# Determination of Molecular Mass

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### Method – I (Molecular Mass by Molar Volume Method)

#### Principle:

• In this method we find a known volume of a gas at S.T.P. and using the concept that one mole of every gas occupies 22.4 dm3 by volume, we calculate the molar mass of the gas.

#### Procedure:

• Find the volume of the gas at STP, from given data.
• Find molecular mass, using the formula #### Example – 1:

• 10 dm3 of gas at 14 °C and 729 mm pressure has a mass of 17.925 g. Calculate the relative molecular mass of the gas.
• Solution:
• Given: V = 10 dm3, t = 14 °C, T = 14 + 273 = 287 K, P = 729 mm of Hg , W = 17.925 g Thus the mass of 9.125  dm3 of a gas is 17.925 g. Hence mass of 22.4 dm3 of a gas Ans: The relative molecular mass of the gas is 44 g

#### Example – 2:

• 5 litres of a gas at NTP has a mass of 14.4 g. Find the relative molecular mass of the gas.
• Solution:
• Given: Vo = 5 litres = 5 dm3, W = 14.4 g

The mass of 5  dmof a gas at NTP is 14.4 g. Hence mass of 22.4 dm3 of a gas Ans: The molecular mass of the gas is 64.51 g.

#### Example – 3:

• 240 ml of a dry gas measured at 300 K and 750 mm of mercury has a mass of 0.42336 g. calculate the relative molecular mass of the gas.
• Solution:
• Given: V = 240 ml = 0.240 dm3, T = 300 K, P = 750 mm of Hg , W = 0.42336 g The mass of 0.2155  dm3 of a gas is 0.42336 g. Hence mass of 22.4 dm3 of a gas Ans: The molecular mass of the gas is 44 g.

#### Example – 4:

• 1.25 g of a pure carbonate on ignition leave a residue of 0.70 g and evolved is 312 mm at 27 °C and 755 mm of mercury. Calculate the molecular mass of the gas.
• Solution:
• Given: Mass of carbonate = 1.25 g, Mass of residue = 0.70 g, Mass of the gas = 1.25 – 0.70 = 0.55 g, V = 312 ml = 0.312 dm3    T =  27 °C = 27 + 273 = 300 K, P = 755 mm of Hg , W = 0.55 g The mass of 0.2821  dm3 of a gas is 0.42336 g. Hence mass of 22.4 dm3 of a gas Ans: The molecular mass of the gas is 43.67 g.

### Method – II (Molecular Mass by Regnault’s Method):

#### Principle:

• In this method vapour densities of gases are determined by direct weighing. The vapour density of a gas is the ratio of the mass of a certain volume of a gas to the mass of the same volume of hydrogen at the same temperature & pressure.

#### Procedure:

• In this method, two hollow glass globes of the same capacity, same mass and same size are taken. They are evacuated and suspended by two sides of a physical balance.
• Now one of the glass globes is filled with a gas whose vapour density is to be found. then the mass of the globe is measured. the difference in the filled globe and empty globe gives the mass of the gas in the globe,
• Now the globe is evacuated again and filled with hydrogen gas at same temperature and pressure. The mass of globe filled with hydrogen is measured again. The difference in the empty globe and hydrogen filled globe gives the mass of hydrogen.
• Vapour density is found by using the formula. Now, Molecular Mass = 2 x Vapour density

#### Example – 5:

• 1 dm3 of hydrogen at S.T.P. has a mass of 0.09 g. If 2 dm3 of a gas at S.T.P. has a mass of 2.880 g. Calculate the vapour density and molecular mass of the gas.
• Solution:

1 dm3 of hydrogen has a mass of 0.09 g.

2 dm3 of gas has a mass of 2.880 g.

Hence, 1 dm3 of gas has a mass of 1.440 g. Now, Molecular Mass = 2 x Vapour density = = 2 x 16 = 32

Ans: The molecular mass of the gas is 32.

