What is the percentage of calcium in an ionic compound if a 1.23 g sample completely dissolves in water and yields 0.893 g of calcium carbonate upon addition of excess sodium carbonate solution?

Question

Quizplus · Accepted Answer

First, we need to write the balanced chemical equation for what is happening when the sample dissolves in water and then reacts with excess sodium carbonate:

CaX + Na2CO3 → CaCO3 + 2NaX

where CaX represents the calcium-containing compound that we are trying to analyze.

We know that the mass of CaCO3 produced is 0.893 g, which means that we can use stoichiometry to determine the amount (in moles) of CaX that was originally in the sample. The molar mass of CaCO3 is 100.09 g/mol, so:

0.893 g CaCO3 × (1 mol CaCO3/100.09 g CaCO3) × (1 mol CaX/1 mol CaCO3) = 0.008923 mol CaX

This represents the amount (in moles) of CaX that was originally in the 1.23 g sample. To convert this to a percentage, we need to know the molar mass of CaX. Let's call this M. Then:

1.23 g CaX × (1 mol CaX/M g CaX) = 0.008923 mol CaX

Solving for M:

M = 138.24 g/mol

So, the percentage of calcium in the compound is:

(1 mol Ca/1 mol CaX) × (40.08 g Ca/mol Ca) ÷ (138.24 g/mol CaX) × 100% = 29.1%

Therefore, answer B) is the best choice.

What Is the Percentage of Calcium in an Ionic Compound