In the last article, we have studied the concept of atomic mass and to calculate the average atomic mass. In this article, we shall study the Cannizzaro’s method for determination of atomic mass. 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 – 01:

For series of volatile compounds following data is obtained. Using it calculate atomic mass of carbon.

Solution:

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

Example – 02:

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:

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

Example – 03:

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:

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

Example – 04:

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:

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

Example – 05:

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:

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 the 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 – 06:

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:

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

And formulae of chlorides are MCl, MCl₂, MCl₃ respectively.

In the next article, we shall study to determine atomic mass using the law of isomprphism.

Previous Topic: The Concept of Atomic Mass

Next Topic: Atomic Mass Using the Law of Isomorphism

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Cannizzaro’s Method

The post Cannizzaro’s Method of Determination of Atomic Mass appeared first on The Fact Factor.

The smallest particle of an element which can take part in a chemical reaction is called an atom. According to Dalton’s atomic theory atom is the smallest particle of an element which is indivisible. In modern research, it is proved that the atom is divisible into its constituent particles like electrons, protons, and neutrons. In this article, we shall understand the concept of atomic mass and gram atomic mass (GAM).

Atomic Mass:

Water contains 11.19 % of hydrogen and 88.89% of oxygen. Thus hydrogen and oxygen combine with each other in a ratio 1 : 8 by mass. Besides in water, there are 2 atoms of hydrogen and 1 atom of oxygen. From these two observations, it follows that the mass of oxygen atom is 16 times that of the hydrogen atom. In this hydrogen based system mass of hydrogen is taken as unity and masses of other element were determined relative to the mass of an atom of hydrogen.

A later atom of oxygen was chosen as reference atom because by taking its mass as 16 units, the relative atomic masses of other elements were very close to whole numbers. Oxygen has 3 isotopes. Hence the standard of oxygen was considered as inappropriate. Hence instead of taking the average atomic mass of the mixture of oxygen, stable isotope of carbon (C-12) was taken as standard. Hence in 1961 International Union of Chemists selected the most stable isotope of carbon (C – 12) as a standard atom to compare masses of various elements.

The relative atomic mass of an element is a mass of one atom of the element compared with the mass of an atom of ⁶C₁₂ isotope taken as 12000 units.

Average Atomic Mass:

Isotopes are the atoms of the same element having the same atomic number containing the same number of protons and electrons but different numbers of neutrons hence they possess different mass numbers.

The observed atomic mass of the atom of the element is the average atomic mass of the element taking into consideration the natural abundance of the element. For example, atomic masses of chlorine’s two isotopes are 36 u and 37 u. u stands for unified mass. They are found in the ratio 3: 4 in nature. Hence average atomic mass of chlorine is

The gram atomic mass of an element is atomic mass expressed in grams (GAM). e.g. The gram atomic mass of chlorine is 35.5 g

Thus one gram hydrogen atom means 1.008 g of hydrogen. one gram atom of carbon means 12 g of carbon. 2 gram atom of chlorine means 2 × 35.5 = 71 g of chlorine.

Number of Atoms in Gram Atom:

By Avogadro’s law, one gram atom of an element contains 6.023 × 10²³ atoms. Thus 1 gram atom (i.e. 1.008 g) of hydrogen contains 6.023 × 10²³ atoms. Thus the mass of each atom of hydrogen is

Mass in gram = number of gram atom ×  gram atomic mass

Definition of Valency:

The valency of an element is the number of electrons an atom of the element can accept or donate in the formation of molecule of a compound. OR the number of a hydrogen atom, which can combine with or displaced by one atom of an element is called the valency of the element.

Relation Between Atomic Mass, equivalent Mass, and Valency:

Let A, E, and v be the atomic mass, equivalent mass, and valency of element X. Then the formula of its compound with hydrogen is XH_v.

Thus v parts by mass of hydrogen will combine with A parts by mass of X.

i.e. 1 part by mass of hydrogen will combine with A/v parts by mass of X.

By definition A/v is equivalent mass of the element.

