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Mathematical Skills in Chemistry

Units, standard form and proportion in chemical calculation: volume, equilibrium, rate and energy units

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Overview

Converting volumes for c = n/V and pV = nRT, deducing the units of Kc and of rate constants, matching kJ and J in Gibbs calculations, and the units of Kw and Ka.

What you need to know

Answer each one before you reveal it.

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Numbers

Calculate the amount, in moles, of solute in 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution.

Key term

A student writes that 250 cm³ of solution is 0.250 m³. What has gone wrong, and what are the correct values in dm³ and m³?

Numbers

Why must volume be in m³ rather than dm³ when the ideal gas equation is used with pressure in pascals?

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    A pipette delivers 25.0 cm³ of a solution. What is this volume in dm³, and why does the conversion matter before using c = n/V?

    Model answer

    25.0 cm³ is 25.0 ÷ 1000 = 0.0250 dm³, or 2.50 × 10⁻² dm³. The conversion matters because concentration in chemistry is quoted in mol dm⁻³, so the volume substituted into n = c × V must be in dm³; leaving it in cm³ inflates the number of moles by a factor of 1000.

  2. 2

    Calculate the amount, in moles, of solute in 25.0 cm³ of 0.100 mol dm⁻³ sodium hydroxide solution.

    Model answer

    V = 25.0 cm³ = 0.0250 dm³. n = c × V = 0.100 mol dm⁻³ × 0.0250 dm³ = 2.50 × 10⁻³ mol. The units cancel to leave mol, which confirms the conversion was made correctly.

  3. 3

    A student writes that 250 cm³ of solution is 0.250 m³. What has gone wrong, and what are the correct values in dm³ and m³?

    Model answer

    The student has used the cm³ to dm³ factor but written the answer as m³. 250 cm³ = 0.250 dm³, and since 1 dm³ = 1 × 10⁻³ m³, the volume is 2.50 × 10⁻⁴ m³. Three different volume units are in routine use and each conversion step must be tracked separately.

  4. 4

    Why must volume be in m³ rather than dm³ when the ideal gas equation is used with pressure in pascals?

    Model answer

    R = 8.31 J K⁻¹ mol⁻¹ is expressed in SI base units, and the product pV must come out in joules. One pascal is one newton per square metre, so Pa × m³ gives N m, which is a joule. Using dm³ with pascals gives an answer 1000 times too large.

  5. 5

    Calculate the volume occupied by 0.0100 mol of an ideal gas at 101 000 Pa and 298 K. Give your answer in m³ and in cm³. (R = 8.31 J K⁻¹ mol⁻¹)

    Model answer

    V = nRT/p = (0.0100 × 8.31 × 298) ÷ 101 000 = 24.76 ÷ 101 000 = 2.45 × 10⁻⁴ m³. Since 1 m³ = 1 × 10⁶ cm³, this is 245 cm³. The data are given to three significant figures, so the answer is given to three.

  6. 6

    Deduce the units of the equilibrium constant Kc for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).

    Model answer

    Kc = [NH₃]² ÷ ([N₂][H₂]³). Substituting mol dm⁻³ for every concentration gives (mol dm⁻³)² ÷ (mol dm⁻³)⁴ = (mol dm⁻³)⁻², so the units are mol⁻² dm⁶. The units must be derived from the expression each time; they are not fixed for all equilibria.

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