Mass and Mole Calculations

The mole is a unit in chemistry that tells us the amount of a substance. It gives an indication of how many particles are present in a substance. If we know the mass of a substance, we can calculate the moles by dividing the mass by the Mr.

Moles

A ‘mole’ simply refers to the amount of a certain substance. Just like asking your butcher for a pound of bacon, one mole of an element refers to a particular amount of that element. It is equivalent to 6.02 x 10²³ atoms, which chemists like to refer to as Avagadro's number, funnily enough because a guy called Amedeo Avogadro came up with it.

One mole of any element will always weigh the same as its mass number (or RAM). This means that one mole of carbon will weigh 12 g and one mole of calcium will weigh 40 g.

The same thing is true for molecules. Let’s say we have exactly one mole of carbon dioxide gas, CO₂. We can determine how much this weighs by working out the relative formula mass (12 + (16 x 2) = 44). So one mole of CO₂ weighs 44 g.

Moles in equations

If we have a balanced symbol equation then we can work out the moles of each substance that are reacting. For example, look at the equation below where magnesium reacts with hydrochloric acid.

From the numbers written in front of each substance in our equation, we can see that every one mole of magnesium will react with two moles of hydrochloric acid to form one mole of magnesium chloride and one mole of hydrogen gas.

This doesn’t mean that we’ll always have 1 mole of magnesium in a particular reaction. It just means that a certain number of moles of magnesium needs to react with double the amount of hydrochloric acid, and will form the same number of moles of each product.

Let’s say we start with 6 grams of magnesium. We can work out how many moles of magnesium we actually have by dividing the mass by the Mr. The Mr of magnesium is 24 which means that we have 6/24 = 0.25 moles of magnesium. Since there is a 1:2 ratio of Mg : HCl, we know that 0.25 moles of magnesium will react with 0.5 moles of hydrochloric acid.

We can also work out the masses of products formed because we know the moles that will be formed and we can work out their formula masses (Mr). There is a 1:1 ratio between Mg and MgCl₂, so we know that the moles of MgCl₂ is also 0.25. The mass can be calculated by multiplying the moles with the formula mass (24 + 35.5 + 35.5 = 95). 0.25 x 95 = 23.75 g. This means that whenever 6 grams of magnesium reacts with plenty of hydrochloric acid, 23.75 g of magnesium chloride is formed.

Using moles to balance equations

You can also use the masses of the substances in a reaction to write a balanced symbol equation. To do this, you just need to convert the masses in grams to amounts in moles then converts the number of moles to simple whole number ratios. Have a look at the worked example below to see the steps involved.

Worked example

In the production of ammonia for fertiliser, a chemist adds 5.6 g of nitrogen gas (N₂) to 1.2 g of hydrogen gas (H₂) to form 6.8 g of ammonia (NH₃). Write a balanced symbol equation for this reaction.

First we use the masses to work out the number of moles of nitrogen, hydrogen and ammonia.

o Moles of nitrogen = 5.6/28 = 0.2 mol
o Moles of hydrogen = 1.2/2 = 0.6 mol
o Moles of ammonia = 6.8/17 = 0.4 mol

You can then divide these amounts by the smallest to convert it into a ratio. Dividing all the moles by 0.2 gives us a 1 : 3 : 2 ratio, which are the numbers we need to write in front of each substance in our balanced equation:

Limiting reactants

In a chemical reaction involving two reactants, it is common to use an excess of one of the reactants to ensure that all of the other reactant is used. The reactant that is completely used up is called the limiting reactant because it limits the amount of products.

You may be asked to use this idea of ‘limiting reactants’ to work out how many moles (or how much mass) of a product is formed. Have a look at the worked example below to see how you’d do this.

Worked example

7.95 grams of copper oxide reacts with 7.2 grams of carbon to form carbon dioxide and pure copper, as shown in the balanced equation below:

We can work out which reactant is limiting by using the masses to work out the number of moles we have of each reactant.

7.95 g / 79.5 = 0.1 mol of CuO
7.2 g / 12 = 0.6 mol of C

We know from our balanced equation that copper oxide reacts with carbon in a 2 : 1 ratio. This means that 0.1 moles of copper oxide will react with just 0.05 moles of carbon. We have way more than that, which means it is the copper oxide which is limiting the amount of products that will form, not the carbon.

Since copper oxide is the limiting reactant, this is the substance which will determine how much products are formed. By looking at the ratios in the balanced equation, we know that 0.1 moles of copper oxide will form 0.1 moles of copper and 0.05 moles of carbon dioxide. We can also work out the masses of the products using the mass = moles x Mr equation.

0.1 x 63.5 = 6.35 g of copper
0.05 x 44 = 2.2 g of carbon dioxide

Concentration of solutions

The concentration of a solution gives us an idea of how much solute is dissolved in a given volume of solvent. A concentrated salt solution will contain a larger mass of salt dissolved in a particular volume of water compared to a dilute salt solution. Concentration can be measured in grams per dm³ (g/dm³). This means that a concentration of 1 g/dm³ means that 1 gram of solute is dissolved in every dm³ of solvent. Use the equation concentration = mass/volume to work out the concentration of any given solution.

For example, if I dissolve 25 grams of sodium chloride into a 1000 dm³ of solution, I have made a solution with a concentration of 0.025 g/dm³.

Did you know…

If you had the same number of rice grains as the number of particles in one mole of a substance (i.e. 6 x 10 to the power of 23), all the land regions on Earth would be completely covered in rice. In fact, there would be so many rice grains that they would pile up to a depth of around 75 meters.

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