Chemical Formulae, Equations and Calculations

Chemistry revision notes · Edexcel · 4CH1

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1. Word equations and balanced symbol equations

Conservation of mass

During a chemical reaction, atoms are rearranged but never created or destroyed
This is the law of conservation of mass: the total mass of reactants equals the total mass of products
It is why every chemical equation must be balanced

Word equations

A word equation uses the full chemical names of the substances involved:

reactants → products

The left of the arrow shows the reactants (the starting substances)
The right of the arrow shows the products (the new substances formed)
The arrow (→) is read as "produces" or "reacts to give"
Reaction conditions (e.g. heat, the name of a catalyst) can be written above the arrow

Example. Magnesium burns in oxygen to form magnesium oxide:

magnesium + oxygen → magnesium oxide

Symbol (chemical) equations

A symbol equation uses chemical formulae instead of full names
Three rules to remember:
1. The diatomic non-metals must be written as molecules: H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂
2. can be written in brackets after each formula:
  - (s) = solid
  - (l) = liquid
  - (g) = gas
  - (aq) = aqueous (dissolved in water)
3. The equation must be balanced — the same number of each kind of atom must appear on both sides

Example (combustion of sulfur):

S (s) + O₂ (g) → SO₂ (g)

Balancing equations

Work across the equation from left to right, checking each element in turn
Adjust the coefficient (the number written in front of each formula) until the counts match on both sides
Never change the subscripts inside a formula — that would change what the substance is. Adding a 2 to make H₂O into H₂O₂ turns water into hydrogen peroxide
Treat polyatomic groups that survive the reaction unchanged (NO₃⁻, SO₄²⁻, CO₃²⁻, etc.) as single units rather than counting their individual atoms
Balance any element that appears on its own last

Example. Balancing the combustion of methane:

CH₄ + O₂ → CO₂ + H₂O

Carbon: 1 on each side ✓
Hydrogen: 4 on the left, only 2 on the right → put a 2 in front of H₂O:

CH₄ + O₂ → CO₂ + 2H₂O

Oxygen: 2 on the left, now 4 on the right (2 from CO₂ + 2 from 2H₂O) → put a 2 in front of O₂:

CH₄ + 2O₂ → CO₂ + 2H₂O

Re-check: C 1 = 1, H 4 = 4, O 4 = 4 ✓

Common exam question

Writing a chemical equation and its state symbols

Question: Write, complete or balance the chemical equation for a reaction described in words, or add the state symbols to a printed equation (1–2 marks).

An equation to write or balance is set on 19 of the 23 papers. For two marks, the first is for the correct formula of every substance and the second for balancing, and the second depends on the first, so check each formula (diatomic O₂, H₂ and Cl₂; the ion ratio in a salt) before the numbers in front. Multiples and fractions, such as 3½O₂, are accepted. State symbols that were not asked for are ignored even when wrong.

State symbols are asked for on 12 of the 23 papers, for one or two marks. Every solution, including a dilute acid, is (aq); water is (l), though (g) is also accepted for steam leaving a hot reaction; capital letters are allowed. Give (l) only to a pure liquid, never to a solution.

2. Relative formula mass (Mr)

The (M_r) of a substance is the sum of the relative atomic masses of every atom in its chemical formula
The relative atomic mass (A_r) of each element is read from the periodic table (it's the larger of the two numbers in each box)
For an element written as just its symbol (e.g. Fe), M_r = A_r
For a compound, multiply each A_r by the number of atoms of that element in the formula, then add the products

Substance	Atoms present	Calculation	M_r
H₂O	2 H + 1 O	(2 × 1) + 16	18
CO₂	1 C + 2 O	12 + (2 × 16)	44
CaCO₃	1 Ca + 1 C + 3 O	40 + 12 + (3 × 16)	100
H₂SO₄	2 H + 1 S + 4 O	(2 × 1) + 32 + (4 × 16)	98
Mg(NO₃)₂	1 Mg + 2 N + 6 O	24 + (2 × 14) + (6 × 16)	148

By the law of conservation of mass, the total M_r on the reactant side (with each M_r multiplied by its coefficient) equals the total M_r on the product side

Common exam question

Counting atoms and calculating a relative formula mass

Question: Calculate the relative formula mass (Mr) of a compound from its formula, or give the number of atoms, or of different elements, in the formula (1–2 marks).

