Answer to Question #83795 in Organic Chemistry for Ghalib

Question

1.0gm of a sample of an organic compound was burnt in excess of oxygen to yeld 3.03gm of co2 and 1.55gm of h2o if the molecular mass of the compound is 58 , find it's molecular formulaC4H10

Answer 1

Answer on Question #83795, Chemistry/ Organic Chemistry

1.0gm of a sample of an organic compound was burnt in excess of oxygen to yeld 3.03gm of co2 and 1.55gm of h2o if the molecular mass of the compound is 58 , find it's molecular formulaC4H10

Solution

An unknown organoc compound may or may not contain oxygen. Let the formula to be C_xH_yO_z. Then the scheme of the combustion process is:

C_xH_yO_z + O₂ -> CO₂ + H₂O

1)Find amount of carbon atoms in the organic compound.

n(CO₂) = m(CO₂)/M(CO₂) = 3.03 g/ 44 g/mol = 0.0689 moles

As one molecule of CO₂ conatins 1 atom of C, then n(C) = n(CO₂) = 0.0689 moles

2) Find amount of hydrogen atoms in the organic compound.

n(H₂O) = m(H₂O)/ M(H₂O) = 1.55 g/18g/mol = 0.0861 moles

As one molecule of H₂O contains 2 atoms of hydrogen, then n(H) = 2*n(H₂O)= 2*0.0861 = 0.172 moles

3) Check up if an organic compound contains oxygen:

m(C_xH_yO_z) = m(C) + m(H) + m(O)

m(C) = n(C)*M(C) = 0.0689*12 = 0.827 g

m(H) = n(H)*M(H) = 0.172*1 = 0.172 g

m(C_xH_yO_z) = 1.0 g

1.0 g = 0.827 g + 0.172 g + m(O)

m(O ) = 0

So, an organic compound contains only atoms of hydrogen and carbon.

4) Find ratio n(C):n(H):

n(C):n(H) = 0.0689:0.172 = 1:2.5|x2 = 2:5 => empirical formula C₂H₅

5) Find molecular formula

M(C_xH_y)/ M(C₂H₅) = 58/29 =2

So indexes of the empirical formula should be multipled by 2 => molecular formula C₄H₁₀

Answer: C₄H₁₀