A-Level Chemistry October/November 2024 Q1(d)(i): Hydrated rhodium(III) chloride, RhCl3•xH2O, catalyses the conversion of ethene to but-2…

1 answer

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  accepted ✓

  Assume 100 g of the hydrated salt. Mass of $\text{H}_2\text{O} = 20.5 \text{ g}$ Mass of $\text{RhCl}_3 = 100 - 20.5 = 79.5 \text{ g}$

  $M_r(\text{RhCl}_3) = 102.9 + (3 \times 35.5) = 209.4$ $M_r(\text{H}_2\text{O}) = (2 \times 1.0) + 16.0 = 18.0$

  Ratio of moles of $\text{RhCl}_3$ : $\text{H}_2\text{O}$ $= \frac{\text{mass}}{M_r} : \frac{\text{mass}}{M_r}$ $= \frac{79.5}{209.4} : \frac{20.5}{18.0}$ $= 0.3796... : 1.1388...$

  Divide by the smallest value (0.3796...) to find the simplest whole number ratio: $= \frac{0.3796...}{0.3796...} : \frac{1.1388...}{0.3796...}$ $= 1 : 3.00$

  Therefore, the integer value of $x$ is 3.

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  How the marks are awarded

  • M1 — Awarded for the correct method to find the molar ratio. This is shown by setting up the calculation rac{79.5}{209.4} : rac{20.5}{18.0} or an equivalent algebraic expression as described in the mark scheme.
  • M2 — Awarded for the correct final integer value of $x = 3$, derived from the correct working.

  Common mistakes

  • Using incorrect relative atomic masses from the Periodic Table, for example using the mass number for Cl (35) instead of the relative atomic mass (35.5).
  • Inverting the ratio calculation, for example, calculating rac{M_r}{\text{mass}} instead of rac{\text{mass}}{M_r}.
  • Rounding intermediate values too early (e.g., rounding 0.3796 to 0.4), which can lead to an incorrect final integer ratio.
  • Making a calculation error, such as incorrectly calculating the percentage of RhCl3 as 100 - 20.5 = 89.5.

  Examiner tip: For percentage composition problems, always set up a clear ratio of moles by dividing the percentage (or mass) of each component by its respective molar mass.

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