Chemistry problem - percentage of sample in mixture?

Question

A sample of cocaine, C17H21O4N, is diluted with sugar, C12H22O11. When a 1.00-mg sample of this mixture is burned, 1.00mL of carbon dioxide (d=1.80 g/L) is formed. What is the percentage of cocaine in this mixture?

The answer, according to the text, is 28%, but my friend and I have no idea how this answer was obtained. If you can offer us any insights, we would greatly appreciate it!

Answer 1

To answer this question, recall the atomic masses of: oxygen (16), carbon (12), nitrogen (14) and hydrogen (1). It follows that, by mass, the fraction of carbon in cocaine (C17H21O4N) is

17*12/(17*12 + 21 + 4*16 + 14) = 68/101

(i.e., cocaine is 67.33% carbon). And the fraction of carbon in sugar (C12H22O11) is

12*12/(12*12 + 22 + 11*16) = 8/19

(i.e., sugar is 42.11% carbon).
The fraction of carbon in carbon dioxide (CO2) is

12/(12+32) = 3/11

(i.e., carbon dioxide is 27.27% carbon).

So the total mass of carbon in the CO2 gas produced is

(3/11)*1.00 mL*1.80 g/L = 27/55 mg (= 0.49091 mg).

On the other hand, let x be the fraction (by mass) of cocaine in the mixture. The mass of carbon in the 1 mg of mixture is

[68x/101 + 8(1-x)/19] mg.

This equals the mass of carbon in the CO2. So

68x/101 + 8(1-x)/19 = 27/55

This solves to

x = 7373/26620 = 27.70% = 28% approx. (by mass)

In other words, 1 mg of the mixture contains 0.28 mg of cocaine.

(Of course, you don't have to work out the percentage to 2 d.p., esp. that the atomic masses are also rounded off above. You can also switch to percentages earlier, instead of the fractions used above.)
NB: Could you use oxygen balance for the calculation? No, because the O in the reactants goes also into H2O (and we are not told how much water was produced). So you have to balance C.
Also, it'd be INCORRECT to divide, as someone else proposed, the number of moles of cocaine by the moles of cocaine + moles of sugar. This would give the % of molecules, NOT the % of mass (and cocaine and sugar do have different molecular masses). The question clearly asks for % by mass.

Answer 2

OK, I'll answer this off the cuff generally... w/o doing the calculation.

Two of the products of cocaine and sugar combusting to completion are going to be CO2 and H2O.

First balance the equation :
C17H21O4N + C12H22O11 ---> CO2 + H20 (It seems as if you do not have to worry about balancing the N in this problem, given the way it is asked, so forget about it... but in real life, you would need to know where the N ends up in the products of this reaction)

Knowing 1 mL of CO2 is evolved, and dCO2 = 1.8 g/L, you can calculate that .0018 g of CO2 is obtained. Calculate the molecular weight of CO2 (Sorry... I don't have a periodic table at hand), and calculate the number of moles of CO2 evolved.

If your equation is balanced, you then do the simple addition/subtraction/multiplication division math and calculate the number of moles of Cocaine and the number of moles of Sugar. Then divide the number of moles of Cocaine by the (moles of cocaine + moles of sugar) = your answer.

Sorry I don't have a periodic table at hand... or I'd do it for you.

Answer 3

Check the sources for some examples