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Stoichiometry / TitrationFRQ

Compound X contains 2.239% hydrogen, 26.681% carbon and 71.080 % oxygen by mass. The titration of 0.Stoichiometry / Titration Chemistry Question

Problem Context

Compound X contains 2.239% hydrogen, 26.681% carbon and 71.080 % oxygen by mass. The titration of 0.154 g of this compound with 0.3351 M KOH produces the curve shown.

a.

Determine the empirical formula of the compound.

Model Answer

convert masses to moles:
2.239 g H × (1 mol / 1.008 g) = 2.221 mol (÷2.157) = 1.03
26.681 g C × (1 mol / 12.011 g) = 2.157 mol (÷2.157) = 1
71.08 g O × (1 mol / 16.00 g) = 4.443 mol (÷2.157) = 2.06
These numbers are close enough to whole numbers that the empirical formula must be CHO2

b.

Calculate its molar mass and give its molecular formula.

Model Answer

Obtain molar mass from titration (estimate endpoint at 10.4 mL)
Mol NaOH = 0.3351 mol/L × 0.0104 L = 0.00348 mol
molar mass = 0.154 g ÷ 0.00348 mol = 44.2 g/mol (× 2 because titration curve is diprotic) = 88.4 g/mol
The molar mass is 88.4 g/mol, the empirical formula molar mass is 45.02. This value is close to half the value of the experimentally determined molar mass, so the molecular formula must be C2H2O4.

c.

When K2Cr2O7 is reacted with X in acidic solution the products are chromium(III) ions and carbon dioxide. Describe the color change that accompanies this reaction.

Model Answer

The color change will be from orange for Cr2O72– to green for Cr3+.

d.

Write a balanced ionic equation for this reaction.

Model Answer

The balanced ionic equation is:
Cr2O72– + 3H2C2O4 + 8H+2Cr3+ + 6CO2 + 7H2O

e.

Find the volume of dry carbon dioxide that could be collected at 22 ˚C and 738 mm Hg when 0.839 g of compound X is reacted with an excess of K2Cr2O7.

Model Answer

Do the stoichiometry for oxalic acid to carbon dioxide, then calculate volume using ideal gas law.
0.839 g H2C2O4 × (1 mol H2C2O4 / 90.04 g H2C2O4) × (6 mol CO2 / 3 mol H2C2O4) = 0.0186 mol CO2
V = nRT / P = (0.0186 mol)(0.0821 L·atm·mol–1·K–1)(295 K) / (738 mmHg × (1 atm / 760 mmHg)) = 0.464 L

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