Practice Exam—Stoichiometry
Show all steps for complete credit. Report answers to three digits (beyond leading zeroes). 1. Which of the following measurements represents the most atoms? a. 0.500 mol H2
b.
0.0800 mol B12
c.
0.900 mol C
d. 0.300 mol N4
e. 0.150 mol S6
in each example, the number of atoms per sample is equal to the number of atoms per molecule multiplied by the number of molecules in the sample. For simplicity (to avoid unnecessary use of Avogadro’s Number), we can multiply moles of atoms per mole of molecules by the number of moles:
𝑚𝑜𝑙 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 ⋅
𝑚𝑜𝑙 𝑎𝑡𝑜𝑚𝑠 𝑚𝑜𝑙 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒
The most atoms are therefore represented by 0.300 mol N4 : (0.300 mol molecules) ∙ (4 mol atoms / mole molecules) = 1.2 mol atoms 2. Rank the following measurements in order of increasing mass:
Ne: 1.5 mol ∙ 20.18 g / mol = 30.27 g Ne
1.5 moles of Ne, 0.30 moles of Ag, 0.20 moles of Pb, and 0.20 moles of U.
Ag: 0.30 mol ∙ 107.47 g / mol = 32.24 g Ag Pb: 0.20 mol ∙ 207.2 g / mol = 41.44 g Pb U: 0.20 mol ∙ 238.03 g / mol = 47.61 g U 1.5 mol Ne < 0.30 mol Ag < 0.20 mol Pb < 0.20 mol U 3. To three digits beyond “leading zeroes”, what is the mass, in grams, of 0.800 mol of ammonium chloride, NH4Cl?
𝟎. 𝟖𝟎𝟎 𝒎𝒐𝒍 𝑵𝑯𝟒 𝑪𝒍 ⋅
𝟓𝟑.𝟒𝟗 𝒈 𝑵𝑯𝟒 𝑪𝒍 𝟏 𝒎𝒐𝒍 𝑵𝑯𝟒 𝑪𝒍
= 42.8 g NH4Cl
4. To three digits beyond “leading zeroes”, how many moles are there in 1.50 g of propanone, (CH3)2CO?
𝟏. 𝟓𝟎 𝒈 (𝑪𝑯𝟑 )𝟐 𝑪𝑶 ⋅
𝟏 𝒎𝒐𝒍 (𝑪𝑯𝟑 )𝟐 𝑪𝑶 = 0.0258 g (CH3)2CO 𝟓𝟖.𝟎𝟖 𝒈 (𝑪𝑯𝟑 )𝟐 𝑪𝑶
5. Which one of the following cannot be the empirical formula of a chemical compound? a. C5H5S2
b. C10H12S
c.
C10H13S
d. C10H15S2
e. C10H15S5
An empirical formula represents the smallest whole-number ratio of atoms in a compound. The subscripts of C10H15S5 have a common whole-number divisor—5—and C10H15S5 cannot be the simplest whole-number ratio of elements. 6. Which cannot be a molecular formula of the compound that has the empirical formula C4O3Br? a. C4O3Br
b. C6O5Br3
c.
C8O6Br2
d. C16O12Br4
e. C40O30Br10
A molecular formula must be a whole-number multiple (1x, 2x, 3x, etc.) of the empirical formula. All of the listed options, apart from C6O5Br3, are whole-number multiples of C6O5Br.
7. A compound of chlorine and fluorine is used in some rocket fuels. A sample of the compound consists of 0.497 g chlorine and 0.801 g fluorine. What is the empirical formula of the compound?
𝟎. 𝟒𝟗𝟕 𝒈 𝑪𝒍 ⋅ 𝟎. 𝟖𝟎𝟏 𝒈 𝑭 ⋅
𝟏 𝒎𝒐𝒍 𝑪𝒍 𝟑𝟓.𝟒𝟓 𝒈 𝑪𝒍 𝟏 𝒎𝒐𝒍 𝑭 𝟏𝟗.𝟎𝟎 𝒈 𝑭
= 0.01402 mol Cl = 0.04216 mol F
Dividing each by the smaller, we obtain 0.01402 mol Cl 0.01402
= 1
0.04216 mol F 0.01402
= 3.01 → 3
The empirical formula is ClF3 8. A compound used as an anti-seizure medication has a molecular weight of 132 g/mol. A 0.750 g sample of the compound contains 0.409 g of carbon, 0.0686 g of hydrogen, and 0.272 g of oxygen. What is the molecular formula of the compound?
