Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 5.50-g sample is burned, and 3.10 g of CO2(g) is produced. What was the mass percentage of the table salt in the mixture?

Answer 1

C12H22O11 + 12O2 ==> 12CO2 + 11H2O
So convert 3.10g CO2 to mols C12H22O11.
mols CO2 = 3.10/44 = about 0.07 but you should do it more accurately.
0.07 x (1 mol C12H22O11/12 mol CO2) = about 0.006 (again, an estimate).
Then mols C12H22O11 x molar mass = about 2.0 grams.
sugar + salt = 5.50g
sugar = about 2.0
salt = about 3.0 g
%NaCl = (mass NaCl/mass sample)*100 = ?