0.1152 g of a compound containing carbon, hydrogen, nitrogen and oxygen are burned in excess oxygen. — Stoichiometry / Gas Laws Chemistry Question
Problem Context
0.1152 g of a compound containing carbon, hydrogen, nitrogen and oxygen are burned in excess oxygen. The gases produced are treated further to convert nitrogen-containing products into N2. The resulting mixture of CO2, H2O and N2 and excess O2 is passed through a CaCl2 drying tube, which gains 0.09912 g. The gas stream is bubbled through water where the CO2 forms H2CO3. Titration of this solution to the second endpoint with 0.3283 M NaOH requires 28.81 mL. The excess O2 is removed by reaction with copper metal and the N2 is collected in a 225.0 mL measuring bulb where it exerts a pressure of 65.12 mmHg at 25 ˚C. In a separate experiment the molar mass of this compound is found to be approximately 150 g.mol-1.
Calculate the number of moles of i. H2O
Model Answer
moles H2O = 0.09912 g × (1 mol H2O / 18.02 g H2O) = 5.501 × 10–3 mol H2O
ii. CO2
Model Answer
moles CO2 = 0.3283 mol NaOH / L soln × 0.02881 L soln × (1 mol CO2 / 2 mol NaOH) = 4.729 × 10–3 mol CO2
iii. N2
Model Answer
moles N2 = pV / RT = (65.12 torr)(0.2250 L) / ((62.4 L·torr·mol-1·K-1)(298.15 K)) = 7.879 × 10–4 mol N2
Determine the mass in the original compound of i. C
Model Answer
g C = 4.729 × 10–3 mol CO2 × (1 mol C / 1 mol CO2) × (12.01 g C / 1 mol C) = 0.05680 g C
ii. H
Model Answer
g H = 5.501 × 10–3 mol H2O × (2 mol H / 1 mol H2O) × (1.008 g H / 1 mol H) = 0.01109 g H
iii. N
Model Answer
g N = 7.879 × 10–4 mol N2 × (2 mol N / 1 mol N2) × (14.01 g N / 1 mol N) = 0.02208 g N
iv. O
Model Answer
g O = g sample – (g C + g H + g N) = 0.1152 g – (0.05680 g + 0.01109 g + 0.02208 g) = 0.1152 g – 0.08997 g = 0.02523 g O
Find the empirical formula of the compound.
Model Answer
C : 0.05680 g C / 12.01 g C/mol = 4.729 × 10–3 / 1.576 × 10–3 = 3.001 ≈ 3
H : 0.01109 g H / 1.008 g H/mol = 1.100 × 10–2 / 1.576 × 10–3 = 6.980 ≈ 7
N : 0.02208 g N / 14.01 g N/mol = 1.576 × 10–3 / 1.576 × 10–3 = 1.000 = 1
O : 0.02523 g O / 16.00 g O/mol = 1.577 × 10–3 / 1.576 × 10–3 = 1.001 ≈ 1
Therefore the empirical formula is C3H7NO
Find the molecular formula.
Model Answer
The molar mass of the empirical formula is 73.10 g/mol.
Compare this molar mass to the measured molar mass.
Therefore the molecular formula is C6H14N2O2