Question

answer me Q 27 plzz plzzz plzzz

Answer 1

Heyy !
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percentage of substances :
carbon = 40.687 %
hydrogen = 5.085 %
Oxygen = 54.228%
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In 100 g 
40.687 g = C 
5.085 g  = H 
54.228 g = O
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No. Moles of each element 
Moles of C = 40.687 / 12.01     =  3.38
 
Moles of H = 5.085 / 1.008       =  5.04

Moles of O = 54.228 / 15.9994 =  3.38 
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Divide the mole value by smallest number 

3.38 / 3.38 = 1 X 2   = 2

5.04 / 3.38 = 1.5 X 2 = 3

3.33/ 3.38  = 1  X 2   = 2
As we need whole number ratio so multiplied by 2

empirical formula ratio = 2:3:2 for C: H : O 
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Empirical formula -  C₂H₃O₂
Emirical formula mass = 12.1X2 + 1.008 X 3 + 15.9994 X 2
                                    = 24.2 +3.04+ 31.998
                                    = 59.222 g ≡ 59 g
now we have to determine the value of n 

n = molar mass / empirical formula mass 

but we don't know molar mass but we know vapor density 

vapour density = 59 

Also molar mass = 2 x vapour density 
                           = 2 X 59 
                           = 118 g 
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Now n = 118/59  = 2 
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molecular formula = n(empirical formula )
                             = 2 (C₂H₃O₂)
                             = C₄H₆O₄
So the given compound is Dimethyl Oxalate 
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Answer 2

percentage of substances :
carbon = 40.687 %
hydrogen = 5.085 %
Oxygen = 54.228%
___________________________________________________________
In 100 g 
40.687 g = C 
5.085 g  = H 
54.228 g = O
___________________________________________________________
No. Moles of each element 
Moles of C = 40.687 / 12.01     =  3.38
 
Moles of H = 5.085 / 1.008       =  5.04

Moles of O = 54.228 / 15.9994 =  3.38 
____________________________________________________________
Divide the mole value by smallest number 

3.38 / 3.38 = 1 X 2   = 2

5.04 / 3.38 = 1.5 X 2 = 3

3.33/ 3.38  = 1  X 2   = 2
As we need whole number ratio so multiplied by 2

empirical formula ratio = 2:3:2 for C: H : O 
_______________________________________________________
Empirical formula -  C₂H₃O₂
Emirical formula mass = 12.1X2 + 1.008 X 3 + 15.9994 X 2
                                    = 24.2 +3.04+ 31.998
                                    = 59.222 g ≡ 59 g
now we have to determine the value of n 

n = molar mass / empirical formula mass 

but we don't know molar mass but we know vapor density 

vapour density = 59 

Also molar mass = 2 x vapour density 
                           = 2 X 59 
                           = 118 g 
__________________________________________________________
Now n = 118/59  = 2 
__________________________________________________________
molecular formula = n(empirical formula )
                             = 2 (C₂H₃O₂)
                             = C₄H₆O₄
So the given compound is Dimethyl Oxalate

Hope it helps