Chemistry – Unit 5 Review
1. Definitions
a. mole- the number equal to the number of carbon atoms in exactly 12 g of pure C12; Avogadro’s number. One mole represents 6.022 x 1023 representative units.
b. molar mass- the mass in grams of 1 mole of a substance
c. Avogadro’s number- the number of atoms in exactly 12 grams of pure C-12,
equal to 6.022 x 1023.
d. empirical formula - the lowest whole # ratio of the elements in a compound.
e. molecular formula- the exact formula of a molecule; giving the types of atoms
and the # of each type.
2. Find the molar mass of the following:
a. KNO3 = 101.1 g/mol
d. oxygen gas= O2; 32.0 g/mol
b. (NH4)2CO3= 96.0 g/mole
e. calcium nitrate = Ca(NO3)2; 164.1 g/mol
c. Ag2CrO4 = 331.8 g/mol
f. lead(II) nitrate = Pb(NO3)2; 331.2 g/mol
3. Consider the masses of various hardware below.
Type
Mass (g)
Relative mass
Washer
1.74
1.00
Hex nut
3.16
1.82
Anchor
5.22
3.00
4.39
Bolt
7.64
a. Do the calculations necessary to complete the table.
Divide all masses by the smallest mass.
1.74/1.74 = 1
3.16/1.74 = 1.82
3 x 1.74 = 5.22
7.64/1.74 = 4.39
b. Explain the connection between these calculations and the atomic masses in the
Periodic Table.
Originally, the atomic masses were relative to the lightest element, hydrogen,
which was assigned a molar mass of 1.0 g. Modern masses are based on assigning
C-12 a molar mass of exactly 12.000 g/mol.
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4. Convert from g  moles or from moles  g. Show units.
a. 12.0 g Fe  1 mole Fe = 0.215 moles
55.8 g
b. 25.0 g of chlorine gas
 1 mol Cl2 = 0.352 moles
71.0 g
c. 0.476 g of (NH4)2SO4  1 mole (NH4)2SO4 = 3.60 x 10-3 moles
132.1 g
d. 0.15 moles NaNO3 
85.0 g =
12.8 g
1 mol NaNO3
e. 0.0280 moles NO2 
46.0 g
= 1.29 g
1 mole NO2
f. 0.64 moles aluminum chloride 
5.
133.5 g AlCl3
1 mol AlCl3
= 85.4 g
Use Avogadro’s number to do the following. Show work, use labels.
a. How many atoms are there in 0.00150 moles Zn?
1.50 x 10-3 mol Zn  6.02 x 1023 atoms Zn
1 mol Zn
= 9.03 x 1020 atoms
b. If you had 2.50 moles of oxygen gas, what mass of the gas would be in the
sample?
2.50 mol O2  32.0 g = 80.0 g O2
1 mol
c. A 4.07 g sample of NaI contains how many atoms of Na?
4.07 g NaI 
1 mol  6.02  1023 form. units
 1 atom Na =
1.63 x 1022
149.9 g
1 mol
1 form. unit NaI atoms Na
d. How many atoms of chlorine are there in 16.5 g of iron (III) chloride?
16.5 g FeCl3  1 mol  6.02 x 1023 form. units  3 atoms Cl =
1.84 x 1023
162.3 g
1 mol
1 form.unit FeCl3
atoms Cl
*e. What is the mass of 100 million atoms of gold? Could you mass this on a balance?
1.00  108 atoms Au  1 mol

197.0 g
= 3.27 x 10-14 g Au
23
6.02 x 10 atoms
1 mol
No, this mass is much too small to be detected by a balance.
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6. Calculate the empirical formula of a compound that contains 4.20 g of nitrogen and
12.0 g of oxygen.
4.20 g N  1 mol = 0.300mol N
14.0 g
0.750 mol O = 2.5 = 5 O
0.300 mol N
2N
12.0 g O  1 mol = 0.750 mol O
16.0 g
E.F. = N2O5
7. When 20.16 g of magnesium oxide reacts with carbon, carbon monoxide forms and
12.16 g of Mg metal remains. What is the empirical formula of magnesium oxide?
20.16 g
12.16 g
magnesium oxide + carbon  carbon monoxide + magnesium
12.16 Mg  1 mol = 0.5000 mol Mg
24.3 g
8.00 g O2  1 mol
16.0 g
= 0.5000 mol O
1:1 ratio therefore EF = MgO
8.
What is the molecular formula of each compound?
Empirical Formula
Actual Molar Mass of
Compound
Molecular Formula
CH = EFM = 13.0 g/mol 78 g/mole = 78 g/mol= 6
13 g/mol
C6H6
NO2 = EFM = 46 g/mol
N2O4
92 g/mole= 92 g/mol = 2
46 g/mol
9.
A compound is composed of 7.20 g of carbon, 1.20 g of hydrogen and 9.60 g of
oxygen. The molar mass of the compound is 180 g/mole. Determine the empirical and
molecular formulas of this compound.
7.20 g C  1 mol
12.0 g
= 0.600 mol C
1.20 g H  1 mol
1.0 g
= 1.200 mol H
Dividing each of these values by 0.600, one
obtains a 1 : 2 : 1 ratio. The ef is CH2O. This
has a mass of 30 g/mol.
Since the molar mass is 6 times as large,
the molecular formula must be C6H12O6
9.60 g O  1 mol
= 0.600 mol O
16.0 g
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10. What is the % by mass of oxygen in water?
16.0 g O
= 88.9% oxygen
18.0 g H2O
11. A compound of iron and oxygen is found to contain 28 g of Fe and 8.0 g of O. What
is the % by mass of each element in the compound?
28 g Fe = 77.8% iron
36 g
8.0 g O = 22.2% oxygen
36 g
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