A.Preparation of Vinegar aprox. mass ofvinegar 3g Trial1 Trial 2 1. Mass ofweigh

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A.Preparation of Vinegar aprox. mass ofvinegar       3g                                                                                                Trial1                               Trial 2 1. Mass ofweighing boat(g)                                                     102.301g                   100.306g
2. Mass of weighing boat + vinegar sample(g)                          105.117g                    103.127  g [3] Mass ofvinegar(g)                                                               2.816g                        2.821g B. Analysis of Vinegar Sample
                                                                                               Trial1                           Trial 2 1. Buret (NaOH):Initial reading(mL)                                       7.81mL                7.15 mL
2. Buret (NaOH): Final reading(mL)                                        31.81mL               32.50mL [3] Volume of NaOHused(mL)                                                24.00mL              25.35 mL 4. MolarConcentration of NaOH (mol/L)**                             0.0024mol/L        0.0025mol/L [5] Moles of NaOHadded(mol)                                             ____________       ___________    [6] Moles of CH3COOH in vinegar(mol)                                ____________      ____________
[7] Mass of CH3COOH in vinegar(g)                                     ____________       ___________
[8] Percent by Mass of CH3COOH in vinegar(%)                  ____________       ___________
[9] Avg Percent by Mass of CH3COOH in vinegar(%)            ____________      ___________ A.Preparation of Vinegar aprox. mass ofvinegar       3g                                                                                                Trial1                               Trial 2 aprox. mass ofvinegar       3g 1. Mass ofweighing boat(g)                                                     102.301g                   100.306g
2. Mass of weighing boat + vinegar sample(g)                          105.117g                    103.127  g [3] Mass ofvinegar(g)                                                               2.816g                        2.821g B. Analysis of Vinegar Sample
                                                                                               Trial1                           Trial 2 1. Buret (NaOH):Initial reading(mL)                                       7.81mL                7.15 mL
2. Buret (NaOH): Final reading(mL)                                        31.81mL               32.50mL [3] Volume of NaOHused(mL)                                                24.00mL              25.35 mL 4. MolarConcentration of NaOH (mol/L)**                             0.0024mol/L        0.0025mol/L [5] Moles of NaOHadded(mol)                                             ____________       ___________    [6] Moles of CH3COOH in vinegar(mol)                                ____________      ____________
[7] Mass of CH3COOH in vinegar(g)                                     ____________       ___________
[8] Percent by Mass of CH3COOH in vinegar(%)                  ____________       ___________
[9] Avg Percent by Mass of CH3COOH in vinegar(%)            ____________      ___________

Explanation / Answer

                                                                                               Trial1                           Trial 2 1. Buret (NaOH):Initial reading(mL)                                       7.81mL                  7.15 mL
2. Buret (NaOH): Final reading(mL)                                        31.81mL                32.50 mL [3] Volume of NaOHused(mL)                                                24.00mL               25.35 mL 4. MolarConcentration of NaOH (mol/L)**                             0.0024mol/L         0.0025mol/L We know that,moles = concentration * volume so moles ofNaOH in Trail 1 = 0.0024mol /L * 24.00mL =0.0576mol       moles of NaOH inTrail 1 = 0.0025mol /L * 25.35mL =0.0634mol [5] Moles of NaOHadded(mol)                                             ____________       ___________    CH3COOH (Vinegar ) + NaOH...........> CH3COONa (aq) + H2O (l) Atequilivelence point, moles of acid = moles of base ; Therefore,moles of CH3COOH required in Trail 1 =0.0576mol and Trail 2 = 0.0634mol [6] Moles of CH3COOH in vinegar(mol)                                ____________      ____________


Mass = moles * molar mass
So mass of Vinegar in Trail 1 = 0.0576mol * 60.05g/mol =3.4589g
                             in Trail 2 = 0.0634mol * 60.05g /mol = 3.8072g
[7] Mass of CH3COOH in vinegar(g)                                     ____________       ___________

Percent by Mass of CH3COOH in vinegar = Mass of CH3COOH in vinegar/ (Mass of CH3COOH in vinegar + Mass of NaOH )
In Trail 1 , % Vinegar = { 3.4589g / (3.4589g + 2.3038g)}*100 = 60.02%
In Trail 2 , % Vinegar = { 3.8072g / (3.8072g + 2.5358g)}*100 = 60.02%

[8] Percent by Mass of CH3COOH in vinegar(%)                  ____________       ___________



Average % = (60.02% + 60.02% ) /2 = 30.01%
[9] Avg Percent by Mass of CH3COOH in vinegar(%)            ____________      ___________

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