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Mathematical Skills in Chemistry

Symbols, rearranging equations and logarithms: rearranging an equation

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Overview

Fourteen recall questions on symbols, rearranging equations and logarithms: rearranging an equation.

What you need to know

Answer each one before you reveal it.

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Key term

Why is the equilibrium sign drawn with two half-arrows of equal length in most equations, and what would unequal arrows suggest?

Numbers

Rearrange c = n/V to find the volume of 0.200 mol dm⁻³ solution that contains 0.0500 mol of solute.

Numbers

A first-order reaction has a rate of 4.50 × 10⁻⁴ mol dm⁻³ s⁻¹ when [A] = 0.150 mol dm⁻³. Calculate the rate constant.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    Why is the equilibrium sign drawn with two half-arrows of equal length in most equations, and what would unequal arrows suggest?

    Model answer

    Two equal half-arrows show that both the forward and reverse processes occur and that neither is being singled out as dominant. Drawing one half-arrow longer is sometimes used informally to suggest that the equilibrium position lies to that side, but the position of equilibrium is properly conveyed by the value of the equilibrium constant, not by the drawing.

  2. 2

    Rearrange n = m/M to make M the subject, and use it to find the molar mass of a substance of which 3.20 g is 0.100 mol.

    Model answer

    Multiplying both sides by M gives nM = m, and dividing by n gives M = m/n. Substituting, M = 3.20 ÷ 0.100 = 32.0 g mol⁻¹.

  3. 3

    Make m the subject of n = m/M and calculate the mass of 0.250 mol of sodium chloride. (M = 58.5 g mol⁻¹)

    Model answer

    Multiplying through by M gives m = nM. So m = 0.250 × 58.5 = 14.6 g to three significant figures.

  4. 4

    Rearrange c = n/V to find the volume of 0.200 mol dm⁻³ solution that contains 0.0500 mol of solute.

    Model answer

    V = n/c = 0.0500 ÷ 0.200 = 0.250 dm³, which is 250 cm³.

  5. 5

    Make T the subject of pV = nRT and calculate the temperature of 0.500 mol of gas occupying 1.20 × 10⁻² m³ at 1.01 × 10⁵ Pa. (R = 8.31 J K⁻¹ mol⁻¹)

    Model answer

    Dividing both sides by nR gives T = pV/(nR). T = (1.01 × 10⁵ × 1.20 × 10⁻²) ÷ (0.500 × 8.31) = 1212 ÷ 4.155 = 292 K.

  6. 6

    Show how pV = nRT can be rearranged to give molar mass, and calculate M for 0.320 g of a gas occupying 250 cm³ at 1.00 × 10⁵ Pa and 298 K.

    Model answer

    Since n = m/M, substituting gives pV = mRT/M and therefore M = mRT/(pV). Converting the volume, 250 cm³ = 2.50 × 10⁻⁴ m³. M = (0.320 × 8.31 × 298) ÷ (1.00 × 10⁵ × 2.50 × 10⁻⁴) = 792.4 ÷ 25.0 = 31.7 g mol⁻¹.

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