# Question 532b9

Oct 21, 2017

$\text{15.7 g}$

#### Explanation:

My approach here would be to convert the mass of acetic acid to moles first, then use the chemical formula of the acid to find the number of moles of oxygen present in the sample, then use the molar mass of atomic oxygen to convert that to grams of oxygen.

So, you know that acetic acid has a molar mass of ${\text{60.05 g mol}}^{- 1}$, which means that you need $\text{60.05 g}$ of acetic acid in order to have $1$ mole of acetic acid.

You can thus say that your sample contains

29.4 color(red)(cancel(color(black)("g"))) * ("1 mole CH"_3"COOH")/(60.05color(red)(cancel(color(black)("g")))) = "0.4896 moles CH"_3"COOH"

Now, the chemical formula of the acid tells you that $1$ mole of acetic acid contains $2$ moles of oxygen.

This means that your sample contains

0.4896 color(red)(cancel(color(black)("moles CH"_3"COOH"))) * "2 moles O"/(1color(red)(cancel(color(black)("mole CH"_3"COOH")))) = "0.9792 moles O"#

Finally, use the molar mass of atomic oxygen, ${\text{15.9994 g mol}}^{- 1}$, to calculate the mass of oxygen present in the sample.

$0.9792 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles O"))) * "15.9994 g"/(1color(red)(cancel(color(black)("mole O")))) = color(darkgreen)(ul(color(black)("15.7 g}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of acetic acid.