# 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, 2018 at 20:42

$$1000$$ mL of vinegar will weigh $$1106$$ g and contain $$1000 · 0.017078$$ 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 $$4$$%).