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JEE Advanced 2018 Paper 1 · Chemistry · Qualitative and Practical Chemistry · medium

An organic compound of molar mass 93 g mol^-1 contains only carbon, hydrogen and nitrogen. In a Kjeldahl determination 0.930 g of the compound…

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An organic compound of molar mass 93 g mol−193\ \mathrm{g\,mol^{-1}} contains only carbon, hydrogen and nitrogen. In a Kjeldahl determination 0.9300.930 g of the compound liberates ammonia that is absorbed in 30.030.0 mL of 1.00 N1.00\ \mathrm{N} H2SO4\mathrm{H_2SO_4}; the unreacted acid needs 20.020.0 mL of 1.00 N1.00\ \mathrm{N} NaOH. A separate 0.9300.930 g sample on complete combustion gives 2.642.64 g of CO2\mathrm{CO_2}. Taking C=12, H=1, N=14\mathrm{C=12,\ H=1,\ N=14}, the number of hydrogen atoms in one molecule of the compound is ______.
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Answer: 7

Kjeldahl: acid consumed by ammonia =30.0−20.0=10.0=30.0-20.0=10.0 meq, so the sample contains 10.010.0 mmol of N (all nitrogen converted to NH3\mathrm{NH_3}), i.e. 0.1400.140 g N. Combustion: 2.64 g CO2/44=0.06002.64\ \mathrm{g\ CO_2}/44=0.0600 mol C, i.e. 0.7200.720 g C. The sample mass is 0.9300.930 g, so hydrogen is 0.930−0.720−0.140=0.0700.930-0.720-0.140=0.070 g, i.e. 0.0700.070 mol H. Atom ratio C : H : N =0.060:0.070:0.010=6:7:1=0.060:0.070:0.010=6:7:1, giving empirical formula C6H7N\mathrm{C_6H_7N} with mass 72+7+14=9372+7+14=93. The molar mass equals the empirical mass, so the molecular formula is C6H7N\mathrm{C_6H_7N} (aniline-type) and contains 7 hydrogen atoms.
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