Fractional Atomic Masses

Atomic masses of most elements are not whole number, i.e., these are fractional.

It is because most elements in nature occur as a mixture of two or more isotopes in certain fixed proportions.

This is illustrated below.

Calculating the atomic mass of an element from the mass numbers of its isotopes

The atomic mass of an element is the weighted arithmetic mean of the atomic masses of its isotopes present in the sample of the element.

Let us consider a sample of an element X containing its two isotopes X1 and X2. Then

This method is illustrated by taking the case of chlorine.

The two isotopes of chlorine, ³⁵₁₇Cl and ³⁷₁₇Cl occur in the ratio 3 : 1. Then,

Atomic mass of chlorine = (35 u × 3) + (37 u × 1) / 3 + 1

 = (105 u + 37u) / 4 = 142 u / 4      =      35.5 u

Mass numbers are whole numbers,

whereas atomic masses are generally fractional.

EXAMPLE 1  

If bromine occurs in the form of say two isotope ⁷⁹₃₅Br (49.7%) and ⁸¹₃₅Br (50.3%), then calculate the atomic mass of bromine atom.

Solution:

Atomic mass of bromine = (79 u × 49.7) + (81 u × 50.3) / (49.7 + 50.3) = 80.0 u

So, the atomic mass of bromine is 80.0 u.

EXAMPLE 2

The atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotope ¹⁶₈X and ¹⁸₈X in the sample?

Solution:

Atomic mass of X = 16.2 u

Let,

Percentage of the isotope ¹⁶₈X be a

Then,

Percentage of the isotope ¹⁸₈X = (100 – a)

Atomic mass of the sample = (% ¹⁶₈X × 16 u) + ((100 – %¹⁶₈X) × 18 u) / 100

      16.2 u = (a × 16 u) + (100 – a) × 18 u / 100

                             1620 u = 16 a u + 1800 u – 18 a u

This gives,

        (18 a u – 16 a u) = 1800 u – 1620 u

                                2 a u = 180 u

                                     a = 180 / 2 = 90

 Thus,      Percentage of ¹⁶₈X = 90%

 and        Percentage of ¹⁸₈X = (100 – 90)% = 10%

Examples based on Avogadro’s Law

EXAMPLE 1.

100 mL of a gas ‘A’ contains x molecules. How many molecules of the gas ‘B’ will be present in 50 mL of ‘B’ under same conditions of temperature and pressure?

Solution:                       A          :         B

 Volume:             100 mL               50 mL

 No. of molecules:              x                     ?

According to the Avogadro’s law, equal volumes of two gases under similar conditions would contain the same number of molecules. Therefore,  100 mL of gas B would contain x molecules.

Hence, 50 mL of gas B would contain = x × 50 / 100 molecules = x / 2 molecules

 

EXAMPLE 2.

Samples of the gases, oxygen, nitrogen, carbon dioxide and carbon monoxide under the same conditions of temperature and pressure contain the same number of molecules, (say X). The molecules of oxygen (O₂) occupy V litres and have a mass of 8 g.

(a) What is the volume occupied by

                       i.        X molecules of nitrogen (N₂)?

                      ii.        2X molecules of carbon monoxide (CO)?

 (b) What is the mass of CO₂ in grams?

 (Atomic masses are : C = 12, N = 14, O = 16)

Solution: The given data are rearranged as follows:

Gas:           Oxygen      Nitrogen    Carbon      Carbon

dioxide      monoxide

No. of molecules:             X             X             X                X

Volume:                        V L

Mass of gas:                  8 g

Equal volumes of gases under similar conditions of temperature and pressure contain the same number of molecules. Therefore,

 (a)    (i) Volume occupied by X molecules of N₂ = V L

        (ii) Volume occupied by 2X molecules of CO = 2 × V L = 2V L

 (b) On molar basis;          O₂         º               CO₂

 1 mol                              1 mol

 2 × 16 g = 32 g               (12 g + 2 × 16 g) = 44 g

Therefore,           32 g of O₂           º       44 g of CO₂

So,                    8 g of O₂ = (44 g / 32 g) × 8 g of CO₂ = 11 g

 

EXAMPLE 3.

