Question

45. Upon combustion, a 0.8233 g sample of a compound containing only carbon, hydrogen, and oxygen produced 2.445 g CO2 and 0.6003 g H2O. Find the empirical formula of the compound.

PLEASE EXPLAIN VERY SINGLE STEP CLEARLY AS TO HOW YOU GOT THERE

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

= 2.445/44

= 0.055568

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

= 0.6003/18

= 0.03335

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.055568

so, x = 0.055568

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.03335 = 0.0667

mass O = total mass - mass of C and H

= 0.8233 - 0.055568*12 - 0.0667*1

= 0.089782

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

= 0.089782/16

= 0.005611

so, z = 0.005611

Divide by smallest to get simplest whole number ratio:

C: 0.055568/0.005611 = 10

H: 0.0667/0.005611 = 12

O: 0.005611/0.005611 = 1

So empirical formula is:C10H12O