Question

A 0.1156g of an unknown compound C, H, and N compnd is analyzed by comubustion analysis and produces a 0.1638g CO2 and 0.1676g of water. In a separate experiment, the molar mass was determined to be 31 g/mol. What is its empirical formula and molecular formula?

Answer 1

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 0.1638/44

= 0.003723

Number of moles of H2O = mass of H2O / molar mass H2O

= 0.1676/18

= 0.009311

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.003723

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2* 0.009311 = 0.018622

mass N = total mass - mass of C and H

= 0.1156 - 0.0037*12 - 0.0186*1

= 0.0523

number of mol of N = mass of O / molar mass of O

= 0.0523/14

= 0.003736

So, we have:

C : 0.003723 mol

H : 0.018622 mol

N : 0.003736 mol

divide by smallest to get whole number:

C : 0.003723 / 0.003723 = 1

H : 0.018622 / 0.003723 = 5

N : 0.003736 / 0.003723 = 1

So, the empirical formula is CH5N

Molar mass of CH5N,

MM = 1*MM(C) + 5*MM(H) + 1*MM(N)

= 1*12.01 + 5*1.008 + 1*14.01

= 31.06 g/mol

Now we have:

Molar mass = 31.0 g/mol

Empirical formula mass = 31.06 g/mol

Multiplying factor = molar mass / empirical formula mass

= 31.0/31.06

= 1

Hence the molecular formula is : CH5N

Answer: CH5N is both empirical and molecular formula