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

1. Calculate the moles of carbon from carbon dioxide 1 mark :

$$\begin{aligned} \text { Moles of } \mathrm{CO}_2= & \frac{0.4490}{44.01}=0.01020 \mathrm{~mol}\left(1 \mathrm{~mol} \mathrm{CO}_2 \text { contains } 1 \mathrm{~mol} \mathrm{C}\right) \\ & \text { Moles of carbon }(\mathrm{C})=0.01020 \mathrm{~mol} \end{aligned}$$

2. Calculate the moles of hydrogen from water 1 mark :

$$\begin{gathered} \text { Moles of } \mathrm{H}_2 \mathrm{O}=\frac{0.1840}{18.02}=0.01021 \mathrm{~mol} \\ \text { Moles of hydrogen }(\mathrm{H})=2 \times 0.01021=0.02042 \mathrm{~mol} \end{gathered}$$

3. Calculate the mass of oxygen in menthol by difference 1 mark :

• Total mass of menthol $=0.1595 \mathrm{~g}$.
• Mass of carbon $=0.01020 \times 12.01=0.1225 \mathrm{~g}$.
• Mass of hydrogen $=0.02042 \times 1.008=0.02057 \mathrm{~g}$.
• Mass of oxygen $=0.1595-(0.1225+0.02057)=0.01643 \mathrm{~g}$.

4. Calculate the moles of oxygen 1 mark :

$$\text { Moles of oxygen }(O)=\frac{0.01643}{16.00}=0.001027 \mathrm{~mol}$$

5. Find the simplest mole ratio 1 mark :

• Divide each mole quantity by the smallest number of moles (oxygen, 0.001027 ):

$$\begin{gathered} \mathrm{C}: \frac{0.01020}{0.001027} \approx 9.93(\approx 10) \\ \mathrm{H}: \frac{0.02042}{0.001027} \approx 19.88(\approx 20) \\ \mathrm{O}: \frac{0.001027}{0.001027}=1 \end{gathered}$$

6. Write the empirical formula 1 mark :

• The empirical formula is $\mathrm{C}_{10} \mathrm{H}_{20} \mathrm{O}$.

temperature = 423 K

$M = \frac{mRT}{pV}$ $M = \frac{0.150\text{ g} \times 8.31\text{ J K}^{-1}\text{ mol}^{-1} \times 423\text{ K}}{100.2\text{ kPa} \times 0.0337\text{ dm}^3}$ $M = 156\text{ g mol}^{-1}$ 1 mark

Accept "pV = nRT AND n = m/M" for M1.

1 mark for correct answer with no working shown.