Question

A gaseous hydrogen and carbon containing compound is decomposed and formed to contain 82.66% carbon and 17.34% hydrogen by mass. The mass of 158 mL of the gas, measured at 556 mmHg and 25C, is 275g. What is the molecular formula of the compound?

Answer 1

Explanation:

Mole percentage of carbon = $\frac{percentage given}{molar mass of carbon}$

                                              = $\frac{82.66}{12}$

                                              = 6.89

Mole percentage of hydrogen = $\frac{percentage given}{molar mass of hydrogen}$

                                               = $\frac{17.34}{1}$

                                               = 17.34

Now, dividing mole percentage of both the atoms by 6.89.

Then,     C = 1 and H = $\frac{17.34}{6.89}$ = 2.5

Hence, empirical formula is $C_{2}H_{5}$.

As, it is given that P = 556 mm Hg. Convert mm Hg into atm as follows.

                        $\frac{556 mm Hg \times atm}{760 mm Hg}$

                             = 0.7316 atm

Volume is given as 158 mL. So, in liter volume is $\frac{158}{1000}$ equals 0.158 L.

According to ideal gas equation, PV = nRT

                      $0.7316 atm \times 0.158 L = \frac{mass}{molar mass} \times 0.082 atm L/mol K \times 298 K$          

                       $0.7316 atm \times 0.158 L = \frac{0.275 g}{molar mass} \times 0.082 atm L/mol K \times 298 K$        

                   molar mass = 58.2 g

Hence, molecular weight of $C_{2}H_{5}$ is $12 \times 2 + 5$ = 29.

Therefore, $(C_{2}H_{5})_{n}$ = 58

                               29 × n = 58

                                   n = 2

Thus, molecular formula of the compound is $C_{4}H_{10}$.