A sample of cocaine, C17H21O4N, is diluted with sugar, C12H22O11. When a 1.00 mg sample of this mixture is burned, 2.00 mg CO2 is formed. What is the percentage in this mixture? A good question A good question mg of C = 2.00mg of CO2 * 1mmolCO2/44mg CO2 * 1mmolC/1mmol CO2 * 12mg C /1mmol CO2 = 0.545 mg of C If we assume X mg of cocaine mass of sugar will be 1mg- Xmg of C in X mg of cocaine = Xmg *1mmol cocaine /303.3mg * 17mmol C /1mmol cocaine * 12mg C /1mmol C mg of C in 1- X mg of sugar= 1-Xmg *1mmol sugar /342.3mg * 12 mmol C /1mmol sugar * 12mg C /1mmol C Solve it for X get the % cocaine = Xmg / 1mg *100% Log in or register to post comments
A good question A good question mg of C = 2.00mg of CO2 * 1mmolCO2/44mg CO2 * 1mmolC/1mmol CO2 * 12mg C /1mmol CO2 = 0.545 mg of C If we assume X mg of cocaine mass of sugar will be 1mg- Xmg of C in X mg of cocaine = Xmg *1mmol cocaine /303.3mg * 17mmol C /1mmol cocaine * 12mg C /1mmol C mg of C in 1- X mg of sugar= 1-Xmg *1mmol sugar /342.3mg * 12 mmol C /1mmol sugar * 12mg C /1mmol C Solve it for X get the % cocaine = Xmg / 1mg *100%
A good questionÂ
A good question
mg of C = 2.00mg of CO2 * 1mmolCO2/44mg CO2 * 1mmolC/1mmol CO2 * 12mg C /1mmol CO2 = 0.545 mg of C
If we assume X mg of cocaine
mass of sugar will be 1mg- X
mg of C in X mg of cocaine = Xmg *1mmol cocaine /303.3mg * 17mmol C /1mmol cocaine * 12mg C /1mmol C
mg of C in 1- X mg of sugar= 1-Xmg *1mmol sugar /342.3mg * 12 mmol C /1mmol sugar * 12mg C /1mmol C
Solve it for X