Immiscible Liquids

Consider the immiscible pair of components A and B in which A denser than B

The immiscible pair is kept in a separating funnel in order to see them clearly.

The distribution of solute in a pair of immiscible liquid is governed by the Partition law or Distribution law.

Immiscible liquids are separated by the process of steam distillation.

Steam Distillation

This is the process of separating immiscible liquids of different boiling points by passing super heated steam through.

Condition Necessary for Steam Distillation

In order for steam distillation to be feasible, the following conditions are;

1. The two liquids should have different B.P.
2. The two liquids should be immiscible.
3. There should be no volume change.
4. The total vapor pressure of the liquids should be equal to the sum of the components vapor pressure.

Steam distillation can be used to determine the molar mass of unknown liquid.

The conditions necessary for steam distillation are;

1. The liquids must form immiscible solution.
2. The total vapor pressure is equal to the sum of the vapor pressure of the components.
3. There should be no change in volume.
4. The liquids should have different boiling point.

Let the two liquids A and B form an immiscible pair And

nA = a

nB = b

Since the two liquids are immiscible then, their distillation process can be explained in terms of their proportions or compositions (mole fraction).

From the number of moles of components the total number of moles can be obtained

i.e nT = nA + nB

nT = a + b

If nT is known then mole fraction or composition can be calculated i.e

$X_A = \frac{n_A}{n_T}$

$X = \frac{a}{a+b}$

Also

$X_B = \frac{n_B}{n_T}$

$= \frac{a}{a+b}$

For immiscible liquids, the ratio of their compositions is equal to the ratio of their vapor pressure.

$\frac{X_A}{X_B} = \frac{P_A}{P_B}$

$\frac{a}{a + b} \times \frac{b + a}{b} = \frac{P_A}{P_B}$

$\frac{a}{b} = \frac{P_A}{P_B}$

But

$a = \frac{\text{mass of } A}{\text{Molar mass of } A}$

$a = \frac{m_A}{M_A}$

and

$b = \frac{\text{mass of } B}{\text{Molar mass of } B}$

$b = \frac{m_B}{M_B}$

$\frac{m_A}{M_A} \div \frac{m_B}{M_B}$

$\frac{m_A}{M_A} \times \frac{M_B}{m_B} = \frac{P_A}{P_B}$

$\boxed{\frac{M_B m_A}{M_A m_B} = \frac{P_A}{P_B}}$

Where:

MB is the molar mass of B

MA is the molar mass of A

mA is the mass of A

mB is the mass of B

PA is the vapor pressure of A

PB is the vapor pressure of B

A solution of 6gm of substance X in 50cm3 of aqueous solution is in equilibrium at room temperature with an ether solution of X containing 108gm of X in 100cm3 . Calculate what weight of X could be extracted by shaking 100cm3 of an aqueous liquids containing 10gm of X with;

i) 100cm3 ether ii) 50cm3 of ether twice at room temperature.

Solution

Concentration of X in H2O = 6 g/50 cm3 Concentration of X in ethereal = 108/100

$Kd = \frac{[x]_{in~extracting}}{[x]_{in~extracted}}$

$= \frac{(108/100)}{(6/50)}$

K = 9

Now

(i) $A_{ex} = w_o - \left(\frac{V_a}{V_b + KV_b}\right)^n \times w_o$

$= \left(\frac{100}{1000+9\times100}\right)^1 \times 10$

$= 9g$

$\therefore$ The amount extracted is $9g$

(ii) $Aex = 10 + \left(\frac{100}{100+9\times50}\right)^2 \times 10$

$= 9.7g$

$\therefore$ The amount extracted is $9.7g$

A organic liquid distills in steam, the partial pressure of the two liquids at the boiling point are 5.3 k pa for organic liquid and 96 k pa for water. The distillate contains the liquids in the ratio of 0.48g organic liquid to 1g of water. Calculate the molar mass of organic liquid.

Solution

Po = 5.3kpa

mo = 0.48

Mo = ?

Pw = 96 kpa

Mw = 18

mw = 1g

$\frac{M_o m_w}{M_w m_o} = \frac{P_w}{P_o}$

$M_o = \frac{M_w m_o P_w}{P_o m_w}$

$= \frac{18 \times 0.48 \times 96}{5.3 \times 1}$

$= 156.498$

$\therefore \text{The molar mass of organic compound is } 156 \text{ g/mol}$

Note: Unit conversion

1. i) 1 N/m2 = 1 Pa ii) 1 atm = 760 mmHg iii) 1 atm = 1.01 × 10^5 Pa iv) 1 atm = 1.01 × 10^5 N/m2

1. a) Differentiate between thermal distillation and steam distillation. b) Bronobenzene (C6H5Br) distills in steam at 95°C the vapor of Bromobenzene and water are 1.39 × 10^4 N/m2 and 8.5 × 10^4 N/m2. Calculate the percentage by mass of bromobenzene.

