Question

Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 5.50-g sample is burned, and 2.30 g of CO2(g) is produced. What was the mass percentage of the table salt in the mixture?

Answer 1

Moles of CO2 = mass/molar mass of CO2

= 2.30 g/44.0 g/mol = 0.052 mol

C12H22O11 + 12 O2 ----------> 12 CO2 + 11 H2O

Moles of C12H22O11 = 1/12 x moles of CO2

= 1/12 x 0.052 = 0.0043 mol

Mass of C12H22O11 = moles x molar mass of C12H22O11

= 0.0043 mol x 342.3 g/mol = 1.472 g

Mass of NaCl = mass of mixture - mass of C12H22O11

= 5.50 - 1.472 g = 4.028 g

Mass percent of NaCl = mass of NaCl/mass of mixture x 100

= 4.028/5.50 x 100

= 73.23 %

Therefore, mass percentage of the table salt in the mixture = 73.23 %