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

##### 1 Answer

Jun 6, 2016

The **percent by mass** (also written as

Note that you have to account for the *number of equivalents* of that atom in that compound, not just its molar mass.

So, the

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

#M_"H"# is themolar mass of hydrogen,#"1.0079 g/mol"# , and#M_("H","tot")# is thetotalmass 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 themolecular mass of ethane, which is the sum of the atomic masses of carbon and hydrogen.

To get

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

Hence, the

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

#= color(blue)(20.11%)#