# Why is it inaccurate to say "CO is 50 percent oxygen"?

Mar 5, 2017

As a binary compound it is 50% oxygen by composition.........

#### Explanation:

$\text{......but it is not 50% by mass, i.e.}$

$\text{% composition of oxygen by mass in carbon monoxide} =$

"Mass of oxygen"/"Mass of carbon monixide"xx100%

=(15.999*g*mol^-1)/((15.999+12.011)*g*mol^-1)xx100%=??%

Mar 5, 2017

Because it would imply that carbon has the same molar mass as oxygen... That is, when we assume "50%" means "50% w/w".

${M}_{\text{CO" = M_"C" + M_"O}}$

$= \left(12.011 + 15.999\right)$ $\text{g/mol}$

Clearly, 15.999/(12.011 + 15.999) xx 100% ne 50%. It's actually 57.12%.

So, it would be correct to say that $\text{CO}$ is about 57.12% oxygen atom by weight.

Mar 5, 2017

We use percentage by mass in chemistry, not percentage of atoms.

The percentage by mass of oxygen in carbon monoxide is about 57.14%.

#### Explanation:

Saying that $C O$ is 50% oxygen is only counting by numbers of atoms. Mostly in chemistry, we count things by mass, and carbon and oxygen have very different masses.

Percentage by mass is what percentage overall mass of a substance is made up by a particular element.

To calculate percentage by mass, use the formula

(M_r[O])/(M_r[CO]) xx 100%

where ${M}_{r} \left[X\right]$ is the relative mass of substance $X$.

Using the periodic table, we know that

${M}_{r} \left[O\right] = 16$

${M}_{r} \left[C O\right] = {M}_{r} \left[C\right] + {M}_{r} \left[O\right] = 12 + 16 = 28$

so

16/28 xx 100% ~~ 57.14%