Question

A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume 10.0 L (measured at NTP) of this welding gas is found to weigh 11.6 g. Calculate
(i) Empirical formula
(ii)- Molar Mass of the gas
(iii)- Molecular formula

Answer 1

Hint First find the percentage composition of carbon and hydrogen by finding the mass of carbon in carbon dioxide and hydrogen in water. Then find the number of moles of carbon and hydrogen, and take the ratio. This will give the empirical formula. Empirical is the simplest form of the molecule but the molecular formula is the exact formula.

Complete answer:
In the question the mass of carbon dioxide formed is 3.38g, so the mass of carbon present in the sample will be:
$\dfrac{12}{44}\text{ x 3}\text{.38 = 0}\text{.92 g}$
And the mass of water formed is 0.690 g, so the mass of hydrogen present in the sample will be:
$\dfrac{2}{18}\text{ x 0}\text{.690 = 0}\text{.077 g}$
So, the percentage of carbon will be:
$C%=\dfrac{0.92}{0.92+0.077}\text{ x 100 = 92}\text{.3 }\!\!%\!\!\text{ }$
The percentage of hydrogen will be:
$H%=\dfrac{0.077}{0.92+0.077}\text{ x 100 = 7}\text{.7 }\!\!%\!\!\text{ }$
(i)- To calculate the empirical formula, we have to calculate the number of moles. Moles of carbon will be:
$=\dfrac{92.3}{12}=7.7$
So, there are 7.7 moles of carbon.
Moles of hydrogen will be:
$=\dfrac{7.7}{1}=7.7$
So, there are 7.7 moles of hydrogen.
Now taking the ratio we get: 7.7 : 7.7 = 1: 1
The empirical formula will be $CH$.
(ii)- We are given that volume 10.0 L (measured at NTP) of this welding gas is found to weigh 11.6 g. So, 22.4 L of gas at N.T.P will have:
$=\dfrac{11.6}{10.0}\text{ x 22 = 26 g/mol}$
Therefore, the molar mass will be 26 g/mol.
(iii)- The molar mass of the empirical formula of the compound is
$12+1=13$
And the actual molar mass is 26 g/mol.
So, the ratio will be = $\dfrac{26}{13}=2$
The molecular formula will be:
$2(CH)={{C}_{2}}{{H}_{2}}$
Therefore, the gas is ethyne.

Note: When we get the ratio of the molar mass of the empirical formula and actual molar mass, then we get the actual number of atoms in the compound. If there were other products then the percentage composition of carbon and hydrogen would be different.