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CHEMISTRY 111 Unit 3 Practice Problems
Empirical Formula Calculations and Stoichiometry Problems (Solutions)
1. Determine for the compound adiponitrile (C 6 N 2 H 8 ):
a) the mass containing 4.50 x 10^23 C atoms
First, calculate the number of moles required to have 4.50 x 10^23 C atoms in the sample. Be sure to use the mole ratios in the chemical formula itselfโฆi.e., there are 6 C atoms in every ONE C 6 N 2 H 8 molecule.
๐. ๐๐ ร ๐๐๐๐^ ๐ ๐๐ญ๐จ๐ฆ๐ฌ (
๐. ๐๐๐ ร ๐๐๐๐^ ๐ฆ๐จ๐ฅ๐ ๐๐๐๐๐๐
Then, convert the moles to mass as usual. ๏
b) the mass of carbon in 83.5 g
Remember that there are 6 mol of C for every 1 mole C 6 N 2 H 8 ๏ How much does one mol of C weigh? (Hintโฆperiodic table!)
c) the mass of carbon in a sample containing as many N atoms as 25.0 g of ammonia.
This problem requires you to use the same conversions as in parts a and b in a little different way. Remember that there are 6 mol of C for every 1 mole C 6 N 2 H 8 , 1 mol N for 1 mol NH 3 , and 2 mol N for every 1 mole C 6 N 2 H 8. ๏ How much does one mol of C weigh? (Hintโฆperiodic table!)
2. a) Osmium forms a number of molecular compounds with CO. One yellow compound gave the following elemental composition:
15.89% C, 21.18% O, and 62.93% Os. What is the empirical formula of this compound?
Need moles; assume 100 g sample.
C: 100 g sample x 1. 12.011gC
1 molC x 100 gsample
15.89 gC ๏ฝ
O: 21.18 g O x 1. 15.999gO
1 molO ๏ฝ
Os: 62.93 g Os x 0. 190.23gOs
1 molOs ๏ฝ
Find mole ratio:
C: 3. 99
O: 3. 99
331
32 ๏ฝ
Os: 1
- 331
Therefore, empirical formula is C 4 O 4 Os.
2. b) The molecular mass of the compound in #7 above was determined to be 907 g/mol. What is its molecular formula?
What is x in (C 4 O 4 Os)x?
907 g/mol = {4(12.011 g/mol) + 4(16.00 g/mol) + 190.23 g/mol}x
x = 3.
Therefore, the molecular formula is C 12 O 12 Os 3.
4. Using the following equation: 2 NaOH + H 2 SO 4 ๏ 2 H 2 O + Na 2 SO 4
How many grams of sodium sulfate will be formed if you start with 200.0 grams of sodium hydroxide and you have an excess of sulfuric acid?
Solve using dimensional analysis:
200.0 g NaOH (
1 mol NaOH 40.0 g NaOH
1 mol Na 2 SO 4 2 mol NaOH
142.1 g Na 2 SO 4 1 mol Na 2 SO 4
) = 355.3 g Na 2 SO 4
5. Using the following equation:
Pb(SO 4 ) 2 + 4 LiNO 3 ๏ Pb(NO 3 ) 4 + 2 Li 2 SO 4
How many grams of lithium nitrate will be needed to make 250. grams of lithium sulfate, assuming that you have an adequate amount of lead (IV) sulfate to do the reaction?
Solve using dimensional analysis:
6. The combustion of ammonia in the presence of excess oxygen yields NO2 and H2O, according to
the balanced chemical equation below:
4 NH3 (g) + 7 O2 (g) โ 4 NO2 (g) + 6 H2O (g)
How many grams of NO 2 are produced in the combustion of 43.9 g ammonia (NH 3 )?
43.9 g NH 3 (
1 mol NH 3 17.024 g NH 3
4 mol NO 2 4 mol NH 3
46 g NO 2 1 mol NO 2
) = 118.62 g NO 2
119 g NO 2
3 3
3 2 4
3 2 4
2 4 (^2 4) 1molLiNO^313 gLiNO
68.9gLiNO 2molLiSO
4molLiNO 109.9gLiSO
1molLiSO
- gLi SO ๏ด ๏ด ๏ด ๏ฝ
7. a) Write the balanced equation for the reaction of acetic acid with aluminum hydroxide to form water and aluminum acetate:
3 C 2 H 3 O 2 H + Al(OH) 3 ๏ Al(C 2 H 3 O 2 ) 3 + 3 H 2 O
b) Using the equation from part a, determine the mass of aluminum acetate that can be made if I do this reaction with 125 grams of acetic acid and 275 grams of aluminum hydroxide.
Two calculations are required. One determines the moles of aluminum acetate that can be
made with 125 grams of acetic acid and the other determines the moles of aluminum
acetate that can be made using 275 grams of aluminum hydroxide. The smaller of these
two answers is the limiting reactant.
125 g C 2 H 3 O 2 H (
1 mol C 2 H 3 O 2 H 60.05 g C 2 H 3 O 2 H
1 mol Al(C 2 H 3 O 2 ) 3 3 mol C 2 H 3 O 2 H
) = 0.694 moles Al(C 2 H 3 O 2 ) 3
275 g Al(OH) 3 (
1 mol Al(OH) 3 78.00 g Al(OH) 3
1 mol Al(C 2 H 3 O 2 ) 3 1 mol Al(OH) 3
) = 3.53 moles Al(C 2 H 3 O 2 ) 3
C 2 H 3 O 2 H is limiting
0.694 moles Al(C 2 H 3 O 2 ) 3 (
204.11 g Al(C 2 H 3 O 2 ) 3 1 mol Al(C 2 H 3 O 2 ) 3
) = 142 g Al(C 2 H 3 O 2 ) 3
c) What is the limiting reagent in part b?
Acetic acid
d) How much of the excess reagent will be left over after the reaction is complete?
142 g Al(C 2 H 3 O 2 ) 3 (
1 mol Al(C 2 H 3 O 2 ) 3 204.11 g Al(C 2 H 3 O 2 ) 3 ) (
1 mol Al(OH) 3 1 mol Al(C 2 H 3 O 2 ) 3 ) (
78.00 g Al(OH) 3 1 mol Al(OH) 3 ) = 54.3 g used
275 g Al(OH) 3 โ 54.3 g = 221 g Al(OH) 3 e) If the actual yield of aluminum acetate is 101.4 g, what is the percentage yield?
101.4 / 142 x 100% = 71.4%
