Question

please complete the missing boxes

DER NO. A. Preparation of Vinegar Sample Calculate the approximate volume of the vinegar sample needed for the analyses (Part A.1). Brand of vinegar or unknown no. Trial 1 Trial 2 1. Mass of flask (8) Trial I Trial 2 52.086 62015 57332 550) 66.001 6o. 159 3.015 2.989 2.822 2. Mass of flask + vinegar (8) 3. Mass of vinegar (8) B. Analysis of Vinegar Sample 1. Buret reading of NaOH, initial (mL) 0.02 23.35 23.35 47.27 2. Buret reading of NaOH, final (mL) 3. Volume of NaOH used (mL) 4. Molar concentration of NaOH (moll) 0163M 0.103 5. Moles of NaOH added (mol) 6. Moles of CH,COOH in vinegar (mol) 7. Mass of CH,COOH in vinegar (8) 8. Percent by mass of CH,COOH in vinegar (%) 9. Average percent by mass of CH,COOH in vinegar (%) *Calculations for Trial 1 of the first vinegar sample on next page.

Answer 1

TRIAL 1

Volume of NaOH = FINAL –INITIAL READING

= 23.35-0.02

= 23.33 ml

CH3COOH + NaOH = CH3COONa+ H2O

Moles of NaOH = Molarity * volume in L

= 0.103* 23.33 /1000

= 0.0024 Moles NaOH

From reaction mole of CH3COOH in sample =

0.0024 Moles NaOH *1 mole CH3COOH/ 1 Moles NaOH

=0.0024 Moles CH3COOH

Mass in g = number of moles * molar mass

= 0.0024 Moles CH3COOH*60.052 g/mol

= 0.144 g

Percentage of CH3COOH = amount of CH3COOH/ sample mass *100

= 0.144 g/3.015 g*100

= 4.79 %

Trail 2

Volume of NaOH = FINAL –INITIAL READING

= 47.27 -23.35

= 23.92 ml

CH3COOH + NaOH = CH3COONa+ H2O

Moles of NaOH = Molarity * volume in L

= 0.103* 23.92 /1000

= 0.0025 Moles NaOH

From reaction mole of CH3COOH in sample =

0.0025Moles NaOH *1 mole CH3COOH/ 1 Moles NaOH

=0.0025 Moles CH3COOH

Mass in g = number of moles * molar mass

= 0.0025 Moles CH3COOH*60.052 g/mol

= 0.148 g

Percentage of CH3COOH = amount of CH3COOH/ sample mass *100

= 0.148g/2.989 g*100

= 4.95 %

Average percentage = 4.79+4.95/2

= 4.87%