Chemistry › Maths skills › Rearranging equations and units

Rearranging equations and units

Rearrange before substituting, convert every value into the units the equation expects, and check the units of the answer. Most wrong answers in a mole or gas calculation are a correct method with a units error.

Rearrange first, then substitute

An equation is a balance, and whatever is done to one side is done to the other. To make n the subject of pV = nRT, divide both sides by RT: n = pV ÷ RT. To make V the subject of c = n ÷ V, multiply both sides by V and then divide by c: V = n ÷ c. Rearranging in symbols first and substituting numbers last keeps the working readable, and a reader marking it can see where a slip happened.

A formula triangle does the same job for a three-term equation. For n = c × V, the n sits on top and c and V underneath, and covering the one wanted shows the rest: cover c and n sits over V. It works for n = m ÷ M and for n = V ÷ 24.0 as well. It does not work for pV = nRT, which has four symbols and a constant, so the both-sides method is the one to be sure of.

Units go through the same rearrangement as the symbols. If n = pV ÷ RT then the units are Pa × m³ ÷ (J K⁻¹ mol⁻¹ × K) = Pa m³ ÷ (J mol⁻¹), and since a joule is a pascal cubic metre, that is mol. Checking the units of an answer is the quickest check that a rearrangement is right.

The ideal gas equation and its units

pV = nRT with R = 8.31 J K⁻¹ mol⁻¹ needs p in Pa, V in m³ and T in K. Questions give pressure in kPa, volume in cm³ or dm³ and temperature in °C, so three conversions come before any arithmetic: multiply kPa by 1000; divide cm³ by 10⁶ or dm³ by 1000; add 273 to °C. A volume in cm³ left as it is gives an answer a million times too large, which is the error a units check catches at once.

Mr of a volatile liquid

0.250 g of a volatile liquid is vaporised in a gas syringe and occupies 95.0 cm³ at 100 °C and 101 kPa. Find its relative molecular mass.

Show the working

Convert first. p = 101 × 10³ = 1.01 × 10⁵ Pa. V = 95.0 ÷ 10⁶ = 9.50 × 10⁻⁵ m³. T = 100 + 273 = 373 K.

Rearrange: n = pV ÷ RT = (1.01 × 10⁵ × 9.50 × 10⁻⁵) ÷ (8.31 × 373) = 9.595 ÷ 3100 = 3.10 × 10⁻³ mol.

M = m ÷ n = 0.250 ÷ 3.10 × 10⁻³ = 80.8 g mol⁻¹, so Mr = 80.8, to the three figures the data allow; 81 only if the question asks for a whole number.

Moles, concentration and volume, and the two kinds of dm⁻³

n = c × V holds with V in dm³, and a titre comes in cm³, so divide by 1000: 23.50 cm³ of 0.100 mol dm⁻³ acid is 0.100 × 0.02350 = 2.35 × 10⁻³ mol. Writing n = c × V ÷ 1000 with V in cm³ is the same thing and is what most working shows. Going the other way, a concentration from moles and a volume in cm³ needs the volume divided by 1000 first, and a mark scheme awards the conversion as a step of its own.

A concentration can be in mol dm⁻³ or in g dm⁻³, and the molar mass converts between them: g dm⁻³ = mol dm⁻³ × M. 5.85 g dm⁻³ of sodium chloride, M = 58.5 g mol⁻¹, is 5.85 ÷ 58.5 = 0.100 mol dm⁻³. The units say which way to go: grams per litre divided by grams per mole leaves moles per litre.

Energy is the other place two units meet. Enthalpy changes are quoted in kJ mol⁻¹, q = mcΔT comes out in J, and entropy is in J K⁻¹ mol⁻¹. Before ΔG = ΔH − TΔS can be used, ΔS has to be divided by 1000 to put it in kJ K⁻¹ mol⁻¹, or ΔH multiplied by 1000 to put it in J. Either works; mixing them gives a ΔG that is wrong by a factor of a thousand and a feasibility conclusion that is wrong with it.

Two unit conversions inside one calculation

For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.2 kJ mol⁻¹ and ΔS = −199 J K⁻¹ mol⁻¹. Find ΔG at 298 K, and state whether the reaction is feasible at that temperature.

Show the working

ΔS in kJ: −199 ÷ 1000 = −0.199 kJ K⁻¹ mol⁻¹.

TΔS = 298 × (−0.199) = −59.3 kJ mol⁻¹.

ΔG = ΔH − TΔS = −92.2 − (−59.3) = −32.9 kJ mol⁻¹. Negative, so the reaction is feasible at 298 K.

Leaving ΔS in J gives ΔG = −92.2 − (−59 300), a nonsense number of +59 000 that would be read as not feasible. The sign of the answer turned on the conversion.

TRY IT: Four rearrangements with units

(a) The volume, in cm³, of 0.0200 mol of a gas at 298 K and 100 kPa.

(b) 0.250 mol dm⁻³ sulfuric acid, M = 98.1 g mol⁻¹, in g dm⁻³.

(d) −125 J K⁻¹ mol⁻¹ in kJ K⁻¹ mol⁻¹.

Check your answer

(a) V = nRT ÷ p = 0.0200 × 8.31 × 298 ÷ 100 000 = 4.95 × 10⁻⁴ m³ = 495 cm³.

(b) 0.250 × 98.1 = 24.5 g dm⁻³.

(d) −125 ÷ 1000 = −0.125 kJ K⁻¹ mol⁻¹.

The lessons that use this: Gases and the ideal gas equation · Concentration, volumetric solutions and titration calculations · The mole, the Avogadro constant and molar mass · Entropy and Gibbs free-energy change. Every skill the boards number, and where each is taught: maths skills.