#### Example – 6:

• The capacity of a glass bulb is 30.5 ml, 0.146 g of a gas is filled in a bulb at 22 °C and 755 mm of Hg pressure.  1 dm3 of hydrogen at S.T.P. have mass of 0.09 g. Calculate the vapour density and molecular mass of the gas.
• Solution:
• Given: V = 30.5 ml, t = 22°C, T = 22 + 273 = 295 K, P = 755 mm of Hg , W = 0.146 g, Po = 760 mm of Hg, To = 273 K Po  = 0.02804 litre = 0.02804 dm3 Now, Molecular Mass = 2 x Vapour density = 2 x 57.85 = 115.7

Ans: The molecular mass of the gas is 115.7.

#### Example – 7:

• The capacity of a glass bulb is 127 ml, 0.4524 g of a gas is filled in a bulb at 409 K and 758 mm of Hg pressure. 1 ml of hydrogen at S.T.P. have the mass of 0.00009 g. Calculate the vapour density and molecular mass of the gas.
• Solutions:
• Given: V = 127 ml, T = 409 K, P = 758 mm of Hg , W = 0.4524 g, Po = 760 mm of Hg, To = 273 K  Now, Molecular Mass = 2 x Vapour density = 2 x 59.45 = 118.90

Ans: The molecular mass of the gas is 118.90

#### Example – 8:

• In Regnault’s method, 27.32 ml of a gas were found to have a mass of 0.1008 g. The experiment was performed at 16.5 °C and 694 mm of mercury. Calculate the molecular mass of the gas.
• Solution:
• Given: V = 27.32 ml, t = 16.5 °C, T = 16.5 + 273 = 289.5 K, P = 694 mm of Hg , W = 0.1008 g, Po = 760 mm of Hg, To = 273 K  Now, Molecular Mass = 2 x Vapour density = 2 x 47.6 = 95.2

Ans: The molecular mass of the gas is 95.2

#### Example – 9:

• When a glass bulb with a stop cock is evacuated, weighed and then filled with oxygen , the weight increases by 0.25 g. When the same bulb is evacuated and filled with another unknown gas under the same conditions of temperature and pressure, the mass increases by 0.5525 g. If molecular mass of oxygen is 32, calculate the molecular mass of the unknown gas. Calculate the volume of the glass bulb assuming that it does not change appreciably with the change in pressure and temperature.
• Solution:

The molecular mass of oxygen = 32

Vapour density of oxygen = 32 / 2 = 16

Mass of oxygen = 0.25 g Now, the vapour density of the gas Now, Molecular Mass = 2 x Vapour density = 2 x 35.36 = 70.72

Hence, the molecular mass of the gas is 70.72 Ans: The volume of the bulb is 0.1736 dm3.

### Method – III (Molecular Mass by Graham’s Law Diffusion Method):

#### Principle:

• Graham’s law of diffusion states that “The rate of diffusion of different gases under similar conditions of temperature and pressure are inversely proportional to square root of their densities”.
• But the densities are directly proportional to their molar masses. Hence the law can be restated as “The rate of diffusion of different gases under similar conditions of temperature and pressure are inversely proportional to square root of their molar masses”. Mathematically, #### Example – 10:

• 300 c.c. of oxygen gas takes 50 seconds to effuse through an aperture while the same volume of unknown gas takes 75 seconds. If the relative molecular mass of oxygen is 32, find the relative molecular mass of the unknown gas.
• Solution:
• Given: Rate of diffusion of oxygen ro = 300/50 = 6 c.c./s, Rate of diffusion of gas g = 300/75 = 4 c.c./s, The molecular mass of oxygen MO = 32,
• To Find: Molecular mass of gas Mg  =?

By Graham’s law of diffusion Ans: The molecular mass of the gas is 72.

