Define relative atomic and molecular mass and the mole, and use moles to calculate reacting masses and amounts from balanced equations

What this dot point is asking

SEAB wants you to define relative atomic mass and relative molecular mass, to define the mole as a counting unit for particles, and to use the moles equation with balanced equations to calculate reacting masses. The calculations are short, but the marks come from a reliable routine: find moles, use the equation ratio, convert back to mass.

The answer

Relative atomic and molecular mass

The relative atomic mass ($A_r$) of an element is the average mass of its atoms compared with a standard, on a scale where carbon-12 is exactly $12$. The relative molecular mass ($M_r$) of a compound is the sum of the relative atomic masses of all the atoms in its formula. For example, water $\text{H}_2\text{O}$ has $M_r = (2 \times 1) + 16 = 18$.

The mole

A mole is the amount of substance that contains the same number of particles as there are atoms in $12\ \text{g}$ of carbon-12. That number, the Avogadro constant, is $6.0 \times 10^{23}$ particles per mole. The key point for calculations is that one mole of any substance has a mass in grams equal to its relative atomic or molecular mass.

The moles equation

The link between mass, moles and relative mass is:

so $\text{mass} = \text{moles} \times \text{relative mass}$. This lets you switch between a measured mass and a number of moles.

Reacting masses from equations

A balanced equation gives the ratio in which substances react and form, in moles. The routine for a reacting-mass problem is:

1. write the balanced equation,
2. convert the known mass to moles,
3. use the mole ratio from the equation to find moles of the wanted substance,
4. convert that back to a mass.

Examples in context

Example 1. Working out a fertiliser's nitrogen content. Farmers compare fertilisers by the mass of nitrogen they supply. Using relative masses, you can calculate what fraction of ammonium nitrate is nitrogen, and so how much usable nitrogen a given mass delivers to the soil.

Example 2. Scaling up a reaction. A chemist who knows the reacting masses for a small test can scale them up in proportion to make a kilogram of product, because the mole ratios stay the same. The mole turns a balanced equation into a practical recipe of masses.

Try this

Q1. Define the term mole. [2 marks]

• Cue. A mole is the amount of substance containing $6.0 \times 10^{23}$ particles, the same number as there are atoms in $12\ \text{g}$ of carbon-12.

Q2. Calculate the relative molecular mass of carbon dioxide, $\text{CO}_2$. ($A_r$: C $= 12$, O $= 16$.) [2 marks]

• Cue. $M_r = 12 + (2 \times 16) = 12 + 32 = 44$.

Q3. How many moles are there in $8.0\ \text{g}$ of methane, $\text{CH}_4$? ($M_r = 16$.) [2 marks]

• Cue. $\text{moles} = \dfrac{\text{mass}}{M_r} = \dfrac{8.0}{16} = 0.50\ \text{mol}$.