What is the mass percent?

The reaction between acetic acid in the vinegar sample against sodium hydroxide.
CH₃COOH + NaOH → CH₃COONa + H₂O
OR: Mole ratio CH₃COOH : NaOH = 1 : 1

In the titration :
Volume of NaOH used = (28.71 - 2.34) mL = 26.37 mL = 0.02637 L
No. of moles of NaOH reacted = (0.396 mol/L) × 0.02637 = 0.01044 mol
No. of moles of CH₃COOH in the vinegar sample = 0.01044 mol

Molar mass of CH₃COOH = (12.0×2 + 1.0×4 + 16.0×2) g/mol = 60.0 g/mol
Mass of CH₃COOH in the vinegar sample = (60.0 g/mol) × (0.01044 mol) = 0.6264
Mass percent of CH₃COOH in the vinegar sample = (0.6264/10.56) × 100% = 5.93%

First, calculate moles of NaOH used:

Volume NaOH used = 28.71 - 2.34 = 26.73 mL NaOH
Moles NaOH = 0.02673 L X 0.396 mol/L = 0.0104 moles NaOH

NaOH and acetic acid react is a 1:1 ratio, so the sample of vinegar contained 0.0104 moles Acetic acid.

Molar mass acetic acid = 60.05 g/mol. So,
Mass acetic acid in sample = 0.0104 mol X 60.05 g/mol = 0.6271 grams acetic acid

% Acetic acid by mass = (0.6271 g / 10.56 g) X 100 = 5.94% acetic acid