Atomic, Molecular and Formula Units

Atomic Weight

Recall that in chapter 2 we discussed atomic weights, and found that the average mass of a given type of atom is equal to the weighted average of its isotopes (the weighting is proportional to the relative abundance of each isotope).

Example

There are three isotopes of Si, which have the following masses and abundances:

Isotope

Mass (amu)

Abundance

28Si

27.977

92.21%

29Si

28.976

4.70%

30Si

29.974

3.09%

Therefore, the atomic weight of silicon is

(27.977)(0.9221) + (28.976)(0.0470) + (29.974)(0.0309)

= 28.086 amu

It is important to note that if we were to pick out a silicon atom at random and measure its mass, it would have a mass of either 27.977 amu, 28.976 amu or 29.974 amu, not 28.086 amu. However, if you were to measure the masses of a billion silicon atoms (a quantity probably to tiny to see with the naked eye) individually and calculate the average mass you would find the average mass of a silicon atom is 28.086 amu.

Molecular Weight

The molecular weight (MW) of a compound is simply the sum of the atomic weights (AW) of all atoms in the molecule.

Example

Determine the molecular weight of ethanol, C2H6O.

MW = 2 (AW of C) + 6 (AW of H) + 1 (AW of O)

MW = 2 (12.01 amu) + 6 (1.01 amu) + 1 (16.00 amu) = 46.08 amu

Formula Weight

Unlike molecular compounds, ionic compounds do not contain discreet molecules. Therefore, we use the term formula weight instead of molecular weight when referring to ionic compounds.

A formula weight is the sum of the atomic weights of all atoms in the formula unit.

Example

Calculate the formula weight of Na2CO3

MW = 2(AW of Na) + 1(AW of C) + 3(AW of O)

MW = 2(22.99 amu) + 1(12.01 amu) + 3(16.00 amu) = 105.99 amu

Note that we can use the same principles starting from either molecular or formula weights, that I used earlier with atomic weights, to calculate numbers of moles and molecules.

Example

How many moles of Na2CO3 are contained in a 10.0 g sample? How many sodium ions are present in this sample?

10.0 g Na2CO3 (1 mol Na2CO3/105.99 g Na2CO3) = 9.43 10-2 mol Na2CO3

9.43 10-2 mol Na2CO3 (2 mol Na ions/1 mol Na2CO3) (6.02 1023 ions/1 mol Na+ ions) = 1.14 1023 Na+ ions