# How do you calculate the percentage composition of Carbon in C_2H_6?

Mar 10, 2016

It really is just asking you what a percent is, and you know that it is just a ratio of $\text{C}$ to ${\text{C"_2"H}}_{6}$.

The molar mass of $\text{C}$ is $\text{12.011 g/mol}$ and the molar mass of ${\text{C"_2"H}}_{6}$ is $12.011 \cdot 2 + 1.0079 \cdot 6 = \text{30.0694 g/mol}$.

So, knowing that ${\text{C}}_{2}$ in ${\text{C"_2"H}}_{6}$ means there are two carbons in the compound, we get:

(2xx"12.011 g/mol C")/("30.0694 g/mol C"_2"H"_6)xx100%

= ("24.022 g/mol C"_2)/("30.0694 g/mol C"_2"H"_6)xx100%

= color(blue)(79.89% "C in C"_2"H"_6)

CHALLENGE: What is the percent composition of $H$ in the compound?