Question

A compound contains only C, H, and N. Combustion of 28.0 mg of the compound produces 26.8 mg CO2 and 32.9 mg H2O. What is the empirical formula of the compound? (Type your answer using the format CO2 for CO2.)

Answer 1

moles = mass / molar mass

molar mass of CO2 = 44 g/ mole

so 0.02680 g of CO2 has 0.027 /44 = 0.0006 moles of CO2
there is 1 mole of C in CO2 and all the C from the compound becomes CO2 so moles of C in the compound = 0.0006 moles
mass of C = 0.0006 x 12 = 0.0073 g

molar mass of H2O = 18 g/ mole
0.0329 g of H2O has 0.0329 / 18 = 0.0018 moles of H2O
there are 2 moles of H in H2O so moles of H in the compound = 0.0018*2=0.0037 moles
mass of H = 0.0037 x 1.0079 = 0.0037 g

mass of H + C = 0.0110 g
mass of sample = 0.0280 g
mass of N by difference = 0.0170 g
moles of N = 0.0170 /14 = 0.0012 moles

molar ratio of C : H : N = 0.0006 : 0.0037 : 0.0012
smallest number 0.0006
divide the ratio by the smallest number we get
molar ratio of C : H : N = 1.00 6.00 1.99

the empirical formula is CH6N2.