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