Question

Vinegar is a solution ofacetic acid, CH3COOH, dissolved in water. A 5.54-gsample of vinegar was neutralized by 30.10 mL of 0.100 M NaOH. Whatis the percent by weight of acetic acid in the vinegar?

Why is the answer3.26%?

Answer 1

Answer:

The mass percentage is 3.26 %.

Explanation:

The reaction between acetic acid and NaOH will be:

$CH_{3}COOH + NaOH--->CH{3}COONa+H_{2}O$

Thus one mole of acetic acid will react with one mole of NaOH.

The moles of NaOH used = molarity X volume = 0.1 X 30.10 mL = 3.010 mmol

The moles of acetic acid reacted = 3.010 mmol

The moles of acetic acid will be used to calculate the mass of acetic acid used

Mass of acetic acid used = moles X molar mass = 3.010mmol X 60

mass of acetic acid used = 180.6 mg = 0.1806 g

The mass of vinegar used = 5.54 g

The percentage of acetic acid in vinegar will be =$\frac{massofaceticacidX100}{massofvinegar}=\frac{0.1806X100}{5.54}=3.26$%

Answer 2

The mass percentage is 3.26 %.

Balanced chemical reaction:

$CH_3COOH+NaOH--->CH_3COONa+H_2O$

One mole of acetic acid reacts with one mole of sodium hydroxide to produce one mole of sodium acetate.

• The moles of NaOH used = molarity * volume

$\text{Number of moles}= 0.1 * 30.10 mL = 3.010 mmol$

The moles of acetic acid reacted = 3.010 mmol

In order to calculate Mass of acetic acid we will use moles of NaOH.

• Mass of acetic acid used = moles * molar mass

$\text{Mass of acetic acid used}= 3.010mmol * 60$

Mass of acetic acid used = 180.6 mg = 0.1806 g

• The mass of vinegar used = 5.54 g

The percentage of acetic acid in vinegar will be:

$\frac{\text{Mass of acetic acid*100}}{\text{Mass of vinegar}}=\frac{0.1806*100}{5.54} =3.26\%$

Thus, percentage will be 3.26%.

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