# 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?

Jul 3, 2017

"%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 $100 \cdot g$ of unknown compound, and get the so-called $\text{empirical formula}$, which is the simplest whole number ratio defining constituent elements in a species.

And thus for $100 \cdot g$ of this compound..........there are....

$\frac{42.9 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 3.57 \cdot m o l \cdot C$. And........

$\frac{57.1 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1} = 3.57 \cdot m o l \cdot O$.

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

This could be $C O$ or cyclopropanetrione, ${C}_{3} {O}_{3}$, or ${C}_{4} {O}_{4}$; these beasts are fiercely reactive, and it would be hard to put them in a bottle.