An Mr calculation is on 11 of the 23 papers and an atom or element count on 9. For Mr, multiply each relative atomic mass by the number of that atom and add them: a one-mark version wants only the total; a two-mark version gives one mark for the written sum and one for the total; one scheme followed a wrong sum through to the total mark and another did not, so get the sum right. Units are ignored. Brackets are the trap: in Mg(NO₃)₂ everything inside the bracket is doubled, and a sum that leaves the bracket undoubled loses the first mark and keeps the second only where the total is followed through from your own sum.

For the counts, add every subscript for the atoms (a dot formula's water counts too: CuSO₄·5H₂O has 21 atoms) and count each element symbol once for the elements (four in that formula). "Different elements" and "atoms" are different numbers.

Worked example

Mr of a compound containing brackets

A student needs the relative formula mass of calcium hydroxide, Ca(OH)₂. Calculate it.

Solution:

Identify all atoms: 1 Ca, and the bracket (OH)₂ means 2 O and 2 H
A_r values: Ca = 40, O = 16, H = 1
M_r = (1 × 40) + (2 × 16) + (2 × 1)
M_r = 40 + 32 + 2 = 74

3. The mole and mass conversions

What a mole is

A (symbol mol) is the chemist's unit for counting chemical particles
One mole of any substance always contains exactly the same number of particles — the Avogadro constant:

$N_\text{A} = 6.02 \times 10^{23}$

The "particles" can be atoms, molecules, ions, electrons or formula units — whichever the substance is made of
- 1 mole of helium (He) → 6.02 × 10²³ atoms
- 1 mole of water (H₂O) → 6.02 × 10²³ molecules
- 1 mole of sodium chloride (NaCl) → 6.02 × 10²³ formula units

Molar mass

The molar mass is the mass, in grams, of one mole of a substance
Numerically:
- For an element, molar mass = A_r in grams
- For a compound, molar mass = M_r in grams
Examples:
- 1 mol of carbon (A_r = 12) weighs 12 g
- 1 mol of water (M_r = 18) weighs 18 g
- 1 mol of sodium chloride (M_r = 58.5) weighs 58.5 g

Converting between mass and moles

The key relationship is:

$n = \frac{m}{M_r}$

where n = number of moles, m = mass in grams, and M_r = molar mass in g/mol
Rearranged:
- mass = moles × M_r
- M_r = mass ÷ moles

Formula triangle linking mass, moles and molar mass: mass sits on top, with moles and M below, so moles = mass ÷ M and mass = moles × M — Source: Calculating moles and mass by Save My Exams

Example. How many moles are in 5.0 g of NaOH?

M_r(NaOH) = 23 + 16 + 1 = 40
n = 5.0 / 40 = 0.125 mol

Example. What is the mass of 2.5 mol of glucose, C₆H₁₂O₆?

M_r(C₆H₁₂O₆) = (6 × 12) + (12 × 1) + (6 × 16) = 180
m = 2.5 × 180 = 450 g

Common exam question

Counting particles with the Avogadro constant

Question: Given that one mole contains 6.0 × 10²³ particles, calculate the number of atoms, molecules or electrons in a stated mass or amount, or the mass of one atom (2–3 marks).

Set in 6 of the 23 papers. Find the moles first, from mass ÷ Mr or from the equation, then multiply by the value of the constant the question quotes, which may be 6.0, 6.02 or 6.022 × 10²³. The last mark is often for the form of the answer: standard form, or the number of significant figures the question names. A correct answer with no working scores in full. The mark most often lost is a factor that comes from the formula or the equation: five waters in each formula unit, sixty atoms in each molecule, an Mr of 2 (not 1) for hydrogen gas. Going the other way, the mass of one atom is the molar mass divided by the constant.

