# Answer to Question #83795 in Organic Chemistry for Ghalib

Question #83795
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
1
2018-12-17T06:27:04-0500

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 CxHyOz. Then the scheme of the combustion process is:

CxHyOz + 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(CxHyOz) = 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(CxHyOz) = 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(CxHy)/ M(C2H5) = 58/29 =2

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

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