Question

fill in 5-9

A. Preparation of Vinegar Sample Brand of vinegar Trial 1 Trial 2 1. Volume of vinegar used (mL) 2. Density of vinegar (g/mL) 3. Mass of vinegar used (g) solution - P.V. 5.0 1002 5.01 5.0 1.000 5.01 B. Analysis of Vinegar Sample 1. Buret reading of NaOH, initial (mL) 2. Buret reading of NaOH, final (mL) 3. Volume of NaOH used (mL) 0 0 43.60 46,90 42.60 41.90 1005 4. Molar concentration of NaOH (mol/L) 5. Moles of NaOH added (mol) 6. Moles of CH3COOH in vinegar (mol) 7. Mass of CH3COOH in vinegar (9) 8. Percent by mass of CH3COOH in vinegar (%) 9. Average percent by mass of CH3COOH in vinegar (%)

Answer 1

Considering the data of trial 1:

volume of NaOH added is 42.60 ml = 0.04260L

As the molar concentration of NaOH solution is 0.1005 mol/L

Moles of NaOH added= (0.1005×0.04260) mol = 0.0043 mol.

From the balanced chemical equation:
Moles of CH3COOH(vinegar) = Moles of NaOH (titrant)

Moles of CH3COOH(vinegar) = 0.0043 moles

Now you need to convert the moles of acetic acid to a mass of acetic acid. You will do this by multiplying the moles of acetic acid with the molecular weight of acetic acid which is 60.0 g of acetic acid in one mole of acetic acid.

Mass of acetic acid = 0.0043 mol of acetic acid × (60 g of acetic acid/1 mol of acetic acid) = 0.258 g

percentage by mass of acetic acid in vinegar = (0.258 g of acetic acid/5.01 g of vinegar) × 100% =5.1497%

Following the same procedure we can calculate the 2nd set using the data of trial 2:

Moles of NaOH added= (0.1005×0.0419) mol = 0.0042 mol.

Mass of acetic acid = 0.252 g

percentage by mass of acetic acid in vinegar = 5.0299%

Av percentage by mass of CH3COOH in vinegar = (5.1497%+5.0299%)/2 = 5.0898%