Question

Table salt, nacl(s), and sugar, c12h22o11(s), are accidentally mixed. a 3.50-g sample is burned, and 2.40 g of co2(g) is produced. what was the mass percentage of the table salt in the mixture?

Answer 1

First we will write the formula for the combustion of the sugar.

C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂O

We have produced 2.40 g of CO₂. We will convert this to moles of CO₂ and then convert that value to moles of sugar.

2.40 g / 44.0 g/mol = 0.0545 moles CO₂ x 1 mole C₁₂H₂₂O₁₁/12 mole CO₂ = 0.00455 moles of C₁₂H₂₂O₁₁

We can now convert the moles of sugar to a mass of sugar.

0.00455 moles C₁₂H₂₂O₁₁ x 342 g/mol = 1.56 g C₁₂H₂₂O₁₁

We want to determine the mass percentage of salt in the 3.5 g mixture of sugar and salt.

3.5 g mixture - 1.56 g sugar = 1.94 g salt

(1.94 g salt/ 3.5g mixture) x 100% = 55%

Therefore, the mass percentage of table salt in the mixture is 55%.