Exam tip

Working, significant figures and standard form

In every calculation on these papers a correct final answer scores full marks even without working, but the working is what earns credit when the answer is wrong, and a slip can only be carried forward if the later steps are on the page. Round nothing until the end: a value given to one significant figure is penalised, and so is incorrect rounding. When the question says "to three significant figures" or "in standard form", that instruction is the final mark, and an answer written out in full instead of in standard form has stopped at two of three marks.

4. Reacting masses and percentage yield

Reacting masses

A balanced equation tells you the ratio in which substances react
Using that ratio, you can work out the mass of any product (or reactant) from the mass of any other species in the reaction

The three-step recipe:

Find the moles of the substance you are given (mass ÷ M_r)
Use the equation's mole ratio to convert to moles of the substance you want
Convert moles back to mass (moles × M_r)

Example. What mass of magnesium oxide forms when 2.4 g of magnesium burns completely in oxygen?

2Mg + O₂ → 2MgO

Moles of Mg = 2.4 / 24 = 0.10 mol
Mole ratio Mg : MgO = 2 : 2 = 1 : 1, so 0.10 mol of MgO is formed
M_r(MgO) = 24 + 16 = 40, so mass of MgO = 0.10 × 40 = 4.0 g

Percentage yield

The actual yield is the mass of product you actually recover in the lab
The theoretical yield is the mass that would form if the reaction went perfectly with no loss — calculated from the balanced equation and reacting masses
The compares the two:

$\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

Yields are never 100 % in practice. Common reasons:
- Some of the reactant is left unreacted
- The reaction is a reversible reaction so the products partially turn back into reactants
- Product is lost during filtration, washing, drying or transfer between vessels
- Side reactions form other products you didn't want

Example. A student decomposes 10.0 g of calcium carbonate by heating:

CaCO₃ (s) → CaO (s) + CO₂ (g)

They recover 4.5 g of calcium oxide. What is the percentage yield?

M_r(CaCO₃) = 100, M_r(CaO) = 56
Theoretical: moles of CaCO₃ = 10.0 / 100 = 0.10 mol → 0.10 mol of CaO → 0.10 × 56 = 5.6 g theoretical
Percentage yield = (4.5 / 5.6) × 100 = 80.4 %

Common exam question

Calculating a maximum mass and a percentage yield

Question: Use the equation to calculate the maximum mass of a product from a given mass or amount of reactant, sometimes in tonnes, and then the percentage yield from the mass actually obtained (2–4 marks).

A maximum-mass calculation is on 7 of the 23 papers and a percentage yield on 5. The marks follow the three steps: moles of the given substance (mass ÷ Mr), the mole ratio from the equation, then moles × Mr of the product. Skipping the ratio, the commonest slip, costs exactly one mark: an answer double or half the right value keeps the rest. Any number of significant figures is accepted except one. Tonnes can be converted to grams (× 10⁶) or kept throughout.

For the yield, one mark is for the theoretical mass (or moles) and one for actual ÷ theoretical × 100. Dividing the actual mass by the moles of reactant instead of by the theoretical mass scores nothing.

Worked example

Showing that a reactant is in excess

A mass of 2.0 g of calcium carbonate is added to 50 cm³ of dilute hydrochloric acid containing 0.050 mol of HCl. Show by calculation that the acid is in excess.

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ (M_r of CaCO₃ = 100)

Solution:

Moles of CaCO₃ = 2.0 ÷ 100 = 0.020 mol
The equation needs 2 mol of HCl for every 1 mol of CaCO₃, so 0.020 mol of CaCO₃ needs 2 × 0.020 = 0.040 mol of HCl
0.050 mol of HCl is present, which is more than 0.040 mol, so the acid is in excess

The other route scores equally: 0.050 mol of HCl could react with 0.025 mol of CaCO₃, which is 2.5 g, more than the 2.0 g used. Whichever you choose, the second mark is the comparison, so compare the same kind of quantity (moles with moles, or grams with grams) and end by stating which reactant is in excess.

