An impure sample of (COOH)2 · 2 H2O that has a mass of 4.3 g was dissolved in water and titrated with standard NaOH solution. The titration required 44.5 mL of 0.11 mo- lar NaOH solution. Calculate the percent (COOH)2 · 2 H2O in the sample.
Answer in units of %.
H2C2O4.2H2O + 2NaOH ==>Na2C2O4 + 4H2O
mols NaOH = M x L = 0.11 x 0.0445 = ?
mols H2C2O4.2H2O used = 1/ 2 x mols NaOH = ? (note that 2 mols NaOH = 1 mol H2C2O4.2H2O)
grams H2C2O4.2H2O in sample = mols H2C2O4.2H2O x molar mass = ?
% H2C2O4.2H2O = (grams H2C2O4.2H2O/4.3 g)*100 = ?
Post your work if you get stuck.
Perfect, thank you for your help, DrBob222!
To calculate the percent (COOH)2 · 2 H2O in the sample, we need to determine the amount of (COOH)2 · 2 H2O in the solution and the total mass of the impure sample.
First, let's calculate the number of moles of NaOH used in the titration. We can use the molarity and volume of NaOH solution:
moles of NaOH = molarity × volume
= 0.11 mol/L × 0.0445 L
= 0.004895 mol (rounded to 4 decimal places)
Next, we can use the balanced equation for the reaction between NaOH and (COOH)2 · 2 H2O to determine the mole ratio:
2 moles of NaOH react with 1 mole of (COOH)2 · 2 H2O
Therefore, the moles of (COOH)2 · 2 H2O in the solution are:
moles of (COOH)2 · 2 H2O = 0.004895 mol ÷ 2
= 0.0024475 mol (rounded to 4 decimal places)
Now, let's calculate the molar mass of (COOH)2 · 2 H2O:
The molar mass of C is 12.01 g/mol.
The molar mass of O is 16.00 g/mol.
The molar mass of H is 1.01 g/mol.
(COOH)2 · 2 H2O contains 2 carbon (C) atoms, 4 oxygen (O) atoms, 4 hydrogen (H) atoms, and 2 water (H2O) molecules.
Molar mass of (COOH)2 · 2 H2O = (2 × 12.01 g/mol) + (4 × 16.00 g/mol) + (4 × 1.01 g/mol) + (2 × (2 × 16.00 g/mol + 2 × 1.01 g/mol))
= 88.08 g/mol (rounded to 2 decimal places)
Next, we can calculate the mass of (COOH)2 · 2 H2O in the solution:
mass of (COOH)2 · 2 H2O = moles of (COOH)2 · 2 H2O × molar mass of (COOH)2 · 2 H2O
= 0.0024475 mol × 88.08 g/mol
= 0.214 g (rounded to 3 decimal places)
Finally, we can calculate the percentage of (COOH)2 · 2 H2O in the impure sample:
percentage = (mass of (COOH)2 · 2 H2O ÷ mass of impure sample) × 100
percentage = (0.214 g ÷ 4.3 g) × 100
= 4.98% (rounded to 2 decimal places)
Therefore, the percent (COOH)2 · 2 H2O in the sample is approximately 4.98%.