DP Chemistry (last assessment 2024)

Question 17N.2.sl.TZ0.4

Date	November 2017	Marks available	[Maximum mark: 5]	Reference code	17N.2.sl.TZ0.4
Level	sl	Paper	2	Time zone	TZ0
Command term	Calculate, Determine	Question number	4	Adapted from	N/A

[Maximum mark: 5]

17N.2.sl.TZ0.4

Menthol is an organic compound containing carbon, hydrogen and oxygen.

(a)

Complete combustion of 0.1595 g of menthol produces 0.4490 g of carbon dioxide and 0.1840 g of water. Determine the empirical formula of the compound showing your working.

[3]

Markscheme

carbon: « $\frac{{0.4490\,{\text{g}}}}{{44.01\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}$ » = 0.01020 «mol» / 0.1225 «g»

hydrogen: « $\frac{{0.1840 \times 2}}{{18.02}}$ » = 0.02042 «mol» / 0.0206 «g»

oxygen: «0.1595 – (0.1225 + 0.0206)» = 0.0164 «g» / 0.001025 «mol»

empirical formula: C₁₀H₂₀O

Award [3] for correct final answer.

(b)

0.150 g sample of menthol, when vaporized, had a volume of 0.0337 dm³ at 150 °C and 100.2 kPa. Calculate its molar mass showing your working.

[2]

Markscheme

temperature = 423 K

«M $= \frac{{mRT}}{{pV}}$

«M $= \frac{{0.150\,{\text{g}} \times 8.31\,{\text{J}} {{\text{ K}}^{ - 1}}\,{\text{mol}}{}^{ - 1} \times 423\,{\text{K}}}}{{100.2\,{\text{kPa}} \times 0.0337\,{\text{d}}{{\text{m}}^3}}} =$ » 156 «g mol^–1»

Award [1] for correct answer with no working shown.

Accept “pV = nRT AND n = $\frac{m}{M}$ ” for M1.

Syllabus sections

Core

Core » Topic 1: Stoichiometric relationships » 1.2 The mole concept

Core » Topic 1: Stoichiometric relationships

Core » Topic 1: Stoichiometric relationships » 1.3 Reacting masses and volumes

Question 17N.2.sl.TZ0.4

Select a Test

Syllabus sections

Confirm action