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Example-1

Example-1
An organic compound contains 49.3% carbon, 6.84% hydrogen, and its vapor density is 73. Molecular formula of the compound is:
  1. Description: 51999.png
  2. Description: 52007.png
  3. Description: 52015.png
  4. Description: 52025.png
Solution (B)
Element % Number of moles Simple ratio
C 12 49.3/12 = 4.1 4.1/2.7 = 1.3 × 2 = 2.6 = 3
H 1 6.84/1= 6.84 6.84/2.7 = 2.5 × 2 = 5
O 16 43.86/16 = 2.7 2.7/2.7 = 1 × 2 = 2
 
Empirical formula = Description: 52050.png
 
Empirical formula weight = 12 × 3 + 1 × 5 + 16 × 2 = 73
 
Molecular weight = Vapor density × 2 = 73 × 2 = 146
 
Description: 52058.png
 
Molecular formula = (Empirical formula)n Description: 52065.png.
 

Example-2

Example-2
If 0.228 g of silver salt of dibasic acid gave a residue of 0.162 g of silver on ignition, then the molecular weight of the acid is:
  1. 70
  2. 80
  3. 90
  4. 100
Solution (C)
Mass of silver salt taken = 0.228 g
 
Mass of silver left = 0.162 g
 
Basicity of acid = 2
 
Step 1: To calculate the equivalent mass of the silver salt (E)
 
Description: 52076.png
 
Description: 52084.png
 
Description: 52093.png
 
Step 2: To calculate the equivalent mass of acid.
 
Equivalent mass of acid
 
= Equivalent mass of silver salt – Equivalent mass of Ag + Basicity
 
= 152 – 108 + 1 = 152 – 109 = 43 (equivalent mass of acid)
 
Step 3: To determine the molecular mass of acid.
 
Molecular mass of the acid = Equivalent mass of acid × Basicity = 45 × 2 = 90.
 

Example-3

Example-3
About Description: 52100.png of carbohydrate of empirical formula Description: 52110.png contain 1 g of hydrogen. The molecular formula of the carbohydrate is:
  1. Description: 52131.png
  2. Description: 52140.png
  3. Description: 52148.png
  4. Description: 52155.png
Solution (D)
∵ 0.0833 mol carbohydrate has hydrogen = 1 g
 
∴ 1 mol carbohydrate has hydrogen Description: 52176.png
 
Empirical formula Description: 52186.png has hydrogen = 2 g
 
Hence Description: 52201.png
 
Hence molecular formula of carbohydrate Description: 52210.png
Description: 52220.png
 

Example-4

Example-4
A hydrocarbon contains 10.5 g carbon and 1 g hydrogen. Its 2.4 g has 1 L volume at 1 atm and Description: 52229.png. The hydrocarbon is:
  1. Description: 52238.png
  2. Description: 52246.png
  3. Description: 52255.png
  4. None of these
Solution (A)
Description: 52263.png
Description: 52273.png
∴ Description: 52281.png
Description: 52289.pngDescription: 52296.png
Description: 52310.png
Description: 52320.png
 

Example-5

Example-5
An organic compound on analysis gave C = 48 g, H = 8 g, and N = 56 g. Volume of 1.0 g of the compound was found to be 200 mL at NTP. Molecular formula of the compound is:
  1. Description: 52328.png
  2. Description: 52338.png
  3. Description: 52346.png
  4. Description: 52353.png
Solution (A)
Element % Number of moles Simple ratio
C 48 48/12 = 4 1
H 8 8/1 = 8 2
N 56 56/14 = 4 1
 
 
Empirical formula = Description: 52383.png
 
Empirical formula mass = 28
 
Now, 200 mL of compound = 1 g
 
22400 mL of compound = Description: 52392.png
 
Description: 52403.png
 
∴ Molecular formula Description: 52411.png.
 




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