Asample of lsd (d-lysergic acid diethylamide, c24h30n3o) is added to some table salt (sodium chloride) to form a mixture. given that a 1.00-g sample of the mixture undergoes combustion to produce 1.20 g of co2, what is the mass percent of lsd in the mixture?

1) All the CO2 comes from the C24 H30 N3 O

2) To balance C from CO2 with C from C24, the ratio of cC24 H30 N3 O to CO2 is 24 / 1

3) Convert 1.20 g of CO2 to number of moles

number of moles = mass in grams / molar mass

mass = 1.20 g

molar mass = 12.01 g/mol + 32.00 g/mol = 44.01 g/mol

number of moles = 1.2 g / 44.01 g/mol = 0.0273 mol

4) proportion

1 mol C24 H30 N3 O / 24 mol CO2 = x / 0.0273 mol CO2

=> x = 0.0273 mol CO2 * 1 mol C24H30N3O / 24 mol CO2 = 0.00114 mol C24H30N3O

5) Convert 0.00114 mol C24H30N3O to grams

molar mass = 376.5 g/mol

mass = 0.00114 mol * 376.5 g/mol = 0.429 g

6) mass percent in the mixture

mass percent = (mass of lsd / mass of mixture) * 100 = (0.429g / 1.0g) * 100 = 42.9%

42.9%

Sorry i think the answer is b