Question

Furnace H,O absorber CO, absorber Sample A 9.314 gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and 13.21 of CO2 and 8.112 grams of H2O are produced. grams In a separate experiment, the molar mass is found to be 62.07 g/mol. Determine the empirical formula and the molecular formula of the organic compound. Enter the elements in the order C, H, O empirical formula- molecular formula

Answer 1

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

= 13.21/44

= 0.3002

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

= 8.112/18

= 0.4507

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.3002

so, x = 0.3002

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.4507 = 0.9013

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 9.314 - 0.3002*12 - 0.9013*1

= 4.81

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

= 4.81/16.0

= 0.3006

so, z = 0.3006

Divide by smallest to get simplest whole number ratio:

C: 0.3002/0.3002 = 1

H: 0.9013/0.3002 = 3

O: 0.3006/0.3002 = 1

So empirical formula is:CH3O

Answer: CH3O

2)

Molar mass of CH3O,

MM = 1*MM(C) + 3*MM(H) + 1*MM(O)

= 1*12.01 + 3*1.008 + 1*16.0

= 31.034 g/mol

Now we have:

Molar mass = 62.07 g/mol

Empirical formula mass = 31.034 g/mol

Multiplying factor = molar mass / empirical formula mass

= 62.07/31.034

= 2

So molecular formula is:C2H6O2

Answer: C2H6O2