How to Find Product Mass from a Balanced Equation
Learn how to calculate product mass from a balanced chemical equation using the mass-to-moles-to-moles-to-mass method. Follow worked examples covering mole ratios, limiting reagents, and theoretical yield, then test your understanding with HSC-style questions.
A reaction equation can tell you exactly how much product should form from a known mass of reactant. The catch is that the numbers in a balanced equation are not mass ratios. They’re mole ratios.
To find the mass of a product, follow one reliable route:
Mass of reactant → moles of reactant → moles of product → mass of product.
You cannot usually jump straight from the starting mass to the product mass using the coefficients. First, convert the mass into moles. Then use the balanced equation to work out how many moles of product form. Finally, convert those moles back into grams.
Let’s work through why this method works, when it needs an extra step, and how to avoid the mistakes that cost marks in HSC Chemistry.
01The method: mass to moles to mass
Imagine reacting 5.60 g of iron with excess oxygen. The reaction produces iron(III) oxide:
\[ 4\mathrm{Fe} + 3\mathrm{O_2} \rightarrow 2\mathrm{Fe_2O_3} \]
You know the starting mass of iron. You want the mass of iron(III) oxide.
The equation tells you that four moles of iron produce two moles of iron(III) oxide. It does not say that four grams of iron produce two grams of iron(III) oxide.
Why not? Because an iron atom and an iron(III) oxide formula unit have different masses. You need to account for that before calculating the product mass.
Step 1: Convert the reactant mass into moles
The amount of a substance, measured in moles, is calculated using:
\[ n = \frac{m}{M} \]
Where:
- \(n\) = amount, in mol
- \(m\) = mass, in g
- \(M\) = molar mass, in g mol\(^{-1}\)
For iron, use \(M(\mathrm{Fe}) = 55.85\ \mathrm{g\,mol^{-1}}\).
\[ n(\mathrm{Fe}) = \frac{5.60}{55.85} \]
\[ n(\mathrm{Fe}) = 0.1003\ \mathrm{mol} \]
You now know the amount of iron available to react.
Step 2: Use the balanced equation to find the product moles
Look at the coefficients in the balanced equation:
\[ 4\mathrm{Fe} + 3\mathrm{O_2} \rightarrow 2\mathrm{Fe_2O_3} \]
The mole ratio is:
\[ \frac{n(\mathrm{Fe_2O_3})} {n(\mathrm{Fe})} = \frac{2}{4} \]
Therefore:
\[ n(\mathrm{Fe_2O_3}) = 0.1003 \times \frac{2}{4} \]
\[ n(\mathrm{Fe_2O_3}) = 0.0501\ \mathrm{mol} \]
Notice that we multiplied by \(2/4\), not \(4/2\). The product has fewer moles than the iron because the equation requires two moles of iron(III) oxide for every four moles of iron.
Step 3: Convert product moles into product mass
The molar mass of iron(III) oxide is:
\[ M(\mathrm{Fe_2O_3}) = 2(55.85) + 3(16.00) \]
\[ = 159.70\ \mathrm{g\,mol^{-1}} \]
Rearrange the mole formula:
\[ m = nM \]
Substitute the product amount and molar mass:
\[ m(\mathrm{Fe_2O_3}) = 0.0501 \times 159.70 \]
\[ \boxed{m(\mathrm{Fe_2O_3}) = 8.00\ \mathrm{g}} \]
So, 5.60 g of iron produces approximately 8.00 g of iron(III) oxide, assuming complete reaction and sufficient oxygen.
The product is heavier than the starting iron because oxygen contributes additional mass. That mass has not appeared from nowhere. It has come from the oxygen gas.