Unit 1: Back to the Basics

Section 5: The Mole And Its Uses

Introduction

Atomic Weight

The Mole

Molecular Mass

Percent Composition

Introduction

The previous unit, was about naming compounds, but this unit takes that information one step further. In this unit the objective is to not only understand how things react, but how much of these things will react. This information helps chemists work backwards to figure out how much of the compounds that they started with and/or how much of the substance was used up.

Atomic Weight

Atomic Weight is defined as the mass of one atom of an element. The first element which was assigned a mass was the carbon 12 atom, and it was said to be exactly 12 amu (atomic mass units). Using the carbon 12 atom as a basis for all the other atoms, proportions were set up and atomic weights were assigned to each of the other atoms.

Atomic weights can be looked up on the periodic table and they are the larger numbers in each box. It is important to note that the atomic weights are averages of all the isotopes of that particular element. For example, Carbon exists in the form of a carbon 12 atom, carbon 13 atom, and carbon 14 atom, and when the atomic weights were calculated, an average was taken. Below is an example in which the atomic weight of carbon is solved for.

98.89% of all carbon atoms are carbon 12

1.11% of all carbon atoms are carbon 13

The percentage of carbon 14 atoms present in all carbon atoms is so small that it plays no part in this calculation.

.9889 (12 amu) + .0111 (13 amu) = 12.01

This is how the atomic weight is calculated.

The Mole

The mole is defined as the number of carbon 12 atoms in 12 grams of carbon 12. The number of atoms in 12 grams of carbon 12 atoms corresponds to Avogadro's Number, which is 6.02 * 10²³. This may seem confusing, but just understand that:

• 1 mole (of any substance) = 6.02*10²³ atoms, molecules, or ions (depending on how that substance exists at that time.) (For example, 1 mole of Na⁺ ions has 6.02*10²³ Na⁺ ions in it.)
• 1 mole (of any substance) = the total of the elements atomic masses expresses in grams. (For example, 1 mole of Na weighs 23 grams.)

The mole plays many valuable roles in Chemistry because it allows Chemists to weigh substances and tell how many particals are in that substance. For instance, if there are 12 grams of Carbon 12 then there are also 6.02 * 10²³ atoms of carbon 12, but this ratio also applies to all elements. For instance, if a student has 6.941 grams of lithium, then that same student has 6.02 * 10²³ atoms of lithium. Because the mole is used so often in Chemistry, an abbreviation has been created as mol.

TRY IT YOURSELF: Examples and Practice Problems

Molecular Mass (Molecular Weight)

If someone asked right now how much one mole of Helium weighs, the answer would be 4.003 grams, but if someone asked how much one mole of Al₂S₃ (Aluminum Sulfide) weighs, uh??? To do this there is a basic assumption that has to be put into place. This assumption is that in one mole of Al₂S₃, there are two moles of Al and three moles of S.

This assumption might seem strange, but look at it like this. If Bob has 3 pens, and he proceedes to pull off the caps, how many caps does he have? Of course he has 3! Well, how many pens (without the caps) does he have? Of course he has 3! The same principle applies to molecules and compounds. If Bob has 1 mole of Al₂S₃, and he decides to break it up into Al and S, then he would have 2 moles of Al and 3 moles of S.

So using this assumption, there are 2 moles of Al and 3 moles of S, so take 26.98 grams (which is the weight of 1 mole of Al) and multiply it by 2 (because there are two moles of Al in Al₂S₃). Then take 32.06 grams (which is the weight of 1 mole of S) and multiply it by 3 (because there are three moles of S in Al₂S₃). Then add the two amounts up.

Al⁺³ = 2 mol Al⁺³ = 2 (26.98) = 53.96 g

S^-2 = 3 mol S^-2 = 3 (32.06) = 96.18 g

Total - 150.14 g

TRY IT YOURSELF: Examples and Practice Problems

Percent Composition

Percent composition is defined as the mass of an element in a compound divided by the total weight of the compound. So if Bob wanted to find the percent composition of Aluminum and Sulfur in Aluminum Sulfide, he would:

% Al = (g of Al/g of Al2S3)

% S = (g of S/g of Al2S3)

% Al = (53.96 g / 150.14 g) * 100 = 35.94% Al⁺³

% S = (96.18 g / 150.14 g) * 100 = 64.06% S^-2

So, Aluminum is 35.94 percent by mass of Aluminum and Sulfur is 64.06 percent by mass.

TRY IT YOURSELF: Examples and Practice Problems