 3.26 𝑚𝑜𝑙 𝑂
*Note: Use three significant figures for moles
Step 3: Determine whole number ratios by dividing each by the smallest number of moles.
0.818 𝑚𝑜𝑙
0.818 𝑚𝑜𝑙
= 1 𝑁𝑎
0.818 𝑚𝑜𝑙
= 1 𝐶𝑙
0.818 𝑚𝑜𝑙
3.26 𝑚𝑜𝑙
=4𝑂
0.818 𝑚𝑜𝑙
* Note: If one ratio is not a whole number or more than 0.06 from a whole number then you may
have to multiply to get a whole number ratio. Example if one of your ratios was 1.5, then multiply
all ratios by 2 to make a whole number.
Step 4: Write metal or most metallic element first (These are elements furthest to left on periodic table).
Follow this order to the least metallic. Oxygen is always last. If carbon is present, write hydrogen directly
after carbon.
NaClO4
* Note: For an ionic compound, the empirical formula is always the chemical formula
Steps for determining the chemical formula (molecular formula) from the empirical formula:
Problem: Find the molecular formula for a compound that has a molecular weight of 180.18 g/mol with
an empirical formula of CH2O.
Step 1: Find molar mass of empirical formula.
CH2O has a molar mass of 30.03 g/mol
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Step 2: Divide molecular weight of compound by molar mass of the empirical formula.
180.18 𝑔/𝑚𝑜𝑙
30.03 𝑔/𝑚𝑜𝑙
=6
Step 3: Multiply all subscripts in empirical formula by this number.
1C=6
2 H = 12
1O=6
Step 4: Write correct molecular or chemical formula
C6H12O6
*Note: You always need the molar mass and empirical formula in order to determine correct chemical or
molecular formula.
Empirical Formula Introduction
Write the Empirical Formula (lowest whole number ratio) for each of the following:
a. P4O6
_________________
b. C6H9
_________________
c. CH2OHCH2OH
_________________
d. BrCl2
_________________
e. C6H8O6
_________________
f. C10H22
_________________
g. Cu2C2O4
_________________
h. Hg2F2
_________________
Calculating Empirical and
Molecular Formula from
percentages
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EF and MF Calculations from
percentages
Calculate the Empirical Formula for all of the Following (show your work):
a. A compound composed of: 72% iron (Fe) and 27.6% oxygen (O) by mass.
b. A compound composed of: 9.93% carbon (C), 58.6% chlorine (Cl), and 31.4% fluorine (F). (This
compound is commonly known as Freon)
c. A compound composed of: 0.556g carbon (C) and 0.0933g hydrogen (H).
Empirical Formula Worksheet 1:
1. Find the empirical formula of a compound that is 48.38% carbon, 8.12% hydrogen, and 43.5%
oxygen by mass.
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2. C.I.Pigment Yellow 45 ("sideran yellow") is a pigment used in ceramics, glass, and enamel.
When analyzed, a 2.164 grams sample of this substance was found to contain 0.5259 grams of Fe
and 0.7345 grams of Cr. The remainder was oxygen. Calculate the empirical formula of this
pigment. Answer: Fe2Cr3O12
3. The composition of nicotine is 74.0% C, 8.7% H, and 17.3% N. The molecular mass of nicotine
is 162. What is its molecular formula? Answer: C10H14N2
4. One of the most deadly poisons, strychnine, has a formula weight of 334 and the composition
75.42% C, 6.63% H, 8.38% N; the rest is oxygen. Calculate the empirical and molecular formulas
of strychnine, arranging the atomic symbols in alphabetical order. Answer: C21H22O2N2
Empirical and Molecular formula worksheet 2
1. Find the empirical formula for a compound which contains 0.463 g Tl (#81), 0.0544 g of carbon,
0.00685 g of hydrogen and 0.0725 g oxygen by finding its empirical formula.
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2. What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?
3. The characteristic odor of pineapple is due to ethyl butyrate, an organic compound which contains
only carbon, hydrogen and oxygen. If a sample of ethyl butyrate is known to contain 0.62069 g of
carbon, 0.103448 g of hydrogen and 0.275862 g of oxygen, what is the empirical formula for ethyl
butyrate?
