Chem 201X In-Class Worksheet 4: Stoichiometry Problems Answer Key
Dr. Lara Baxley
1. Caporic acid, which is responsible for the smell of dirty socks, contains atoms of
C, H, and O. Combustion analysis of 0.225 g of caporic acid produced 0.512 g of CO2 and
0.209 g of H2O.
What is the empirical formula for caporic acid? [C3H6O]
calculate moles of C and H
0.512 g CO2 | mol CO2
| 1 mol C = 0.011634 mol C (keeping 2 more digits than are sig)
| 44.01 g CO2 | 1 mol CO2
0.209 g H2O | mol H2O | 2 mol H = 0.023196 mol H
| 18.02 g H2| 1 mol H2O
calculate mass of C and H
0.01164 mol C | 12.01 g C = 0.139796 g C
| mol C
0.023196 mol H | 1.008 g H = 0.0233816 g H
| mol H
use mass of caporic acid in problem and mass of C and H to calculate mass of O
0.225 g – 0.139796 g – 0.0233816 g = 0.06182 g O
calculate moles O
0.06182 g O | mol O = 0.0038638 mol O
|16.00 g O
divide each moles by smallest number to get some whole numbers
C: 0.011634 mol/0.0038638 mol = 3.01 rounds to 3
H: 0.023196 mol H/0.0038638 mol = 6.00 = 6
O: 0.0038638 mol/0.0038638 mol = 1
values are all whole numbers, so there is no need to multiply. Formula is:
C3H6O
a. Caporic acid has a molar mass of 116 g/mol. What is the molecular formula? [C6H12O2]
empirical mass = 3(12.01) + 6(1.008) + 16.00 = 58.08 g/mol
n=
116 g/mol
58.08 g/mol
= 1.997 = 2
multiply all of the subscripts in the empirical formula by 2
molecular formula is:
C6H12O2
2. How many grams of Cu(NO3)2 will be produced if 46.97 g of HNO3 reacts according to the
following reaction. [52.42 g Cu(NO3)2]
3 Cu + 8 HNO3 → 3 Cu(NO3)2 + 2 NO + 4 H2O
46.97 g HNO3 | mol HNO3 | 3 mol Cu(NO3)2 | 187.57 g Cu(NO3)2 = 52.42 g Cu(NO3)2
| 63.02 g HNO3 | 8 mol HNO3 | mol Cu(NO3)2
Chem 201X In-Class Worksheet 4: Stoichiometry Problems Answer Key
Dr. Lara Baxley
3. Bornite (Cu3FeS3) is a copper ore used in the production of copper. When heated, the
following reaction shown below occurs. If 2.50 metric tons of bornite is reacted with excess
O2 and the process has an 86.3 % yield, what mass (in kg) of copper is produced?
(metric ton = 1,000 kg) [1.20 x 103 kg]
2 Cu3FeS3(s) + 7 O2(g) → 6 Cu(s) + 2 FeO(s) + 6 SO2(g)
2.50 mtons | 1,000 kg | 103 g | mol Cu3FS3 | 6 mol Cu | 63.55 g Cu
| 1 mton | kg | 342.71 g Cu3FS3 | 2 mol Cu3FS3 | mol Cu
= 1.39 x 106 g = 1390.7 kg = theoretical yield
actual yield = 1390.7 kg · 86.3 = 1.20 x 103 kg
100
4. If 245 g of chlorine dioxide is combined with 60.2 g of water, how many grams of chloric
acid will be produced? [256 g HClO3]
6 ClO2 + 3 H2O → 5 HClO3 + HCl
245 g ClO2 | mol ClO2 | 5 mol HClO3 | 84.46 g HClO3 = 255.6548 g HClO3
|67.45 g ClO2 | 6 mol ClO2 | mol HClO3
60.2 g H2O | mol H2O | 5 mol HClO3 | 84.46 g HClO3 = 470.26 g HClO3
| 18.02 g H2O | 3 mol H2O | mol HClO3
smaller amt is all that can be produced: 256 g HClO3
5. For the above problem, after the reaction, how many grams of each reactant will be left after
the reaction is complete? [0 g ClO2 and 27.5 g H2O]
0 g ClO2 remains because it was the limiting reactant
255.6548 g HClO3 | mol HClO3 | 3 mol H2O | 18.02 g H2O = 32.727 g H2O reacted
| 84.46 g HClO3 | 5 mol HClO3 | mol H2O
60.2 g H2O initial – 32.727 g H2O reacted = 27.4727 g = 27.5 g H2O remains