Question 6.Int.1: Combustion of 1.110 g of a gaseous hydrocarbon yields 3.613 ...
Combustion of 1.110 g of a gaseous hydrocarbon yields 3.613 g CO_2 and 1.109 g H_2 O, and no other products. A 0.288 g sample of the hydrocarbon occupies a volume of 131 mL at 24.8 °C and 753 mmHg. Write a plausible structural formula for a hydrocarbon corresponding to these data.
Analyze
Use the combustion data for the 1.110 g sample of hydrocarbon and the method of Example 3-6 on page 83 to determine the empirical formula. Use the P – V – T data in equation (6.13) for the 0.288 g sample to determine the molar mass and molecular mass of the hydrocarbon. By comparing the empirical formula mass and the molecular mass, establish the molecular formula. Now write a structural formula consistent with the molecular formula.
PV = \frac{mRT}{M} (6.13)
Calculate the number of moles of C and H in the 1.110 g sample of hydrocarbon based on the masses of CO_2 and H_2 O obtained in its combustion.
? mol C = 3.613 g CO_2 \times \frac{1 mol CO_2}{44.01 g CO_2} \times \frac{1 mol C}{1 mol CO_2} = 0.08209 mol C
? mol H = 1.109 g H_2 O \times \frac{1 mol H_2 O}{18.02 g H_2 O} \times \frac{2 mol H}{1 mol H_2 O} = 0.1231 mol H
Use these numbers of moles as the provisional subscripts in the formula.
C_{0.08209}H_{0.1231}
Divide each provisional subscript by the smaller of the two to obtain the empirical formula.
C_{\frac{0.08209}{0.08209}}H_{\frac{0.1231}{0.08209}}
CH_{1.500} = C_{2}H_{3}
To determine the molar mass, use a modified form of equation (6.13).
M = \frac{mRT}{PV} = \frac{0.288 g \times 0.08206 atm L K^{-1} mol^{-1} \times (24.8 + 273.2) K}{(753 mmHg \times 1 atm/760 mmHg) \times 0.131 L}
= 54.3 g mol^{-1}
The empirical formula mass is
\left(2 C \text { atoms } \times \frac{12.0 u }{1 C \text { atom }}\right) + \left(3 H \text { atoms } \times \frac{1.01 u }{1 H \text { atom }}\right) = 27.0 u
The molar mass based on the empirical formula 27.0 g mol^{-1}, is almost exactly onehalf the observed molar mass of 54.3 g mol^{-1}. The molecular formula of the hydrocarbon is
C_{2 \times 2}H_{2 \times 3} = C_{4}H_{6}
The four-carbon alkane is butane, C_4 H_{10}. Removal of 4 H atoms to obtain the formula C_4 H_6 is achieved by inserting two C-to-C double bonds.
\begin{array}{r c}\begin{matrix} H_2 C=CH-CH=CH_2 \text{or} H_2 C=C=CH-CH_3 \end{matrix} \end{array}
Two other possibilities involve the presence of a C-to-C triple bond.
H_3 C-C \equiv C-CH_3 \text{or} H_3 C-CH_2-C \equiv CH
Assess
The combination of combustion data and gas-law data yields a molecular formula with certainty. However, because of isomerism the exact structural formula cannot be pinpointed. All that we can say is that the hydrocarbon might have any one of the four structures shown, but it might be still another structure, for example, based on a ring of C atoms rather than a straight chain.