The volumes of gases A, B, C and D are in the ratio of 1 : 2 : 2 : 4     under the same conditions of temperature and pressure.

a.    Which sample of gas contains the maximum number of molecules?

b.   If the temperature and pressure of gas A are kept constant, then what will happen to the volume of A, when the number of its molecules is doubled?

c.    If the volume of A is actually 5.6 L at STP, calculate the number of molecules in the actual volume of D at STP. [Avogadro’s number: 6.02 × 10²³]

d.   Using your answer in (d), state the mass of D, if the gas is dinitrogen oxide (N₂O). (N = 14, O =16).

Solution: The given data are written as follows:

             Gas:      A      B      C      D

 Volume ratio:      1   :   2   :   2   :   4

a.    The number of molecules is proportional to the volume.

Hence, gas D will contain the maximum number of molecules.

b.   Volume of A will get doubled since the number of molecules depends directly on the volume.

c.    Volume of A at STP = 5.6 L

So, Volume of D at STP = 4 × 5.6 L = 22.4 L

22.4 L of any gas at STP will contain one Avogadro’s number of molecules. So, No. of molecules in the actual volume of D = 6.02 × 10²³

d.   The mass of any gas occupying a volume of 22.4 at STP is equal to its molar mass. Hence

Mass of gas D = Mass of 1 mole of N₂O  

    = (2 × 14 + 1 × 16) g = (28 + 16) g = 44 g

 

EXAMPLE 4.

The gases chlorine, nitrogen, ammonia and sulphur dioxide are collected under the same conditions of temperature and pressure. The following table gives volume of gases collected and the number of molecules x in 20 L of nitrogen. Complete the table giving the number of molecules in the other gases in terms of x.

Gas                   Volume/L           No. of molecules

Chlorine                10                          —

Nitrogen                20                          x

Ammonia               20                          —

Sulphur dioxide        5                           —

Solution: The number of molecules in any sample of a gas is proportional to its volume.

So,

Gas                             Volume/L                 No. of molecules

Chlorine                       10                                         x / 2

Nitrogen                       20                                         x

Ammonia                     20                                         x

Sulphur dioxide            5                                        x / 4

Applications of Avogadro’s Law

Avogadro’s law has been found very useful in chemistry. The usefulness of the Avogadro’s law lies in the fact that it differentiates between an atom and a molecule.

Some important applications of the Avogadro’s law are described below.

1. Explaining Gay-Lussac’s Law of Combining Volumes

Consider this reaction,

a A    +     b B    ®    c C   +     d D

Let, a molecules of A react with b molecules of B, and there be n molecules per unit volume of A.

Under similar conditions of temperature and pressure, B should also have n molecules per unit volume.

Then,

Volume of A involved in the reaction = 1 / n × a units

and,   Volume of B involved in the reaction = 1 / n × b units

Therefore, the ratio of volumes of the reacting gases A and B is, a/n : b/n or a : b

Since a and b are whole numbers (molecules cannot be in fractions), hence the volumes of the reacting gases bear a simple whole number ratio to each other.

This is the Gay-Lussac’s law of combining volumes.

2. Determining the atomicity of an elementary gas

To illustrate the determination of the atomicity of an elementary gas, let us consider the reaction between hydrogen and chlorine to form hydrogen chloride.

