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A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas ?

A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas ?

Grade:12

6 Answers

Gaurav
askIITians Faculty 164 Points
8 years ago
Number of Moles of H = 4.07/1 = 4 (almost)
Number of Moles of C = 24.27/12 = 2(almost)
Number of Moles of Cl = 71/35.5 = 2​(almost)
Ratio
C:H:Cl = 2:4:2 = 1:2:1
Empirical Formula = CH2Cl
Empirical Formula Mass =12+2+35 = 49
n = 98.96/46 = 2(almost)
Molecular formula = n x empirical formula = C2H5Cl2
saksham
13 Points
6 years ago
In this question  Gaurav answered right but in last steo i.e.  [molecular formula = n x empirical formula = c(2)H(4)CL(2),   
           thank u
Mayank Tiwari
15 Points
5 years ago
He had divided it by 46.  HE has to divided  it  by  49  n=98.96\49=2 appox. 
Molecular  formula= n * emperical formula 
=C2hcl2
Khadeeja Salih
13 Points
5 years ago
No: of moles of H=4.07÷2.01=2
No: of moles of C=2.02÷2.01=1
No: of moles of Cl=2.02÷2.01=1
Therefore EF=CH2Cl
n=Mm÷Ef
MF=n(CH2Cl)
Therefore MF=C2H4Cl2
Therefore MF=n[CH2Cl)
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Hello student
Let us assume 100 g of sample of given compound. Thus the percentage of each element will represent its mass content in sample.
The number of moles of
Hydrogen, H = 4.07/1 = 4 (approx)
Carbon, C = 24.27/12 = 2 ( approx)
Chlorine, Cl = 71/35.5 = 2
 
The ratio of moles and thus of elements in the given compound, C:H:Cl = 2:4:2 = 1:2:1
Hence, Empirical formula = CH2Cl
The molecular formula will be of the form, (CH2Cl)n
Empirical formula mass = 12 + 2 x 1 + 35.5
                                      =  49.5 g
Now, molar mass of Compound = 98.96 g
 
Hence, n = 98.96/49.5 = 1.999 = 2
Hence the molecular formula is : C2H4Cl2
 
Hope it helps
Regards,
Kushagra
nisha bajaj
13 Points
3 years ago
atomic mass of H C and cl are 1, 12 and 35.5 respectively
let compound form is 100g, so the given percentages of each element represent its mass.
so. the
no. of moles of hydrogen= prcentage of hydrogen/atomic mass of hydrogen 
=4.07/1=4 (approx because AM of H is 1.008)
no.of moles of H = 4
 
No. of moles of  C is 24/12=2 [approx because carbon prcent is 24.7]
No. of moles of  C=2
 
no. of moles of Cl = 71\35.5=2
no. of moles of Cl = 2
 
thus the ratio of given compound = C:H:Cl
= 2:4:2
=1:2:1
so Emperical formula of this CH2Cl
emperical mass will be = mass of C + mass of H +mass of CL 
= 12+2*1+35.5
=49.5 g
so, the Emperical formula mass = 49.5 g
 
molar mass of compound=98.96 g
:.n= 68.96/49.5 = 1.99 =2 
 
hence the molecular formula = C2H2*2Cl2
=C2H4Cl2
So the EF = CHCl
         MF= C2H4Cl2
 
 

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