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
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