# Determination of Atomic Mass: Cannizzaro’s Method

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• This method was proposed by Cannizzaro, Stanislao (1826-1910), an Italian chemist.

#### Principle of Cannizzaro’s Method:

• An atom is the smallest part of an element that can be present in a molecule of a compound. Hence the smallest weight of an element contained in the molecular mass of its compounds shall be the atomic mass of that element.

#### Steps Involved in Cannizzaro’s Method:

1. Collect as many compounds of the element as possible.
2. Determine molecular mass of each compound.
3. Determine percentage composition of these compounds.
4. Calculate the relative mass of that particular element in the molecular mass of each compound from the molecular mass of the compound and its percentage composition.
5. The highest common factor (HCF) of the values obtained gives atomic mass.

### Problems Based on Cannizzaro’s Method:

#### Example – 1:

• For series of volatile compounds following data is obtained. Using it calculate atomic mass of carbon.
 Substance Vapour Density Percentage Carbon Carbondioxide 22 27.3 Methane 8 75.3 Acetone 29 62.1 Ethyl acetate 44 54.5 Pyridine 39.5 76.0 Benzene 39.0 92,3
• Solution:
 Substance Molecular Mass = 2 × Vapour Density mass of carbon in one gram mole       of compound Carbondioxide 2 × 22 g = 44 g (27.3 / 100) × 44 g = 12 g Methane 2 × 8 g = 16 g (75.3 / 100) × 16 g = 12 g Acetone 2 × 29 g = 58 g (62.1 / 100) × 58 g = 36 g Ethyl acetate 2 × 44 g = 88 g (54.6 / 100) × 88 g = 48 g Pyridine 2 × 39.5 g = 79 g (76.0 / 100) × 79 g = 60 g Benzene 2 × 39 g = 78 g (92.3 / 100) × 78 g = 72 g

By Cannizzaro’s method HCF of the numbers in last column is 12. Hence the atomic mass of carbon is 12 g

#### Example – 2:

• The pecentage of carbon in its four compounds is 92.2; 62.0; 40.0 and 15.8 respectively. The vapour densities of these compounds are 39; 29; 30 and 38 respectively. Deduce atomic mass of the carbon.
• Solution:
 Substance Molecular Mass = 2 × Vapour Density mass of carbon in one gram mole       of compound Compound – I 2 × 39 g = 78 g (92.2 / 100) × 78 g = 72 g Compound – II 2 × 29 g = 58 g (62.0 / 100) × 58 g = 36 g Compound – III 2 × 30 g = 60 g (40.0 / 100) × 60 g = 24 g Compound – IV 2 × 38 g = 76 g (15.8 / 100) × 76 g = 12 g

HCF of the numbers in last column is 12. Hence the atomic mass of carbon is 12 g

#### Example – 3:

• The vapour densities of five compounds of a certain element are 23, 26, 22, 8.5 and 24 respectively. The percentage of the same element in these compounds are 91.3, 53.8, 63.7; 82.4; and 97.7 respectively. Find atomic mass of the element.
• Solution:
 Substance Molecular Mass = 2 × Vapour Density mass of carbon in one gram mole       of compound Compound – I 2 × 23 g = 46 g (91.3 / 100) × 46 g = 42 g Compound – II 2 × 26 g = 52 g (53.8 / 100) × 52 g = 28 g Compound – III 2 × 22 g = 44 g (63.7 / 100) × 44 g = 28 g Compound – IV 2 × 8.5 g = 17 g (82.4 / 100) × 17 g = 14 g Compound – V 2 × 24 g = 48 g (97.7 / 100) × 48 g = 56 g

By Cannizzaro’s method HCF of the numbers in last column is 14. Hence the atomic mass ofthe element is 14 g

#### Example – 4:

• Vapour densities of three substances referred to hydrogen as unity were 45, 70 and 25 and percent mass of certain element contained in each were 22.22, 42.86 and 40 respectively. Find the probable atomic mass of the element.
• Solution:
 Substance Molecular Mass = 2 × Vapour Density mass of carbon in one gram mole       of compound Compound – I 2 × 45 g = 90 g (22.22 / 100) × 90 g = 20 g Compound – II 2 × 70 g = 140 g (42.86 / 100) × 140 g = 60 g Compound – III 2 × 25 g = 50 g (40 / 100) × 50 g = 20 g

By Cannizzaro’s method  HCF of the numbers in last column is 20. Hence the atomic mass ofthe element is 20 g

