Question

you are analyzing an unknown organic compound in the lab. the sample of unknown organic compound has a mass of 18.8 grams. you decide to determine the emprirical formula of the unknown compound using combustion analysis. after combustion, you find that 27.6 grams of carbon dioxide and 11.3 grams of waer have been produced. what is the empirical formula for the unknown organic compound?

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

= 27.6/44

= 0.6273

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

= 11.3/18

= 0.6278

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.6273

so, x = 0.6273

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.6278 = 1.2556

mass o = total mass - mass of C and H

= 18.8 - 0.6273*12 - 1.2556*1

= 10.0172

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

= 10.0172/16

= 0.6261

so, z = 0.6261

Divide by smallest to get simplest whole number ratio:

C: 0.6273/0.6261 = 1

H: 1.2556/0.6261 = 2

O: 0.6261/0.6261 = 1

So empirical formula is:CH2O