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

  • $\begingroup$ HINT: You want: mass(vinegar)/mass(solution) $\endgroup$
    – MaxW
    Oct 15, 2018 at 20:42

1 Answer 1


$1000$ 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 $4$%).

  • $\begingroup$ According to Wikipedia, vinegar has typically 5-8%. I do not know common values found in stores now, but for some reasons I remember it had 8% in early 80s when I was a young chemist. $\endgroup$
    – Poutnik
    Sep 27, 2020 at 5:46

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