A 3.455 g sample of vinegar is titrated. It requires 16.83 mL of 0.2145 M NaOH to get to the phenolphthalein end point. Calculate the weight percent of acetic acid in this sample of vinegar. ____ % Note: calculate the molecular weight to one decimal place, so the answer should be 3 significant figures.
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To calculate the weight percent of acetic acid in the vinegar sample, we need to determine the amount of acetic acid present in grams and divide it by the total sample weight (3.455 g), then multiply by 100 to get the percentage.
1. Calculate the amount of acetic acid reacted with NaOH:
We know that the molar ratio between acetic acid (CH3COOH) and NaOH (sodium hydroxide) is 1:1. Therefore, the number of moles of CH3COOH can be calculated using the volume and concentration of NaOH used.
Number of moles of NaOH = volume (L) × concentration (mol/L)
= (16.83 mL / 1000 mL/L) × 0.2145 mol/L
= 0.003603495 mol
Since the stoichiometry is 1:1, this is also the number of moles of acetic acid (CH3COOH) reacted.
2. Calculate the molecular weight of acetic acid (CH3COOH):
The molecular weight of acetic acid is the sum of the atomic weights of each element in the formula:
C = 12.01 g/mol
H = 1.01 g/mol (3 hydrogen atoms)
O = 16.00 g/mol
Molecular weight of acetic acid = (12.01 g/mol) + (1.01 g/mol × 3) + 16.00 g/mol
= 60.05 g/mol
3. Calculate the amount of acetic acid in grams:
Amount of acetic acid (g) = number of moles × molecular weight
= 0.003603495 mol × 60.05 g/mol
= 0.216309870 g
4. Calculate the weight percent of acetic acid in the vinegar sample:
Weight percent = (Amount of acetic acid / Sample weight) × 100
= (0.216309870 g / 3.455 g) × 100
= 6.263806 % (rounded to 3 significant figures)
Therefore, the weight percent of acetic acid in the sample of vinegar is approximately 6.26%.