Capstone 4 Review
Moles
1. 1 mole = 6.02 x 1023 molecules/particles/things
2. How many moles of H2 are needed in the reaction 2H2 + O2  2H2O
2
Molar Mass
For each of the following, calculate the molar mass.
1. BF3
B: 1 x 11 = 11
F: 3 x 19 = +57
68 g/mol
2. CCl2F2
C: 1 x 12 = 12
Cl: 2 x 35 = 70
F: 2 x 19 = +38
120 g/mol
3. Ca(C2H3O2)2
Ca: 1 x 40 = 40
C: 4 x 12 = 48
H: 6 x 1 = 6
O: 4 x 16 = +64
158 g/mol
Molar Conversions
1. What is the mass of 1 mole of Barium acetate, Ba(C2H3O2)2?
255 g/mol
2. What is the molar mass (g/mol) of cyclohexanol, C6H11OH?
100 g/mol
3. How many moles are in 2.35 g of H2O?
2.35 g H2O x 1 mol H2O = 0.13 mol H2O
18 g H2O
4. If we have 0.072 g of FeCl3 then how many moles are there?
0.072 g FeCl3 x 1 mol FeCl3 = 0.00045 mol FeCl3
161 g FeCl3
Percent Composition
Molar Mass:
1. CuBr2
Cu: 1 x 64 = 64
Total molar mass:
Br: 2 x 80 = 160
Total:
224
Molar mass:
Fe: 1 x 56 = 56
Cl: 3 x 35 = 105
Total:
161 g/mol
Mass due to Cu:
64/224 x 100 = 28.57%
Cu: 28.57%
Mass due to Br:
160/224 x 100 = 71.43%
Br: 71.43%
Want to check your work? Add the percentages together – they should be close to 100%
2. NaOH (solve like #1)
Na: 57.5%
O:40%
H: 2.5%
H:11.76%
S: 47.06%
3. (NH4)2S (solve like #1)
N: 41.18 %
Empirical Formulas
1. Dr. Barry Um has a sample of a compound, which weighs 200 grams and contains only carbon, hydrogen,
oxygen and nitrogen. By analysis, he finds that it contains 97.56 grams of carbon, 4.878 g of hydrogen, 52.03 g
of oxygen and 45.53 g of nitrogen. Find its empirical formula.
Convert grams to moles for each element in the compound:
C: 97.56 g x 1 mol = 8.13 mol = 2.500096 ≈ 3
12 g 3.251875 mol
O: 52.03 g x 1 mol = 3.251875 mol = 1
16 g
3.251875 mol
H: 4.878 x 1 mol = 4.878 mol = 4.878 = 1.5 ≈ 2
1g
3.251875 mol
N: 45.53 g x 1 mol = 3.2521428 mol = 1
14 g
3.251875 mol
Find the smallest number of moles:
Divide all molar amounts by the smallest number of moles:
Round:
Use the rounded numbers as coefficients: C3 H2 O1 N1
2. 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?
C: 0.62069 g x 1 mol = 0.05172 mol = 3.0
12 g
0.01724 mol
H: 0.013448 g x 1 mol = 0.103448 mol = 6.0
1g
0.01724 mol
O: 0.275862 g x 1 mol = 0.01724 mol = 1
16 g
0.01724 mol
Formula = C3H6O
3. 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.
100% - 5.98823% = 94.01177%
94.01177% ÷ 2 = 47.01% (for Carbon and Oxygen)
Use g to calculate #s; e.g. C: 47.01 g x 1 mol = 3.92 mol = 1.33 ≈ 1
12 g
2.94 mol (from Oxygen)
FORMULA: CH2O
Molecular Formulas – solve for the molecular formula and/or empirical formula as needed.
1. A compound with an empirical formula of CFBrO and a molecular mass of 254.7 grams per mole.
Molar mass of the empirical formula:
C: 1 x 12 = 12
F: 1 x 19 = 19
Br: 1 x 80 = 80
O: 1 x 16 = 16
Total:
127 g/mol
Molecular mass/molar mass of the empirical formula: 254.7/127 = 2.0055 ≈ 2
Round: 2
Multiply by the subscripts: (example: C subscript is 1, so multiply x 2)
Molecular Formula: C2F2Br2O2
2. A compound with an empirical formula of C2H8N and a molar mass of 46 grams per mole.
C: 2 x 12 = 24
H: 8 x 1 = 8
N: 1 x 14 = 14
Total:
46
Molecular Formula = C2H8N
46/46 = 1
3. A well-known reagent in analytical chemistry, dimethylglyoxime, has the empirical formula C2H4NO. If its molar
mass is 116.1 g/mol, what is the molecular formula of the compound?
