1. The manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0% acetic acid. What is the percent error between your result and the manufacturer statement

72.8 % (But verify explanation).

Explanation:

Hello,

In this case, with the following obtained results, the percent error is computed as follows:

Volume of vinegar= 7.0 mL

Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL

Used concentration of NaOH= 1.5M

Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M

Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol

Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g

% of acetic acid in vinegar=8.64 %

% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %

Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.

Regards.

hi my dude or dudette