Numericals Based on Mole Concept

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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

• Molecular weight = 2 x Vapor density

• 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 = 22.4L

2.8 g of N₂ at STP = ?

28 g of N₂ = 22.4L

2.8g of N₂ = ?

2.8 g of N₂ at STP occupies a volume of 2.24L.

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 32g of sulphur dioxide measured at STP?

Solution

Molecular formula = SO₂

Molecular weight = 1 x 32 + 2 x 16 = 64g

64g of SO₂ occupies 22.4L

32 g of SO₂ = ?

Volume of 32 g of SO₂ is 11.2 liters.

Example: 4

Calculate the volume at S.T.P. of 7.1g of chlorine.

Solution:

Cl = 35.5

1 mole of a substance = 22.4L

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 liters.

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 = 18g.

18g = 1 mole

Xg = 0.4 moles

0.4 moles of water weighs 7.2g.

Example:7

Calculate the vapor density and molecular weight of CO₂ gas occupying 200 mL and weighing 0.40 g at STP.

Solution:

200 mL of CO₂ = 0.40 g

22.4 L of CO₂ = 12 + 32 (44g)

Molecular weight = 2 x vapor density

= 2 x 22.22 = 44.44

Note

Molecular weight has no units.

Example: 8

Calculate the gram atoms present in 8g of oxygen.

Solution

Example: 9

Calculate the gram molecules present in 45g of water.

Solution

1 molecule = 2 + 16g = 18g of H₂O

? = 45 g of H₂O

Example: 10

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 is

34.2 x 6.023 x²³ molecules

Example: 11

0.48 g of a gas forms 100 cm³ of vapors 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.

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.

Percentage composition of an element

Numericals based on percentage composition

Example:1

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: 2

Calculate the percentage by weight of potassium in potassium dichromate.

Solution

Potassium dichromate = K₂Cr₂ O₇

GMM = (2 x 39) + (2 x 52) + (7 x 16)

= 78 + 104 + 112 g = 294 g