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The Power of Algebraic Rearrangement

Imagine you are an environmental scientist testing river water for toxic pollutants. You know the concentration you detected and the volume of your sample, but you need to find the exact mass of the toxin. To do this, you cannot just plug numbers into a standard formula; you must flip the equation around to make the unknown value the star of the show. This is called changing the subject of an equation.

Core Stoichiometry Formulas

In OCR Chemistry B, you will frequently use two main relationships to solve chemical analysis problems. The first links the amount of substance ( $n$ ) to mass ( $m$ ) and molar mass ( $M$ ):

n = \frac{m}{M}

The second links molar concentration ( $c$ ) to the amount of substance and volume ( $V$ ):

c = \frac{n}{V}

$c$ is measured in moles per decimetre cubed ( $\text{mol/dm}^3$ )
$V$ is measured in decimetres cubed ( $\text{dm}^3$ )

How to Rearrange Equations

While you might have used a formula triangle in earlier studies, examiners expect higher-tier students to rely on algebraic deduction. The golden rule is simple: whatever mathematical operation is happening to your target variable, apply the inverse operation to both sides of the equation.

If your target variable is in the denominator, multiply both sides by that variable to bring it to the top.
If a variable is multiplied by your target, divide both sides by that variable to isolate your target.

Worked Example: Calculating Mass

Calculate the mass of 0.45 moles of calcium carbonate ( $\text{CaCO}_3$ ). The molar mass ( $M$ ) of calcium carbonate is 100 g/mol.

Step 1: State the core formula.

n = \frac{m}{M}

Step 2: Rearrange to make mass ( $m$ ) the subject.

Multiply both sides by $M$ to isolate $m$ :

m = n \times M

Step 3: Substitute the given values.

$m = 0.45 \text{ mol} \times 100 \text{ g/mol}$

Step 4: Calculate the final answer with units.

$m = 45 \text{ g}$

Worked Example: Calculating Volume

A quality control chemist dissolves 0.150 moles of sodium hydroxide to make a $0.500 \text{ mol/dm}^3$ solution. Calculate the volume of the solution in $\text{dm}^3$ .

Step 1: State the core formula.

c = \frac{n}{V}

Step 2: Rearrange to make volume ( $V$ ) the subject.

First, multiply both sides by $V$ to remove it from the denominator:

c \times V = n

Next, divide both sides by $c$ to isolate $V$ :

V = \frac{n}{c}

Step 3: Substitute the given values.

$V = \frac{0.150 \text{ mol}}{0.500 \text{ mol/dm}^3}$

Step 4: Calculate the final answer with units.

$V = 0.300 \text{ dm}^3$

Precision in Calculations

When dealing with real-world context questions, your final calculated answer should reflect the accuracy of your instruments. Always round your final result to the same number of significant figures as the least precise piece of data provided in the question (typically 2 or 3 significant figures).

Students frequently forget to convert volume from cm³ to dm³ before substituting it into the concentration equation; always divide your cm³ value by 1000 first.

In 'Calculate' questions, you are awarded a method mark for showing the correct algebraic rearrangement or substitution, meaning you can still score marks even if you make a final calculator error.

Always use the precise terminology 'amount of substance' rather than just writing 'moles' when describing the physical quantity n in your written explanations for OCR B.

When rearranging, it is safer to isolate the letter variable first before substituting the numbers in, as this prevents transcription errors during multi-step calculations.

The relationship between the relative quantities of substances taking part in a reaction, typically as a ratio of whole integers.

The single variable that is isolated on one side of the equals sign in a mathematical formula.

A physical quantity representing the number of specified elementary entities (such as atoms or molecules), measured in moles (mol).

The mass of one mole of a specific substance, usually expressed in grams per mole (g/mol).

The amount of solute dissolved per decimetre cubed of solution, expressed in mol/dm³.

A visual aid used to help rearrange simple three-variable equations, though algebraic deduction is preferred at the higher tier.