Question

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

Answer 1

`%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 `"mass of all elements in the compound"` is simply equal to the sum of the atomic masses: `(6xx12.011(C)+8xx1.00794(H)+7xx15.999(O))*g*mol^-1` `=` `192.12*g*mol^-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.