A-level Chemistry/WJEC/Module 1/Equations

In chemistry, we have many shorthand ways to write about chemicals and how they react. It is important to understand - fluently - what a chemical formula means.

Examples:

H₂O(l) - a molecule of liquid water, containing two hydrogen atoms and one oxygen atom.

• Each element is written using its symbol; H, O, C, Al, Cu, etc.
• If there is more than one atom then we write a subscript after the symbol; H₂, S₈, etc

Al₂O₃(s) - the formula of solid aluminium oxide, which has two aluminium ions for every three oxygen ions.

• Not all chemicals are made of molecules. Ionic compounds are made of positive ions and negative ions. The structures of ionic compounds are huge and repetitive, so the formula only records the proportions of the ions, not the actual numbers.

(CH₃)₃COH(l) - a molecule of the liquid MethylPropan-2-ol, which has four carbon atoms, ten hydrogen atoms and an oxygen atom.

• Brackets help us to use a familiar formula (CH₃) even when it occurs three times. Writing C₃H₉COH would be less helpful.

CuSO₄ ⋅ 5 H₂O(s) - the formula of solid copper(II) sulfate pentahydrate, which has one copper ion and one sulfate ion (SO₄^2-) for every two water molecules. Sulfate and water are not elements. A sulfate ion has one sulfur atom, four oxygen atoms and a -2 electrical charge.

• Many ionic solids contain extra molecules of water, which are written after a dot (⋅) and with a coefficient to show how many extra molecules are in the structure (the 5 in CuSO₄ ⋅ 5 H₂O(s)).

Most ions can be deduced from the element's Group in the Periodic Table:

Na⁺ Mg²⁺ Al³⁺ ... P^3- S^2- Cl^-

Sometimes an element forms more than one ion. If so, the name will indicate what the ion is:

The iron(II) ion is Fe²⁺; The iron(III) ion is Fe³⁺

But there are some ions, with multiple atoms, that you need to learn. Looking at Past Papers, the most common formulae you need to learn are:

• hydroxide OH^-
• carbonate CO₃^2-
• sulfuric acid/sulfate H₂SO₄ / SO₄^2-
• ethanoic acid/ethanoate CH₃COOH / CH₃COO^-
• nitric acid/nitrate HNO₃ / NO₃^-
• ammonia/ammonium NH₃ / NH₄⁺

Ionic compounds are usually made of one positive ion (a "cation") and one negative ion (an "anion"). The positive and negative charges cancel out. This is easily done with some pairs of ions:

• Sodium chloride, Na⁺ and Cl^- ions make a compound with the formula NaCl.
• Iron(II) oxide, Fe²⁺ and O^2- ions make a compound with the formula FeO.
• Aluminium phosphate, Al³⁺ and PO₄^3- ions make a compound with the formula AlPO₄.

What if the cation and anion have different charges? To balance the charges we need to find the lowest common multiple of the two charges.

• Calcium chloride, Ca²⁺ and Cl^- ions make a compound with the formula CaCl₂.
• Iron(III) oxide, Fe³⁺ and O^2- ions make a compound with the formula Fe₂O₃.
• Magnesium phosphate, Mg²⁺ and PO₄^3- ions make a compound with the formula Mg₃(PO₄)₂.

Try some yourself:

• Potassium bromide
• Barium nitrate
• Vanadium(V) oxide
• Chromium(III) sulfate

The formulae of molecules are more difficult to predict.

Several non-metal elements exist as diatomic molecules: H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂

You need to learn these. This only happens in the pure elements. An easy way to remember the diatomic elements is that, by coincidence, their names either end with "-gen" or "-ine".

For many non-metal elements we can assume that they bond to 4, 3, 2 or 1 other atoms if the element is in Group 4, 5, 6 or 7.

Examples: H₂O, H₂S, NH₃, CH₄, CO₂, PCl₃, SiO₂

There are many exceptions though: CO, PCl₅, SO₂, SO₃, NO, NO₂.

You don't need to learn all the exceptions - there are too many. Expect the formula to be predictable but don't be surprised if you're wrong. Learn the ones you get wrong!

Molecules with several central atoms get even more unpredictable: N₂O, C₂H₆, H₂O₂

A chemical equation describes a chemical reaction, using the formula of the chemicals involved.

Example: CH₄ + Cl₂ → CH₃Cl + HCl

If you count all the atoms of each element in the left hand side of the equation, the numbers are equal to the numbers of atoms of those elements on the right hand side. The equation is "balanced" with equal numbers of reactant atoms (before the reaction, written on the left) and product atoms (after the reaction, written on the right).

Sometimes the equation is not balanced:

Example: CH₄ + Cl₂ → CH₂Cl₂ + HCl

What has happened here is that two Cl₂ have reacted and replaced two H atoms on the CH₄.

CH₄ + 2 Cl₂ → CH₂Cl₂ + 2 HCl

The large "2" values in "2 Cl₂" and "2 HCl" are called "coefficients" and multiply every atom in the formula that follows.

A big part of dealing with chemical equations in the exam is working out how to balance them.

HNO₃ + CaCO₃ → Ca(NO₃)₂ + CO₂ + H₂O is unbalanced.

