Vinegar is a solution of acetic acid, CH3COOH, dissolved in water. A 4.69 g sample of vinegar was neutralized by 32.97 mL of 0.100 M NaOH. What is the percent by weight of acetic acid in the vinegar?
CH3COOH + NaOH ==> CH3COONa + H2O
mols NaOH = M x L = ?
mols CH3COOH = mols NaOH (because of the 1:1 coefficients in the balanced equation.)
g CH3COOH = mols CH3COOH x molar mass CH3COOH.
Then % CH3COOH = (g CH3COOH/mass sample)*100 = ?
To find the percent by weight of acetic acid in the vinegar, we need to calculate the mass of acetic acid in the vinegar.
Step 1: Calculate the number of moles of NaOH used.
Using the formula Molarity = Moles/Volume, we can rearrange the formula to find the number of moles.
0.100 M = Moles/0.03297 L
Moles of NaOH = 0.100 M x 0.03297 L = 0.003297 moles
Step 2: Write the balanced chemical equation for the reaction.
CH3COOH + NaOH -> CH3COONa + H2O
Step 3: Determine the mole ratio between acetic acid and NaOH from the balanced equation.
From the balanced equation, we can see that 1 mole of acetic acid reacts with 1 mole of NaOH.
Step 4: Calculate the number of moles of acetic acid.
The number of moles of acetic acid is the same as the number of moles of NaOH since the mole ratio is 1:1.
Number of moles of acetic acid = 0.003297 moles
Step 5: Calculate the molar mass of acetic acid.
The molar mass of acetic acid (CH3COOH) can be calculated by multiplying the atomic masses of each element and adding them together.
Molar mass of C = 12.01 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of acetic acid = (12.01 x 2) + (1.01 x 4) + 16.00 + 1.01 = 60.05 g/mol
Step 6: Calculate the mass of acetic acid.
Mass of acetic acid = number of moles x molar mass
Mass of acetic acid = 0.003297 moles x 60.05 g/mol = 0.1976 g
Step 7: Calculate the percent by weight.
Percent by weight = (mass of acetic acid / mass of vinegar) x 100%
Mass of vinegar = 4.69 g
Percent by weight = (0.1976 g / 4.69 g) x 100% ≈ 4.22%
The percent by weight of acetic acid in the vinegar is approximately 4.22%.
To find the percent by weight of acetic acid in vinegar, we need to calculate the amount of acetic acid that was present in the vinegar sample.
First, let's find the number of moles of NaOH used in the neutralization reaction. We can use the formula:
moles of substance = concentration × volume (in liters)
Given:
Concentration of NaOH (CNaOH) = 0.100 M
Volume of NaOH (VNaOH) = 32.97 mL = 32.97 cm³ = 32.97 × 10⁻³ L
Using the formula: moles of NaOH = CNaOH × VNaOH
moles of NaOH = 0.100 M × 32.97 × 10⁻³ L = 0.003297 moles
According to the balanced chemical equation for the neutralization reaction between acetic acid and sodium hydroxide:
CH3COOH + NaOH → CH3COONa + H2O
The stoichiometric ratio between acetic acid (CH3COOH) and sodium hydroxide (NaOH) is 1:1. Therefore, the number of moles of acetic acid present in the vinegar sample is also 0.003297 moles.
Next, let's calculate the molar mass of acetic acid (CH3COOH) using the atomic masses of each element:
Molar mass of carbon (C) = 12.01 g/mol
Molar mass of hydrogen (H) = 1.01 g/mol
Molar mass of oxygen (O) = 16.00 g/mol
Molar mass of CH3COOH = (4 × Molar mass of hydrogen) + Molar mass of carbon + (2 × Molar mass of oxygen)
= (4 × 1.01 g/mol) + 12.01 g/mol + (2 × 16.00 g/mol)
= 60.05 g/mol
Now, let's calculate the mass of acetic acid in the vinegar sample using the number of moles and molar mass:
Mass of acetic acid = number of moles × molar mass
= 0.003297 moles × 60.05 g/mol
= 0.1976 g
Finally, let's calculate the percent by weight of acetic acid in the vinegar sample using the mass of acetic acid and the mass of the vinegar sample:
Percent by weight of acetic acid = (mass of acetic acid / mass of vinegar sample) × 100
= (0.1976 g / 4.69 g) × 100
= 4.21%
Therefore, the percent by weight of acetic acid in the vinegar is approximately 4.21%.