the manufacturer of the vinegar used in this experiment claims that the vinegar contains 5% acetic acid by weight. Use your results and density of 1.0 g/mL to determine if this claim is true or false
To determine if the claim that the vinegar contains 5% acetic acid by weight is true or false, we need to calculate the actual percentage of acetic acid in the vinegar based on the results of the experiment.
First, we need to calculate the mass of acetic acid in the 25.0 mL sample of vinegar. From the titration results, we found that it took 21.6 mL of 0.100 M NaOH to neutralize the acetic acid in the vinegar sample.
The balanced chemical equation for the reaction between acetic acid and sodium hydroxide is:
CH3COOH + NaOH --> CH3COONa + H2O
From the balanced equation, we can see that 1 mole of acetic acid reacts with 1 mole of sodium hydroxide. Since the molarity of the NaOH solution is 0.100 M, it means that 0.100 moles of NaOH were used in the titration.
Therefore, there are also 0.100 moles of acetic acid in the sample of vinegar.
Now, we need to calculate the mass of acetic acid in the sample. The molar mass of acetic acid (CH3COOH) is 60.05 g/mol.
The mass of acetic acid in the sample is:
0.100 moles x 60.05 g/mol = 6.005 g
Now, we can calculate the actual percentage of acetic acid in the vinegar sample:
(6.005 g / 25.0 mL *1.0g/mL) x 100 = 24.02%
Since we found that the actual percentage of acetic acid in the vinegar sample is 24.02% and not 5% as claimed by the manufacturer, we can conclude that the manufacturer's claim is false.