# How many grams of acetic acid are present in 2.80 L of this vinegar?

A vinegar sample is found to have a density of $1.006 \mathrm{\frac{g}{mL}}$ and to contain $8.7~\%$ acetic acid by mass. How many grams of acetic acid are present in $\pu{2.80L}$ of this vinegar?

I multiplied: $$\left(\frac{1.006~\mathrm{g}}{\mathrm{ml}}\right)\cdot \left(\frac{1000~\mathrm{ml}}{1~\mathrm{L}}\right)\cdot 2.80~\mathrm{L} = 2816.8~\mathrm{g}$$

I then converted $8.7~\%$ into $0.087$ and multiplied it into $2816.8$, to get the answer $245~\mathrm{g}$.

Am I correct?

• Please add what you have attempted towards solving the problem into the body of your question. For more information, see the site's homework policy for how to ask homework questions. Thanks! – jonsca Sep 12 '14 at 21:13
• (you've done this very well in your other questions so far, so if you could do the same with this one, that would be great) – jonsca Sep 12 '14 at 21:18
• Can you tell me if I am on the right track? – Cetshwayo Sep 12 '14 at 22:19
• Yes, I get the same answer as you. – ron Sep 12 '14 at 22:45
• @JohnSnow I appreciate you working hard in editing the questions with MathJax. I would appreciate it even more, if you would have a look at this meta post, specifically the section about units and variables. – Martin - マーチン Sep 13 '14 at 6:41

Your math and stoichiometry are correct. $2.80L*(\frac{1000 ml}{1 L})*(\frac{1.006 g}{ ml})*8.7$%$=245g \checkmark$