4. 300 grams of a compound which contains only carbon, hydrogen and oxygen is analyzed and found to
contain the exact same percentage of carbon as it has oxygen. The percentage of hydrogen is known to be
5.98823%. Find the empirical formula of the compound.
5. 200.00 grams of an organic compound is known to contain 83.884 grams of carbon, 10.486 grams of
hydrogen, 18.640 grams of oxygen and the rest is nitrogen. What is the empirical formula of the
compound?
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6. A certain compound contains 4.0 g of calcium and 7.1 g of chlorine. Is relative molecular mass is
111 g/mol. Find its empirical and molecular formulas.
7. A certain compound was found to contain 54.0 g of carbon and 10.5 grams of hydrogen. Its relative
molecular mass is 86.0 g/mol. Find the empirical and the molecular formulas.
8. A certain compound was found to contain 26.4 g of carbon, 4.4 grams of hydrogen and 35.2 grams of
oxygen. Its relative molecular mass is 60.0 g/mol. Find the empirical and the molecular formula.
9. A certain compound was found to contain 78.2 % Boron and 21.8 % hydrogen. Its relative molecular
mass is 27.7 g/mol. Find the empirical and the molecular formula.
15. A certain compound contains 7.3%Carbon, 4.5 % hydrogen, 36.4% oxygen, and 31.8% nitrogen. Its
relative molecular mass is 176.0. Find its empirical and molecular formulas.
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Empirical and Molecular Formula Worksheet 3
Show ALL your work for credit!
1. Identify the following as molecular formulas, empirical formulas or both.
a. Ribose, C5H10O5, a sugar molecule in RNA.
b. Ethyl butanoate, C6H12O2, a cmpd w/ the odor of pineapple.
c. Chlorophyll, C55H72MgN4O5, part of photosynthesis.
d. DEET, C12H17ON, an insect repellent.
e. Oxalic acid H2C2O4, found in spinach and tea.
2. Calculate the empirical formula of each compound with the following percent composition.
a. 94.1% O, 5.9% H
b. 79.9% C, 20.1% H
3. The compound meythl butanoate smells like apples. Its percent composition is 58.8% C, 9.8% H, and
31.4% O. If its gram molecular mass is 102 g/mol, what is its molecular formula?
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4. a. A compound of carbon and hydrogen has the composition of 92.25% carbon and
7.75% hydrogen by mass. What is the empirical formula of this composition?
b. If the compound has a mass of 52.03 g/mol, what is the molecular formula of the compound?
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The Mole Diagram
Volume
of Gas at
STP
22.4 L / mol at STP
÷
X
÷
Molar
Mass
(g/mol))
÷
÷
Moles
Mass
X
6.02 x 1023 particles/ 1mol
÷
X
÷
Number of
Particles
(atoms, ions,
molecules)
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Mole-Particle Conversions
Moles to Particle Conversions
1. How many moles of magnesium is 3.01 x 1022 atoms of magnesium?

1 mole

 = 5.00 x 10-2 moles
23
 6.02 x10 atoms 
3.01 x 1022 atoms 
2. How many molecules are there in 4.00 moles of glucose, C6H12O6?
3. How many moles are 1.20 x 1025 atoms of phosphorous?
4. How many atoms are in 0.750 moles of zinc?
5. How many molecules are in 0.400 moles of N2O5?
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Mole-Mass Conversions
Mass to Mole Conversions
1. How many moles in 28 grams of CO2?
Molar mass of CO2
1 C = 1 x 12.01 g = 12.01 g
2 O = 2 x 16.00 g = 32.00 g
44.01 g/mol.
 1 mole 
 = 0.64 moles CO2
 44.01 g 
28 g CO2 
2. What is the mass of 5.0 moles of Fe2O3?
3. Find the number of moles of argon in 452 g of argon.
4. Find the grams in 1.26 x 10-4 mol. of HC2H3O2.
5. Find the mass in 2.60 mol. of lithium bromide.
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Mole-Volume Conversions
Moles and Molar
Volume
1. Determine the volume, in liters, occupied by 0.030 moles of a gas at STP.
 22.4 L 
 = 0.67 L
 1 mole 
0.030 mol 
2. How many moles of argon atoms are present in 11.2 L of argon gas at STP?
3. What is the volume of 0.05 mol of neon gas at STP?
4. What is the volume of 1.2 moles of water vapor at STP?
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Mixed Practice
1. You have 1.20 x 10-2 moles of Tantalum (Ta). How many grams is this?
2. You discover that the head of a match contains 1.66 g of Sulfur, S. How many atoms of S does a
match contain?
