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A compound on analysis gave the following percentage composition C = 54.54%, H, 9.09%, O = 36.36 %. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound. Can anyone give me questions like this

  1. A compound on analysis gave the following percentage composition C = 54.54%, H, 9.09%, O = 36.36 %. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound.
 
Can anyone give me questions like this

Grade:9

3 Answers

A M S ARUN KRISHNA
216 Points
8 years ago
for the first question
1)A compound on analysis gave the following percentage composition: Na=14.31%, S = 9.97%,              H = 6.22%, O = 69.5%, calculate the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallisation. Molecular mass of the compound is 322 [Na = 23, S = 32, H = 1, 0 = 16].
Solution :- Calculation of empirical formula
 
Element
Percentage
Relative number of moles
Simple ratio moles
Simplest whole number ratio
Na
14.31
14.31
------- = 0.62
  23
0.62
------  = 2
0.31
2
S
9.97                      
19.97
-------- = 0.31
  32
0.31
------  = 1
0.31
1
H
6.22
6.22
------ = 6.22
  1
6.22
-----  = 20
0.31
20
O
69.5
  69.5
-------- = 4.34
   16
4.34
------ = 14
0.31
14
 
The empirical formula is Na2SH20O14
Calculation of Molecular formula
Empirical formula mass = (23 x 2) + 32 + (20 x 1) + (16 x 14) = 322
Molecular mass                                322
n = ----------------------------------------   = ------------  = 1
            Empirical formula mass                   322
 
Hence molecular formula = Na2SH20O14
Since all hydrogens are present as H2O in the compound, it means 20 hydrogen atoms must have combined. It means 20 hydrogen atoms must have combined with 10 atoms of oxygen to form 10 molecules of water of crystallisation. The remaining (14 - 10 = 4) atoms of oxygen should be present with the rest of the compound.
Hence, molecular formula = Na2SO4.10H2O.
 
 
  1. A compound on analysis gave the following percentage composition C = 54.54%, H, 9.09%, O = 36.36 %. The vapour density of the compound was found to be 44. Find out the molecular formula of the compound.
Solution
Calculation of empirical formula
Element
Percentage
Relative number of moles
Simple ratio moles
Simplest whole number ratio
C
54.54
  54.54
------- -  =  4.53
    12
  4.53
--------  = 2 
  2.27
2
H
9.09
   9.09
--------    =  9.09
     1
  9.09
--------  = 4
  2.27
4
O
36.36
  36.36
----------  =  2.27
    16
  2.27
---------  =  1
  2.27
1
 
Calculation of Molecular formula
Empirical formula mass = 12 x 2 + 1 x 4 + 16 x 1 = 44
Molecular mass =  2 x   Vapour density
 = 2 x 44 = 88
             Molecular mass                       88
n = ------------------------------------  = ---------   = 2
        Empirical Formula mass              44
 
= C2H4O        x   2
= C4H8O2
 
A M S ARUN KRISHNA
216 Points
8 years ago
  1. An organic compound was found to have contained carbon = 40.65%, hydrogen = 8.55% and Nitrogen = 23.7%. Its vapour - density was found to be 29.5. What is the molecular formula of the compound?
Ans:- C2H5NO
  1.  A compound contains 32% carbon, 4% hydrogen and rest oxygen. Its vapour density is 75. Calculate the empirical and molecular formula.
Ans:- C2H3O3, C4H6O6
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  3. A compound analyzes as 79.08% C,  5.54% H and 15.38%N. what is the molecular formula if the molar mass is 273.36g/mol?
  4. Determine the molecular formula
  5. A compound has an empirical formula of NO2 and a molar mass of 92g/mol
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Raja
21 Points
6 years ago
What is the simplest formula of the compound which has the following percentage composition: carbon 80%, Hydrogen 20%, If the molecular mass is 30, calculate its molecular formula.

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