Atoms and Molecules


   
 
Numericals Based on Mole Concept
Important Relationships
1 mole of an atom = 1 gram atomic weight of an atom
1 mole of a molecule = 1 gram molecular weight of molecule
1 mole of a gas = 22.4 liters of gas at STP
1 mole of a substance = 6.023 x 1023, atoms, molecules or ions
1 molar volume = 22.4 dm3 /L at STP
 
Example: 1
Calculate the volume occupied by 2.8 g of N2 at STP.
 
Solution:
Molecular weight of N2 = 2 x 14 = 28 g
28 g of N2 at STP occupies = 22.4 L
2.8 g of N2 at STP = ?
28 g of N2 = 22.4 L
 2.8g of N2 = ?
 
2.8 g of N2 at STP occupies a volume of 2.24 L.
 
Example: 2
Calculate gram molecular weights of the following gases:
a. N2 (if 360 cm3 at STP weighs 0.45g)
b. Cl2 (if 308 cm3 at STP weighs 0.97g)
 
Solution:
a. 360 cm3 of N2 = 0.45g
22.4L of gas = 1 gram molecular weight
22.4L = 22,400 cm3. (1L = 1000 cm3)
360 cm3 of N2 = 0.45g
 22,400 cm3 of N2 = ?
Gram molecular weight of N2 is 28 g.
 b. 308 cm3 Cl2 = 0.979 g
 22.400 cm3 of Cl2 = ?
 
Molecular weight of Cl2 = 71.9 g
 
Example: 3
What is the volume of 32 g of sulphur dioxide measured at STP?
 
Solution:
Molecular formula = SO2
Molecular weight = 1 x 32 + 2 x 16 = 64 g
64g of SO2 occupies 22.4 L
32 g of SO2 = ?
Volume of 32 g of SO2 is 11.2 litres.
 
Example: 4
Calculate the volume at S.T.P. of 7.1g of chlorine.
 
Solution:
Cl = 35.5
1 mole of a substance = 22.4 L
1 Mole of a substance = 1 GMM
1 GMM of Cl2 = 71 g
71 g of Cl2 = 22.4 L
7.1 g of Cl2 = ?
7.1 g of Cl2 will occupy a volume of 2.24 litres.
 
Example: 5
Calculate the number of moles of nitrogen in 7g of nitrogen.
 
Solution:
1 mole of N2 = 1 GMM
1 mole of N2 = 2 x 14g = 28g
1 mole = 28 g
? = 7 g
7g of nitrogen is equal to 0.25 moles.
 
Example: 6
Calculate the mass of 0.4 moles of water.
 
Solution:
1 GMM of water (H2O) = 2 x 1 + 16 = 18 g.
18 g = 1 mole
Xg = 0.4 moles
0.4 moles of water weigh 7.2 g.
Example: 7
Calculate the gram atoms present in 8g of oxygen.
 
Solution:
 
Example: 8
Calculate the gram molecules present in 45 g of water.
 
Solution:
1 molecule = 2 + 16g = 18g of H2O
 ? = 45 g of H2O
 
Example: 9
Calculate the number of molecules in 500g of sodium chloride.
 
Solution:
1 GMM = 6.023 x 1023 molecules
23 + 35.5 g = 58.5g = 1 GMM of NaCl
58.5g = 6.023 x 1023 molecules
500g = ?
The number of molecules in 500g of sodium chloride
  = 34.2 x 6.023 x1023 molecules
 
Example: 10
About 0.48 g of a gas forms 100 cm3 of vapours at STP. Calculate the gram molecular weight of the gas.
 
Solution:
22.4L of a gas = 1 GMM
100 cm3 of gas = 0.48 g
22.4 x 1000 of gas = ?
 
Gram molecular weight of the gas is 107.52 g.
 
Numericals Based on Percentage Composition
All elements are represented by symbols and all compounds represented by chemical formulae indicating the number of atoms of elements and also the proportion of the atoms in the compound.
Example: Hydrogen atom is represented as H.
Hydrogen molecule is represented as H2.
A compound of hydrogen, water is represented as H2O.
In H2O - proportion of atoms H : O = 2 : 1
 
Knowing the proportion of atoms in a compound, the percentage composition can be calculated. Percentage composition of a compound is the percent by weight of each element present in it. 
 
 
 
Example: 11
Calculate the percentage by weight of all the elements present in calcium carbonate.
 
Solution:
Calcium carbonate = CaCO3 Ca = 40, C = 12, O = 16
GMM = 1 x 40 + 1 x 12 + 3 x 16
 = 40 + 12 + 48 = 100 g
 
 
 
 
 
Example: 12
Calculate the percentage by weight of potassium in potassium dichromate.
 
Solution:
Potassium dichromate = K2Cr2O7
GMM = (2 x 39) + (2 x 52) + (7 x 16)
 = 78 + 104 + 112 g = 294 g
 
Empirical and Molecular Formula of a Compound
 
Numericals Based on Empirical Formula
Empirical formula is the formula of a compound, which shows the simplest whole number ratio between the atoms of the elements in the compound. It does not indicate the actual number of atoms of the elements present but the simplest whole number ratio.
Example:
 
How can we differentiate a molecular formula from an empirical formula? If the subscripts in the formula have a common divisor, it is usually a molecular formula. Generally the empirical formula is multiplied by this common divisor to get the molecular formula.
Example:
Empirical formula of acetic acid is CH2O
 Molecular formula is CH3COOH = C2H4O2
 
 C1H2O1 x 2 = C2H4O2 [Molecular formula]
 
 Steps for Calculation
Calculate the percentage by weight of each element.
Find out relative number of atoms by dividing percentage of weight by atomic weight.
Choose the simplest and the smallest ratio; divide all the ratios by it
If whole numbers are not obtained, then multiply it by a smallest integer to make it whole.
 
Example: 13
An oxide of iron contains 72.41% of iron. Calculate the empirical formula for the oxide of iron [Fe = 56; O=16].
 
 Solution: 
 Therefore simple ratio = Fe3O4.
 Empirical formula = Fe3O4.
 
Determination of Molecular Formula from Empirical Formula
Molecular formula is the chemical formula, which represents the actual numbers of atoms of each element present in a compound.
 
Steps for Calculation
Calculate empirical formula
Use vapour density if given
If molecular weight given, calculate 'n' using this formula
 
Molecular formula = n x empirical formula
 
Example: 14
Calculate the molecular formula of a compound with vapour density of 30 having 40% carbon; 6.67% of hydrogen and the rest is oxygen.
 
Solution:
 
 
Empirical formula = C1H2O1
Empirical formula weight = 12 x 1 + 2 x 1 + 1 x 16
  = 12 + 2 + 16
  = 30 g
 
Molecular weight = 2 x vapour density
 = 2 x 30
  = 60
Molecular weight = n x empirical weight
  60 = n x 30
 
Molecular formula = n x empirical formula = 2 x CH2O = C2H4O2
 
Example: 15
A compound has molecular formula C5H10., what is its empirical formula?
 
Solution:
Ratio of C atoms to H atoms is 5:10 = 1:2
Empirical formula is C1H2
 
  
     
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