The ratio in a balanced equation

The numbers in front of formulae in a balanced equation are called coefficients. They show the reacting ratio. For mass calculations, multiply each coefficient by the relative formula mass, $M_{r}$, of that substance.

In this lesson, equation mass is a shortcut for setting up the ratio; Edexcel questions may simply ask you to use a balanced equation and relative formula masses.

For example:

$\ce{2Mg + O2 -> 2MgO}$

Using $A_{r}$: $\ce{Mg} = 24$, $O = 16$

So the balanced equation tells you that 48 g of magnesium reacts with 32 g of oxygen to form 80 g of magnesium oxide. The same ratio works at any scale: half those masses, ten times those masses, or any other proportional amount.

Do not assume the product mass equals the mass of the one reactant you were given. The other reactant may also add mass to the product.

The mass-ratio method

Once you have the two equation masses you need, use them as a ratio.

For a theoretical mass, assume the given reactant reacts completely, any other required reactant is present in sufficient quantity, and the stated product is formed without loss.

Use this route:

1. Check the equation is balanced.
2. Calculate the $M_{r}$ of the given substance and the wanted substance.
3. Multiply each $M_{r}$ by its coefficient in the equation.
4. Use the equation-mass ratio to calculate the wanted mass.
5. Give the answer with a suitable unit.

Worked example:

$\ce{Zn + S -> ZnS}$

Calculate the mass of zinc sulfide made from 6.50 g of zinc. Use $A_{r}$: $\ce{Zn} = 65$, $S = 32$.

$M_{r}$ of $\ce{Zn} = 65$

$M_{r}$ of $\ce{ZnS} = 65 + 32 = 97$

The coefficients are both 1, so the equation masses are 65 for $\ce{Zn}$ and 97 for $\ce{ZnS}$.

mass of ZnS = 6.50 × 97 / 65

mass of ZnS = 9.70 g

The answer is larger than 6.50 g because sulfur also becomes part of the zinc sulfide.

Working backwards from a product

The same method works if the given mass is a product and the wanted mass is a reactant. The fraction changes direction because the wanted substance changes.

Worked example:

$\ce{Fe + S -> FeS}$

Calculate the mass of iron needed to produce 8.80 g of iron sulfide. Use $A_{r}$: $\ce{Fe} = 56$, $S = 32$.

$M_{r}$ of $\ce{Fe} = 56$

$M_{r}$ of $\ce{FeS} = 56 + 32 = 88$

The coefficients are both 1, so the equation masses are 56 for $\ce{Fe}$ and 88 for $\ce{FeS}$.

mass of Fe = 8.80 × 56 / 88

mass of Fe = 5.60 g

This is not a new method. It is the same proportional reasoning, but the product mass is the starting point.

Coefficients and excess substances

Coefficients matter. If the balanced equation has a 2, 3 or another number in front of a formula, include it in the equation mass.

Sometimes a question says one reactant is in excess. This means there is more than enough of that reactant, so it does not limit the amount of product. For this lesson, use the substance whose mass is given and the substance whose mass is wanted.

Worked example:

$\ce{N2 + 3H2 -> 2NH3}$

Calculate the mass of ammonia made from 5.60 kg of nitrogen, with hydrogen in excess. Use $A_{r}$: $N = 14$, $H = 1$.

$M_{r}$ of $\ce{N2} = 2 \times 14 = 28$

$M_{r}$ of $\ce{NH3} = 14 + (3 \times 1) = 17$

Equation mass of $2\ce{NH3} = 2 \times 17 = 34$

The ratio needed is 28 of $\ce{N2}$ to 34 of $\ce{NH3}$.

mass of NH3 = 5.60 × 34 / 28

mass of NH3 = 6.80 kg

The unit stays as kg because the calculation is a ratio of masses. If the question asks for a different unit, convert at the end unless the conversion is needed earlier.

Now use the same idea in a shorter calculation where sodium is the given substance and chlorine is present in excess.

Checking exam answers

Show enough working for the method to be clear. A valid alternative is to calculate moles, use the coefficient ratio, then convert back to mass. Equivalent simplified mass ratios also work: for example, 128:160 and 64:80 give the same copper-to-copper-oxide ratio.

Check these points before you finish:

• The equation used is balanced.
• The $M_{r}$ values use all atoms in the formula, such as $\ce{O2}$ rather than just O.
• Coefficients have been included in the equation masses.
• The fraction has the wanted substance on top and the given substance on the bottom.
• The answer has a mass unit, such as g or kg.
• You have not rounded intermediate values too early.

If your answer seems unexpected, use chemistry to sense-check it. A product can have a larger mass than the given reactant if another reactant contributes atoms. A product can have a smaller mass than the starting compound if another product also forms and takes away some of the atoms.

That final example has all the pieces an examiner expects to see: relative formula mass, coefficient ratio, substitution, answer and unit.

Link it together

You have now worked through every section of this lesson. This last exercise checks how well you can link those ideas: explain your answer in your own words, as if to a classmate, using what you learned above.