Question

What is the percent by mass of hydrogen in the compound \[{C_2}{H_6}\]\[?\]

Answer 1

Hint: Here we have to calculate mass percent. First, we know that \[1.0079\;g/mol\] is the mass of hydrogen in one mole. Then to calculate the mass percent of an element in a compound, we divide the mass of the element in \[1\] mole of the compound by the compound's molar mass and multiply the result by \[100\].

Complete answer:
The percent by mass is basically the fraction or percent of the total molecular mass that is contributed to by the atom(s) of choice.
Note that you have to account for the number of equivalents of that atom in that compound, not just its molar mass.
So, the percent by mass of \[H\] in \[{C_2}{H_6}\] \[ = \dfrac{{Total\;molar\;mass\;of\;hydrogen\operatorname{in} \;{C_2}{H_6}}}{{Total\;molar\;mass\;of\;\;{C_2}{H_6}}} \times 100\].
Since, the molar mass of hydrogen is \[1.0079\;g/mol\]. Also, there are \[6\] equivalents of hydrogen in ethane. Hence, the total mass contributed by hydrogen in the compound \[{C_2}{H_6}\] is equal to \[6\] times the molar mass of hydrogen. i.e., the total mass or atomic mass of hydrogen is \[6 \times 1.0079\;g/mole\] in the compound \[{C_2}{H_6}\].
 The molecular mass of ethane \[{C_2}{H_6}\], which is the sum of the atomic masses of carbon and hydrogen. We know that the molar mass of carbon per one mole is \[12.011\;g/mole\].
Hence molar mass of \[{C_2}{H_6}\]\[ = 2 \times 12.011 + 6 \times 1.0079 = 30.069\;g/mole\]
Hence, the percent by mass of hydrogen in \[{C_2}{H_6}\]\[ = \dfrac{{6 \times 1.0079}}{{30.069}} \times 100 = 20.11\% \]

Note:
Note that the ethane is a colourless, odourless, gaseous hydrocarbon (compound containing only hydrogen and carbon). Ethane is the simplest hydrocarbon since it contains only one carbon–carbon bond in its structure. It has a tetrahedral molecular geometry