3. While cleaning a cut, you spill a bottle of Iodine. The label says that the bottle holds 500. grams of I2.
How many moles of I2 are there? How many I atoms are present?
4. Your silver watchband masses out at 326 g. How many moles of Ag do you have?
5. EXTRA STEP HERE! Can you catch it? While dropping off you recycling, you are overcome by the
urge to weigh the tin cans you brought in. You find that the mass of cans in the box you brought
massed out at 23.0 kg. How many moles do you have?
6. Water has a molar mass of 18.02 grams (that’s 18.02 grams per mole…). You drink a 2-liter bottle of
water every day, and you are such a smarty that you know that 1-ml of H2O weighs 1 g. Can you tell
me how many moles of water you consume a day?
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7. Your toothpaste probably contains around 62.0 g of fluorine per tube. How many molecules and
atoms of fluorine are in one tube of toothpaste?
8. The head of a golf club might contain 250.0 grams of titanium. How many atoms is this?
10. The shaft of that same golf club probably contains around 35.0 moles of graphite, a natural form of
carbon. What is the mass of the shaft of the club? Also how many C atoms are present?
Multiple Variable Conversions
1. How many oxygen molecules are in 3.36 L of oxygen gas at STP?
 1 mole   6.02 x1023 molecules 
 = 9.03 x 1022 molecules
 
1
mole
22
.
4
L



3.36 L 
23
2. Find the mass in grams of 2.00 x 1023 molecules of F2.
3. Determine the volume in liters occupied by 14.0 g of nitrogen gas at STP.
4. Find the mass, in grams, of 1.00 x 1023 molecules of N2.
5. How many particles are there in 1.43 g of a molecular compound with a molecular mass of 233 g/mol?
6. Aspartame is an artificial sweetener that is 160 times sweeter than sucrose (table sugar) when
dissolved in water. It is marketed by G.D. Searle as Nutra Sweet. The molecular formula of aspartame is
C14H18N2O5 .
a) Calculate the gram-formula-mass of aspartame.
b) How many moles of molecules are in 10 g of aspartame?
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c) What is the mass in grams of 1.56 moles of aspartame?
d) How many molecules are in 5 mg of aspartame?
e) How many atoms of nitrogen are in 1.2 grams of aspartame?
Volume-Volume Worksheet
If volumes of reactants or products are provided you can use the ratios of the coefficients from the
balanced equation to work out the answer
1. N2 + 3H2 2NH3
What volume of Hydrogen is necessary to react with 5.78 L of nitrogen?
What volume of ammonia is produced?
2. C3H8 + 5O2  3CO2 + 4H2O
If 20.5 L of Oxygen are consumed in the above reaction, how many litres of carbon dioxide are produced?
3. 2H2O  2H2 +O2
If 30.0 ml of hydrogen are produced in the above reaction, how many milliliters of oxygen are produced?
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4. 2CO + O2  2CO2
How many litres of carbon dioxide are produced if 75.5 L of CO2 are combusted in oxygen? How many
litres of Oxygen are needed?
Mole-Mole Problems
From the balanced equation the coefficients are also used to predict the moles of reactants/products
needed or produced in a chemical reaction. Use the ratios given from the balanced equation.
1. N2 + 3H2  2NH3
How many moles of hydrogen are needed to completely react with two moles of nitrogen?
2. 2KClO3  2KCl + 3O2
How many moles of oxygen are produced by the decomposition of six moles of potassium chlorate?
3. Zn + 2HCl  ZnCl2 + H2
How many moles of hydrogen are produced from the reaction of three moles of zinc with an excess of
hydrochloric acid.
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Three step mole conversions
Mass to Mass Conversions
Normally you will not have nice whole number of moles in your problems but the method you use to
change the number of moles will be the same. For the next set of problems you cannot just use the
coefficients to change the amounts. When converting from the mass of a reactant/product to the mass of
another reactant/product you need to do three steps, these are
Mass  Moles
Moles  Moles (using coefficients from the balanced equation)
Moles  Mass
Three steps will be required for anything on the mole diagram. The mole diagram below shows the steps
involved to convert between the different quantities you may be given in these types of reactions.
