An organic compound with a vapour density 56.5 has the following percentage composition:C =53.1% N = 12.4%, O = 28.3%, H = 6.2%. The molecular formula of the compound is
Relative atomic masses: N = 14, O = 16, C = 12, H = 1 )

olyezema

4 years ago

first you would look for the empirical formula
C
53.1%/12= 4.425 mole
N
12.4%/14= 0.886 mole
O
28.3%/16= 1.769mole
H
6.2%/1= 6.2 mole
divide by the smallest number of mole
C
4.425/0.886= 5
N
0.886/0.886= 1
O
1.769/0.886= 2
H
6.2/0.886= 7
empirical formula= C5H7O2N
V.D= ½×M.m
m.m= V.D×2
56.5×2=113g
molecular formula=
n(C5H7O2N) = 113
n(12×5+1×7+16×2+14)=113
113n=113
n=1
: . the molecular formula is equal to the empirical formula.
D is the correct answer.

Atere Jr

10 years ago

Here is an explanation:

Go about it in a similar fashion that u do with empirical formulae.

First, divide the percentage composition of each elements by their respective atomic mass.

Then, whatever value obtained should be divided by the element with the smallest ratio. For example, when i divide hydrogen's percentage (6.2%) by its atomic mass (1), i obtain 6.2, if it is the smallest value i have among others, i'll divide every other value by it (i.e 6.2) including the 6.2 itself.

After which arrangements are made according to the organic compound.

NB: if u obtain decimals, then u must correct to 1SF.

REF: Go to the topic empirical formula/molecular formula (New System Chemistry, New School Chemistry, Calculations in Chemistry, and virtually any other chemistry text book.

Clinton14589

1 year ago

I need explanation