(e) Chemical formulae, equations and calculations

Learning objectives

1.25Chemical equationsWrite word equations and balanced chemical equations (including state symbols):• for reactions studied in this specification• for unfamiliar reactions where suitable information is provided.1.26Relative formula massCalculate relative formula masses (including relative molecular masses) (Mr) from relative atomic masses (Ar)1.27The moleKnow that the mole (mol) is the unit for the amount of a substance 1.28Amount of substance calculationsUnderstand how to carry out calculations involving amount of substance, relative atomic mass (Ar) and relative formula mass (Mr)1.29Reacting massesCalculate reacting masses using experimental data and chemical equations 1.30Percentage yieldCalculate percentage yield 1.31Experimental formulaeUnderstand how the formulae of simple compounds can be obtained experimentally, including metal oxides, water and salts containing water of crystallisation 1.32Empirical and molecular formulaeKnow what is meant by the terms empirical formula and molecular formula 1.33Formula calculationsCalculate empirical and molecular formulae from experimental data 134C Solution concentration calculationsCarry out calculations involving amount of substance, solution volume and concentration in mol/dm³.135C Gas volume calculationsCarry out gas-volume calculations using a molar gas volume of 24 dm³ mol⁻¹ (24 000 cm³ mol⁻¹) at room temperature and pressure.1.36Metal oxide formula practicalPractical: know how to determine the formula of a metal oxide by combustion (e.g. magnesium oxide) or by reduction (e.g. copper(II) oxide)

A word equation names the reactants and products. A symbol equation replaces each name with the correct chemical formula. Balancing then changes coefficients only, so the number of atoms of every element is the same on both sides.

\ce{2Mg(s) + O2(g) -> 2MgO(s)}

Write correct formulae first; count each element; change the coefficient before a whole formula; recount; then add state symbols from the information given: (s), (l), (g) or (aq). For an unfamiliar reaction, use the supplied names, formulae and conditions rather than inventing products.

Never alter a subscript to make an equation balance: changing $\ce{H2O}$ to $\ce{H2O2}$ changes the substance. State symbols describe physical state; they do not balance atoms.

Relative formula mass, $M_r$ , is the sum of the relative atomic masses, $A_r$ , of every atom shown in a formula. The same calculation is called relative molecular mass when the substance consists of molecules.

M_r = \sum (\text{number of each atom} \times A_r)

For $\ce{Mg(NO3)2.6H2O}$ , count brackets and water separately: $24 + 2(14) + 6(16) + 6[2(1)+16] = 256$ . A coefficient before a formula is not part of one formula unit and must not be included in its $M_r$ .

$A_r$ and $M_r$ are relative values and have no unit. In mass calculations, molar mass has the same numerical value but the unit $\mathrm{g\,mol^{-1}}$ .

Amount of substance is measured in moles. Its unit name is mole and its symbol is mol. The symbol for amount of substance is $n$ .

Quantity	Symbol	Unit
Amount of substance	$n$	mol
Mass	$m$	g
Molar mass	$M$	$\mathrm{g\,mol^{-1}}$

The mole lets chemical amounts be compared using the coefficients in a balanced equation. For $\ce{N2 + 3H2 -> 2NH3}$ , the amount ratio is 1 mol : 3 mol : 2 mol.

A mole is a unit of amount, not a unit of mass. Different substances can have the same amount in moles but different masses because their molar masses differ.

Convert mass to amount by dividing by molar mass; convert amount to mass by multiplying. For an element use its $A_r$ as the numerical molar mass, and for a compound use its $M_r$ .

n = \frac{m}{M} \qquad m = nM

For $2.00\,\mathrm{g}$ of $\ce{CuSO4}$ with $M_r=159.5$ , $n=2.00/159.5=0.0125\,\mathrm{mol}$ . Check that division makes sense: the mass is much less than one molar mass, so the amount must be less than 1 mol.

Use the mass of the stated substance, not a mass copied from another stage. Keep grams with $\mathrm{g\,mol^{-1}}$ , and do not round intermediate results so early that the final answer changes.

A balanced equation gives mole ratios, not mass ratios. Convert the known mass to moles, apply the coefficient ratio, then convert the required moles back to mass.

Step	Operation
1	Balance the equation and identify known and required substances.
2	Known moles = known mass ÷ known molar mass.
3	Required moles = known moles × required coefficient ÷ known coefficient.
4	Required mass = required moles × required molar mass.

For $\ce{CH4 + 2O2 -> CO2 + 2H2O}$ , $32\,\mathrm{g}$ of methane is $32/16=2$ mol. The 1:2 ratio requires 4 mol of oxygen, so its mass is $4\times32=128\,\mathrm{g}$ .

Do not compare masses directly from equation coefficients. Coefficients compare amounts in moles; $M_r$ is needed on both sides of the mole-ratio step.

The theoretical yield is the maximum product predicted by the balanced equation. The actual yield is the product obtained in the experiment. Percentage yield compares the actual amount with that maximum.

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

Make sure actual and theoretical yields refer to the same product and use the same unit. If the theoretical yield is $29.9\,\mathrm{g}$ and the actual yield is $23.92\,\mathrm{g}$ , the yield is $23.92/29.9\times100=80.0\%$ .

The denominator is theoretical yield, so reversing the fraction is incorrect. A value above 100% usually signals wet or impure product, a measurement problem, or the wrong theoretical yield—not extra successful reaction.

Experimental formulae come from the mole ratio of the elements or components present. First use mass differences to isolate each component, then convert every component mass to moles and simplify the ratio to whole numbers.

