Question

Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 4.00-g sample is burned, and 3.40 g of CO2(g) is produced. What was the mass percentage of the table salt in the mixture?

Answer 1

Since carbon only present in sugar molecule so carbon in the carbon dioxide would be come from the sugar, so

First we find moles of CO2:

3.40 g / (44.02 g/mol) = 0.077 mol CO2 = 0.077 mol C

ie CO2 contains 0.077 moles of carbon.

Since every sugar molecule contains 12 carbon molecule , th so 0.077 / 12 = 0.00641 mol of sugar.

Now, multiply the moles of sugar by its molar mass:

0.00641 mol * 342.3 g/mol = 2.19 g sugar

So sugar% in 4.0 g of mixture = (2.19 g / 4.0 g) *100 = 54.7%

So, salt = (100 - 54.7) = 45.3%

answered by: anonymous