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Expanded Mole Diagram
* Use the conversion factor formed from the coefficients of A and B in the balanced equation.
Volume
of A at
STP
÷
÷
Volume
of B at
STP
*
X
÷
X
÷
22.4 L / mol at STP
22.4 L / mol at STP
÷
Mass of
A
÷
÷
Moles
of A
X
Molar
Mass of
A (g/mol)
÷
23
6.02 x 10 particles/mol
X
*
X
X
÷
÷
6.02 x 1023 particles/mol
*
Number of
particles
of A
Number of
particles
of B
28
Mass of
B
Molar
Mass of
B (g/mol)
X
÷
÷
Moles
of B
The Mole and Chemical Equations
1. A student weighed out 2.30 g of magnesium and burned it in air. Magnesium burns in air to form
magnesium oxide. The equation for the reaction is:
Mg(s) + O2(g)  MgO(s) (unbalanced)
Calculate the mass of magnesium oxide produced in the reaction.
2. The reaction for the decomposition of calcium carbonate is:
CaCO3(s)  CaO(s) + CO2(g)
If 100.0 g of calcium carbonate is heated, what mass of calcium oxide will form?
3. Calcium burns in air according to the equation:
Ca(s) + O2(g)  CaO(s) (unbalanced)
How much calcium is needed when 8.00 g of oxygen is used up?
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4. Iron oxide is converted into iron by carbon monoxide according to the equation:
Fe2O3 + CO  Fe + CO2 (unbalanced)
Calculate the mass of iron, which could be obtained from 1.60 tonnes (1 tonne = 1000 kg) of iron oxide.
5. Calculate the mass of water that will react completely with 4.00 g of pure calcium metal
according to the following equation:
Ca(s) + H2O(l)  Ca(OH)2(s) + H2(g) (unbalanced)
6. Calculate the mass of ammonia is that is required to produce 182.0 kg of urea, CO(NH2)2,
according to the following equation:
CO2(g) + NH3(g)  CO(NH2)2(s) + H2O (unbalanced)
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Mixed Stoichiometry Problems
For the problems involving gases, assume that the reactions are being performed at STP
1)
Given the reaction:
4NH3(g) + 5O2(g)

4NO(g) + 6H2O(l)
What is the total number of molecules of water formed when 1.20 L of ammonia reacts with
excess oxygen?
2)
Ethylene burns in oxygen to form carbon dioxide and water vapor:
C2H4(g) + 3O2(g)  2CO2(g) + 2H2O(g)
How many liters of carbon dioxide can be formed if 1.25 x 1024 molecules of oxygen are consumed in this
reaction?
3)
Calcium carbonate decomposes at high temperatures to form carbon dioxide and calcium oxide:
CaCO3(s)  CO2(g) + CaO(s)
How many grams of calcium carbonate will I need to form 3.45 liters of carbon dioxide?
31
4)
Given the following reaction:
H2SO4 (aq) + 2NaOH(aq)  H2O(l) + Na2SO4(aq)
How many molecules of water are produced if 2.00 grams of sodium sulfate are also produced?
5)
A car air bag requires 70.0 L of Nitrogen to inflate properly. The following reaction represents
the reaction:
2NaN3(s)
→
2Na(s) + 3N2(g)
What mass of sodium azide (NaN3) is needed to generate the proper amount of nitrogen gas to fill
the air bag? Assume that the reaction is carried out at STP
6)
When chlorine is added to acetylene, 1,1,2,2-tetrachloroethane is formed:
2Cl2(g) + C2H2(g)  C2H2Cl4(l)
How many molecules of chlorine will be needed to make 75.0 grams of C2H2Cl4?
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Limiting Reactants and Theoretical Yield
Tutorial on Limiting Reactant
Another Limiting Reactant
Tutorial
In real reactions, reactants are not always present in exact stoichiometric ratios. Usually one reactant will
be present in excess and will not be completely consumed. The reactant that is completely consumed is
called the limiting reactant. The reactant that is not completely used up is called the excess reactant.
The amount of the limiting reactant determines how much of the excess reactant is consumed and how
much of each product is produced. The amount of product that is calculated based on the limiting reactant
is called the theoretical yield.