Thus E = A/v

At. Mass (A) = Equ. Mass (E) x Valency (v)

Calculation of Average Atomic Mass by Relative Abundance Method:

Example – 01:

Naturally, occurring lead is found to contain four isotopes 1.40 % ⁸²Pb₂₀₄ isotope with isotopic mass 203.973,  24.10 % ⁸²Pb₂₀₆ isotope with isotopic mass 205.974, 22.10 % ⁸²Pb₂₀₇ isotope with isotopic mass 206.976 and 52.40 % ⁸²Pb₂₀₈ isotope with isotopic mass 207.977. Calculate the average atomic mass of lead.

Solution:

Hence average atomic mass of lead is 207.2 u.

Example – 02:

Naturally, occurring neon is found to contain three isotopes 90.92 % ¹⁰Ne₂₀ isotope with isotopic mass 9.9924, 8.82 % ¹⁰Ne₂₂ isotope with isotopic mass 21.9914, 0.26 % ¹⁰Ne₂₁ isotope with isotopic mass 20.9940 Calculate the average atomic mass of neon.

Solution:

Hence average atomic mass of neon is 20.17 u.

Example – 03:

Naturally, occurring lithium is found to contain two isotopes 8.24 % ³Li₆ isotope with isotopic mass 6.0151 and 91.76 % ³Li₇ isotope with isotopic mass 7.0160 Calculate the average atomic mass of lithium.

Solution:

Hence average atomic mass of lithium is 6.934 u.

Example – 04:

Naturally, occurring silicon is found to contain three isotopes 92.23 % ¹⁴Si₂₈ , 4.67 % ¹⁴Si₂₉, 3.10 % ¹⁴Si₃₀ Calculate the average atomic mass of silicon.

Solution:

Hence average atomic mass of silicon is 28.1 u

Example – 05:

In naturally occurring neon the fractional abundance of various isotopes is as follows 0.9051 of ¹⁰Ne₂₀, 0.0027 of ¹⁰Ne₂₁, 0.0922 of ¹⁰Ne₂₂. Calculate the average atomic mass of neon.

Solution:

Average atomic mass =0.9051 × 20  + 0.0027 ×  21  + 0.0922 × 22

= 18.102 + 0.057 + 2.028 = 20.187 u

Hence average atomic mass of neon is 20.187 u.

Example – 06:

Nitrogen occurs in nature in the form of two isotopes with atomic mass 14 and 15 respectively. If the average atomic mass of nitrogen is 14.0067. What is the percentage abundance of the two isotopes?

Solution:

Let % abundance of the isotope with atomic mass 14 be ‘x’. Hence that of the isotope with atomic mass 15 will be (100 -x).

% abundance of isotope of nitrogen with atomic mass 14 is 99.33 and  with atomic mass 15 is 0.67

Example – 07:

Boron occurs in nature in the form of two isotopes with atomic mass 10 and 11 respectively. If the average atomic mass of boron is 10.80 u. What is the percentage abundance of the two isotopes?

Solution:

Let % abundance of the isotope with atomic mass 10 be ‘x’. Hence that of isotope with atomic mass 11 will be (100 -x).

% abundance of isotope of boron with atomic mass 10 is 20 and 

% abundance of the isotope of boron with atomic mass 11 is 80.

Example – 08:

Chlorine has two stable isotopes Cl – 35 and Cl -37, with atomic masses 34.968 u and 36.956 u respectively. If the average atomic mass is 35.452 u, calculate the % abundance of isotopes.

Solution:

Let % abundance of the isotope with atomic mass 35 be ‘x’. Hence that of the isotope with atomic mass 37 will be (100 -x).

% abundance of isotope of chlorine with atomic mass 35 is 75.65 and 

% abundance of isotope of chlorine with atomic mass 37 is 24.35

In the next article, we shall study Cannizzaro’s method to determine atomic mass.

Previous Chapter: Laws of Chemical Combinations

Next Topic: Atomic Mass by Cannizzaro’s Method

Science > Chemistry > Concept of Atomic Mass and Equivalent Mass > Concept of Atomic Mass

The post Concept of Atomic Mass appeared first on The Fact Factor.