Question #120317

When 0.72g of a liquid vaporized at 110.0oC and 0.967atm, the gas occupies a volume of 0.417L. The empirical formula of the gas is CH2. What is the molecular formula of the gas?

Expert's answer

To find the molecular formula of the gas, its molecular weight must be calculated.

According to the ideal gas law:

*pV = nRT or pV = mRT / Mr*

where p - pressure, V - volume, n - number of moles, R - gas constant, T - temperature, m - mass, Mr - molecular weight.

From here:

*Mr = mRT / pV*

As *R = 0.082057 L atm / K mol* and *T = 110.0 °C = 383.15 K*:

*Mr(gas) = (0.72 g × 0.082057 L atm / K mol × 383.15 K) / (0.967 atm × 0.417 L) = 56 g/mol*

As a result, molecular weight of the gas is 56 g/mol and its empirical formula is CH_{2}. From here:

*(12g/mol × x) + (2 × 1 g/mol × x) = 56 g/mol*

where x - index coeeficient, 12g/mol and 1 g/mol - molecular weight of carbon and hydrogn respectively. From here:

*12x + 2x = 56*

*14x = 56*

*x = 4*

As a result, the molecular formula of the gas is C_{4}H_{8}

**Answer: C**_{4}**H**_{8}

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