Equations and Calculations
Conservation of Mass
In any chemical reaction, atoms are neither created nor destroyed — they are simply rearranged. This is the law of conservation of mass. It means a correctly written symbol equation must have the same number of each type of atom on both sides of the arrow. An equation with unequal atom counts is unbalanced and does not represent a real reaction.
In a chemical reaction, total mass is conserved. A balanced symbol equation has the same number of atoms of each element on both sides of the arrow. You balance equations by adding large numbers (coefficients) in front of formulas — never by changing the subscripts inside a formula.
Balancing Symbol Equations
To balance a symbol equation, count the atoms on each side and adjust the large numbers (coefficients) in front of formulas until both sides match. You must never change the subscripts inside a formula (that would change the substance itself). Work through one element at a time, leaving hydrogen and oxygen until last when they appear in water or acids.
Balance the equation for the combustion of propane: \(C_3H_8 + O_2 \rightarrow CO_2 + H_2O\)
Step 1: Count carbons. Left: 3; Right: 1. Put a 3 in front of \(CO_2\):
\(C_3H_8 + O_2 \rightarrow 3CO_2 + H_2O\)
Step 2: Count hydrogens. Left: 8; Right: 2. Put a 4 in front of \(H_2O\):
\(C_3H_8 + O_2 \rightarrow 3CO_2 + 4H_2O\)
Step 3: Count oxygens. Right: 3×2 + 4×1 = 10. Left needs 5 \(O_2\):
\(C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O\)
Step 4: Check all atoms. C: 3=3 ✓ H: 8=8 ✓ O: 10=10 ✓. The equation is balanced.
State symbols can be added in brackets after each formula to show the physical state: (s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous (dissolved in water). For example: \(Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)\)
Relative Formula Mass
The relative formula mass (\(M_r\)) of a compound is calculated by adding together the relative atomic masses (\(A_r\)) of all the atoms shown in its formula. For example, for water (\(H_2O\)):
$$M_r(H_2O) = 2 \times A_r(H) + 1 \times A_r(O) = 2 \times 1 + 16 = 18$$
For sulfuric acid (\(H_2SO_4\)): \(M_r = 2(1) + 32 + 4(16) = 98\). You need the periodic table to look up \(A_r\) values. The relative formula mass tells you the mass of one formula unit in atomic mass units, and is numerically equal to the mass of one mole in grams.
Relative formula mass (Mr): add up the Ar of every atom in the formula, accounting for subscripts. Mr of CaCO₃ = 40 + 12 + 3×16 = 100. No units — it is a ratio relative to ¹²C.
The Mole
Atoms and molecules are far too small to count individually, so chemists use the mole as a counting unit. One mole of any substance contains the same enormous number of particles — approximately \(6.02 \times 10^{23}\) — known as Avogadro's constant. The genius of the mole is that one mole of an element has a mass in grams exactly equal to its relative atomic mass.
The key equation linking moles, mass, and molar mass is:
$$\text{number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$$
Rearranged: mass = moles × molar mass, or molar mass = mass ÷ moles.
Worked example: How many moles are in 44 g of carbon dioxide (\(CO_2\), \(M_r = 44\))?
Moles = 44 ÷ 44 = 1 mol.
Second example: What mass of sodium hydroxide (NaOH, \(M_r = 40\)) is needed to make 0.5 mol?
Mass = 0.5 × 40 = 20 g.
Always show your working in full and include units at every step — a final answer with no unit usually loses a mark.
Using Equations to Calculate Reacting Masses
A balanced symbol equation shows the molar ratios of reactants and products — the large numbers (coefficients) tell you how many moles of each substance react. Combined with molar masses, this allows calculation of the mass of any reactant or product from a known mass of another.
Calculate the mass of iron produced when 160 g of iron(III) oxide (\(Fe_2O_3\)) reacts with excess carbon:
\(2Fe_2O_3 + 3C \rightarrow 4Fe + 3CO_2\)
Step 1: Find \(M_r\) of \(Fe_2O_3\): \(2 \times 56 + 3 \times 16 = 160\) g/mol.
Step 2: Calculate moles of \(Fe_2O_3\): \(\frac{160}{160} = 1 \text{ mol}\).
Step 3: Use the molar ratio. From the equation, 2 mol \(Fe_2O_3\) → 4 mol Fe, so 1 mol \(Fe_2O_3\) → 2 mol Fe.
Step 4: Convert moles of Fe to mass: \(2 \times 56 = \mathbf{112 \text{ g}}\).
Reacting mass calculation (Higher): (1) write balanced equation, (2) calculate Mr of known substance and find moles, (3) use molar ratio from equation to find moles of unknown substance, (4) convert moles to mass. Always show every step and include units.