Writing balanced equations

For covalent molecules, the molecular formula represents the actual number of atoms in one single molecule. To deduce these, we can cross over their valency, which is the "combining power" or the number of electrons an atom needs to share to achieve a full outer shell. For example, to deduce the formula of methane: Carbon (Group 4) has a valency of 4, and Hydrogen has a valency of 1. Swapping these values gives one Carbon atom for every four Hydrogen atoms, resulting in the formula $CH_4$. Alternatively, formulae can be deduced using prefixes in the name like "mono-" (1), "di-" (2), and "tetra-" (4). For example, dinitrogen tetroxide contains two nitrogen atoms and four oxygen atoms, giving the formula $N_2O_4$.

Certain non-metal elements do not exist as single atoms in nature. These diatomic elements always travel in pairs when appearing alone in an equation: $H_2$, $N_2$, $O_2$, $F_2$, $Cl_2$, $Br_2$, and $I_2$.

For ionic compounds, the formula represents the empirical formula, which is the simplest whole-number ratio of ions in the giant lattice. All ionic compounds obey the principle of electrical neutrality. This means the total positive ionic charge from the cations must exactly balance the total negative charge from the anions to equal zero.

The charges of ions can usually be predicted from their periodic table group (e.g., Group 1 is $1+$, Group 6 is $2-$). Transition metals do not have fixed charges; their charge is given by a Roman numeral, such as Iron(III) for $Fe^{3+}$. You must memorise two exceptions: Zinc is always fixed at $Zn^{2+}$ and Silver is always $Ag^+$.

You must also memorise common polyatomic ions, which are groups of covalently bonded atoms that act as a single charged unit. These include hydroxide ($OH^-$), nitrate ($NO_3^-$), sulfate ($SO_4^{2-}$), carbonate ($CO_3^{2-}$), and ammonium ($NH_4^+$).

To translate chemical names into formulae, you must apply standard naming conventions. The metal (cation) always comes first. Single non-metals change their suffix to "-ide" (e.g., Chlorine becomes Chloride). Polyatomic ions containing oxygen generally end in "-ate" (e.g., Sulfate, Nitrate).

Worked Example: Writing Ionic Formulae ("Swap and Drop")

Deduce the formula for Calcium Hydroxide.

Step 1: Identify the ions and their charges.

• Calcium (Group 2) = $Ca^{2+}$
• Hydroxide (Polyatomic ion) = $OH^-$

Step 2: Swap the charge numbers and drop them to the subscript position.

• The "2" from Calcium applies to the whole Hydroxide group.
• The "1" from Hydroxide goes to Calcium.

Step 3: Apply brackets and simplify.

• Because there are multiple polyatomic ions, brackets are required.
• Final Formula: $Ca(OH)_2$
• Proof of neutrality: $(1 \times 2+) + (2 \times 1-) = 0$.

The Law of Conservation of Mass

If you burn a massive log, the tiny pile of ash left behind might look like mass has vanished into thin air—but it hasn't. The Law of Conservation of Mass states that no matter is lost or gained during a chemical reaction. The total mass of the reactants always exactly equals the total mass of the products.

In a chemical reaction, existing bonds are broken and atoms are simply rearranged. The type and number of atoms remain completely constant. We can prove this mathematically by showing that the sum of the relative formula mass ($M_r$) of the reactants equals the sum of the products.

Sometimes, mass appears to change if the reaction occurs in a non-enclosed (open) system. If a product is a gas (like $CO_2$), it escapes into the atmosphere and cannot be weighed, causing an apparent decrease in mass. If a reactant is a gas drawn from the air (like $O_2$ reacting with solid metal), the mass appears to increase. Only in a closed system (like a sealed flask) will the measured mass remain perfectly constant.

Worked Example: Mass Verification Calculation

Prove that mass is conserved in the reaction: $2H_2(g) + O_2(g) \rightarrow 2H_2O(l)$.

Step 1: Calculate the total $M_r$ of the reactants.

• $2H_2 = 2 \times (1 \times 2) = 4$
• $O_2 = (16 \times 2) = 32$
• Total reactant mass = $4 + 32 = 36$

Step 2: Calculate the total $M_r$ of the products.

• $2H_2O = 2 \times [(1 \times 2) + 16] = 2 \times 18 = 36$

Step 3: Compare the totals.

• Total reactant $M_r$ (36) = Total product $M_r$ (36). Mass is fully conserved.

Balancing Chemical Equations

Understanding chemical equations explains why factory chemists must calculate exact reactant amounts—otherwise, expensive leftover ingredients are wasted. Chemical equations must be balanced to obey the Law of Conservation of Mass and show accurate stoichiometry, which is the exact ratio of reacting substances.

The golden rule of balancing is that you can only change the coefficient (the large number placed in front of a formula). You must never change or add subscripts (the small numbers within a formula), as doing so completely changes the chemical identity of the substance.

OCR requires the use of standard state symbols to describe the physical states of substances in a balanced equation: $(s)$ for solid, $(l)$ for liquid, $(g)$ for gas, and $(aq)$ for aqueous (dissolved in water).

Worked Example: Balancing by Atom Counting (Standard Tier)

Balance the equation: $Al(s) + CuO(s) \rightarrow Al_2O_3(s) + Cu(s)$

Step 1: Count the atoms on both sides of the unbalanced equation.

• Left side: $Al=1$, $Cu=1$, $O=1$
• Right side: $Al=2$, $Cu=1$, $O=3$

Step 2: Balance the Oxygen atoms.

• Place a coefficient of '3' in front of $CuO(s)$.
• $Al(s) + 3CuO(s) \rightarrow Al_2O_3(s) + Cu(s)$
• New count: Left ($O=3, Cu=3$), Right ($O=3$).

Step 3: Balance the Aluminium atoms.

• Place a coefficient of '2' in front of $Al(s)$.
• $2Al(s) + 3CuO(s) \rightarrow Al_2O_3(s) + Cu(s)$

Step 4: Balance the Copper atoms (leave lone elements until last).

• Place a coefficient of '3' in front of $Cu(s)$.
• Balanced Equation: $2Al(s) + 3CuO(s) \rightarrow Al_2O_3(s) + 3Cu(s)$

Step 5: Verify that the total atom count remains constant.

• Left side: $Al=2$, $Cu=3$, $O=3$
• Right side: $Al=2$, $Cu=3$, $O=3$
• Proof of conservation: The total number of atoms of each element is equal on both sides, confirming that mass is conserved.

Worked Example: Deducing Coefficients from Masses (Higher Tier)

$64g$ of liquid methanol ($CH_3OH$) reacts with $96g$ of $O_2$ gas to produce $88g$ of $CO_2$ gas and $72g$ of liquid $H_2O$. Deduce the balanced equation, including state symbols.

Step 1: Find the relative formula mass ($M_r$) for each substance.

• $CH_3OH = 32$, $O_2 = 32$, $CO_2 = 44$, $H_2O = 18$

Step 2: Convert each given mass into moles (Mass / $M_r$).

• $CH_3OH$: $64 / 32 = 2$ mol
• $O_2$: $96 / 32 = 3$ mol
• $CO_2$: $88 / 44 = 2$ mol
• $H_2O$: $72 / 18 = 4$ mol

Step 3: Apply the mole ratio as coefficients and add state symbols.

• The simplest mole ratio is $2 : 3 : 2 : 4$.
• Final Equation: $2CH_3OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 4H_2O(l)$