Question

A scientist decides to determine the molecular formula of a compound containing only carbon and hydrogen. The compound is first analyzed by combustion analysis. 2.257 g of the compound was found to produce 4.21 L of carbon dioxide gas at 1.0 bar, 298.15 K, and 1.92 g of water. Determine the empirical formula. The molecular formula is determined using data from the combustion analysis and the Dumas method. Given the Dumas method data below, determine the molecular formula of the compound. In this question the masses of carbon dioxide and water are used to determine the empirical formula. The Dumas method data is used to determine the molecular formula. Volume of flask = 247.0 mL temperature of water = 100.0 °C Mass of empty flask = 71.8148 atmospheric pressure = 813.0 torr Mass of empty flask and compound vapour = 73.188 g Molecular formula of compound:

Answer 1

Total mass of the hydrocarbon taken = 2.257 g

Amount of carbon dioxide formed = V = 4.21 L

Pressure , P = 1.0 bar

Temperature = T = 298.15 K

Assuming the CO2 gas to be ideal, we can calculate the number of moles of CO2 formed as follows:

$PV= nRT \\ \Rightarrow n = \frac{PV}{RT} = \frac{1.0 \ bar \times 4.21 \ L}{8.314 \times 10^{-2} \ L \cdot bar \cdot K^{-1}\cdot mol^{-1} \times 298.15\ K } \approx 0.1698 \ mol$

Each mole of CO2 contains 1 mole of C.

Since all the C in CO2 must come from the hydrocarbon, the number of moles of C in the hydrocarbon must be 0.1698 mol.

Now, the amount of water formed = 1.92 g

Molar mass of water = 18.0 g/mol

Hence, number of moles of water formed is

$\frac{ mass}{molar\ mass} = \frac{1.92 \ g}{ 18.0 \ g/mol} \approx 0.1067 \ mol$

Note that each mole of water contains 2 mole of H.

Hence, the number of mole of H in the hydrocarbon is $2 \times 0.1067 \ mol = 0.2133 \ mol$

Hence, we can find the empirical formula of the compound by finding the whole number ratio of moles of C and H in the hydrocarbon.

$\frac{C}{H} = \frac{0.1698}{0.2133} = \frac{0.1698/0.1698}{0.2133/0.1698} = \frac{1}{1.25} = \frac{1 \times 4}{1.25 \times 4} = \frac{4}{5}$

Hence, the empirical formula of the compound is $C_4H_5$ .

Now, we will use the Dumas method data to calculate the molecular weight and then we will use the molecular weight to calculate the molecular formula.

Mass of empty flask = 71.814 g

Mass of flask and compound vapour = 73.188 g

Hence, mass of the compound vapour, m = 73.188 g - 71.814 g = 1.374 g

Volume of the flask i.e. Volume of the compound vapor = V = 247.0 mL = 0.2470 L

Temperature of the water bath i.e. temperature of the vapor , T= 100.0 C = 100+273.15 K = 373.15 K

Pressure = P = 813.0 torr

Using Ideal gas law to the vapour we can write

Replacing the number of moles , n by $\frac{m}{M}$

where m = mass of vapour and M = molar mass of vapour, we get

$PV = \frac{m}{M}RT$

Hence, putting in the value of all the known quantities, we can calculate the molecular weight of the hydrocarbon as follows:

$M = \frac{mRT}{PV} = \frac{1.374 \ g \times 62.363 \ L \ torr \ K^{-1} \ mol^{-1} \times 373.15 \ K}{813.0 \ torr \times 0.2470 \ L} \approx 159.48 \ g/mol$

Hence, the molecular weight of the hydrocarbon is 159 g/mol.

One unit of empirical formula $C_4H_5$ has a mass of

$12.01 \times 4 + 1.00 \times 5 = 53.04 \ g/mol$

Hence, the number of empirical units in the molecular formula is

$\frac{ 159.48 \ g/mol}{53.04 \ g/mol} = 3.00$

Hence, the molecular formula has 3 times more carbon and hydrogen than the empirical formula.

Hence, the molecular formula is $C_{3 \times 4} H_{3 \times 5} = {\color{Red} C_{12}H_{15}}$ .

Hence, the molecular formula of the compound is $C_{12}H_{15}$ .