Question

01. An organic compound has the following percentage composition by me
C=40 % H=6.7 %and 0-53.3%
If vapour density of the compound is 30.find
(a) the empirical formula of the compound
(b) the empirical formula mass of the compound
(c) the molecular formula of the compound.
(Atomic mass of C=12 H=1.0=16)
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Answer 1

Answer:

for carbon

%comp=40%

at. no.=12

no. of moles= 40/12=3.3

for hydrogen

% comp=6.7%

at. no= 1

no. of moles= 6.7/1=6.7

for oxygen

%comp=53.3%

at. no.=16

no. of moles= 53.3/16=3.3

lowest no. of moles = 3.3

simplest ratio for carbon=3.3/3.3=1

simplest ratio for hydrogen= 6.7/3.3=2

simplest ratio for oxygen=3.3/3.3=1

therefore, empirical formula= $CH_{2} O$  

(ii) empirical formula mass = 12+1+1+16=30

(iii) v.d = 30

    molecular mass = 2x v.d=60

n=$\frac{molecular mass}{empirical formula mass}$

n=60/30=2

molecular formula=$(CH_{2} O)_{2}=C_{2}H_{4}O_{2}$

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