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a compound contains 61%of carbon,11.88% hydrogen and 27.12%oxygen.if the volume density of compound is 59%.calculate the molecular formula.

a compound contains 61%of carbon,11.88% hydrogen and 27.12%oxygen.if the volume density of compound is 59%.calculate the molecular formula.

Grade:11

2 Answers

Vikas TU
14149 Points
4 years ago
Dear student 
C=61/12=5.08
H=11.8/1=11.8
O=27/16=1.68
Lowest ratio value multiple to all value
C=5.08/1.68=3.02
H=11.8/1.68=7.2
O=1.68/1.68=1.68
Empirical formula is =C3H7O1
Empirical formula mass =3×12+7×1+1×16=59
n=density /emyrical formula mass =59/59=1
Molecular formula =n×emyrical formula =1×C3H7O1=C3H7O
askiitians
15 Points
one year ago
Relative atomic weight ratio :
carbon= 61/12
            =5.08
hydrogen= 11.8/1
                     =1
oxygen=27/16
             =1.68
Simple atomic weight ratio ( smallest ratio value multiple to all  value)
carbon =5.08/1.68
             =3.02
hydrogen= 11.8/1.68
                    7.2
oxygen = 1.68/1.68
                 =1
therefore, the ratio is    = 3:7:1
 
so ,Empirical formula is C3H7O
 
Empirical formula weight (mass) = 3×12+7×1+1×16 
                                                             = 59
Now, Molecular formula = n ×Empirical formula (n= density/ empirical formula weight ) 
 
=59/59×C3H7O
= 1×C3H7O 
 = C3H7O   
 therefore , the correct answer is = C3H7O
 
 

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