Earlier, we saw how the mass spectrometer could be used to determine the relative masses of atoms. We determined, for example, that Li had an atomic mass of 7 AMU while C had an atomic mass of 12 AMU. These were considered relative weights because we really had no idea what an AMU was, except that, by definition, whatever H’s mass, it was approximately 1 AMU.

The problem:

Consider the chemical combination of carbon and lithium: xC + yLi => LiyCx.

But, what numbers are x and y? Since the mass spectrometer was not available in the 18th and 19th centuries when molecular formulas were determined leading to the discovery of combining capacity, other methods of determining “x” and “y” were necessary.

Equal numbers:

The first problem scientists needed to solve was “How can we get samples of C and Li that contained equal number of atoms?” So long as Li and C were in the ratio of 7 to 12 mass units, the samples must have equal numbers of particles. For example:

7 AMU of Li and 12 AMU of C would each contain 1 atom (ratio 7:12)

14 AMU of Li and 24 AMU of C would each contain 2 atoms (ratio still 7:12)

70 AMU of Li and 120 AMU of C would each contain 10 atoms (ratio still 7:12)

7n AMU of Li and 12n AMU of C would each contain n atoms (ratio still 7:12)

7 grams of Li and 12 grams of C would contain the same number of atoms,

although we didn’t have any idea how many!

How many?

For convenience, since we could not weigh out AMU, and since our balances measure in grams, 7 g of Li and 12 g of C were selected to contain equal numbers of particles. This was an arbitrary choice with no special meaning whatsoever, except that the gram was a convenient unit of mass and 7 and 12 were the relative masses of Li and C. How many atoms are in 7 g of Li? How many atoms are in 12 g of C? At this stage in history, no one had even the slightest clue, except that there must have been equal numbers of atoms in each sample.

The experiment:

In order to proceed further, experimental results were needed, so an experiment was done. When 7 g of Li were combined with 12 g of C in the laboratory, it was found that 10 g of lithium carbide were formed, with 9 g of C left over. No lithium was left over. What does this mean?

The mole:

If an equal number of Li and C atoms are combined, the entire number of lithium atoms react, but only 1/4th of that same number of carbon atoms combined. The words are beginning to get awkward. Instead of speaking of “that same number”, let’s use the word “mole”. The mole is that number of Li atoms in 7 g of Li. It is also that number of C atoms in 12 g of C. 1 mole of Li is 7 g of Li. 1 mole of C is 12 g of C. 1 mole of H is 1 g of H, etc.

The formula of lithium carbide:

Now, we can state that 1 mole of Li (7 g) combined with .25 moles of C (3 g) to produce 10 g of lithium carbide, with .75 moles (9 g) of carbon left over.

If 1 mole Li combines with .25 moles C, the simplest whole number ratio is: 4:1.

4 Li + 1 C => Li₄C

The mole ratio:

The molecular weight of Li₄C should be 7*4 + 12 = 40. 1 mole of Li₄C should weigh 40g. 10g of of lithium carbide would then be .25 mole.

4 Li + 1 C => Li₄C

1 mole .25 mole .25 mole

Notice that we have determined the chemical formula, the molecular weight, the balanced equation, and we have shown that the mole ratio is the same as the coefficients of the balanced equation. This reasoning was greatly simplified by introducing the concept of “mole”.

Why is a mole?

Experiments in chemistry, at least in its early days usually resulted in data in the form of masses. We determine the masses of the reactants and the masses of the products. Yet the information we were seeking from these experiments invariably involved numbers of particles. When we determine a chemical formula we discover the number of each type of atom in the molecule. The balanced equation specifies the numbers of each type of molecule taking part in the reaction. How do we get conclusions about number from data about mass? The gram is our preferred unit of mass. The mole is our preferred unit of number.

What is a mole?

It is

• “that equal number” of atoms arbitrarily chosen.
• the number of atoms in the atomic weight in g of any element.
• the number of atoms in 16 g of oxygen, in 4 grams of He, in 32 g of sulfur, etc.
• the number of molecules in the molecular weight, in grams, of any compound.
• the number of molecules in 18 g of water, in 40 g of lithium carbide.
• Much later, a mole, that equal number, was found to be 6.022 x 10²³.

Test your understanding with the following questions:

1. How many bananas are in a mole of bananas?

2. How many kisses is a half mole of kisses?

3. What is the mass of 1.0 mole of Mg atoms?

4. What is the mass of 2.0 moles of NH₃ molecules?

Answers:

1. 6.022 x 10²³ bananas

2. 3.011 x 10²³ kisses

3. 24 grams

4. 34 grams

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