Question

A sample of a hydrocarbon produced 3.14 grams of CO2 and 1.28 grams of H2O during combustion analysis. If the hydrocarbon has a molar mass between 50 and 60 g/mol, what is its molecular formula? (1) C3H6 (2) C3H8 (3) C4H4 (4) C4H8 (5) C4H10

Answer 1

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

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

= 3.14/44

= 0.0714

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

= 1.28/18

= 0.0711

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.0714

so, x = 0.0714

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.0711 = 0.1422

Divide by smallest to get simplest whole number ratio:

C: 0.0714/0.0714 = 1

H: 0.1422/0.0714 = 2

So empirical formula is:CH2

option 1 and option 4 has empirical formula of CH2

molar mass of C3H6 = 3*12 + 1*6 = 42 g/mol

molar mass of C4H8 = 4*12 + 1*8 = 56 g/mol

So, the molecular formula is C4H8

Answer: (4)C4H8