The mass of an alkane with a volume of 11.2 liters under normal conditions is 15 g. What is this alkane?

univerkov | May 4, 2021 | education

Let’s find the amount of the substance of the unknown alkane:

v (CnH2n + 2) = V (CnH2n + 2) / Vm = 11.2 / 22.4 = 0.5 (mol).

On the other hand, the amount of substance of the unknown alkane:

v (CnH2n + 2) = m (CnH2n + 2) / M (CnH2n + 2) = 15 / (12 * n + 1 * (2n + 2)) = 15 / (14n + 2).

Let’s compose and solve the equation:

15 / (14n + 2) = 0.5, n = 2, whence the alkane formula is C2H6 (ethane).

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