Amounts of substances in equations (HT only)

AQA·Chemistry (Combined Science)·5.3.2.2·Trilogy — Higher

Mole ratios in balanced equations

A balanced chemical equation shows the fixed ratio in which substances react and are formed. The numbers in front of the formulae, called coefficients, count particles, but they also represent amounts in moles. This is because one mole of any substance contains the same number of particles.

For example, the balanced equation for magnesium reacting with hydrochloric acid is:

Mg+2 HCl→MgClX2+HX2\ce{Mg + 2HCl -> MgCl2 + H2}

A formula with no written coefficient has a coefficient of 1. The equation therefore gives the mole ratio of magnesium : hydrochloric acid : magnesium chloride : hydrogen as 1 : 2 : 1 : 1.

This means that 1 mole of magnesium reacts with 2 moles of hydrochloric acid to produce 1 mole of magnesium chloride and 1 mole of hydrogen. The same ratio applies to other amounts: 3 moles of magnesium require 6 moles of hydrochloric acid and, if the reaction goes to completion, produce 3 moles of hydrogen.

Using a mole ratio

When the amount of one substance is known, the coefficients show how to find the amount of another substance needed or produced. Select the two coefficients from the balanced equation and use their ratio to scale from the known amount to the unknown amount.

The ratio comes from the balanced equation, so the equation must be balanced before using it for an amount calculation.

Stoichiometric mass calculations

A balanced chemical equation shows the fixed mole ratio in which substances react and form. The coefficients give the ratio of amounts in moles, not directly the ratio of masses. To find masses, link those mole amounts to the molar mass of each substance.

A stoichiometric mass calculation follows three steps:

  1. Convert the known mass to moles.
  2. Use the coefficient ratio in the balanced equation to find the required number of moles.
  3. Convert the required number of moles to mass.

Relating mass and moles

The molar mass of a substance is the mass of one mole, usually expressed in $$\mathrm{g,mol^{-1}}$$. Its numerical value is found from the formula mass: add the relative atomic masses of all the atoms in the formula. For example, using relative atomic masses of $$24$$ for magnesium and $$16$$ for oxygen, $$\ce{MgO}$$ has a formula mass of $$40$$ and a molar mass of $$40\ \mathrm{g,mol^{-1}}$$.

To convert a mass to moles, divide by the molar mass:

amount in moles=mass in gmolar mass in g mol−1\text{amount in moles} = \frac{\text{mass in g}}{\text{molar mass in g mol}^{-1}}

To convert moles back to mass, multiply by the molar mass:

mass in g=amount in moles×molar mass in g mol−1\text{mass in g} = \text{amount in moles} \times \text{molar mass in g mol}^{-1}

Mass relationships in a balanced equation

Magnesium reacts with oxygen to form magnesium oxide:

2 Mg+OX2→2 MgO\ce{2Mg + O2 -> 2MgO}

The coefficients show that $$2$$ moles of $$\ce{Mg}$$ react with $$1$$ mole of $$\ce{O2}$$ to form $$2$$ moles of $$\ce{MgO}$$. Using molar masses of $$24$$, $$32$$ and $$40\ \mathrm{g,mol^{-1}}$$ respectively, these amounts correspond to:

  • $$2\ \mathrm{mol}$$ of $$\ce{Mg}$$: $$48\ \mathrm{g}$$
  • $$1\ \mathrm{mol}$$ of $$\ce{O2}$$: $$32\ \mathrm{g}$$
  • $$2\ \mathrm{mol}$$ of $$\ce{MgO}$$: $$80\ \mathrm{g}$$

Thus the equation implies a mass relationship of $$48\ \mathrm{g}$$ of magnesium reacting with $$32\ \mathrm{g}$$ of oxygen to form $$80\ \mathrm{g}$$ of magnesium oxide. The mass ratio is different from the mole ratio because the substances have different molar masses.

Calculating mass from a known mass

Suppose $$12\ \mathrm{g}$$ of magnesium reacts completely. To find the mass of oxygen needed, first convert the magnesium mass to moles:

n(Mg)=12 g24 g mol−1=0.50 moln(\ce{Mg}) = \frac{12\ \mathrm{g}}{24\ \mathrm{g\,mol^{-1}}} = 0.50\ \mathrm{mol}

The equation shows that $$2$$ moles of $$\ce{Mg}$$ react with $$1$$ mole of $$\ce{O2}$$. Therefore, the amount of oxygen is half the amount of magnesium:

n(OX2)=0.50 mol×12=0.25 moln(\ce{O2}) = 0.50\ \mathrm{mol} \times \frac{1}{2} = 0.25\ \mathrm{mol}

Finally, convert the oxygen amount to mass using the molar mass of $$\ce{O2}$$:

m(OX2)=0.25 mol×32 g mol−1=8.0 gm(\ce{O2}) = 0.25\ \mathrm{mol} \times 32\ \mathrm{g\,mol^{-1}} = 8.0\ \mathrm{g}

So $$12\ \mathrm{g}$$ of magnesium requires $$8.0\ \mathrm{g}$$ of oxygen. The same mass-to-moles, ratio, then moles-to-mass method can be used whether the known substance is a reactant or a product.

Amounts of substances in equations (HT only) | AQA GCSE Combined Science: Trilogy (8464) — Chemistry — Higher Revision Notes | KnowledgeSet