### Learning outcomes

– calculate the percentage composition of a compound– use the terms relative atomic mass, relative molecular mass and relative formula mass correctly– calculate the percentage yield of a chemical reaction– calculate the formula of a compound from the masses used in a reaction– calculate the masses of reactants or products using a balanced chemical equation– state that a mole of an element or compound contains the same number of particles and that this is the Avogadro constant– find the mass of one mole of an element or compound using the Periodic table– use the Avogadro constant and moles in chemical calculations.

### What you should already know

– How to write a chemical formula– How to balance a chemical equation– How to find the mass number and atomic number of an element from the Periodic table– How to work out a percentage

### Chemistry calculations

If I put a small piece of sodium metal in some water, a reaction occurs. The sodium fizzes and races around and there is an orange flame.

The reaction in words is:

sodium + water → sodium hydroxide + hydrogen

In symbols it is:

$$2 \mathrm{Na}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2}$$

Water and sodium hydroxide are both compounds. If you look at the Periodic table, you will see that sodium has a mass number of 23, oxygen has a mass number of 16 and hydrogen has a mass number of 1. These are called relative atomic masses; they are the ratio of the masses to an atom Carbon-12, which has mass 12. Each atom only weighs a very tiny amount; you would get around $$6 \times 10^{23}$$ atoms of hydrogen in 1g.

We can use the relative atomic mass (RAM) of an element to work out the percentage of each element in a compound by weight.

Water has a molecular mass of 16 + 1 + 1 = 18.

It is 16/18 x 100% = 89% oxygen, and 2/18 x 100% = 11% hydrogen

How many atoms are there in 2.8g of iron? There are a lot, but how many?

The number $$6 \times 10^{23}$$ is called Avogadro’s number. It is the number of atoms in 1 mole of an element. A mole is a convenient way of working without using the very big and very small numbers you would otherwise have to use. The relative atomic mass (RAM) is the number of neutrons and protons in the nucleus of the atom. You get $$6 \times 10^{23}$$ atoms in a mole of any element, whether it is a tiny atom like hydrogen with only 1 proton in it or a huge one like uranium-238 with 92 protons and 146 neutrons. Because the hydrogen atoms only have a RAM of 1 and the Uranium-238 has a RAM of 238, 1g of hydrogen and 238g of uranium both have the same number of atoms in them – $$6 \times 10^{23}$$. So we can use moles and g instead of numbers of atoms when we are doing chemical calculations.

A mole of an element weighs the RAM in g.

A mole of a compound or molecule weighs the relative formula mass in g.

A mole of an element contains $$6 \times 10^{23}$$ atoms.

I can now calculate how many atoms are in 2.8g of iron.

• Iron has RAM of 56.
• 2.8g is 2.8/56 = 0.05 moles.
• Each mole contains $$6 \times 10^{23}$$ atoms.
• So there are 0.05 x $$6 \times 10^{23}$$ = $$3 \times 10^{22}$$ atoms in 2.8g of iron.

Be careful with things which come in molecules, like hydrogen, H2.

• There are $$6 \times 10^{23}$$ atoms of hydrogen in 1 mole of hydrogen atoms which weighs 1g because the RAM of one hydrogen atom is 1.
• There are $$6 \times 10^{23}$$ molecules of hydrogen in 1 mole of hydrogen gas which weighs 2g because its relative molecular mass is 2. It is $$\mathrm{H}_{2}$$.
• Gases often come in molecules like this: $$\mathrm{O}_{2}, \mathrm{~N}_{2}, \mathrm{~F}_{2}, \mathrm{Cl}_{2}$$ and $$\mathrm{H}_{2}$$. So do the other halogens you learn about: $$\mathrm{Br}_{2}$$, which is a liquid, and $$\mathrm{I}_{2}$$, which is a solid.

We can use moles to work out how much product we will get in a chemical reaction. These can be the sort of reaction you do in chemistry lessons or industrial reactions.