5. Empirical and molecular formulae

Definitions

The gives the actual number of each kind of atom in one molecule of the substance
- e.g. ethanoic acid: C₂H₄O₂
The gives the simplest whole-number ratio of the atoms
- e.g. ethanoic acid: CH₂O
For organic molecules, the empirical formula and the molecular formula are often different
For ionic compounds, the formula given (e.g. NaCl, MgCl₂) is always the empirical formula

Common exam question

Empirical formula from a molecular formula

Question: Give the empirical formula of a compound whose molecular formula is shown, or explain what the term means using that compound as the example (1–2 marks).

Set in 6 of the 23 papers. Divide every subscript by the highest common factor: C₆H₁₂O₆ becomes CH₂O, and the symbols may be written in any order. When the subscripts share no factor, as in C₃H₈, the empirical and molecular formulae are the same. For the definition, the mark is for the simplest whole-number ratio of the atoms of each element, and a second mark, when asked, is for applying it to the compound named. A related one-mark question asks why every alkene has the empirical formula CH₂: because the number of hydrogen atoms is always double the number of carbon atoms.

Calculating an empirical formula

Use a table to keep the working tidy:

List each element in its own column
Write the mass (or % by mass) of each
Write each element's A_r
Calculate moles = mass ÷ A_r
Divide every mole value by the smallest — this gives the simplest ratio
If the ratio isn't whole numbers, multiply all values by the same factor (×2, ×3, etc.) until they are

Example. A 24 g sample of magnesium burns fully and the magnesium oxide formed weighs 40 g. Find its empirical formula.

	Mg	O
Mass / g	24	40 − 24 = 16
A_r	24	16
Moles	24 / 24 = 1	16 / 16 = 1
Divide by smallest	1	1

Empirical formula: MgO

Example. A compound contains 40.0 % carbon, 6.7 % hydrogen and 53.3 % oxygen by mass. Find its empirical formula.

	C	H	O
% mass	40.0	6.7	53.3
A_r	12	1	16
Moles	40.0 / 12 = 3.33	6.7 / 1 = 6.7	53.3 / 16 = 3.33
Divide by smallest (3.33)	1	2	1

Empirical formula: CH₂O

Common exam question

Empirical formula from percentages or masses

Question: Given the percentage composition by mass, or the masses of each element from an experiment, calculate the empirical formula, sometimes then the molecular formula from a given Mr (2–4 marks).

The most common calculation in the topic, on 12 of the 23 papers. Divide each mass or percentage by the element's relative atomic mass (first mark), divide the results by the smallest (second) and write the formula (third). Dividing by atomic numbers, or the Ar by the mass, scores nothing for those marks; in the practical version only the first mark, for the masses from the weighings, survives. When the ratio is not whole, multiply through: 1 : 1.33 becomes 3 : 4. Keep at least two significant figures in the mole values. A correct formula alone scores full marks. For the molecular formula, divide the given Mr by the empirical formula's Mr and multiply every subscript by the result (one or two further marks).

Calculating a molecular formula

A molecular formula is always a whole-number multiple of the empirical formula
To find it, the question must also give the actual M_r of the compound

Procedure:

Calculate M_r of the empirical formula
Divide: n = (real M_r) ÷ (empirical M_r)
Multiply every subscript in the empirical formula by n

Example. A compound has empirical formula CH₂O and M_r = 180. Find its molecular formula.

M_r(CH₂O) = 12 + 2 + 16 = 30
Multiplier: 180 / 30 = 6
Molecular formula: C₆H₁₂O₆ (glucose)

Practical: formula of magnesium oxide

Aim. Find the empirical formula of magnesium oxide by burning a known mass of magnesium in air.

