Question

You are given a sample of limestone, which is mostly CaCO3, to determine the mass percentage of Ca in the rock. You dissolve the limestone in hydrochloric acid, which gives a solution of calcium chloride. Then you precipitate the calcium ion in solution by adding sodium oxalate, Na2C2O4. The precipitate is calcium oxalate, CaC2O4. You find that a sample of limestone weighing 128.3 mg gives 140.2 mg of CaC2O4. What is the mass percentage of calcium in the limestone

Answer 1

Answer:

34.15% is the mass percentage of calcium in the limestone.

Explanation:

Mass of precipitate that is calcium oxalate = 140.2 mg = 0.1402 g

1 mg = 0.001 g

Moles of calcium oxalate = $\frac{0.1402 g}{128 g/mol}=0.001095 mol$

1 mole of calcium oxalate have 1 mole of calcium atom.

Then 0.001095 moles of calcium oxalate will have 0.001095 moles of calcium atom.

Mass of 0.001095 moles of calcium :

0.001095 mol × 40 g/mol = 0.04381 g

Mass of sample of limestone = 128.3 mg = 0.1283 g

Percentage of calcium in limestone:

$\frac{0.04381 g}{0.1283 g}\times 100=34.15\%$

34.15% is the mass percentage of calcium in the limestone.

Answer 2

Answer:

here are the percentages I got, I know this question is old but for others, for future reference, I figured I would add my answer. Hope this helps :)