Molar Mass and Mole Conversions
Molar Mass:
Molar mass is the mass of one mole of a substance. It is usually expressed in grams and is very easy to
calculate. For ALL molar mass calculations in this class I want you to round the atomic masses from the
periodic table to the nearest tenth.
1.
The molar mass of any element is simply the atomic mass of that element expressed in grams.
 the molar mass of Na is 23.0 grams (22.989770 rounds up to 23.0)
 the molar mass of Cl is 35.5 grams (35.453 rounds up to 35.5)
 the molar mass of B is 10.8 grams (10.811 rounds down to 10.8)
NOTE: Sometimes you see the formula of an element that shows it is a diatomic molecule. There IS a
difference between the molar mass of H and the molar mass of H2
 the molar mass of H is 1.0 g The molar mass of molecular H2 will be covered in #2
below (see the last example in bold text).
2.
The molar mass of any compound or molecular element is calculated by adding the atomic masses of
each atom in the formula.
 In H2O there are two hydrogen atoms and one oxygen atom
o 2 hydrogens:
2 x 1.0 g
=
2.0 grams
o 1 oxygen:
1 x 16.0 g
=
16.0 grams
o the molar mass of water is the sum: 18.0 grams

In C6H12O6 there are six carbon atoms, twelve hydrogen atoms and six oxygen atoms
o 6 carbons:
6 x 12.0 g
=
72.0 grams
o 12 hydrogens:
12 x 1.0 g
=
12.0 grams
o 6 oxygen:
6 x 16.0 g
=
96.0 grams
o the molar mass of C6H12O6 is the sum: 180.0 grams

in Mg3(PO4)2 there are three Mg, two P and six O atoms, so the molar mass would be:
o (3 x 24.3) + (2 x 31.0) + (6 x 16.0) = 230.9 grams

in Na2CrO4 there are two Na, one Cr and four O atoms, so the molar mass would be:
o (2 x 23.0) + (1 x 52.0) + (4 x 16.0) = 162.0 grams

in BaCl2 there is one Ba, and two Cl atoms, so the molar mass would be:
o (1 x 137.3) + (2 x 35.5) = 208.3 grams
o
in H2 there are two atoms of hydrogen, so the molar mass is 2 x 1.0 = 2.0 g

Mole Conversions:
Quantities of items can be converted from one unit to another as long as you have “statements of equality”
which provide information that may be used in conversion factors
Here are a few “statements of equality” that you are familiar with:
1 dollar = 4 quarters = 10 dimes = 20 nickels = 100 pennies
1 yard = 3 feet = 36 inches
So you can convert quarters to dollars, for example:
How many dollars are there in 81 quarters?
æ 1 dollar ö
81 quarters ç
÷ = 20.25 dollars
è 4 quarters ø
Or you can convert 925 inches to yards:
æ 1 yard ö
925 inches ç
÷ = 25.7 yards
è 36 inches ø
In chemistry there is a very important “statement of equality” that is used to convert one unit of measurement
of a substance into another unit of measurement:
1 mole = 6.02x1023 particles = the molar mass (in grams) of the compound = 22.4 L (for gases at STP)
Sometimes we might measure the mass of a solid or liquid substance in grams on a balance. Other times we
might measure its volume (in the case of gases). These are direct methods of measurement that can be done in
any chemistry lab. There is no way to actually measure the number of particles (atoms, molecules or subatomic
particles) because they are too small to be seen. So we have to measure the mass or volume of a substance and
then convert it to atoms or molecules if necessary.
So just as you can convert quarters to dollars or inches to yards, you can also convert between any of the units
listed above (moles, particles, grams and liters).
The molar mass (mass of one mole of a substance) varies in the same sort of way that the mass of a dozen
objects may vary. On the first page you learned how to find the molar mass of any substance.
Avogadro’s Hypothesis (page 320) states that equal volumes of different gases at the same temperature and
pressure will have the same number of particles. One application of this is that when gases are at standard
temperature and pressure (STP) one mole of any gas will have the same volume, which is 22.4 L (equal number
of particles, 1 mole, and equal volume, 22.4 liters)
So here are some sample molar conversions:
How many molecules of water are in 10 grams of H2O ?
æ 6.02x10 23 molecules H 2O ö
23
10 g H 2O ç
÷ = 3.34x10 molecules H 2 O
18
g
H
O
è
ø
2
How many grams of CO2 are in a 60 liter sample of the gas?
æ 44.0 grams CO2 ö
60 L CO2 ç
÷ =117.86 grams CO2
è 22.4 L CO2 ø
Notice that these are just conversion problems and the information in the “statement of equality” listed above
in bold text is being used in the conversion factors. As always, when you set up conversion factors plug in the
units first so that they cancel out correctly and then place the appropriate numbers where they belong.