#### Example – 5:

• Vapour densities of seven compounds of phosphorous phosphoric oxide, phosphorous oxide, phosphorous trichloride, phosphorous pentafluoride, phosphorous oxychloride, phosphorous pentasulphide and tetra phosphorous trisulphide were 150, 110, 70, 63, 77, 111, 114 and percent mass of phosphorous contained in each were 43.7, 56.4, 22.5, 24.8, 20.2, 27.9 and 56.4 respectively. Find the probable and exact atomic mass of phosphorous.
• Solution:
 Substance Molecular Mass = 2 × Vapour Density mass of carbon in one gram mole       of compound Phosphoric acid 2 × 150 g = 300 g (43.7 / 100) × 300 g = 131.1 g Phosphorous oxide 2 × 110 g = 220 g (56.4 / 100) × 220 g = 124.1 g Phosphorous trichloride 2 × 70 g = 140 g (22.5 / 100) × 140 g = 31.1 g Phosphorous pentafluoride 2 × 63 g = 126 g (24.8 / 100) × 126 g = 31.1 g Phosphorous oxichloride 2 × 77 g = 154 g (20.2 / 100) × 154 g = 31.1 g Phosphorous oentasulphide 2 × 111 g = 222 g (27.9 / 100) × 222 g = 61.9 g Tetra Phosphorous trisulphide 2 × 114 g = 228 g (56.4 / 100) × 228 g = 128.6 g

By Cannizzaro’s method  The approximate HCF of the numbers in last column is 31.1. Hence the probable atomic mass of phosphorous is 31.1 g

• To find exact atomic mass we can consider any one compound in the list. Let us consider first compound phosphoric acid.% of phosphorous = 43.7% of oxygen = 100 – 43.7 = 56.3, Mass of phosphorous = 43.7 g Mass of oxygen = 56.3 g

Equivalent mass of an element = (Mass of an element in compound / Mass of oxygen in compound) × 8

∴  Equivalent mass of an element = (43.7 g / 56.3 g) × 8 g = 6.21 g

Now, Valency = Approx atomic mass / Equivalent Mass = 31.1 / 6.21 = 5 (nearest whole number)

Corrected atomic mass = Equivalent mass x valency – 6,21 g x 5 = 31.05 g

Thus the exact atomic mass of phosphorous is 31.05.

#### Example – 6:

• A metal forms three volatile chlorides containing 23.6, 38.2 and 48.3 per cent of chlorine respectively. The vapour densities of chlorides are 74.6, 92.9 and 110.6 respectively. The specific heat of the metal is 0.055. Find the exact atomic mass of the metal and formulae of its chlorides.
• Solution:
 Substance % of Chlorine % of metal Chloride – I 23.6 100 – 23.6 = 76.4 Chloride – II 38.2 100 – 38.2 = 61.8 Chloride – III 48.3 100 – 48.3 = 51.7

 Substance Molecular Mass = 2 × Vapour Density mass of carbon in one gram mole       of compound Chloride – I 2 × 74.6 g = 149.2 g (23.6 / 100) × 149.2 g = 114 g Chloride – II 2 × 92.9 g = 185.8 g (38.2 / 100) × 185.8 g = 144.82 g Chloride – III 2 × 110.6 g = 221.2 g (48.3 / 100) × 221.2 g = 114.36 g

The approximate HCF (The least mass) is 114. Hence probable  atomic mass of metal is 114.

• To find exact atomic mass we can consider any one compound in the list. Let us consider first chloride (100 g).

% of metal = 76.4

% of chlorine = 23.6

Mass of metal = 76.4 g

Mass of chlorine = 23.6 g

Equivalent mass of metal = Mass of metal in chloride x 35.5 / Mass of chlorine in metal chloride

Equivalent mass of metal = 76.4 x 35.5 / 23.6 = 114.9

Valency = Approximate atomiic mass / Equivalent mass = 114 / 114.9 = 1  (Nearest whole number)

Actual atomic mass = Equivalent mass x valency = 114.9 x 1 = 114.9

• To find molecular formulae of the chlorides: Let x be the valency of the metal, hence its molecular formula is MClx.
• First Chloride:

Molecular mass of the first chloride  = 114.9 + 35.5x = 149.2

∴    35.5x = 149.2 – 114.9

∴    35.5x = 34.3

x = 1  (Nearest whole number)

Hence the formula of the first chloride is MCl.

• Second Chloride:

Molecular mass of the second chloride  = 114.9 + 35.5x = 185.8

∴    35.5x = 185.8 – 114.9

∴     35.5x = 70.9

∴       x = 2

Hence the formula of the second chloride is MCl2.

• Third  Chloride:

Molecular mass of the third chloride  = 114.9 + 35.5x = 221.2

∴      35.5x = 221.2 – 114.9

∴      35.5x = 106.3

∴       x = 3

Hence the formula of the third chloride is MCl3.

The exact atomic mass of the metal is 114.9

andformulae of chlorides are MCl, MCl2, MCl3 respectively

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### One Comment

1. aditya kumar patra

its really so nice easy for understanding …….good job…thanx to the company of chemistry