*When given amounts of two reactants use each reactant to calculate the amount of product
that can be produced.
*Only the smallest amount of product will actually be formed because one of the reactants gets
used up and the reaction stops.
*The reactant that forms the least amount of product is the limiting reactant.
*Once you determine the limiting reactant, the amount of limiting reactant given in the
problem is used for all subsequent calculations.
An easy way to find the limiting reactant is to find the number of moles of each reactant and then divide
the number of moles of each of the reactants by the coefficient in the balanced equation. The reactant with
the smallest value will be the limiting reactant. DO NOT USE THIS NUMBER IN THE THREE
STEP MOLE PROBLEM ONLY USE THE MOLES OF THE LIMITING REACTANT.
Example:
2Mg(s) + O2(g)  2MgO(s)

What is the limiting reactant given 0.30 moles of Mg and 0.20 moles oxygen?

What is the limiting reactant given 0.20 moles Mg and 4.48 L oxygen at STP?

How many moles of MgO will form from 4 moles Mg and 10 moles Oxygen?
33
Limiting Reactant Worksheet
1.
Methanol, CH3OH, an excellent fuel, can be made by the reaction of carbon monoxide and
hydrogen.
CO(g) + 2H2(g)  CH3OH(l)
Suppose 356 grams of CO are mixed with 65.0 g of H2.
Which is the limiting reactant? What is the maximum mass of methanol that can be formed? What mass
of the excess reactant remains after the limiting reactant has been consumed?
2.
You have 20.0 grams of elemental sulfur, S8, and 160.0 grams of O2. Which is the limiting
reactant in the combustion of S8 in oxygen to give SO2 gas? What amount of excess reactant (in moles) is
left after complete reaction? What mass of SO2 in grams is formed in the complete reaction?
S8(s) + 8O2(g)  8SO2(g)
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3.
Aluminum chloride, AlCl3, is an expensive reagent used n many industrial processes. It is made
by treating scrap aluminum with chlorine according to the following balanced equation:
2Al(s) + 3Cl2(g)  2AlCl3(s)
Which reactant is limiting if 2.70 grams of Al and 4.05 grams of Cl2 are mixed?
What mass of AlCl3 can be produced?
What mass of the excess reactant remains when the reaction is completed?
4. Disulfur dichloride, S2Cl2, is used to vulcanised rubber. It can be made by
sulfur with gaseous chlorine.
S8(g)
treating molten
+ 4Cl2(g)  4S2Cl2(g)
Starting with a mixture of 32.0 grams of sulfur and 71.0 grams of Cl2, which is the limiting reactant?
What mass of S2Cl2 (in grams) can be produced? What mass of the excess reactant remains when the
limiting reactant is consumed?
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5. If 15.6 grams of copper (II) chloride, CuCl2 react with 20.2 grams of sodium nitrate, NaNO3, how
many grams of sodium chloride, NaCl, can be formed?
CuCl2 + 2NaNO3  Cu(NO3)2 + 2NaCl
6. If 10.4 grams of hydrogen, H2, and 9.14 grams of oxygen, O2, are placed together in a container
and allowed to react according to the equation how many grams of water, H2O, would be
produced by the reaction?
2H2 + O2  2H2O
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The Theoretical Yield of a Chemical Reaction
The theoretical yield is the maximum amount of a product that can form in a chemical reaction. The
theoretical yield is calculated by assuming that all the limiting reagent has reacted to form the product.
During an experiment, why might we discover that this assumption is not true?
(1) Less than perfect collection techniques
(2) Competing reactions e.g. Mg + O2 and Mg + N2
(3) Experimental design and technique
(4) Impure reagents
The percent yield is the amount of a product that is actually obtained from a chemical reaction. For the
reasons that we have given actual yield is almost always less than the theoretical yield.
Theoretical yield is a calculated quantity and the actual yield is an experimentally determined quantity.
The percent yield is the actual yield expressed as a percentage of the theoretical yield by the equation:
% 𝑦𝑖𝑒𝑙𝑑 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑥 100%
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
If impure reactants are used, the percentage purity of the produce may be compromised. The percentage
purity defines what proportion, by mass, of a sample is composed of a specific compound.
e.g. If gold is said to be 98% pure, a 100g sample would contain: 98 g
Example: Suppose 7.00 g of AgNO3 is added to a solution which contains an excess of dissolved KBr. If
7.32 g of AgBr is obtained, what is the percent yield?
