Chemical Reactions, Equations and Moles
Chemical Formulae and Symbols
Every element has a unique chemical symbol shown in the periodic table. Compounds are represented by chemical formulae. For example, \(H_2O\) shows that water contains 2 hydrogen atoms for every 1 oxygen atom. \(CaCO_3\) shows calcium carbonate contains 1 calcium, 1 carbon and 3 oxygen atoms. A subscript number applies to the element immediately before it; a large number placed before a formula (a coefficient) multiplies every atom in that formula. So \(2H_2O\) means two molecules of water, containing a total of 4 hydrogen atoms and 2 oxygen atoms.
You can deduce formulae from ion charges. Because ionic compounds have zero overall charge, the charges of the positive and negative ions must balance. For example, \(Na^+\) and \(Cl^-\) balance 1:1 to give NaCl; \(Ca^{2+}\) and \(Cl^-\) require two chloride ions per calcium ion, giving \(CaCl_2\); \(Al^{3+}\) and \(O^{2-}\) require \(Al_2O_3\) for the charges to cancel (total +6 and total −6).
Chemical formulae show which elements are present and in what ratio. Subscript numbers apply to the preceding element; coefficients multiply the whole formula. Common formulae to know: \(H_2O\), \(CO_2\), \(NH_3\), \(CH_4\), \(HCl\), \(NaCl\), \(MgO\), \(CaCO_3\).
Writing and Balancing Symbol Equations
A chemical equation shows reactants on the left and products on the right, separated by an arrow (\(\rightarrow\)). To be correct, a symbol equation must be balanced — the number of atoms of each element must be the same on both sides. This reflects the conservation of mass: atoms are rearranged during reactions, but none are ever created or destroyed.
To balance an equation: count the atoms of each element on both sides, then add coefficients (large numbers in front of formulae) until both sides are equal. Never change the subscript numbers in a formula — that would change the identity of the substance.
Balance the combustion of methane: \(CH_4 + O_2 \rightarrow CO_2 + H_2O\)
Step 1: Count atoms on each side. Left: 1 C, 4 H, 2 O. Right: 1 C, 2 H, 3 O. Carbon is already balanced. Hydrogen and oxygen are not.
Step 2: Balance hydrogen. There are 4 H on the left and 2 H on the right. Put a 2 in front of \(H_2O\): \(CH_4 + O_2 \rightarrow CO_2 + 2H_2O\). Now right has 4 H. Right now has 4 O total (2 in \(CO_2\) + 2 in \(2H_2O\)).
Step 3: Balance oxygen. Left has 2 O. Right has 4 O. Put a 2 in front of \(O_2\): \(CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O\)
Step 4: Check. Left: 1 C, 4 H, 4 O. Right: 1 C, 4 H, 4 O. Balanced.
A balanced equation has the same number of atoms of each element on both sides. Add coefficients (large numbers) to balance — never change subscripts. Conservation of mass: if reactions occur in sealed containers, total mass before = total mass after.
State Symbols
State symbols are added after each formula to show its physical state: (s) = solid, (l) = liquid, (g) = gas, (aq) = aqueous (dissolved in water). For example:
$$2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)$$
State symbols tell you whether substances are solid, liquid, gaseous or in solution, and help you understand the conditions of the reaction. For example, “(aq)” tells you the substance is dissolved in water, which is relevant when predicting whether ions are free to move.
The Mole and Relative Formula Mass
The relative formula mass (\(M_r\)) of a compound is calculated by adding up the relative atomic masses (\(A_r\)) of all atoms in the formula. For example, for water \(H_2O\): \(M_r = (2 \times 1) + 16 = 18\). For calcium carbonate \(CaCO_3\): \(M_r = 40 + 12 + (3 \times 16) = 100\). You can use \(M_r\) to calculate how much of each substance takes part in a reaction using balanced equations and conservation of mass.
The Mole Concept and Avogadro’s Constant (Higher Tier)
A mole is the unit chemists use for the amount of a substance. One mole of any substance contains \(6.02 \times 10^{23}\) particles — this number is Avogadro’s constant. The relationship between mass, moles and relative formula mass is:
$$\text{amount (mol)} = \frac{\text{mass (g)}}{\text{relative formula mass } (M_r)}$$
How many moles are in 44 g of carbon dioxide (\(CO_2\), \(M_r = 44\))?
Step 1: Apply the formula. $$\text{moles} = \frac{44}{44} = 1 \text{ mol}$$
How many grams of magnesium oxide form when 12 g of magnesium burns completely? The balanced equation is \(2Mg + O_2 \rightarrow 2MgO\). \(A_r\) of Mg = 24. \(M_r\) of MgO = 40.
Step 1: Moles of Mg = \(\frac{12}{24} = 0.5 \text{ mol}\).
Step 2: From the equation, 2 mol Mg forms 2 mol MgO (1:1 ratio). So 0.5 mol Mg forms 0.5 mol MgO.
Step 3: Mass of MgO = \(0.5 \times 40 = 20 \text{ g}\). Check (conservation of mass): 12 g Mg + 8 g O (from \(0.5 \times 0.5 \times 32\)) = 20 g MgO. ✓
When calculating reacting masses, always include units in every step and double-check your answer using conservation of mass. A common error is forgetting to convert between moles and mass in the final step, or failing to use the correct mole ratio from the balanced equation.