Sample Questions – Chapter 7
CP Chemistry
1. Calculate the percent composition of iron (II) nitrate.
Fe(NO3)2
Fe: 1 x 55.85 = 55.85 g
N: 2 x 14.01 = 28.02 g
O: 6 x 16.00 = 96.00 g
% Fe = 55.85/179.87 *100 = 31.05%
% N = 28.02/179.87*100 = 15.58%
% O = 96/179.87*100 = 53.37%
Molar mass = 179.87 g/mol
2. A compound’s empirical formula is NO2. If the formula mass is 138 g/mol, what is
the molecular formula?
NO2 – molar mass = (14.01) + (2*16.00) = 46.01 g/mol
138 g/mol ÷ 46.01 g/mol = 3, so molecular formula is N3O6
3. You have 9.45 g of the compound dinitrogen pentoxide. How many molecules are
contained in this mass?
9.45 g ÷ 108.02 g/mol = 0.0874837 mol * 6.02E23 = 5.27 * 1022 molecules N2O5
4. If you have 3.5 x 1024 formula units of CaI2, how many moles of the compound do
you have?
3.5 * 1024 formula units ÷ 6.02E23 = 5.8 moles
5. Find the empirical formula of a compound that is 41.67% magnesium, 54.87%
oxygen, and 3.46% hydrogen?
41.67/24.31 = 1.714 mol Mg
54.87/16.00 = 3.429 mol O
3.46/1.01 = 3.425 mol H
1.714/1.714 = 1
3.429/1.714 = 2
3.425/1.714 = 2
MgO2H2
6. What is the molar mass (in grams) of Li3PO4?
Li: 3 * 6.94 = 20.82
P: 1 * 30.97 = 30.97
O: 4 * 16.00 = 64.00
Molar mass = 115.79 g/mol
7. Phosgene, a gas used in World War I, consists of 12.41% C, 16.17% O, and 71.69%
Cl.
a. What is the empirical formula for phosgene?
12.41/12.01 = 1.033 mol C
16.17/16.00 = 1.01 mol O
71.69/35.45 = 2.022 mol Cl
1.033/1.01 = 1
1.01/1.01 = 1
2.022/1.01 = 2
COCl2
b. If the molar mass is 99 g/mol, then what is the molecular formula?
Molar mass of COCl2 = (12.01) + (16.00) + (2*35.45) = 98.91 g/mol
99 g/mol ÷ 98.91 g/mol = 1
Molecular formula = COCl2
8. Calculate the percent composition of each element in AgNO3.
Ag: 1 * 107.87 = 107.87
N: 1 * 14.01 = 14.01
O: 3 * 16.00 = 48.00
% Ag = 107.87/169.88 * 100 = 63.49 %
% N = 14.01/169.88 * 100 = 8.25 %
% O = 48.00/169.88 * 100 = 28.25 %
Molar mass = 169.88 g/mol
9. Based on your answer in # 8, how many grams of silver are in a sample of AgNO3
that has a mass of 34.75 g?
34.75 g * 0.6349 = 22.06 g
10. How many moles are in 364 g of thallium? How many atoms?
364 g Tl/204.4 = 1.78 mol Tl
1.78 mol Tl * 6.02E23 = 1.07 x 1024 atoms Tl