Conservation of mass and quantitative reaction relationships

For example, mixing calcium chloride solution with sodium sulfate solution forms an insoluble solid, calcium sulfate. A reaction that forms an insoluble solid from solutions is called a precipitation reaction:

$$\mathrm{CaCl_2(aq) + Na_2SO_4(aq) \rightarrow CaSO_4(s) + 2NaCl(aq)}$$

The solid precipitate is a new substance, not extra matter. In a closed flask, the precipitate, remaining solution and flask together have the same mass as the flask and its contents before mixing.

Why an open flask can appear to gain or lose mass

An open flask allows gases to cross the boundary between the flask and the surrounding air. The balance measures only the flask and what remains inside it, not everything involved in the reaction.

Calcium carbonate reacts with hydrochloric acid to produce carbon dioxide gas:

$$\mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)}$$

In an open flask, carbon dioxide escapes, so the measured mass decreases. If the escaped gas were included, the total mass would still be unchanged.

The opposite happens when a reaction takes in a gas from the air. For example, magnesium gains mass as it reacts with oxygen to form magnesium oxide. The extra mass comes from oxygen that was initially outside the container. Including that oxygen in the starting mass shows that mass is conserved.

Turning a balanced equation into a mass relationship

A balanced equation tells us the relative numbers of particles reacting. Different particles have different masses, so the coefficients are not directly a mass ratio.

To find the mass ratio, multiply each substance’s relative formula mass, $M_r$, by its coefficient. Relative formula mass is the sum of the relative atomic masses in a formula and has no unit. The resulting ratio can be used with grams, tonnes or another consistent mass unit.

Consider magnesium burning in oxygen:

$$\mathrm{2Mg(s) + O_2(g) \rightarrow 2MgO(s)}$$

Using $A_r(\mathrm{Mg})=24$ and $A_r(\mathrm{O})=16$, the relative masses are:

Thus, 48 g of magnesium reacts with 32 g of oxygen to form 80 g of magnesium oxide. Notice that $48+32=80$: the mass relationship also demonstrates conservation of mass.

If 6.0 g of magnesium burns completely with enough oxygen, scale the magnesium-to-magnesium-oxide ratio:

$$m(\mathrm{MgO})=6.0\times\frac{80}{48}=10.0\text{ g}$$

The product is heavier than the magnesium because it also contains oxygen.

The same approach works when finding a reactant mass from a product mass: put the coefficient-weighted relative mass of the substance you want on top of the fraction, and that of the known substance underneath. For aluminium oxide decomposition, $\mathrm{2Al_2O_3\rightarrow4Al+3O_2}$, the aluminium oxide-to-aluminium mass ratio is $204:108$. Therefore, 51 tonnes of aluminium oxide can produce $51\times108/204=27$ tonnes of aluminium.

The reactant that runs out controls the product mass — Higher tier

Reactants must combine in the proportions set by the balanced equation. If one reactant is supplied in excess, there is more of it than the other reactant can use.

The reactant that is not in excess is the limiting reactant. Once it has all reacted, no more product can form, even if another reactant remains. Its starting mass therefore controls the maximum mass of product.

In the magnesium example, 6.0 g of magnesium needs 4.0 g of oxygen to make 10.0 g of magnesium oxide. Supplying more oxygen does not produce more magnesium oxide once all the magnesium has reacted: the surplus oxygen remains unused. This is why a reacting-mass calculation based on magnesium assumes that enough oxygen is available and the magnesium reacts completely.

Working backwards from masses to coefficients — Higher tier

Stoichiometry describes the relative amounts of substances in a reaction. You can deduce the coefficients of an equation from reacting masses, but first you must convert those masses into amounts in moles.

The molar mass, $M$, is the mass of one mole in $\mathrm{g\,mol^{-1}}$; its numerical value equals the relative formula mass. Use:

$$n=\frac{m}{M}$$

Here, $n$ is the amount in mol and $m$ is the mass in g. Divide all the calculated amounts by the smallest to find their ratio. If necessary, multiply every term by the same number to obtain the smallest whole-number ratio.

For example, 64 g of methanol reacts with 96 g of oxygen to produce 88 g of carbon dioxide and 72 g of water. Using $A_r(\mathrm{C})=12$, $A_r(\mathrm{H})=1$ and $A_r(\mathrm{O})=16$:

The amounts have the ratio $2:3:2:4$. Dividing by the smallest gives $1:1.5:1:2$; multiplying every term by two restores the smallest whole-number ratio. These numbers become the coefficients:

$$\mathrm{2CH_3OH + 3O_2 \rightarrow 2CO_2 + 4H_2O}$$

Check the result by counting atoms: both sides contain two carbon atoms, eight hydrogen atoms and eight oxygen atoms. On the left, the oxygen count is $2\times1+3\times2=8$; on the right, it is $2\times2+4\times1=8$. The total masses also match: $64+96=88+72=160$ g.

Reacting masses

1. Write the balanced equation.
2. Calculate each relevant $M_r$ and multiply by its coefficient.
3. Scale the resulting mass ratio:

$$m_{\text{wanted}}=m_{\text{known}}\times\frac{c_{\text{wanted}}M_{r,\text{wanted}}}{c_{\text{known}}M_{r,\text{known}}}$$

Here, $c$ is the equation coefficient. Use consistent mass units and assume sufficient other reactants for complete reaction.

Limiting reactant — Higher tier

The reactant not in excess runs out first and controls the maximum product mass. Excess reactant remains; adding more of it alone cannot make more product.

Deducing stoichiometry — Higher tier

$$n=\frac{m}{M}$$

Convert each reacting mass to moles → divide by the smallest → multiply all terms if needed for whole numbers → use the smallest whole-number ratio as coefficients → check atom balance.