Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 4.50-g s
ID: 741739 • Letter: T
Question
Table salt, NaCl(s), and sugar, C12H22O11(s), are accidentally mixed. A 4.50-g sample is burned, and 1.80 g of CO2(g) is produced. What was the mass percentage of the table salt in the mixture?Explanation / Answer
Salt will not burn under ordinary conditions, so when the sample was "burned", only the sugar was burned, the salt was left behind. So: C12H22O11 + 35/2 O2 ------------------> 12 CO2 + 11 H2O Now 2.90 mol of CO2 is equal to 2.90 / 44.01 g/mol = 0.06589 mol According to the equation the amount of sugar that will produce that number of moles is 1/12 of the number of moles of CO2. So sugar is 1/12 x 0.06589 = 0.0054911 mol of sugar. The mass of that sugar is 0.0054911 mol x 342.34 g/mol = 1.88 g So out of the sample of the mixture 1.88 g of it was sugar. that leaves 4.50 - 1.88 = 2.62 g of salt So the mass percent of salt in the mixture would be 2.62 / 4.50 all x 100 = 58.2%
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