Question

Combustion of a 1.025 g sample of a compound containing only carbon, hydrogen and oxygen produced 2.265 g of CO2 and 1.236 g of H2O. What is the empirical and molecular formulas of the sample compound, if its molecular weight has been roughly determined to be 363 g/mol by mass spectrometer?

Answer 1

1) Moles of CO₂ = Mass / Molar mass

= 2.265 g / 44

= 0.0514 moles of CO₂

If our sample contains 0.0514 moles of CO2 then our sample must contains 0.0514 moles of C, becouse CO2 prepared from 1 moles of carbon & 2 moles of oxygen

Then mass of carbon = Molar mass x moles = 12.011 x 0.0514

= 0.617 g of C.

2) Moles of Water = Mass / Molar mass

= 1.236 g / 18

= 0.0686 moles

One mole water prepared from one mole of oxygen & Two moles hydrogen. Therefore moles of hydrogen

= 2 x 0.0686   

= 0.1372 moles of H2

Now, Mass of H2 = Molar mass x Moles

= 1 x 0.1372

= 0.1372 g of H2

When we add mass of carbon and Hydrogen together then we get,

= 0.617 g C + 0.1372 g H2

= 0.7542 g

3) We cumbust 1.025 g of sample. Then we can find the mass of oxygen as,

= 1.025 - 0.7542

= 0.2708 g of Oxygen.

Moles of O2 = Mass / molar mass

= 0.2708 / 16

= 0.01692 moles

Now, we get,

a) Moles of Carbon = 0.0514 moles

b) Moles of Hydrogen = 0.1372 moles

c) Moles of Oxygen = 0.01692 moles

Dividing by smallest number to above values,

a) Carbon = 0.0514 / 0.01692 = 3

b) Hydrogen = 0.1372 / 0.01692 = 8.1 = 8

c) Oxygen = 0.01692 / 0.01692 = 1

Therefore the empirical formula = C₃H₈O

Empirical formula mass = ( 3 x 12 ) + ( 1 x 8 ) + ( 16 x 1 )

= 36 + 8 + 16

= 60

4) Molecular formula = Molecular wt / 60

   = 363 g/mol / 60

   = 6.05

Multiplying by 6 to empirical formula we get,

  C₁₈H₄₈O_{6 .} This is the Molecular formula.