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

Question

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

3 Answers

Impact of this question

8165 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-05T12:19:06

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

Explanation:

$......but it is not 50% by mass, i.e.$ $"......but it is not 50% by mass, i.e."$

$% composition of oxygen by mass in carbon monoxide =$ $"% composition of oxygen by mass in carbon monoxide"=$

$\frac{Mass of oxygen}{Mass of carbon monixide} \times 100 %$ $"Mass of oxygen"/"Mass of carbon monixide"xx100%$

$= \frac{15.999 \cdot g \cdot m o l^{- 1}}{(15.999 + 12.011) \cdot g \cdot m o l^{- 1}} \times 100 % = ? ? %$ $=(15.999*g*mol^-1)/((15.999+12.011)*g*mol^-1)xx100%=??%$

Answer 2 · 2017-03-05T12:19:56

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

$M_{CO} = M_{C} + M_{O}$ $M_"CO" = M_"C" + M_"O"$

$= (12.011 + 15.999)$ $= (12.011 + 15.999)$ $g/mol$ $"g/mol"$

Clearly, $\frac{15.999}{12.011 + 15.999} \times 100 % \neq 50 %$ $15.999/(12.011 + 15.999) xx 100% ne 50%$ . It's actually $57.12 %$ $57.12%$ .

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

Answer 3 · 2017-03-05T12:20:43

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

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

Explanation:

Saying that $C O$ $CO$ is $50 %$ $50%$ 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

$\frac{M_{r} [O]}{M_{r} [C O]} \times 100 %$ $(M_r[O])/(M_r[CO]) xx 100%$

where $M_{r} [X]$ $M_r[X]$ is the relative mass of substance $X$ $X$ .

Using the periodic table, we know that

$M_{r} [O] = 16$ $M_r[O] = 16$

$M_{r} [C O] = M_{r} [C] + M_{r} [O] = 12 + 16 = 28$ $M_r[CO] = M_r[C] + M_r[O] = 12 + 16 = 28$

so

$\frac{16}{28} \times 100 % \approx 57.14 %$ $16/28 xx 100% ~~ 57.14%$

Science

Math

Humanities

... and beyond

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

3 Answers

Explanation:

Explanation:

Related questions

Impact of this question