Question

What is the percent composition of a carbon-oxygen compound, given that a 95.2 g sample of the compound contains 40.8 g of carbon and 54.4 g of oxygen?

Answer 1

`"%element"="mass of element"/"mass of compound"xx100%`

Explanation:

And thus.........

`"%C"=(40.8*g)/(95.2*g)xx100%=42.9%`.

`"%O"=(54.4*g)/(95.2*g)xx100%=57.1%`.

Why should the individual percentage sum to `100%`? Do they......

We can divide the elemental percentages by the atomic masses if we assume `100*g` of unknown compound, and get the so-called `"empirical formula"`, which is the simplest whole number ratio defining constituent elements in a species.

And thus for `100*g` of this compound..........there are....

`(42.9*g)/(12.011*g*mol^-1)=3.57*mol*C`. And........

`(57.1*g)/(15.999*g*mol^-1)=3.57*mol*O`.

We divide thru by the element with the smallest number of moles, and thus we get an empirical formula of `CO`.........

This could be `CO` or cyclopropanetrione, `C_3O_3`, or `C_4O_4`; these beasts are fiercely reactive, and it would be hard to put them in a bottle.