Method:

Weigh the empty crucible with its lid
Add a coil of clean magnesium ribbon (around 10 cm); reweigh the crucible + lid + ribbon; calculate the mass of magnesium used
Heat strongly with a Bunsen burner. Lift the lid briefly every few seconds so that more air (oxygen) can reach the ribbon, but lower it again quickly so very little white magnesium oxide smoke escapes
Continue heating until no further glow can be produced and the mass has stopped changing
Allow the crucible to cool and reweigh; calculate the mass of magnesium oxide formed

Calculation: apply the empirical-formula procedure with the two columns "Mg" and "O":

mass of Mg from step 2
mass of O = (mass of MgO) − (mass of Mg)
moles of each = mass ÷ A_r
simplest mole ratio → empirical formula

Typical result: ratio Mg : O = 1 : 1, so the formula is MgO.

Apparatus for finding the formula of a metal oxide: a crucible of metal oxide sits on a tripod above a heating Bunsen flame — Source: Investigating metal oxide formulas by Save My Exams

The specification allows the formula of a metal oxide to be found by combustion, as here, or by reduction, for example passing a stream of methane or hydrogen over heated copper(II) oxide so that the oxygen is removed; the mass lost gives the mass of oxygen and the mass remaining gives the mass of copper, which are then turned into a mole ratio in the same way.

Reduction apparatus: methane passes over heated copper(II) oxide in a tube, oxygen is removed leaving copper, and excess gas burns at a small hole in the tube — Source: Investigating metal oxide formulas by Save My Exams

Common exam question

Heating to constant mass

Question: Describe what the student should do to make sure all the water has been driven off (or the oxide fully reduced or oxidised), or state how the results show that the reaction is complete (1–2 marks).

Asked on 9 of the 23 papers, across the hydrated salt, magnesium oxide and metal oxide reduction practicals. The two-mark answer is two actions: heat again and reweigh, then repeat until the mass no longer changes; the phrase "heat to constant mass" earns both marks on its own. If the question asks why the heating is repeated, the reason is to make sure all the water has been given off, or that the reaction is complete. When a results table gives the mass at intervals, the reaction is finished where consecutive readings are the same.

Practical: formula of a hydrated salt

A hydrated salt is a crystalline solid that has fixed numbers of water molecules locked into its structure
- The trapped water is called the water of crystallisation
- The formula is written with a dot: e.g. CuSO₄·5H₂O (copper(II) sulfate pentahydrate)

Aim. Find x in CuSO₄·xH₂O by heating to drive off the water.

Method:

Weigh an empty evaporating dish
Add a known mass of the blue hydrated copper(II) sulfate crystals; reweigh
Heat gently on a tripod and gauze, stirring with a glass rod, until the salt is completely white (all water has been driven off as steam — this is the anhydrous salt)
Cool and reweigh; the mass lost is the mass of water

Calculation: apply the empirical-formula procedure with "anhydrous salt" and "water" as the two components.

M_r(CuSO₄) = 159.5, M_r(H₂O) = 18
Moles of each = mass ÷ M_r
Simplest ratio gives 1 : x

For copper(II) sulfate, the result is CuSO₄·5H₂O (so x = 5).

Apparatus for the hydrated salt experiment: blue hydrated salt in an evaporating dish is heated over a Bunsen flame to drive off the water of crystallisation — Source: Simple compound formulae by Save My Exams

Common exam question

Finding x in a hydrated salt from heating results

Question: Use the masses of the tube empty, with the hydrated salt and with the anhydrous salt to calculate x in a formula such as MgSO₄·xH₂O (3–5 marks).

Set in 7 of the 23 papers, with the Mr of the anhydrous salt and of water given. The marks are the steps: mass of anhydrous salt and mass of water lost by subtraction (one mark each when asked separately), moles of each (mass ÷ Mr), then moles of water ÷ moles of salt = x. A correct x with no working scores full marks. Follow-through from a wrong mass is allowed, but only when the final ratio comes out as a whole number, because x must be an integer: a ratio of 4.12 is given as 4. Do not round the mole values to one significant figure on the way, as that is penalised. In a "show that" version the value is printed, so only the written steps can score.

