Question

0.493g of an organic compound contains carbon, hydrogen and oxygen gave 0.881g of CO2 and 0.359g of H2O. On complete combustion , 0.082g of the compound displaced 27cm3 of air in a victor Mayers ditermination measured at 17deg and 700mmHg... Calculate the molecular formula at stp and determine it's possible isomers naming each..

Answer 1

Solution.

Find the molecular weight of the unknown compound:

$p \times V = \frac{m}{M} \times R \times T$

$M = \frac{m \times R \times T}{p \times V}$

$M = \frac{0.493 \times 0.0821 \times 273}{1 \times 0.027}$

M = 409 g/mole

Now we find the mass of carbon, hydrogen and oxygen in an unknown substance:

m(C) = 0.240 g

m(H) = 0.040 g

m(O) = 0.213 g

Now we find the amount of substance carbon, hydrogen and oxygen:

$n(C):n(H):n(O) = 0.02:0.04:0.01 = 2:4:1$

We found the molecular formula of an unknown substance. Now we find the ratio of the molecular mass of the simplest formula and the mass of the substance that we found through the method of Victor Myers:

$k = \frac{409}{44} = 9$

Formula of the substance:

C18H36O9

Isomers:

[27]-crown-9

m-dPEG.(TM).8 propionaldehyde

Answer:

Formula of the substance:

C18H36O9

Isomers:

[27]-crown-9

m-dPEG.(TM).8 propionaldehyde