Question

Combustion analysis of 3.00g of dietary supplement "X" compromised of C,H, and O yields 8.872 g CO2 and 3.114 g H2O. If this compound has a molar mass of 416.68 g/mol what is the molecular formula?

Answer 1

let in compound number of moles of C, H and O be x, y and z respectively

Number of moles of CO2 = mass of CO2 / molar mass CO2
= 8.872/44
= 0.2016

Number of moles of H2O = mass of H2O / molar mass H2O
= 3.114/18
= 0.173

Since 1 mol of CO2 has 1 mol of C
Number of moles of C in CO2= 0.2016
so, x = 0.2016

Since 1 mol of H2O has 2 mol of H
Number of moles of H = 2*0.173 = 0.346

Molar mass of O = 16 g/mol


mass O = total mass - mass of C and H
= 3.0 - 0.2016*12 - 0.346*1
= 0.2344

number of mol of O = mass of O / molar mass of O
= 0.2344/16.0
= 1.465*10^-2
so, z = 1.465*10^-2
Divide by smallest to get simplest whole number ratio:
C: 0.2016/1.465*10^-2 = 14
H: 0.346/1.465*10^-2 = 24
O: 1.465*10^-2/1.465*10^-2 = 1


So empirical formula is:C14H24O

Molar mass of C14H24O,
MM = 14*MM(C) + 24*MM(H) + 1*MM(O)
= 14*12.01 + 24*1.008 + 1*16.0
= 208.332 g/mol

Now we have:
Molar mass = 416.68 g/mol
Empirical formula mass = 208.332 g/mol
Multiplying factor = molar mass / empirical formula mass
= 416.68/208.332
= 2

So molecular formula is:C28H48O2
Answer: C28H48O2