# Acetic Acid % in vinegar? Given average molarity of acetic acid

Question:

The stockroom claims the percent acetic acid in vinegar to be $$2.0\%$$. The density of vinegar is $$\pu{1.106 g {mL}-1}$$. Using average molarity given ($$\pu{0.2844M}$$) calculate the mass percent acetic acid in vinegar for comparison to the stockroom claim.

Not sure how to approach the problem but this is what I've done:

$$\frac{\pu{0.2844 mol}}{\pu{1L}} \times (\pu{60.05g {mol}-1})\times \left(\frac{\pu{1L}}{\pu{1000mL}}\right) = \pu{ 0.01707822 g {mL}-1}$$

if you have $$\pu{1000mL}$$, you'll have $$\pu{0.001106 g}\text{ Acetic Acid} \div \pu{0.0000170782 g}\text{ Vinegar}$$ so that will give $$\%$$ of acetic acid? I could use some clarification in solving this

• HINT: You want: mass(vinegar)/mass(solution) – MaxW Oct 15 '18 at 20:42

## 1 Answer

1000 mL of vinegar will weigh 1106 g and contain 1000 x 0.017078 g of acetic acid (= 17.078 g). Then 17.078/1106 = 0.0154 = 1.54% acetic acid.

The stockroom manager must be diluting the vinegar, perhaps for a specific experiment. The vinegar I buy is 5% acetic acid (rarely 4%).