# AQA Chemistry Paper 1 Revision: Quantitative Chemistry

Beyond Science returns with another blog in our Chemistry Paper 1 series, focusing on quantitative chemistry revision! Get your notebooks ready folks…there’s a fair few equations in this one.

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## Quantitative Chemistry Revision: Conservation of Mass

No atoms can be created during a chemical reaction, so the mass of the reactants will equal the mass of the product. You will have the same number and types of atoms on both sides of the reaction. This means no mass is lost or gained so we say mass is conserved during a reaction.

Reactions can be shown as a word or symbol equation:

Take the reaction when we burn magnesium, we can write this as:

Magnesium + Oxygen →  Magnesium Oxide

or

Mg + O →  MgO

Symbol equations should also be balanced; they should have the same number of atoms on each side. To balance an equation we can only add numbers in front of the elements’ symbols, we cannot change the formula or symbols for the elements, so: the reaction between magnesium and oxygen would be:

Mg+O2 → MgO

However, this does not balance as we have different amounts of oxygen on each side. So to balance it we have to add numbers to each side.

We currently have 2 molecules of oxygen on the left-hand side of the reaction but only one on the right-hand side, to balance this we put a 2 in front of the MgO.

This gives us:

Mg+O2 → 2MgO

This however then means we have 2 magnesium on the right-hand side. To balance this out with the left-hand side we put a 2 in front of the Mg on the left.

This gives us:

2Mg + O2 → 2MgO

This is now balanced.

## Quantitative Chemistry Revision: Relative Formula Mass

The relative formula mass is the sum of all the relative atomic masses (Ar) of the atoms in the formula added together, for example:

HCl

This contains 1 atom of hydrogen (H) and 1 atom of chlorine(Cl), we firstly work out the relative atomic masses (Ar)of the atoms.

Ar of H = 1

Ar of Cl = 35.5

1 + 35.5 = 36.5

Let’s look at a different example:

H2SO4

We firstly need to find the relative atomic masses (Ar)of the atoms

Ar of H = 1

Ar of S = 32

Ar of O = 16

We have two atoms of hydrogen and 4 of oxygen so we have to include this in our calculations:

(1 × 2) + 32 + (16 × 4)

2 + 32 + 64 = 98

## Concentration of Solutions

Concentration is the amount of a substance in a specific volume of a solution. The more substance that is dissolved, then the more concentrated the solution is.

It is possible to calculate the concentration of a solution with the  following equation:

concentration (g/dm3) = mass of solute (g) ÷ volume of solvent (dm3)

The equation can be rearranged to find the mass of the dissolved substance (solute):

Mass of solute (g) = concentration (g/dm3) × volume of solution (dm3)

## Calculating Percentage Mass of an Element in a Compound

Percentage mass of an element in a compound =

Ar × number of atoms of that element

Mr of the compound this whole equation needs to be multiplied by 100 to make it a percentage

Find the percentage mass of oxygen in magnesium oxide. The formula for magnesium oxide as discussed earlier is MgO.

We firstly need to find the Ar of each atom.

Ar of magnesium = 24                                      Ar of oxygen = 16

We then calculate the relative formula mass Mr  of MgO

Mr of MgO = 24 + 16

= 40

We then use our formula to calculate the percentage mass:

% mass = Ar = 16 = 0.4                                    0.4 x 100 = 40%

Mr   40

## Quantitative Chemistry Revision from Beyond

You can find a whole heap of Quantitative Chemistry revision on the Beyond website, which you can use to bolster your understanding of this tricky topic. Below, you’ll find our AQA GCSE Quantitative Chemistry Lesson Pack Bundle, which is an excellent place to start!

If you liked our Quantitative Chemistry Revision Blog, you can read more of our blogs here!