2 HNO₃ + CaCO₃ → Ca(NO₃)₂ + CO₂ + H₂O is balanced.

There is no single simple method to balance chemical equations (sorry). A system you can follow is:

Balance one element at a time, starting with the first one in the equation:

Example 1: C₃H₈ + O₂ → CO₂ + H₂O

To balance the C, we need 3 CO₂ for every 1 C₃H₈:

1 C₃H₈ + O₂ → 3 CO₂ + H₂O

In the final equation, we won't write any "1" coefficients, but they are useful to include during balancing to keep track of which coefficients have been figured out.

To balance the H, we need 4 H₂O for every 1 C₃H₈:

1 C₃H₈ + O₂ → 3 CO₂ + 4 H₂O

To balance the O, we work backwards. We have 3 x 2 + 4 x 1 = 10 O atoms on the right-hand (product) side. How many O₂ reactant molecules are required to give us 10 product O atoms?

1 C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

We can lose the "1" coefficient now:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Example 2: O₂ + C₃H₆ → CO₂ + H₂O

The first element listed is O, but this is a problem element; It appears in more than one product. If an element appears more than once on either side of the equation, it's best to leave them until last.

Balancing C, and then H, we get:

O₂ + 1 C₃H₆ → 3 CO₂ + 3 H₂O

Now we finally balance the O, and it causes us a second problem. There are 3 x 2 + 3 x 1 = 9 product O atoms. How many O₂ reactant molecules are required? ⁹/₂ is not a whole number!

⁹/₂ O₂ + 1 C₃H₆ → 3 CO₂ + 3 H₂O

Actually, this is sort-of balanced. On special occasions (like when we discuss the reaction of exactly 1 mole of C₃H₆) we would use an equation with fractions. Normally we don't use fractions. The solution is to multiply every coefficient by 2:

9 O₂ + 2 C₃H₆ → 6 CO₂ + 6 H₂O

Example 3: Ba(NO₃)₂ + Al₂(SO₄)₃ → BaSO₄ + Al(NO₃)₃

In this example, we can treat NO₃ as an "element" because every N is part of NO₃. The same is true of the "element" SO₄.

Ba is easily balanced:

1 Ba(NO₃)₂ + Al₂(SO₄)₃ → 1 BaSO₄ + Al(NO₃)₃

Balancing NO₃ is tricky. We need to find the lowest common multiple of 2 and 3...

3 Ba(NO₃)₂ + Al₂(SO₄)₃ → 3 BaSO₄ + 2 Al(NO₃)₃

Notice that both coefficients for Ba are multiplied by 3. Once we've balanced Ba we don't want to balance it again.

We work backwards (from product to reactant) to balance Al:

3 Ba(NO₃)₂ + 1 Al₂(SO₄)₃ → 3 BaSO₄ + 2 Al(NO₃)₃

The SO₄ is also balanced. This often happens - balancing the easy elements automatically balances the rest.

3 Ba(NO₃)₂ + Al₂(SO₄)₃ → 3 BaSO₄ + 2 Al(NO₃)₃

Often we need to know the physical state of a chemical. There are four common "state symbols" used for this purpose:

(s) - a solid

(l) - a liquid

(g) - a gas

(aq) - dissolved in water (an "aqueous" solution)

Examples:

9 O₂(g) + 2 C₃H₆(g) → 6 CO₂(g) + 6 H₂O(l)

3 Ba(NO₃)₂(aq) + Al₂(SO₄)₃(aq) → 3 BaSO₄(s) + 2 Al(NO₃)₃(aq)

People often lose marks in exams because of ionic equations. Ionic equations look difficult but they are actually simplified versions of full equations. They involve ions in aqueous solution:

Example: 3 Ba(NO₃)₂(aq) + Al₂(SO₄)₃(aq) → 3 BaSO₄(s) + 2 Al(NO₃)₃(aq)

For each ionic compound in aqueous solution, we can write the ions separately:

3 Ba²⁺(aq) + 6 NO₃^-(aq) + 2 Al³⁺(aq) + 3 SO₄^2-(aq) → 3 BaSO₄(s) + 2 Al³⁺(aq) + 6 NO₃^-(aq)

Notice how we don't separate BaSO₄(s) into its ions.

The equation looks much more complicated, but wait! The 2 Al³⁺(aq) and 6 NO₃^-(aq) appear as reactants and products. These ions are not doing anything. we call them spectator ions and we can delete them from the equation:

3 Ba²⁺(aq) + 3 SO₄^2-(aq) → 3 BaSO₄(s)

The equation can be simplified even further now:

Ba²⁺(aq) + SO₄^2-(aq) → BaSO₄(s)

Check that the charges are balanced: The reactants are (1 x 2+) + (1 x 2-) = 0 and the products have 0 charge. All good.

The ionic equation shows what happens when any solution containing barium ions is mixed with any solution containing sulfate ions: We get a solid (precipitate) of barium sulfate. This is true of barium nitrate plus aluminium sulfate, or barium chloride plus magnesium sulfate, or barium ethanoate plus potassium sulfate. The list is almost infinite, but the ionic equation shows that it is the same reaction every time.