Use your average AA concentration to calculate the mass percent of acetic acid in vinegar. ResultsQuantityTrial 1Trial 2Trial 3M NaOH0.992 M0.992 M0.992 MV initial buret0 ml15 ml30 mlV Final buret13.1 ml28.2 ml47.9 mlVep = VNAOH added= Vf - Vi(47.9ml-30ml)=(13.1ml-0ml)=13.1 ml(28.2ml-15ml)=13.2ml17.9mlVep = VNAOH in L0.0131 L0.0132 L0.0179 Lmoles NaOH=(0.992M)(0.0131 L)= | (0.992M)(0.0132 L)= | (0.992M)(0.0179 L)=MNAOH x VNaOH0.0129 mol0.0130 mol0.178 molmoles AA = moles0.0129 mol0.0130 mol0.178 molNaOHVsample = Vacid0.0150 L0.0150 L0.0150 LActual molarity of0.0129mol/0.0150L=0.0130mol/0.0150L=0.0178mol/0.0150L=AA0.86 M0.87 M1.186 MAverage Molarity of AA: (0.86M)(0.87M)(1.186M)=0.887 M

Mass percent = Mass of soluteMass of sample×100 Here solute is acetic acid and the sample is vinegar. Mass of solute : average molarity of AA =0.887 M So the number of moles of AA = volume of acetic acid × molarity of AA =0.0150 L × 0.887 moles/L = 0.013305 moles So the mass of AA = moles of the acetic acid ×molar mass of acetic acid ( 60.052 g/mol) =0.013305 moles ×60.052 g/mol = 0.799 grams Mass of sample: Mass of vinegar = volume of vinegar × density of vinegar (Here I am using the density of vinegar as the general value of density of vinegar which is 1.02 g/mL ) Volume of vinegar = volume of sample = 0.0150 L = 0.0150 L ×1000 mL/L = 15 mL so mass of vinegar is =15mL × 1.02 g/mL = 15.3 grams So, Mass percent = 0.799 grams15.3 grams×100 =0.0522 ×100 = 5.22 % So the mass percent of acetic acid in vinegar is 5.22%