Experiment	Mass obtained by difference	Ratio used
Metal heated in oxygen	oxide − metal = oxygen	metal mol : oxygen mol
Metal oxide reduced	oxide − metal = oxygen	metal mol : oxygen mol
Hydrogen and oxygen form water	use measured masses of both elements	hydrogen mol : oxygen mol
Hydrated salt heated	hydrate − anhydrous salt = water	salt mol : water mol

If $12.5\,\mathrm{g}$ of $\ce{CuSO4.xH2O}$ leaves $8.0\,\mathrm{g}$ of $\ce{CuSO4}$ , water lost is $4.5\,\mathrm{g}=0.25$ mol and salt is $8.0/159.5\approx0.050$ mol. The ratio 1:5 gives $\ce{CuSO4.5H2O}$ .

A mass ratio is not yet a formula ratio: divide each mass by the relevant $A_r$ or $M_r$ . For hydrated salts, the water of crystallisation is part of the crystal formula and is separated by a dot.

An empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. A molecular formula gives the actual number of atoms of each element in one molecule.

Molecular formula	Empirical formula	Relationship
$\ce{C2H4}$	$\ce{CH2}$	divide every subscript by 2
$\ce{C6H12O6}$	$\ce{CH2O}$	divide every subscript by 6
$\ce{C3H8}$	$\ce{C3H8}$	already simplest

The molecular formula is a whole-number multiple of the empirical formula. Therefore both formulae have the same elemental ratio, but only the molecular formula states the actual atom count in a molecule.

Simplest means all subscripts have no common whole-number factor greater than 1. Do not reduce one subscript without reducing all of them by the same factor.

Treat percentages as masses out of $100\,\mathrm{g}$ when no sample mass is given. Divide each elemental mass by its $A_r$ , divide all mole values by the smallest, and scale together if needed to obtain the simplest whole-number ratio.

Element	C	H	Cl
Percentage ÷ $A_r$	$38.4/12=3.2$	$4.8/1=4.8$	$56.8/35.5=1.6$
Divide by 1.6	2	3	1

k = \frac{M_r}{\text{empirical formula mass}} \qquad \text{molecular formula}=(\text{empirical formula})_k

Divide by relative atomic masses, not atomic numbers. Round only when the ratio is close to a simple whole number; if a ratio such as 1.5 remains, multiply every ratio by the same integer.

Molar concentration is amount of solute per volume of solution. The equation requires volume in $\mathrm{dm^3}$ when concentration is in $\mathrm{mol\,dm^{-3}}$ .

c = \frac{n}{V} \qquad n=cV \qquad 1,\mathrm{dm^3}=1000,\mathrm{cm^3}

For $25.0\,\mathrm{cm^3}$ of $0.0500\,\mathrm{mol\,dm^{-3}}$ solution, $V=25.0/1000=0.0250\,\mathrm{dm^3}$ , so $n=0.0500\times0.0250=0.00125\,\mathrm{mol}$ .

Do not substitute $\mathrm{cm^3}$ directly into $n=cV$ when $c$ is per $\mathrm{dm^3}$ . Convert the volume first, and distinguish the total solution volume from the volume of solvent alone.

At room temperature and pressure, one mole of gas occupies $24\,\mathrm{dm^3}$ , equivalent to $24{,}000\,\mathrm{cm^3}$ . Use the value whose volume unit matches the question.

V=n\times24,\mathrm{dm^3} \qquad n=\frac{V}{24,\mathrm{dm^3}}

For reacting gases, first convert the known quantity to moles, use the balanced-equation coefficient ratio, then multiply the required gas moles by the molar gas volume. For $60\,\mathrm{cm^3}$ of gas at rtp, $n=60/24000=0.0025\,\mathrm{mol}$ .

The $24\,\mathrm{dm^3\,mol^{-1}}$ value applies at room temperature and pressure. Do not mix $\mathrm{cm^3}$ with 24 or $\mathrm{dm^3}$ with 24,000.

Determine a metal oxide formula by measuring the metal and oxygen masses, converting both to moles, and finding their simplest ratio. Combustion adds oxygen to a metal; reduction removes oxygen from a metal oxide.

Route	Essential measurements and controls
Magnesium combustion	Weigh crucible and lid; add cleaned magnesium and reweigh; heat strongly; lift the lid briefly to admit oxygen while limiting product loss; cool and reweigh; repeat to constant mass.
Copper(II) oxide reduction	Weigh oxide; pass the reducing gas over the heated oxide; continue gas flow while cooling to prevent re-oxidation; reheat and reweigh to constant mass.

Subtract container masses to obtain sample masses. In combustion, oxygen mass is oxide minus metal; in reduction, oxygen mass is oxide minus metal remaining. Divide metal mass by the metal's $A_r$ and oxygen mass by 16, then simplify the mole ratio.

Constant mass shows that further heating causes no measurable change; it is stronger evidence of completion than heating for a fixed time. A closed lid can restrict oxygen, while leaving it off can lose solid product.

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(e) Chemical formulae, equations and calculations

Learning objectives

Write and balance chemical equations

Calculate relative formula mass

Use the mole as the unit of amount

Convert between mass and amount

Calculate reacting masses from an equation

Calculate percentage yield

Obtain formulae from experimental masses

Distinguish empirical and molecular formulae

Calculate empirical and molecular formulae

Calculate solution concentration in mol per dm³

Calculate gas volumes at room conditions

Determine a metal oxide formula experimentally

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