# What is the percentage of carbon by mass in citric acid?

Sep 25, 2016

%C $=$ (6xx12.011*g)/(192.12*g*mol^-1)xx100% $=$ 37.5%.

#### Explanation:

I think you have intuitively realized the correct approach.

%"element" $=$ "mass of element in the compound"/"mass of all elements in the compound"xx100%

The $\text{mass of all elements in the compound}$ is simply equal to the sum of the atomic masses: $\left(6 \times 12.011 \left(C\right) + 8 \times 1.00794 \left(H\right) + 7 \times 15.999 \left(O\right)\right) \cdot g \cdot m o {l}^{-} 1$ $=$ $192.12 \cdot g \cdot m o {l}^{-} 1$ (the which of course is the molecular mass of the compound).

And thus %C $=$ (6xx12.011*g)/(192.12*g*mol^-1)xx100% $=$ 37.5%.

And %H $=$ ??

And %O $=$ ??

Of course all the individual percentages must sum to 100%. Why?

Alternatively see here.