Molecular Formula

source  : quantum mechanics.mc multimedia .com

Q 1– % of C=40%, % of H=6.9% . At 2000C & 76 cm. Pressure 64.7 ml vapours are obtained by heating 0.10 gm. org. compd. calculate mol. formula of compound.

Ans.   % of O = 100–(40+6.9) =53.1 %

 Element % Atomic weight Relative no. of atoms Simplest Ratio C 40 12 40/12=3.33 3.33/3.33=1 H 6.9 1 6.9/1=6.9 6.9/3.33=2 O 53.1 16 53.1/16=3.33 3.33/3.33=1

C:H:O=1:2:1

Empirical formula = CH2O

Empirical formula weight = 12+2+16 =30

Molecular formula =(CH2O)n

Calculation of Molecular weight by victor mayer’s method-

P1 V1/T1 =P2 V2/T2

T1= 200 +273 =473 K; V1 =64.7 ml. ; P1 = 76 cm. =760 mm.

At N.T.P,

P2 =760 mm. ;V2 = ? ml. ; T2= 273 K

P1 V1/T1 =P2 V2/T2

760 x 64.7 /473 =760 x V2 /273

V2 =37.34 ml.

vol. of vapours at N.T.P =37.34 ml.

mol. wt. =wt. of substance x 22400/ vol. of vapours at N.T.P

=0.10 x 22400 /37.34

mol.wt.   =60.0

n = mol.wt. / Empirical formula weight

n =60/30 =2

Molecular formula =(CH2O)n

=(CH2O)2

Molecular formula = C2H4O2 Ans.

Q 2— A compound of C, H & N contain the elements in the ratio of 18:2:7 Calculate its empirical  formula. If mol. wt. is 108. What is its mol. formula

Ans.    C: H: N= 18:2:7

sum of ratio= 18+2+7

= 27

% of C= 18 x 100/27 =66.67 %

% of H=2 x 100 /27 =7.4 %

% of N = 7 x 100 /27 =25.93 %

 Element % Atomic weight Relative no. of atoms Simplest Ratio C 66.67 12 66.67/12=5.55 5.55/1.85=3 H 7.4 1 7.4/1=7.4 7.4/1.85=4 N 25.93 14 25.93/14=1.85 1.85/1.85 =1

C:H:N:O=4:8:2:1

Empirical Formula = C3H4N

Empirical Formula weight = 12 x 3 +1 x 4 +14 x 1= 54

Mol. formula =  (C3H4N )n

mol.wt. = 108

n =mol. wt. / Empirical Formula weight

n = 108 / 54=2

Mol. formula =  (C3H4N )n

=  (C3H4N )2

Mol. formula    = C6H8N2   Ans.

Q 3–The analysis of an organic compound gave the following data ;

(a)- 0.4020 gm. gave 0.6098 gm.& 0.2080 gm. of CO2 &H2O respectively.

(b)-1.01 gm. by kjeldahl’s method produced NH3 which was neutralised by 23.2 ml.N/2  acid.

(c)-0.1033 gm. of  organic compound gave 0.2772 gm.BaSO4.

(d)-0.1015 gm.when vapourised by victor meyer’s  method 27.96 ml.of air at 150C &766 mm. pressure.Calculate molecular formula.( P’ H2O at 150C =16 mm.)

Ans.

(a) wt.of org. compd =0.4020 gm. ; wt. of CO2=0.6098 gm. ; wt. of H2O =0.2080 gm.

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

=12 x 0.6098 x 100 /44 x 0.4020

% of C =41.37 %

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

=2 x 0.2080 x 100/18 x 0.4020

% of H =5.75 %

(b) w =1.01 gm. ;N =1/2 ;V = 23.2 ml.

% of Nitrogen = 1.4 NV/w

=1.4 x 0.5 x 23.2/1.01

% of Nitrogen =16.08 %

(c)     wt.of org. compd =0.1033 gm. ; wt. of BaSO4 = 0.2772 gm.

% of S =32 x  wt. of BaSO4 x 100 /233 x  wt.of org. compd

=32 x  0.2772  x 100 /233 x  0.1033

% of S =36.85 %

 Element % Atomic weight Relative no. of atoms Simplest Ratio C 41.37 12 41.37/12=3.45 3.45 /1.15 =3 H 5.75 1 5.75/1= 5.75 5.75/1.15=5 N 16.08 14 16.08/14=1.15 1.15/1.15 =1 S 36.85 32 36.85/32 =1.15 1.15 /1.15=1

C:H:N:S=3:5:1:1

Empirical Formula = C3H5N S

Empirical Formula weight = 12 x 3 +1 x 5 +14 x 1 +1 x 32= 87

Mol. formula =  (C3H5N S )n

(d)   Calculation of Molecular weight by victor mayer’s method-

P1 V1/T1 =P2 V2/T2

T1= 15 +273 =288 K; V1 =27.96 ml. ; P1  =766 – 16 = 750 mm.

At N.T.P,

P2 =760 mm. ;V2 = ? ml. ; T2= 273 K

P1 V1/T1 =P2 V2/T2

750 x 27.96 /288 =760 x V2 /273

V2 = 26.16  ml.

vol. of vapours at N.T.P = 26.16  ml.

mol. wt. =wt. of substance x 22400/ vol. of vapours at N.T.P

=0.1015 x 22400 /26.16

mol.wt.   =86.9 =87

n = mol.wt. / Empirical formula weight

n =87/87 =1

Molecular formula =(C3H5NS)n

= (C3H5NS)1

Molecular formula = C3H5NS Ans.