An unknown compound X contains only carbon, hydrogen, and oxygen. — Physical Chemistry Chemistry Question
Problem Context
An unknown compound X contains only carbon, hydrogen, and oxygen.
When burned, 1.00 g of X produces 2.00 g of carbon dioxide and 0.818 g of water. What is the empirical formula of X?
Model Answer
2.00 g CO2/(44.01 g mol–1) = 0.0454 mol C = 0.545 g C
0.818 g H2O/(18.02 g mol–1) = 0.0454 mol H2O = 0.0908 mol H = 0.0915 g H
1.00 g X – 0.545 g C – 0.0915 g H = 0.364 g O = 0.0227 mol O
Mol ratio of C : H : O = 2 : 4 : 1, so empirical formula is C2H4O
A 1.68 g sample of X is introduced into a 2.00 L evacuated flask. The flask is slowly heated until the liquid evaporates completely, which occurs at 62.3 °C. At this temperature, the pressure in the flask is 200.0 mmHg. What is the molar mass of X, and what is its molecular formula?
Model Answer
PV = nRT
[(200.0 mmHg)/(760 mmHg/atm)]*(1.013 bar atm–1) = 0.2666 bar
(0.2666 bar)(2.00 L) = n(0.08314 L bar mol–1 K–1)(335.5 K)
n = 0.0191 mol
M = (1.68 g)/(0.0191 mol) = 87.9 g mol–1
Since the molar mass of C2H4O = 44.1 g mol-1, the molecular formula must be double the empirical formula, C4H8O2
The melting point of X is 11.8 °C and its normal boiling point is 101.1 °C. Calculate the standard enthalpy and entropy of vaporization of X.
Model Answer
We have two temperatures at which we know the vapor pressure: 0.2666 bar at 335.5 K (from part b) and 1 atm (1.013 bar) at 374.3 K.
∆H°vap = 35.92 kJ mol–1
At 374.3 K, Keq = 1.013, so
∆G°vap = –RTln(1.013) = –0.0402 kJ mol–1 = ∆H°vap – T∆S°vap
–0.0402 kJ mol–1 = 35.92 kJ mol–1 – (374.3 K)∆S°vap
∆S°vap = 96.1 J mol–1 K–1
What types of intermolecular forces are likely present in liquid X? Justify your answer.
Model Answer
There are of course London dispersion forces present, and these are probably the only forces that are significant, given that the enthalpy of vaporization is modest and the entropy is about what one would expect from Trouton's rule. Hydrogen bonding in particular is unlikely because the boiling point is lower than one would expect for a molecule of this size with hydrogen bonding present. Thus, for example, n-propanol, C3H7OH, has a very similar normal boiling point of 97 °C despite having one fewer carbon and one fewer oxygen atom than this compound.
The unit cell of crystalline X at its melting point has a volume of 241.5 Å3 and the density of the solid is 1.211 g cm-3. How many molecules of X are present in the unit cell?
Model Answer
= 2 molecules/unit cell
The density of liquid X at its melting point is 1.034 g cm–3. Will the melting point increase, decrease, or stay the same as the pressure is increased?
Model Answer
Higher pressures favor the denser phase, which for X is the solid. Therefore the melting point will increase with increasing pressure.