What is the percent composition of the element hydrogen in the compound methane, CH4?
The molar mass of CH4 M=12.011+4×1.008=16.043gmolM=12.011+4\times1.008=16.043\frac{g}{mol}M=12.011+4×1.008=16.043molg
%H=4×Ar(H)M (CH4)×100%=4×1.008 g/mol16.043 g/mol×100%=25.13%\%H=\frac{4\times{A_r(H)}}{M\ (CH_4)}\times100\%=\frac{4\times1.008\ g/mol}{16.043\ g/mol}\times100\%=25.13\%%H=M (CH4)4×Ar(H)×100%=16.043 g/mol4×1.008 g/mol×100%=25.13%
Answer: 25.13%
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