(c =12 Br =80 O = 16 H = 1) Note mass of H2O =24g

Solution

a) Thermal distillation is the process of separating immiscible mixture by the use of thermal energy (heat) while Steam distillation is the process of separating immiscible liquids by having different boiling points by passing super heated steam through it.

Data

PB = 1.39 × 10^4 N/m2

Pw = 8.5 × 10^4 N/m2

MB = 157

Mw = 18

mw = 24g

From

$\frac{M_B m_w}{M_w m_B} = \frac{P_w}{P_B}$

$m_B = \frac{M_B m_w P_B}{M_w P_w}$

$= \frac{157 \times 24 \times 1.39 \times 10^4}{18 \times 8.5 \times 10^4}$

$= 34.23$

Total mass = 34.23 + 24

$= 58.23$

% by mass of $B$ = $\frac{34.23}{58.23}$ × 100

$= 58.78\%$

$\therefore$ The percentage by mass of bromobenzene is 58.78%

An organic liquid Q which do not mix with water distills in steam at 96°C under the pressure of 1.01. The pressure of water at 96°C is 8.77 N/m2. The distillate contains 51% by mass Q. Calculate the molar mass of Q.

Solution

Atmospheric pressure Patm = 1.01 N/m2

PQ = Patm – Pw

Pw = 8.77

Now

PQ = 1.01 – 8.77 × 10^4

= 1033

MQ = ?

Mw = 18

Mass of Q mB = 51g

Mass of W mw = 49

From

$\frac{M_Q m_w}{M_w m_Q} = \frac{P_w}{P_Q}$

$M_Q = \frac{M_w m_Q P_w}{m_w P_Q}$

$= \frac{18 \times 51 \times 8.77 \times 10^4}{49 \times 1.33 \times 10^4}$

$= 123.54 \, \text{g/mol}$

$\therefore \text{The molar mass of } Q \text{ is } 123.54 \, \text{g/mol}$

Calculate the molar mass of the compound B whose mixture with water distills at 95°C. At this temperature the pressure of compound B and water are 119mmHg and 64mmHg. The ratio of B to water is 1.61 : 1

Solution

PB = 119mmHg

PW = 64 mmHg

MB = 1.61g

mw = 1g

MB = ?

Mw = 18g/mol

From

$\frac{M_B m_W}{M_W m_B} = \frac{P_W}{P_B}$

$M_B = \frac{M_W m_B P_W}{P_B m_W}$

$M_B = \frac{18 \times 1.61 \times 64}{119}$

$M_B = 15.59 \, \text{g/mol}$

$\therefore \text{The molar mass of } B \text{ is } 15.59 \, \text{g/mol}$

At a pressure of 760 mmHg, a mixture of nitrobenzene (C6H5NO2) and water boils at 99°C. The vapor pressure at this temperature is 733 mm. Find the proportion of water and nitrobenzene in the distillate obtained by steam distillation of pure C6H5NO2.

Solution

Atmospheric pressure patm = 760

Pressure of H2O Pw = 733

Pressure of nitrobenzene pn = Patm – Pw

= 760 – 733

Pn = 27

Mw = 18

Mn = (12 × 6) + 5 + 14 + (16 × 2)

Mn = 123

From

$\frac{M_w m_n}{M_n m_w} = \frac{P_n}{P_w}$

Multiply by $\frac{M_n}{M_w}$ both sides

$\frac{M_n}{M_w} \times \frac{M_w m_n}{M_n m_w} = \frac{P_n}{P_w} \times \frac{M_n}{M_w}$

$\frac{m_n}{m_w} = \frac{P_n}{P_w} \times \frac{M_n}{M_w}$

$\frac{m_n}{m_w} = \frac{27}{733} \times \frac{123}{18}$

$\frac{m_n}{m_w} = \frac{3321}{13194}$

$\frac{m_n}{m_w} = 0.25$

$\frac{m_n}{m_w} = 1:4$

∴ The proportion of water to nitrobenzene is 1:4

A mixture of water and bromobenzene (C6H5Br) distills at 95°C and the distillate contain 1.6 times as much C6H5Br as water by mass. At 95°C the vapor pressure of water and C6H5Br are 640mmHg and 120mmHg respectively. Calculate the molecular weight of bromobenzene.

Solution

PW = 640mmHg

PB = 120mmHg

Mw = 18g/mol

MB = ?

Let X be the mass of water (Mw).

1.6X will be the mass of C6H5Br (MB).

From

$\frac{M_B m_w}{M_w m_B} = \frac{P_w}{P_B}$

$M_B = \frac{M_w m_B P_w}{P_B m_w}$

$= \frac{18 \times 1.6x \times 640}{120x}$

$= 153.6$

∴ The molecular weight of bromobenzene is $153.6 \text{g/mol}$