What is the percent by weight of acetic acid in the vinegar?

Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 5.54 g sample of vinegar was neutralized by 30.10 mL of 0.100 Molarity NaOH. What is the percent by weight of acetic acid in the vinegar?

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One comment a "What is the percent by weight of acetic acid in the vinegar?"

First, determine the number of moles of NaOH. You could do this by converting the volume of NaOH to liters (0.0301 liter) then multiplying this by ).100 mole. This will yield 0.00301 mole of NaOH. This would also be the number of mole of acetic acid. The next step would be to determine the molar mass of acetic acid (2 C x 12g) + (4 H x 1g) + (2 O x 16g) = 60 grams. Multiplying the number of moles by the molar mass (0.00301 mole x 60 grams per mole of acetic acid) = 0.1806 grams. Dividing 0.1806 grams of acetic acid by the mass of vinergar (5.54 g) then multiplying this by 100% will yield 3.26%.
Hope this helps.

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