Determining Molecular Masses Using Victor Meyer's Method

What is Victor Meyer Method?

Victor Meyer Method is used for volatile organic compounds. A known mass of the compound is vaporised in an instrument called Victor Meyer tube. The vapours obtained, displace an equal volume of air into a graduated tube. The volume of vapours is measured and reduced to STP.

Let the volume of vapours at STP be V mL

22400 mL of vapours are obtained from 1 mole of the compound.

V mL of vapours are obtained from (V/ 22400) mL of the compound

Mole= W/Mw

Therefore,

W/Mw = V/22400

From this, the molecular weight of the compound can be determined.

Also Read

Determination of Empirical and Molecular Formula

If the percentages compositions of the elements are given, then use the following method. For example, if carbon’s composition is 48%, hydrogen 8%, nitrogen 28% and oxygen 16%. Make a table with the given data as follows,

Element Percentage Atomic mass Relative number of atoms Simplest atomic ratio Simplest whole number atomic ratio
Carbon 48.0 12
Hydrogen 8.0 1
Nitrogen 28.0 14
Oxygen 16.0 16

To calculate the relative number of atoms, divide the percentage by the atomic mass of the element

Element Percentage Atomic mass Relative number of atoms Simplest atomic ratio Simplest whole number atomic ratio
Carbon 48.0 12 48/12=4
Hydrogen 8.0 1 8/1=8
Nitrogen 28.0 14 28/14=2
Oxygen 16.0 16 16/16=1

To find the simplest atomic ratio, divide the relative number of atoms, and with the lowest number that is obtained, in this particular case, among 4, 8, 2 and 1, it is 1.

Element Percentage Atomic mass Relative number of atoms Simplest atomic ratio Simplest whole number atomic ratio
Carbon 48.0 12 48/12=4 4/1=4
Hydrogen 8.0 1 8/1=8 8/1=8
Nitrogen 28.0 14 28/14=2 2/1=2
Oxygen 16.0 16 16/16=1 1/1=1

To find the simplest whole number, multiply each simplest atomic ratio with an integer to make it a whole number, in this particular case, we can skip that since all the numbers are whole numbers.

Element Percentage Atomic mass The relative number of atoms Simplest atomic ratio Simplest whole number atomic ratio
Carbon 48.0 12 48/12=4 4/1=4 4
Hydrogen 8.0 1 8/1=8 8/1=8 8
Nitrogen 28.0 14 28/14=2 2/1=2 2
Oxygen 16.0 16 16/16=1 1/1=1 1

Therefore, the empirical formula of the compound is, C4H8N2O

To find out the molecular formula, we need the molecular mass of the compound.

Say, for the above question, the molecular mass of the substance had been given as 200 amu, we need to find out the empirical mass.

Empirical mass= 4×12+8×1+2×14+1×16

=100 amu

The logic here is that we have to find out by how many times, the molecular mass is more than empirical mass, and so we divide them

n= (Molecular mass)/(Empirical mass)

n= 200/100

n=2

Now, we found that molecular mass is twice, so the molecular formula should be = empirical formula x 2

Molecular formula= C4H8N2O x 2 = C8H16N4O2

 


Practise This Question

Which of the following will not under go disproportionation reaction in the presence of dilute (or) Conc NaOH