𝟎. 𝟒𝟎𝟗 𝒈 𝑪 ⋅ 𝟎. 𝟎𝟔𝟖𝟔 𝒈 𝑯 ⋅ 𝟎. 𝟐𝟕𝟐 𝒈 𝑶 ⋅
𝟏 𝒎𝒐𝒍 𝑪 𝟏𝟐.𝟎𝟏 𝒈 𝑪 𝟏 𝒎𝒐𝒍 𝑯 𝟏.𝟎𝟎𝟖 𝒈 𝑯 𝟏 𝒎𝒐𝒍 𝑶 𝟏𝟔.𝟎𝟎 𝒈 𝑶
= 0.03405 mol C = 0.06806 mol H = 0.01700 mol O
Dividing each by the smallest, we obtain 0.03405 mol C 0.01700
= 2
0.06806 mol H 0.01700
= 4
0.01700 mol O 0.01700
= 1
The empirical formula is C2H4O. As the molecular formula must be a whole-number multiple of the empirical formula, we divide the given molecular mass by the empirical formula mass (2 ∙ 12.01 + 4 ∙ 1.008 + 1 ∙ 16 = 44.06 g mol–1): 132 g mol–1 = 3. Each subscript of the empirical formula is to be multiplied by 3, and 44.06 g mol–1 the molecular formula is C6H12O3. 9. Determine the percent composition of Ge3N4 [germanium(IV) nitride].
As the percentage of each element in a compound is the same for any amount of the compound, we can assume that we have one mole of the compound, which allows one to determine the total mass along with the mass of each element toward that total: Ge: 3 ∙ 72.64 = 217.92 + N: 4 ∙ 14.01 = 56.04 273.96
Ge: (217.92 / 273.96) ∙ 100 = 79.5 % N: (56.04 / 273.96) ∙ 100 = 20.5%
10. NaO2 is an ionic compound. The name of NaO2—because the other options cannot be correct—must be a. b. c. d. e.
sodium sodium sodium sodium sodium
ionic compounds are not named using numerical prefixes there is no “dioxygen” ion sodium oxide would have the formula Na2O names of ions, not elements, are used to name ionic compounds must be correct, by elimination. [The uncommon superoxide ion is O2–.]
dioxide dioxygen oxide oxygen superoxide
11. Toluene (C6H5CH3) combusts in oxygen according to the reaction equation C6H5CH3 + 9 O2 → 7 CO2 + 4 H2O. a. If 0.600 mol of oxygen are allowed to react completely with an excess of toluene, how many moles of carbon dioxide would be produced?
𝟎. 𝟔𝟎𝟎 𝒎𝒐𝒍 𝑶𝟐 ⋅
𝟕 𝒎𝒐𝒍 𝑪𝑶𝟐 = 0.467 mol CO2. Coefficients of a balanced reaction equation indicate 𝟗 𝒎𝒐𝒍 𝑶𝟐
the mole ratios in which substances participate.
b. What is the maximum quantity, in moles, of H2O that could be produced from the reaction of 2.50 moles of toluene with 15.0 moles of oxygen?
𝟐. 𝟓𝟎 𝒎𝒐𝒍 𝑪𝟔 𝑯𝟓 𝑪𝑯𝟑 ∙
𝟏𝟓. 𝟎 𝒎𝒐𝒍 𝑶𝟐 ⋅
𝟒 𝒎𝒐𝒍 𝑯𝟐 𝑶 = 10.0 mol H2O if all of the toluene is able to react. 𝟏 𝒎𝒐𝒍 𝑪𝟔 𝑯𝟓 𝑪𝑯𝟑
𝟒 𝒎𝒐𝒍 𝑯𝟐 𝑶 = 6.67 mol H2O if all of the oxygen is able to react.. 𝟗 𝒎𝒐𝒍 𝑶𝟐
The smaller amount of product is the maximum possible, as O2 is the limiting reactant; 6.67 mol H2O is the maximum quantity that can be produced. 12. Acid rain can be neutralized by adding lime to affected lakes. Sulfuric acid in a body of water will react with added calcium hydroxide to form water and calcium sulfate. [It is necessary to write and balance the equation before any stoichiometric calculations can be performed.] H2SO4 + Ca(OH)2
→ 2 H 2O
+ CaSO4
13.
