Molecular Formula-

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Molecular Formula

Q 1–(a)  % of C= 40% ; % of H= 6.7%

(b) 0.9 gm. silver salt of monobasic acid on combustion give 0.582 gm.of silver. calculate  Molecular Formula = ?

Ans.       W= 0.9 gm. , w = 0.582 gm.

% of C= 40% ; % of H= 6.7%

% of O = 100–(40+6.7)=53.3

 Element % Atomic weight Relative no. of atoms Simplest Ratio C 40 12 40/12= 3.3 3,3/3.3= 1 H 6.7 1 6.7 /1 = 6.7 6.7/ 3.3=2 O 53.3 16 53.3 /16= 3.3 3.3/3.3=1

C:H:O=1 : 2 : 1

Empirical formula =CH2O

Empirical formula weight =12+2+16 =30

Calculation of molecular weight by Silver salt method-

E+107/108 =W/w

=0.9/ 0.582

=1.55

E = 167.4 -107 =60.4

molecular weight =E x basicity of acid

=60.4 x 1 = 60.4

n= molecular weight/ Empirical formula weight

n= 60.4 /30 = 2

mol. formula =(  Empirical formula )n

=(CH2O)n

=(CH2O)2

mol. formula = C2H4O2  Ans.

Q 2– (a)% of C= 77.42 %, % of H = 7.53,

(b)  0.465 gm. of organic compd is Kjeldhalised  then evolved NH3 is completely neutralised by 25 ml N/5 H2SO4. 0.596 gm  platini Chloride of a monoacidic base on combustion gave 0.195 gm Pt. calculate the mol. formula of base ?

Ans.  % of nitrogen = 1.4 N V/w

N=1/5  =0.2 ; V=25 ml. ,w =0.465 gm.

% of nitrogen =1.4 x  0.2 x 25 /0.465

= 15.05 %

% of O = 100–(77.42 +7.53 +15.05)= 0 %

 Element % Atomic weight Relative no. of atoms Simplest Ratio C 77.42 12 77.42/12=6.45 6.45/1.07=6 H 7.53 1 7.53/1=7.53 7.53/1.07= 7 N 15.05 14 15.05/14= 1.07 1.07/1.07 =1

C:H:N=6 : 7: 1

Empirical formula =C6H7N

Empirical formula weight =12 x 6 +1 x 7 +14 = 93

Calculation of molecular weight by Platini chloride method-

2E+410/195 =W/w

w=0.195 gm.

W =0.596 gm.

2E+410/195    = 0.596 /0.195

=3.05

2E =(195 x 3.05) -410

E = 184.75 /2 = 92.38

E = 92.38

molecular weight =E x  acidity of  base

=92.38 x 1 = 92.38

n= molecular weight/ Empirical formula weight

n= 92.38 /93 = 0.99 =1.0

mol. formula =(  Empirical formula )n

=(C6H7N)n

=(C6H7N)1

mol. formula = C6H7N  Ans.

Q 3– 0.20 gm of a monobasic acid on combustion give 0.505 gm CO2 & 0.0892 gm. H2O. 0.183 gm. of this acid requires 15 ml of N/10 NaOH for complete neutralisation. Determine molecular formula.

Ans.  % of C =(12 x wt. of CO2 x 100) /(44 x wt. of org.compd)

= (12 x 0.505 x 100) /(44 x 0.20)

% of H =(2 x wt. of H2O x 100) /(18 x wt. of org.compd)

% of H =(2 x 0.0892 x 100) /(18 x wt. of 0.20)

=4.95 %

% of C =68.86%,                              % of H=4.95%

% of O = 100–(68.86+4.95)

=26.19

 Element % Atomic weight Relative no. of atoms Simple Ratio Simplest ratio C 68.86 12 68.86/12= 5.74 5.74/1.64=3.5 3.5 x 2=7 H 4.95 1 4.95/1= 4.95 4.95/1.64=3 3 x 2=6 O 26.19 16 26.19/16=1.64 1.64/1.64=1 1 x 2 =2

C : H : O=7 : 6 : 2

Empirical formula = C7H6O2

Empirical formula weight = 12 x 7 +6 x 1 +2 x 16 = 122

molecular formula =( C7H6O)n

Calculation of Molecular weight by volumetric method-

w/E =N V(l)

w=0.183 gm.,N =1/10 , V =15 ml. =.015 l

0.183 /E = .015/10

E= 122

Mol. wt. = E x basicity of acid = 122 x 1 =122

n= molecular weight / Empirical formula weight

n =122/122 =1

Mol. formula = (C7H6O2)1

molecular formula =C7H6O2 Ans.