• Sodium has a relative atomic mass of 23 and sodium hydroxide has a relative formula mass of 23 + 16 + 1 = 40.
• Using moles tells us that if we react 23g (1 mole) of sodium with lots of water (don’t try this!), we get 40g (1mole) of sodium hydroxide.
• If you look at the section on moles of gases, you will also see that you get 12 litres (½ a mole) of hydrogen gas.

### Exam Questions

Exam Question 1

5.6g of iron wool is reacted with excess iodine. What mass of iron III iodides is produced?

1. Write down a balanced formula equation: $$2 \mathrm{Fe}+3 \mathrm{I}_{2} \rightarrow 2 \mathrm{Fe} \mathrm{I}_{3}$$
2. Work out how many moles of iron you have: mass/RAM = 5.6/56 = 0.1 moles
3. If you have 0.1 moles of iron, the equation tells you you have the same number of moles of iron III iodide.
4. RFM iron III iodide = 56 + 127 + 127 + 127 = 437
5. mass of iron III iodide = RFM x number of moles = 437 x 0.1 = 43.7g

The masses are not always given in g; they may be kg or even Tonnes.

Exam Question 2

Aluminium is extracted from its ore, aluminium oxide, by electrolysis. What is the greatest mass of aluminium that can be extracted from 10200 kg of aluminium oxide?

1. Write the balanced equation: $$2 \mathrm{Al}_{2} \mathrm{O}_{3} \rightarrow 4 \mathrm{Al}+3 \mathrm{O}_{2}$$
2. Work out the formula mass of Aluminium oxide and how many moles you have:27 + 27 + 16 + 16 + 16 = 102. 10200/ 102 = 100 – this isn’t moles because the masses are in kg, but you can leave it like this as long as you remember. (Of course, if you are asked for moles, you need to x$$\mathrm{x} 1 \mathrm{O}^{3}$$.)
3. Now notice that you are getting twice as many moles of aluminium out as aluminium oxide going in, so you are getting 200 x $$\mathrm{x} 1 \mathrm{O}^{3}$$ moles.
4. A mole of aluminium weighs 27g, so $$\mathrm{x} 1 \mathrm{O}^{3}$$ moles weigh 27kg.
5. So, the greatest amount of aluminium that can be extracted is 27 x 200 kg = 5400kg.

You might also be asked to consider the percentage yield.

If the actual amount of aluminium produced by the process in question 2 is 3.6 Tonnes, what is the percentage yield? Give your answer to the nearest 1%.

Once you have the potential yield, this is easy. Remember that 3.6 Tonnes is 3600kg.

Sensible responses would be: the ore is impure; not all the reactants are converted to product.

You can use moles to work out the formula of a compound. You will be given a compound you have not met before. They are usually kind and give you something with only two elements in it.

### Exam Question 3

Manganese sulfide is a compound containing only manganese and sulfur.In an experiment, 4.40g of manganese was reacted with excess sulfur. The manganese sulfide produced weighed 9.52g. What is the simplest formula for manganese sulfide?

1. Look up the RAM of manganese and sulfur in the Periodic table. Manganese has a RAM of 55 and sulfur has a RAM of 32.
2. Work out how many g of sulfur has been used in the reaction. 9.52 – 4.40 = 5.12g
3. Work out how many moles of manganese and how many moles of sulfur were used.Manganese 4.4/55 = 0.08Sulfur 5.12/32 = 0.16
4. Turn 0.08:0.16 into the simplest ratio you can. It is 1:2, so you have 2 sulfurs for each manganese.
5. The formula is $$\mathrm{MnS}_{2}$$.

### Exam Question 4 – a hard one

A hydrocarbon is burned in oxygen and the amounts of carbon dioxide and water produced are weighed. 2.2g of carbon dioxide and 1.8g of water are produced. What is the simplest formula for the hydrocarbon?