Worked example

Finding x in hydrated sodium carbonate

A student heats a sample of hydrated sodium carbonate, Na₂CO₃·xH₂O, in a test tube to constant mass and records the results in the table. Use them to find x. (M_r of Na₂CO₃ = 106; M_r of H₂O = 18)

Measurement	Mass in g
empty test tube	18.25
test tube and Na₂CO₃·xH₂O before heating	21.11
test tube and Na₂CO₃ after heating to constant mass	19.31

Solution:

Mass of anhydrous Na₂CO₃ = 19.31 − 18.25 = 1.06 g
Mass of water lost = 21.11 − 19.31 = 1.80 g
Moles of Na₂CO₃ = 1.06 ÷ 106 = 0.0100 mol
Moles of H₂O = 1.80 ÷ 18 = 0.100 mol
x = 0.100 ÷ 0.0100 = 10, so the formula is Na₂CO₃·10H₂O

The alternative route the mark schemes also accept: the hydrated salt weighs 21.11 − 18.25 = 2.86 g, so its Mr is 2.86 ÷ 0.0100 = 286, and 286 − 106 = 180 is the mass of water in one mole, which is 180 ÷ 18 = 10 waters.

6. Concentration of solutions

Solute, solvent and solution

The solute is the substance that dissolves
The solvent is the liquid in which the solute dissolves
The solution is the homogeneous mixture that results
Most reactions in solution chemistry happen between solutes dissolved in water (an aqueous solution)

What concentration means

Concentration is the amount of solute dissolved in a fixed volume of the solution
The greater the amount of solute in a given volume, the more concentrated the solution
Edexcel uses two units of concentration:
- g/dm³ — grams of solute per dm³ of solution
- mol/dm³ — moles of solute per dm³ of solution

The key formula

For concentration in mol/dm³:

$c = \frac{n}{V}$

where:
- c = concentration in mol/dm³
- n = moles of solute
- V = volume of solution in dm³
Rearranged:
- = × volume (in dm³)
- volume (in dm³) = moles ÷ concentration

Formula triangle linking moles, concentration and volume: moles sits on top with concentration and volume below, so moles = concentration × volume — Source: Solution concentration by Save My Exams

Unit conversion:

cm³ → dm³: divide by 1000
dm³ → cm³: multiply by 1000

Converting between cm³ and dm³: divide by 1000 to go from cm³ to dm³, and multiply by 1000 to go from dm³ to cm³ — Source: Solution concentration by Save My Exams

Example. What is the concentration, in mol/dm³, of a solution made by dissolving 8.0 g of NaOH in water to make 250 cm³ of solution?

M_r(NaOH) = 40, so moles of NaOH = 8.0 / 40 = 0.20 mol
Volume = 250 / 1000 = 0.25 dm³
c = 0.20 / 0.25 = 0.80 mol/dm³

Example. 25.0 cm³ of 0.10 mol/dm³ sodium carbonate is exactly neutralised by dilute hydrochloric acid. The equation is:

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

If 20.0 cm³ of the acid is required, what is its concentration?

Moles of Na₂CO₃ = 0.10 × (25.0 / 1000) = 0.00250 mol
Mole ratio HCl : Na₂CO₃ = 2 : 1, so moles of HCl = 2 × 0.00250 = 0.00500 mol
Volume of HCl = 20.0 / 1000 = 0.020 dm³
c(HCl) = 0.00500 / 0.020 = 0.25 mol/dm³

Common exam question

Concentration from a titration result

Question: Given the volumes of both solutions and the concentration of one, calculate the concentration of the other in mol/dm³, or the volume, mass or amount in moles of one reactant (3–4 marks).