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐴𝑔𝑁𝑂3 =
7.00 𝑔 𝐴𝑔𝑁𝑂3
𝑔
169.9
𝐴𝑔𝑁𝑂3
𝑚𝑜𝑙
= 0.04120 𝑚𝑜𝑙. 𝐴𝑔𝑁𝑂3
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙. 𝐴𝑔𝐵𝑟 = 0.04120 𝑥
1 𝑚𝑜𝑙 𝐴𝑔𝐵𝑟
1 𝑚𝑜𝑙 𝐴𝑔𝑁𝑂3
𝑚𝑎𝑠𝑠 𝐴𝑔𝐵𝑟 = 0.04120 𝑚𝑜𝑙. 𝑥 187.8
= 7.74 𝑔
37
𝑔
𝑚𝑜𝑙
% 𝑦𝑖𝑒𝑙𝑑 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
𝑥 100%
𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑦𝑖𝑒𝑙𝑑
7.32 𝑔
𝑥 100% = 94.6 %
7.74 𝑔
Theoretical and Percent Yield Worksheet
1. Given the following equation:
_____ K2PtCl4 + _____ NH3  _____ Pt(NH3)2Cl2 + _____ KCl
a) Balance the equation.
b) Determine the theoretical yield of KCl if you start with 34.5 grams of NH3 and excess
K2PtCl4. (3 step mole problem to calculate the mass of KCl)
c) If you start with 34.5 g of NH3, and you isolate 76.4 g of Pt(NH3)2Cl2, what is the
percent yield? (3-step mole problem to calculate the theoretical yield and then %
yield calculation)
38
2. CaCO3
→
CaO
+
CO2
Calculate the theoretical yield of Calcium Oxide if 24.8 grams of Calcium Carbonate
decomposes. Calculate the percent yield if you got 13.1 grams in the laboratory.
3. In the reaction between excess K(s) and 4.28 g of O2(g), potassium oxide is formed .
What mass would you expect to form (theoretical yield)? If 17.36 g of K2O is actually
produced, what is the percent yield?
4. Determine the mass of carbon dioxide one could expect to form (and the percent yield)
for the reaction between excess CH4 and 11.6 g of O2 if 5.38 g of carbon dioxide gas is
produced along with some water vapor.
39
5. Determine the mass of water vapor you would expect to form (and the percent yield) in
the reaction between 15.8 g of NH3 and excess oxygen to produce water and nitric oxide
(NO). The mass of water actually formed is 21.8 g.
6.
Al
+
CuSO4
→
Cu
+
Al2(SO4)3
What is the percent yield if you experimentally produce 3.65 grams of copper when 1.87 grams
of Aluminum reacts with 9.65 grams of Copper (II) Sulfate? The equation is unbalanced.
40
Research Chemistry: Percent Yield worksheet 2
1. “Slaked lime,” Ca(OH)2, is produced when water reacts with “quick lime,” CaO. If you start
with 2400. g of quick lime, add excess water, and produce 2060. g of slaked lime, what is
the percent yield of the reaction?
CaO
+ H2O  Ca(OH) 2
2. Some underwater welding is done via the thermite reaction, in which rust (Fe2O3) reacts with
aluminum to produce iron and aluminum oxide (Al2O3). In one such reaction, 258 g of
aluminum and excess rust produced 464 g of iron. What was the percent yield of the
reaction
Fe2O3
+ 2Al
 Al2O3 + 2Fe
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3. Use the balanced equation to find out how many liters of sulfur dioxide are actually produced
at STP if 1.50 x 1027 molecules of zinc sulfide are reacted with excess oxygen and the
percent yield is 75.0%.
2 ZnS(s) + 3 O2(g)  2 ZnO(s) + 2 SO2(g)
4. The Haber process is the conversion of nitrogen and hydrogen at high pressure into ammonia,
as follows:
N2(g) + 3 H2(g)  2 NH3(g)
If you must produce 700. g of ammonia, what mass of nitrogen should you use in the
reaction, assuming that the percent yield of this reaction is 70.0%?
Answers:
L SO2
1. 65%
4. 824 g N2
2. 87%
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3. 4.19 x 104