1. This is not impossible to work out! Hydrocarbons only contain carbon and hydrogen. You need to find how many moles of carbon dioxide and how many moles of water you have and then you can find out how many moles of carbon and hydrogen you have.
2. 1 mole of carbon dioxide = 12 + 16 + 16 = 44 g1 mole of water = 1+ 1+16 = 18g
3. 2.2/44 = 0.05 moles of $$\mathrm{CO}_{2}$$ , 1.8/18 = 0.1 moles of $$\mathrm{H}_{2} \mathrm{O}$$
4. Each mole of $$\mathrm{CO}_{2}$$ contains 1 mole of carbon, so you have 0.05 moles of carbon.Each mole of water contains 2 moles of hydrogen, so you have 0.2 moles of hydrogen.
5. Simplify the ratio. 0.05:0.2 = 1:4.
6. The formula for the hydrocarbon is $$\mathrm{CH}_{4}$$. You might recognise this as methane.

When we are working with gases, we often prefer to use volume rather than mass.

A mole of any gas at ‘room temperature and pressure’ has a volume of 24 litres (or $$\mathrm{dm}^{3}$$). This is the same as 24,000 $$\mathrm{cm}^{3}$$.

This sounds unlikely but if you think about it, it makes sense. A balloon filled with helium floats up to the ceiling while a balloon filled with the same volume of air sits on the floor. They occupy the same volume but the air weighs more than the helium. Air is mostly nitrogen, with a relative molecular mass of 28, while helium has a relative atomic mass of 4. A mole of air weighs a bit more than 28g (because it also contains oxygen and other things), while a mole of helium only weighs 4g. It is less dense.

To see how this works, let’s look at the reaction between propane and oxygen. Propane is a hydrocarbon with the formula $$\mathrm{C}_{3} \mathrm{H}_{8}$$. Oxygen is a molecule, $$\mathrm{O}_{2}$$.The balanced equation for the reaction is $$\mathrm{C}_{3} \mathrm{H}_{8}+5 \mathrm{O}_{2} \rightarrow 3 \mathrm{CO}_{2}+4 \mathrm{H}_{2} \mathrm{O}$$

For this example, I am going to react 1.2 litres of propane with oxygen. What volume of oxygen do I need?

1. Without doing any working out, you can see you need 5 moles of oxygen for each mole of propane.
2. Since each mole of any gas occupies the same volume, I need 5 x 1.2 litres of oxygen = 6 litres.

### Questions for you to try

Assume all gases are measured at room temperature and pressure.

Question 1

a. What is the relative formula mass of calcium carbonate?

b. 20g of calcium carbonate is heated. A gas is produced.Write a balanced formula equation for this reaction.

c. The gas is collected and measured. What volume of gas is produced if all the calcium carbonate reacts?

d. Describe the test you would use for the gas produced in this reaction.

Question 2

A compound is made which contains only cobalt and bromine.1.18g of cobalt reacts with 3.2g of bromine.Find the formula for the compound produced in its simplest form.

Question 3

This is a harder question. Remember oxygen comes as O2 molecules.A compound is made which contains only boron and oxygen.4.4g of boron reacts with 7.2 dm3 of oxygen gas.Find the formula for the compound produced in its simplest form.

Question 4

The most commonly used iron ore is haematite, $$\mathrm{Fe} 2 \mathrm{O}_{3}$$.It is extracted by reducing with carbon monoxide (CO) in a blast furnace.The products of the reaction are iron and carbon dioxide.

a. Write a balanced formula equation for this reaction.

b. 80 Tonnes of haematite is reduced in a blast furnace.What is the maximum amount of iron which could be produced?

c. This process produces 42 Tonnes of iron. What is the percentage yield of iron?

d. Give a reason why the yield in part c is less than the potential yield.

Question 5

Hydrogen and chlorine are reacted together to make hydrogen chloride gas.

a. Write a balanced formula equation for this reaction.

b. 0.36 litres of hydrogen are used. How many molecules of hydrogen are in 0.36 litres?

c. What volume of chlorine is needed to completely react with 0.36 litres of hydrogen

d. How many moles of hydrogen chloride gas is produced?

e. The hydrogen chloride is reacted with magnesium hydroxide.What mass of magnesium hydroxide is needed to react with all the hydrogen chloride?

a. Calcium carbonate is $$\mathrm{Ca} \mathrm{CO}_{3}$$.

