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 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- 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 Submitted by chemtopper on Tue, 09/26/2017 - 00:54 Permalink Log in or register to post comments

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- 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 Submitted by chemtopper on Tue, 09/26/2017 - 00:54 Permalink

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