It has been observed experimentally that hydrogen and chlorine react in equal volumes. Then, one can write,

Atomicity of hydrogen and chlorine

Hydrogen    +      Chlorine     ®     Hydrogen chloride

   1 vol                 1 vol                        2 vol

n molecules         n molecules              2n molecules          (Avogadro’s law)

1/2 molecule       1/2 molecule               1 molecule

Since one molecule of hydrogen chloride contains one atom of hydrogen and one atom of chlorine, hence,

         1/2 molecule of hydrogen = 1 atom of hydrogen

or,         1 molecule of hydrogen = 2 atoms of hydrogen

Therefore, Atomicity of hydrogen = 2

Similarly,  1/2 molecule of chlorine = 1 atom of chlorine

or,             1 molecule of chlorine = 2 atoms of chlorine

Therefore,   Atomicity of chlorine = 2

Thus, the atomicity of both hydrogen and chlorine is two.

Atomicity of nitrogen and oxygen

Similarly, we can determine the atomicity of nitrogen and oxygen.

Nitrogen     +      Oxygen     ®     Nitric oxide

1 vol                    1 vol                    2 vol

n molecules         n molecules         2n molecules                       (Avogadro’s law)

1/2 molecule       1/2 molecule           1 molecule

 1 atom              1 atom                   1 molecule

Thus,    1/2 molecule of nitrogen = 1 atom of nitrogen

              1 molecule of nitrogen = 2 atoms of nitrogen

Therefore, Atomicity of nitrogen = 2

Also,       1/2 molecule of oxygen = 1 atom of oxygen

                1 molecule of oxygen = 2 atoms of oxygen

Therefore,    Atomicity of oxygen = 2

3. Relationship between molecular mass & vapour density

The vapour density (V.D.) of a gas is defined as the ratio between the mass of a certain volume of gas, to the mass of the same volume of hydrogen gas under similar condition of temperature and pressure. Thus,

According to the Avogadro’s law equal volumes of gases under similar conditions should contain equal number of molecules. So, if V mL of any gas contain n molecules, then

Since, one molecule of hydrogen contains 2 atoms, hence

As per definition

Hence,

 Molecular mass of the substance = 2 × V.D. of the substance

In general, therefore,

Molecular mass = 2 × Vapour density

4. Determining the molecular formula of a gaseous compound

The Avogadro’s law can also be used for determining the molecular formula of a gaseous compound, if the volumes of the reactants and products are known. This is illustrated by taking example of the reaction between nitrogen and hydrogen gases to form ammonia. Experimentally, it is found that one volume of nitrogen reacts with three volumes of hydrogen to form two volumes of ammonia. Then, one can write,

 Nitrogen    +      Hydrogen    ®     Ammonia

   1 vol                 3 vol                    2 vol

 n molecules        3n molecules       2n molecules                    (Avogadro’s law)

 1/2 molecule      1/2 molecule       1 molecule

or 1 atom           3 atom               1 molecule (atomicity of nitrogen and hydrogen = 2)

Thus, one molecule of ammonia contains one atom of nitrogen and three atoms of hydrogen.

Therefore, the molecular formula of ammonia is NH₃.

Atomicity

The number of atoms of all the elements present in a molecule of any substance is called its atomicity.

Monatomic molecules.

The molecules which contain only one atom in their molecule are called monatomic molecules.

Examples. Noble gases viz, helium, neon, argon, etc. are monatomic gases. Therefore, the atomicity of helium (He), neon (Ne) and argon (Ar) is one.

Diatomic molecules.

The molecules which contain two atoms in their molecule are called diatomic molecules.

Examples. The molecules of gases like hydrogen, oxygen, nitrogen, chlorine, etc. contain two atoms. Therefore, molecules of hydrogen, oxygen, nitrogen, chlorine are diatomic molecules.

Thus, the atomicity of hydrogen (H₂), oxygen (O₂), nitrogen (N₂) and chlorine

(Cl₂) is two.

Triatomic molecules.

The molecules which contain three atoms in their molecule are called triatomic molecules. Examples. The molecules of water and carbon dioxide are triatomic molecules. Therefore, the atomicity of water (H₂O) and carbon dioxide (CO₂) is three.

The atomicity of

• Ozone is 3

• Phosphorus (yellow) is 4

• Sulphur is 8