Its relative formula mass is 40 + 12 + 3x 16 = 100.

b. $$\mathrm{CaCO}_{3} \rightarrow \mathrm{CaO}+\mathrm{CO}_{2}$$

This is one of those reactions you need to learn. It is used in the manufacture of cement.

c. 20g of calcium carbonate is 20/100 = 0.2 moles.

From the equation, you can see that 0.2 moles of CO3 are produced.

1 mole occupies 24dm3, so there will be $$0.2 \times 24=4.8 \mathrm{dm}^{3}$$ of carbon dioxide produced.

d. You test for carbon dioxide by bubbling the gas through lime water.

If carbon dioxide is present, the lime water turns milky.

(You need both points to answer the question properly, so don’t miss off either of them.)

Cobalt has RAM 59, so 1.18g of cobalt = 1.18/59 = 0.02 Moles of cobalt atoms.

Bromine has RAM 80, so 3.2g bromine = 3.2/80 = 0.04 Moles of bromine atoms.

0.02:0.04 = 1:2, so the formula is $$\mathrm{CoBr}_{2}$$.

Boron has RAM 11, so 4.4g of boron = 4.4/11 = 0.4 moles of boron atoms.

Since the oxygen is given as a volume, we need 1 mole of a gas = $$24 \mathrm{dm}^{3}$$.

$$7.2 \mathrm{dm}^{3}=3.6 / 24=0.3$$ moles of oxygen $$\left(\mathbf{O}_{2}\right)$$ molecules.

Since this is a tricky one to work out, it may be a good idea to write an equation at this point.

$$0.4 \mathrm{~B}+0.3 \mathrm{O}_{2}=4 \mathrm{~B}+3 \mathrm{O}_{2}=4 \mathrm{~B}+6 \mathrm{O}$$

4:6 = 2:3

So the formula is $$\mathrm{B}_{2} \mathrm{O}_{3}$$

a. $$\mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \rightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_{2}$$

b. First find the relative formula mass of haematite.

2 x 56 + 3 x 16 = 160

80 Tonnes = $$80 \times 10^{6} \mathrm{~g}$$

So we have $$80 \times 10^{6} / 160=5 \times 10^{5}$$ moles of haematite.

For each mole of haematite, we get 2 moles of iron, so we have $$10^{6}$$ moles of iron.

Each mole is 56 g, so we have $$56 \times 10^{6} \mathrm{~g}$$ or 56 Tonnes of iron.

The haematite ore is impure/contains impurities.

Some of the haematite is not converted to iron.

a. $$\mathrm{H}_{2}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{HCl}$$

b. 0.36 litres = 0.36/24 = 0.015 moles of hydrogen.

1 mole contains $$6 \times 10^{23}$$ molecules of hydrogen, so 0.015 moles contain

$$0.015 \times 6 \times 10^{23}=9 \times 10^{21}$$ molecules.

c. From the equation, we can see that the same number of moles of chlorine are needed, so we need the same volume; 0.36 litres.

d. Since you get 2 moles of hydrogen chloride for each mole of hydrogen, you will get 0.015 x 2 = 0.03 moles of hydrogen chloride.

e. Begin by writing the balanced formula equation.

$$2 \mathrm{HCl}+\mathrm{Mg}(\mathrm{OH})_{2} \rightarrow \mathrm{Mg} \mathrm{Cl}_{2}+2 \mathrm{H}_{2} \mathrm{O}$$

Since you have 0.03 moles of HCl, you need half as many moles of $$\mathrm{Mg}(\mathrm{OH})_{2}=0.015$$ moles

Mg has a RAM of 24, O has a RAM of 16 and H has a RAM of 1, so the relative formula mass of Mg $$(\mathrm{OH})_{2}=24+16 \times 2+1 \times 2=58$$

mass = moles x formula mass = 0.015 x 58 = 0.87g.