Atoms and Molecules

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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 g28 g of N₂ at STP occupies = 22.4 L2.8 g of N₂ at STP = ?28 g of N₂ = 22.4 L2.8g 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 = 0.45g22.4L of gas = 1 gram molecular weight22.4L = 22,400 cm³. (1L = 1000 cm³)360 cm³ of N₂ = 0.45g22,400 cm³ of N₂ = ?Gram molecular weight of N₂ is 28 g.b. 308 cm³ Cl₂ = 0.979 g22.400 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 = SO₂Molecular weight = 1 x 32 + 2 x 16 = 64 g64g of SO₂ occupies 22.4 L32 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 = 35.51 mole of a substance = 22.4 L1 Mole of a substance = 1 GMM1 GMM of Cl₂ = 71 g71 g of Cl₂ = 22.4 L7.1 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 GMM1 mole of N₂ = 2 x 14g = 28g1 mole = 28 g? = 7 g7g 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 moleXg = 0.4 moles0.4 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 molecules23 + 35.5 g = 58.5g = 1 GMM of NaCl58.5g = 6.023 x 10²³ molecules500g = ?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 GMM100 cm³ of gas = 0.48 g22.4 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 = 16GMM = 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 = K₂Cr₂O₇GMM = (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:

Organic Compound	Empirical Formula	Molecular Formula
Benzene	C1H1	C6H6
Glucose	C1H2O1	C6H12O6

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 CH₂OMolecular 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:

Element	% by wt	At wt	Relative No.of Atoms	Simple Ratio
Fe	72.41	56	72.41/56 = 1.29	1.29/1.29 = 1 x 3 = 3
O	27.59	16	27.59/16 = 1.72	1.72/1.29 = 1.33 x 3 = 4

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:

Element	% by wt	At wt	Rlative No.of Atoms	Simple Ratio
C	40	12	40/12 = 3.33	3.33/3.33 = 1
H	6.67	1	6.67/1 = 6.67	6.67/3.33 = 2
O	100 - 46.67	16	53.33/16 = 3.33	3.33/3.33 = 1

Empirical formula = C₁H₂O₁Empirical 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= 60Molecular weight = n x empirical weight60 = n x 30Molecular 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 = 1:2Empirical formula is C₁H₂