A sample of cocaine, C17H21O4N, is diluted with sugar, C12H22O11. When a 1.00-mg sample of this mixture is burned, 1.00 mL of carbon dioxide (d=1.80g/L) is formed. What is the percentage of cocaine in this mixture? Answer is 28%, but someone show how to get it?
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Since 1.00 mL of CO2 is formed, we know that 1.80 mg of CO2 is formed (1.8 g/L x 1.00 mL).
Since 1.80 mg of CO2 is formed, 0.49 mg of that is carbon (12/(12 + 2(16)) x 1.80 mg)
Each mg of cocaine has (17*12)/(17*12 + 21*1 + 4*16 +1*14) = 0.67 mg of carbon.
Each mg of sugar has (12*12)/(12*12 + 22*1 + 11*16) = 0.42 mg of carbon
Assume we have x mg of cocaine. Then we have (1 - x) mg of sugar. And since all carbon is converted to CO2, we have the equation:
x (0.67 mg carbon per mg of cocaine) + (1 - x) (0.42 mg carbon per mg of sugar) = 0.49 mg of carbon as CO2
(0.67 - 0.42)x = 0.49 - 0.42
(0.25)x = 0.07
x = 0.28 mg cocaine
Fraction of cocaine in mixture = 0.28 mg / 1 mg = 0.28 = 28 %
Since 1.80 mg of CO2 is formed, 0.49 mg of that is carbon (12/(12 + 2(16)) x 1.80 mg)
Each mg of cocaine has (17*12)/(17*12 + 21*1 + 4*16 +1*14) = 0.67 mg of carbon.
Each mg of sugar has (12*12)/(12*12 + 22*1 + 11*16) = 0.42 mg of carbon
Assume we have x mg of cocaine. Then we have (1 - x) mg of sugar. And since all carbon is converted to CO2, we have the equation:
x (0.67 mg carbon per mg of cocaine) + (1 - x) (0.42 mg carbon per mg of sugar) = 0.49 mg of carbon as CO2
(0.67 - 0.42)x = 0.49 - 0.42
(0.25)x = 0.07
x = 0.28 mg cocaine
Fraction of cocaine in mixture = 0.28 mg / 1 mg = 0.28 = 28 %