Question

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

Answer 1

3.26 % of vinegar is acetic acid

Step-by-step explanation:

Step 1: Data given

Mass of the sample = 5.54 grams

Volume of NaOH = 30.10 mL

Molarity of NaOH = 0.100M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles NaOH

Moles NaOH = volume * molarity

Moles NaOH = 0.03010 L * 0.100M

Moles NaOH = 0.00301 moles NaOH

Step 4: Calculate moles CH3COOH

For 1 mol NaOH we need 1 mol CH3COOH

For 0.00301 moles NaOH we nee 0.00301 moles CH3COOH

Step 5: Calculate mass CH3COOH

Mass CH3COOH = moles CH3COOH * molar mass CH3COOH

Mass CH3COOH = 0.00301 moles * 60.05 g/mol

Mass CH3COOH = 0.1808 grams

Step 6: Calculate percent by weight of acetic acid

Mass % = ( 0.1808 / 5.54 ) *100%

Mass % = 3.26 %

3.26 % of vinegar is acetic acid