Question

Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 5.00-g sample is burned, and 2.60 g of CO2(g) is produced. What was the mass percentage of the table salt in the mixture?

Answer 1

no of mole of CO2 = no of mole of C present in sample

= 2.6/44 = 0.06 mole

so that, 1 mole C12H22O11 = 12 mole C

no of mole of sugar present in sample = 0.06/12 = 0.005 mole

mass of sugar present in sample = n*mwt

               = 0.005*342.3 = 1.7115 g

mass of table salt = 5 - 1.7115 = 3.3 g

mass percentage of the table salt in the mixture = wt of sugar / wt of mixer *100

   = 3.3/5*100

   = 66%