#### Example – 11:

• 30 ml of ozone diffuses at the same time as 25 ml of chlorine. If the density of chlorine is 35.4. Find the density and relative molecular mass of ozone.
• Solution:
• Given: Rate of diffusion of ozone rO = 30 / t  ml per second, Rate of diffusion of chlorine rCl = 25 / t ml per second, Density of chlorine ρCl = 35.4
• To Find: Density of ozone ρO  =?, The Molecular mass of ozone MO  =?

By Graham’s law of diffusion Ans: The molecular mass of the gas is 24.58

#### Example – 12:

• In 75 seconds 400 ml of oxygen diffuses through a glass tube. If 25 seconds are taken by 100 ml of another gas ‘X’ to diffuse through the same tube under similar conditions. calculate the relative molecular mass of the gas ‘X’.
• Solution:
• Given: Rate of diffusion of oxygen=  rO = 400 / 75  = 16/3 ml per second, Rate of diffusion of gas ‘X’ = rX = 100 /25 = 4  ml per second

By Graham’s law of diffusion Ans: The molecular mass of the gas is 56.9

### Method – IV (Molecular Mass by Ideal Gas Equation / Law Method):

#### Principle:

The ideal gas equation is PV = nRT

Where P = Pressure of the gas

V = Volume of the gas

n = Number of moles of the gas

R = Universal gas constant

T = absolute temperature of the gas

Using above relation and knowing remaining quantities molecular mass can be calculated.

#### Example – 13:

• What is the relative molecular mass of the gas if 0.866 g sample is 60.0 ml. Bulb has a pressure of 400 mm at 20 °C. R = 0.0821 lit-atm
• Solution:
• Given: w = 0.866 g, V = 60.0 ml  = 60 x 10-3 dm3, P = 400 mm = 400/760  = 0.5263 atm

By ideal gas equation. PV = nRT Ans: The relative molecular mass is 66.00

#### Example – 14:

• 0.1348 g of a gas was found to occupy a volume of 25.80 ml at 0 OC and 760 mm of Hg pressure. calculate the relative molecular mass of the gas.
• Solution:
• Given: w = 0. 1348 g, V = 25.80  ml  = 25.80 x 10-3 dm3, P = 760 mm = 760/760  = 1 atm, T = 0 + 273 = 273 K

By ideal gas equation PV = nRT Ans: The relative molecular mass is 117.1

#### Example – 15:

• 3.895 dm3 of a gas at 293 K and 780 mm pressure were found to have a mass of 2.83 g. Calculate the relative molecular mass of the gas.
• Solution:
• Given: w = 2.83 g, V = 3.895 dm3, P = 780 mm = 780/760   atm, T = 293 K

By ideal gas equation, PV = nRT Ans: The relative molecular mass is 17.03

Example – 16:

• The molecular mass of a gaseous substance is 80. What will be the volume of 1 g of a gas at 0 °C and 720 mm mercury pressure? State whether this gas would diffuse through a porous pot slower or faster than chlorine.
• Solution:
• Given: M = 80, w = 1 g, P = 720 mm = 720/760  atm, T = 0 +273 = 273 k

By ideal gas equation, PV = nRT The volume of the gas is 0.2957 dm3

• By Graham’s diffusion law “The rate of diffusion of different gases under similar conditions of temperature and pressure are inversely proportional to square root of their densities”. In this case the molecular mass of the gas (80) is greater than the molecular mass of chlorine (71). Hence the gas will diffuse slowly.

#### Example – 17:

• What is the relative molecular mass of the gas if 0.866 g sample is 60.0 ml. Bulb has a pressure of 400 mm at 20 °C. R = 0.0821 lit-atm
• Solution:
• Given: w = 0.866 g, V = 60.0 ml  = 60 x 10-3 litres, P = 400 mm = 400/760  = 0.5263 atm

By ideal gas equation, PV = nRT Ans: The relative molecular mass is 66.00

### Method – IV (Molecular Mass by Victor Meyer’s Method):

#### Principle:

• A known mass of volatile substance in Victor Meyer’s tube. As a result, the liquid changes into vapours, the vapours, in turn, displaces an equal volume of air which is collected over water. The volume of air collected is measured at the pressure and temperature of the laboratory. This volume is converted to S.T.P. volume. and using gram molar volume concept molecular mass is calculated.

