A substance X is 18.93% C, 25.21% O, and 55.86% a halogen by mass. X is a gas at 70 °C and 1.00 bar — Stoichiometry and Gases Chemistry Question
Problem Context
A substance X is 18.93% C, 25.21% O, and 55.86% a halogen by mass. X is a gas at 70 °C and 1.00 bar pressure, with a vapor density of 4.45 g/L.
A 1.00 g sample of X is added to an excess of aqueous sodium hydroxide, which causes evolution of a colorless gas. The gas is collected over water at an ambient pressure of 1.00 bar and a temperature of 21 °C, and the volume of the gas is found to be 197 mL. Addition of a dilute solution of barium nitrate to the aqueous solution results in the formation of a white precipitate.
What is the molar mass of X?
Model Answer
At 70 °C and 1.00 bar pressure, a 1-L gas sample would contain: n = PV/RT = (1.00 bar)(1.00 L)/(0.08314 L bar mol-1 K-1)(343.15 K)
n = 0.03505 mol
The molar mass is thus 4.45 g/0.03505 mol = 127.0 g/mol
What is the molecular formula of X?
Model Answer
C: 0.1893 × 127.0 g/mol = 24.0 g/mol C = 2 mol C/mol X
O: 0.2521 × 127.0 g/mol = 32.0 g/mol O = 2 mol O/mol X
Halogen: 0.5586 × 127.0 g/mol = 70.9 g halogen/mol X
This would correspond to 3.73 mol F, 2.00 mol Cl, 0.887 mol Br, or 0.559 mol I. The halogen must therefore be Cl, and the molecular formula is C2O2Cl2.
Draw the Lewis structure of X.
How many moles of gas are evolved in this reaction? (The vapor pressure of water at 21 °C is 18.7 mm Hg.)
Model Answer
P = 1.000 bar – 18.7 mm Hg•(1.013 bar/760 mm Hg) = 0.975 bar
n = PV/RT = (0.975 bar)(0.197 L)/(0.08314 L bar mol-1 K-1)(294.15 K) = 7.85 × 10-3 mol gas
Write a balanced net ionic equation for the reaction of X with aqueous sodium hydroxide, and explain how this reaction is consistent with the number of moles of gas calculated in part (d), and with the observation of a precipitate upon addition of barium nitrate.
Model Answer
The only reasonable colorless gases that could be produced would be CO and CO2; the latter would immediately form carbonate ion, which would explain the white precipitate with Ba2+ (BaCO3). The production of 7.85 × 10-3 mol gas per (1.00 g/127.0 g/mol) = 7.87 × 10-3 mol X is consistent with production of one mol CO per mol of X, suggesting the other C in C2O2Cl2 forms carbonate. In the absence of an oxidant and in basic solution, the only reasonable fate of Cl is Cl–(aq), giving a net balanced reaction of:
C2O2Cl2 + 4 OH–(aq) → 2 Cl–(aq) + CO3 2-(aq) + CO(g) + 2 H2O(l)