C: 2 x 12 = 24
Molecular Formula = C4H8N2O2
H: 4 x 1 = 4
N: 1 x 14 = 14
O: 1 x 16 = 16
Total:
58 g/mol
116.1/58 = 2.0017241 ≈ 2
4. What’s the empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and 29.0% oxygen?
C: 65.5 g x 1 mol = 5.46 = 3.011 ≈ 3
12 g 1.81
H: 5.5 g x 1 mol = 5.5 = 3.03 ≈ 3
1g
1.81
O: 29.0 g x 1 mol = 1.8125 = 1
16 g
1.8125
EMPIRICAL FORMULA: C3H3O
Stoichiometry
1. Mole to Mole You have 30 moles of K2CO3. Given the following equation, how many moles of CO2 could be
produced?
4 FeCr2O7 + 8 K2CO3 + 1 O2  2 Fe2O3 + 8 K2CrO4 + 8 CO2
30 mol K2CO3 x 8 mol CO2 = 30 moles of CO2
1
8 mol K2CO3
2. Mole to Gram You have 7.2 moles of Al. Given the following reaction, how many grams of H2 could be
produced?
6 NaOH + 2 Al  2 Na3AlO3 + 3 H2
7.2 mol Al x 3 mol H2 x 2 g H2 = 21.6 g H2
1
2 mol Al 1 mol H2
3. Gram to Mole You have 250g of FeCr2O7. Given the following equation, how many moles of Fe2O3 would be
formed?
Molar mass:
4 FeCr2O7 + 8 K2CO3 + 1 O2  2 Fe2O3 + 8 K2CrO4 + 8 CO2
250 g FeCr2O7 x 1 mol FeCr2O7 x 2 mol Fe2O3 = 0.98 mol Fe2O3
1
128 g FeCr2O7 4 mol FeCr2O7
Fe: 2 x 56 = 112
O: 3 x 165 = 16
Total:
128 g/mol
4. Gram to Gram You have 16g of NaOH. Given the following reaction, how many grams of Na3AlO3 could you
produce?
Molar masses:
6 NaOH + 2 Al  2 Na3AlO3 + 3 H2
16 g NaOH x 1 mol NaOH x 2 mol Na3AlO3 x 144 g Na3AlO3 = 19.2 g Na3AlO3
1
40 g NaOH
6 mol NaOH
1 mol Na3AlO3
Na: 1 x 23 = 23 Na: 3 x 23 = 69
O: 1 x 16 = 16 Al: 1 x 27 = 27
H: 1 x 1 = 1 O: 3 x 16 = 48
Totals:
40
144
5. Under standard conditions, determine the maximum volume of carbon dioxide, CO2, that could be produced
when 30g of butane (C4H10), completely reacts in an excess amount of oxygen.
[Hint: 1 mole of any gas under standard conditions occupies 22.4L of volume.]
2 C4H10 + 13 O2  10 H2O + 8 CO2
30 g C4H10 x 1 mol C4H10 x 8 mol CO2 x 22.4 L
= 46.34 L
1
58 g C4H10
2 mol C4H10 1 mol CO2
Molar mass:
C: 4 x 12 = 48
H: 10 x 1 = 10
Total:
58
Limiting Reactants
1. What is a limiting reactant?
The substance in a chemical reaction that is completely consumed. When the limiting reagent is gone, the
reaction will stop.
2. Consider the following synthesis reaction: 2H2 + O2  2H2O. If we had 10 moles of H2 and 9 moles of O2, what is
the maximum number of moles of H2O that could be formed?
10 mol H2 x 2 mol H2O = 10 mol H2
1
2 mol H2
Since this reaction produces the least amount, it is the maximum that
can be formed from the reactants that were given.
9 mol O2 x 2 mol H2O = 18 mol H2
1
1 mol O2
3. Consider the following single replacement reaction: 2Al + Fe2O3  Al2O3 + 2Fe. If we had 21 moles of Al and 30
moles of Fe2O3, what is the maximum number of moles of Fe that could be formed?
21 mol Al x 2 mol Fe = 21 mol Fe
1
2 mol Al
30 mol Fe2O3 x 2 mol Fe =
1
1 mol Fe2O3
Since this reaction produces the least amount, it is the maximum that
can be formed from the reactants that were given.
60 mol Fe
4. Consider the following combustion reaction: 2CH3OH + 3O2  2CO2 + 4H2O. What are the maximum number of
moles of CO2 could be formed from 12 moles of O2 and 17 moles of CH3OH?
12 mol O2 x 2 mol CO2 = 8 mol CO2
1
3 mol O2
Since this reaction produces the least amount, it is the maximum that
can be formed from the reactants that were given.
17 mol CH3OH x 2 mol CO2
= 17 mol CO2
1
2 mol CH3OH