Question

What is the percent by mass of hydrogen in the compound `C_2H_6`?

Answer 1

The percent by mass (also written as `"% w/w"`) 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 `"% w/w"` of `"H"` in `"C"_2"H"_6` is:

`\mathbf("% w/w H" = (M_("H","tot"))/(M_("C"_2H"_6))xx100%)`

where:

• `M_"H"` is the molar mass of hydrogen, `"1.0079 g/mol"`, and `M_("H","tot")` is the total mass contributed by hydrogen in the compound. There are `color(red)(6)` equivalents of hydrogen in ethane, so we need to account for that.
• `M_("C"_2"H"_6)` is the molecular mass of ethane, which is the sum of the atomic masses of carbon and hydrogen.

To get `M_("C"_2"H"_6)`, you can either look it up on google, or add it up yourself:

`M_("C"_2"H"_6) = 2xx12.011 + 6xx1.0079 = "30.069 g/mol"`

Hence, the `"% w/w"` of `"H"` is:

`color(blue)("% w/w H") = (color(red)(6)xx1.0079)/(30.069)xx100%`

`= color(blue)(20.11%)`