#### Procedure :

• A round bottom flask filled with a liquid whose B.P. is 10 °C more than the volatile liquid acts as an outer glass jacket.  A Victor Mayor’s tube with an outer tube acts as inner glass jacket. This outer tube is dipped in a trough fitted with water.  Bottom of the Victor Mayor’s tube consists of Hg or asbestos pieces for cushioning. • Outer glass jacket is heated due to which air inside expands and bubbles through the water in the trough. A small. glass tube is known as Hoffman’s bottle. with a stopper is cleaned, washed and dried and weighed.  Volatile Liquid is taken in Hoffman’s bottle and weighed.  A Eudiometer tube filled with water is placed over the tube connected to Victor Mayor apparatus.
• Now Hoffman’s bottle is dropped in the Victor Mayor tube.  Due to heat liquid in the bottle vapourises and blows off the stopper and displace air which corresponds to its own volume which is collected in a eudiometer tube by the downward displacement of water.
• The water temperature and pressure are also noted.  Eudiometer tube is taken into another trough filled with water for equalisation of pressure.  The volume of air displaced is noted, which in fact correspond to the volume of the vapours.

#### Observations:

• The weight of the Hoffman’s Bottle  = W1 gm.
• The weight of the Hoffman’s bottle  +  Liquid  = W2 gm.
• Weight of the volatile liquid     =  W2 – W1 =  W gm.
• Pressure     =  (p – f) mm of Hg.
• Temperature          = t °C = T K
• Volume  of vapours  = V dm3.

#### Calculations:

Now the volume of vapours at S.T.P. from the above data is calculated using following formula. Where the volume of hydrogen at S.T.P. = V0 dm3

Pressure at S.T.P. = P0 mm of Hg = 760 m

Absolute temperature at S.T.P.= T K = 273 K

Now, the molecular weight can be calculated by following formula

#### Merits of Victor Mayer’s Method:

• The method is very simple to carry out. Weight.
• The sample required for the experiment is very small.

#### Demerits of Victor Mayer’s Method:

• This method is applicable to volatile liquid only.
• The method can not be used for the substance which undergoes thermal decomposition.
• The liquid in outer jacket is generally water, hence this method is suitable only for the liquids which have boiling point less than 100oC.
• Due to manual handling, there is the possibility of personal error.
• It is not applicable to volatile substances which are water soluble.

#### Example – 18:

• In the determination of molecular mass by Victor Meyer’s method 0.60 g of volatile substance expelled 123 ml of air measured over water at 20 °C and 757.4 mm pressure. Find the molecular mass of the substance if the aqueous tension at 20 °C is 17.4 mm.
• Solution:
• Given: V = 123 ml = 0.123 dm³, t = 20 °C, T = 20 + 273 = 293 K, P = 757.4 mm of Hg ,  f = 17.4 mm of Hg, P – f = 757.4 – 17.4 = 740 mm of Hg, Po = 760 mm of Hg, To = 273 K Ans: The molecular mass of the substance is 120.43

### Method – V  (Duma’s Method):

#### Procedure:

• In this method a known volume of vapour is heated to some higher temperature and its mass is noted. The calculations are same as Victor Meter’s method.

### Method – VII (Hoffman’s Method):

#### Procedure:

• Some substances are decomposed when they are evaporated at their boiling points.The substance is evaporated at a sufficiently low temperature at reduced pressure. Hoffman’s method is used for calculating molecular mass of such substances.
• A known mass of a substance is vapourised above a mercury column in a barometric tube and the volume of vapour formed is noted, Then the volume is calculated at STP and calculation is done the same way as in Victor Meyer’s method.
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