IB DP Chemistry Structure 1.4 Counting particles by mass: The mole HL Paper 2- Exam Style Questions - New Syllabus
Question
An organic compound, \(A\), has the following composition by mass when its only combustion products, carbon dioxide and water, are analysed.
Element | C / % | H / % |
---|---|---|
Percentage by mass | 71.93 | 12.10 |
(a) Outline why this compound is not a hydrocarbon. [1]
(b) Determine the empirical formula of \(A\). [2]
(c) A sample of the vapour of \(A\) at \(200.0~^\circ\text{C}\) and \(1.00\times10^{5}\ \text{Pa}\) has a density of \(2.544\times10^{3}\ \text{g m}^{-3}\). Determine the molar mass and the molecular formula of \(A\). [2]
(d) Identify the bond responsible for the absorption labelled B in the IR spectrum. [1]

Data booklet actual range used: O–H (alcohols/phenols) broad, strong \(3200\text{–}3600\ \text{cm}^{-1}\).
(e) \(A\) can be converted to compound \(E\), which has a higher molecular mass, by heating under reflux with acidified potassium dichromate(VI), \(\mathrm{K_2Cr_2O_7}\). Identify one functional group present in \(E\) based on this information only. [1]
(f) Deduce a possible structural formula for \(A\) consistent with the evidence presented. [1]
▶️ Answer/Explanation
Markscheme (with detailed working)
(a)
Hydrocarbons contain only C and H. \(71.93\% + 12.10\% = 84.03\%\), so the missing \(15.97\%\) must be another element (O) → not a hydrocarbon. A1
Hydrocarbons contain only C and H. \(71.93\% + 12.10\% = 84.03\%\), so the missing \(15.97\%\) must be another element (O) → not a hydrocarbon. A1
(b)
\(\%O = 100 – 71.93 – 12.10 = 15.97\%\). Assume \(100~\text{g}\):
\[ n(\text{C})=\frac{71.93}{12.01}=5.99,\quad n(\text{H})=\frac{12.10}{1.008}=12.0,\quad n(\text{O})=\frac{15.97}{16.00}=0.998. \] Divide by smallest \(\approx 0.998\): C:H:O \(\approx 6:12:1\).
\(\boxed{\text{Empirical formula }= \mathrm{C_6H_{12}O}}\). M1 A1
\(\%O = 100 – 71.93 – 12.10 = 15.97\%\). Assume \(100~\text{g}\):
\[ n(\text{C})=\frac{71.93}{12.01}=5.99,\quad n(\text{H})=\frac{12.10}{1.008}=12.0,\quad n(\text{O})=\frac{15.97}{16.00}=0.998. \] Divide by smallest \(\approx 0.998\): C:H:O \(\approx 6:12:1\).
\(\boxed{\text{Empirical formula }= \mathrm{C_6H_{12}O}}\). M1 A1
(c)
Use \(M=\dfrac{dRT}{p}\): \(d=2.544\times10^{3}\ \text{g m}^{-3}\), \(R=8.31~\text{J K}^{-1}\text{mol}^{-1}\), \(T=200.0^\circ\text{C}=473~\text{K}\), \(p=1.00\times10^{5}~\text{Pa}\).
\[ M=\frac{(2.544\times10^{3})(8.31)(473)}{1.00\times10^{5}} \approx 1.00\times10^{2}\ \text{g mol}^{-1}. \] \(M_\text{emp}(\mathrm{C_6H_{12}O})\approx 100\ \text{g mol}^{-1}\) ⇒ molecular formula is the same: \(\boxed{\mathrm{C_6H_{12}O}}\). M1 A1
Use \(M=\dfrac{dRT}{p}\): \(d=2.544\times10^{3}\ \text{g m}^{-3}\), \(R=8.31~\text{J K}^{-1}\text{mol}^{-1}\), \(T=200.0^\circ\text{C}=473~\text{K}\), \(p=1.00\times10^{5}~\text{Pa}\).
\[ M=\frac{(2.544\times10^{3})(8.31)(473)}{1.00\times10^{5}} \approx 1.00\times10^{2}\ \text{g mol}^{-1}. \] \(M_\text{emp}(\mathrm{C_6H_{12}O})\approx 100\ \text{g mol}^{-1}\) ⇒ molecular formula is the same: \(\boxed{\mathrm{C_6H_{12}O}}\). M1 A1
(d)
\(\boxed{\text{O–H}}\) (broad stretch \(3200\text{–}3600\ \text{cm}^{-1}\)). A1
\(\boxed{\text{O–H}}\) (broad stretch \(3200\text{–}3600\ \text{cm}^{-1}\)). A1
(e)
Oxidation with acidified \(\mathrm{K_2Cr_2O_7}\) gives higher \(M_r\) (adds O) → consistent with formation of a \(\boxed{\text{carboxylic acid (–COOH)}}\). A1
Oxidation with acidified \(\mathrm{K_2Cr_2O_7}\) gives higher \(M_r\) (adds O) → consistent with formation of a \(\boxed{\text{carboxylic acid (–COOH)}}\). A1
(f)
One valid structure matching all evidence (primary alcohol, \( \mathrm{C_6H_{12}O} \), IR O–H), e.g. \[ \text{HO–CH}_2\text{–CH}_2\text{–CH}_2\text{–CH{=}CH–CH}_3 \] (hex-5-en-1-ol). Alternatives with the same formula and an –OH group also acceptable. A1
One valid structure matching all evidence (primary alcohol, \( \mathrm{C_6H_{12}O} \), IR O–H), e.g. \[ \text{HO–CH}_2\text{–CH}_2\text{–CH}_2\text{–CH{=}CH–CH}_3 \] (hex-5-en-1-ol). Alternatives with the same formula and an –OH group also acceptable. A1
Total Marks: 8