Set in 10 of the 23 papers, every one of them a Paper 2. The three marks are the three steps: moles of the known solution (concentration × volume in dm³, so divide the cm³ by 1000), the mole ratio from the equation, then divide by the other volume in dm³ (or multiply by Mr for a mass). A fourth mark, where there is one, is for rounding to the significant figures asked for. Skipping the ratio, or using it the wrong way round, costs exactly one mark in every scheme: an answer double or half the right one keeps the rest. An answer that is just one volume divided by the other, with the concentration never used, scores nothing. Forgetting the ÷ 1000 costs one mark, and one significant figure is penalised once.

7. Volumes of gases at RTP

Avogadro's law

Avogadro's law states that, at the same temperature and pressure, equal volumes of any gases contain equal numbers of molecules
Consequence: at the same temperature and pressure, the ratio of any two gases is the same as the volume ratio of those gases

Molar gas volume

RTP stands for room temperature and pressure, taken as 20 °C and 1 atm
At RTP, one mole of any gas occupies a volume of 24 dm³ (equivalently 24 000 cm³)
This value is called the molar gas volume at RTP

The key formula

For a gas at RTP:

$n = \frac{V}{24} \qquad (V \text{ in dm}^3)$

or equivalently, when V is given in cm³:

$n = \frac{V}{24\,000} \qquad (V \text{ in cm}^3)$

Rearranged:
- volume (dm³) = moles × 24
- volume (cm³) = moles × 24 000

Formula triangle linking gas volume in dm³, amount of gas in moles and the molar gas volume 24 dm³/mol at RTP — Source: Calculate Gas Volumes by Save My Exams

Example. What volume does 88 g of CO₂ occupy at RTP?

M_r(CO₂) = 44, so moles = 88 / 44 = 2.0 mol
Volume = 2.0 × 24 = 48 dm³

Example. Methane burns completely:

CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (g)

What volume of oxygen reacts with 100 cm³ of methane, and what total volume of gas products forms? All volumes are measured at the same temperature and pressure.

Mole ratio CH₄ : O₂ = 1 : 2, so volume of O₂ needed = 2 × 100 = 200 cm³
Total moles of products per mole of CH₄ = 1 (CO₂) + 2 (H₂O) = 3
Total volume of products = 3 × 100 = 300 cm³

Common exam question

Gas volume at RTP from a mass, or a mass from a volume

Question: Using a molar volume of 24 dm³ (24 000 cm³) at rtp, calculate the volume of gas formed from a given mass of reactant, or the mass or moles of a substance from a measured gas volume (2–4 marks).

Set in 11 of the 23 papers, all of them Paper 2. The molar volume is printed on every paper but one (which instead asks you to show by calculation that a mole of gas occupies about 24 000 cm³), so the marks are for the chain: moles of the given substance, the mole ratio, then moles × 24 dm³ (or volume ÷ 24 000 cm³ going the other way). Match the molar volume to the unit on the answer line: when 24 dm³ is given and cm³ are wanted, the × 1000 is the final mark. Masses in tonnes usually come with "give your answer in standard form", which is again the final mark. The ratio is the step most often skipped and costs one mark; a wrong number of moles is carried forward through the rest.

Worked example

Gas volume at RTP

Calculate the volume of carbon dioxide, measured at RTP, produced when 25 g of calcium carbonate decomposes completely.

CaCO₃ → CaO + CO₂ (M_r of CaCO₃ = 100; molar gas volume = 24 dm³/mol at RTP)

Solution:

Moles of CaCO₃ = mass ÷ M_r = 25 ÷ 100 = 0.25 mol
The mole ratio CaCO₃ : CO₂ is 1 : 1, so moles of CO₂ = 0.25 mol
Volume = moles × 24 = 0.25 × 24 = 6 dm³

Mixing units is the main source of dropped marks: use 24 dm³/mol for an answer in dm³, or 24 000 cm³/mol for cm³. If the question wanted cm³, multiply the dm³ answer by 1000 (6 dm³ = 6000 cm³).