In the previous section, we saw estimation of carbon, hydrogen and nitrogen in organic compounds. In this section, we will see estimation of halogens, sulfur, phosphorus and oxygen.
Estimation of halogens (Cl, Br, I) is done using Carius method. It can be written in 4 steps:
1. A known mass (m) of an organic compound is heated with fuming nitric acid in the presence of silver nitrate.
• The heating is done in a hard glass tube known as Carius tube.
• The heating is done using a furnace.
2. Carbon in the organic compound gets oxidized to CO2 and hydrogen gets oxidized to H2O
• Both will be in gaseous form.
3. The halogen in the organic compound will be converted into the corresponding silver halide (AgX). It will be in solid form.
• It is washed and dried. Then it is weighed and mass mAgX is noted.
4. Using this mAgX, the mass of X and it’s percentage can be calculated. The following solved example shows the procedure.
Solved example 12.24
In Carius method of estimation of halogen, 0.15 grams of an organic compound gave 0.12 grams of AgBr. Find out the percentage of bromine in the compound.
Solution:
1. Molar mass of AgBr is (108 + 80) = 188 grams
• So 188 grams of AgBr will contain 80 grams of Br
⇒ 1 gram of AgBr will contain $\frac{80}{188}$ grams of Br
⇒ 0.12 grams of AgBr will contain $0.12 \times \frac{80}{188}$ grams of Br
2. Original mass of the organic compound was 0.15 grams.
• So the percentage of Br in this compound = $\frac{0.12 \times \frac{80}{188}}{0.15}\times 100$ = 34.04%
Estimation of Sulfur is done using a Carius tube. It can be written in 4 steps:
1. A known mass (m) of an organic compound is heated with sodium peroxide or fuming nitric acid.
• The heating is done in a Carius tube.
• The heating is done using a furnace.
2. Sulfur in the organic compound gets oxidized to sulfuric acid
3. Excess 'barium chloride solution in water' is added.
• The sulfuric acid gets precipitated as barium sulfate.
• The precipitate is washed and dried. Then it is weighed and the mass mB is noted.
4. Using this mB, the mass of sulfur and it’s percentage can be calculated. The following solved example shows the procedure.
Solved example 12.25
In sulfur estimation, 0.157 grams of an organic compound gave 0.4813 grams of barium sulfate. Find
out the percentage of sulfur in the compound.
Solution:
1. Molar mass of BaSO4 is (137 + 32 + 64) = 233 grams
• So 233 grams of BaSO4 will contain 32 grams of S
⇒ 1 gram of BaSO4 will contain $\frac{32}{233}$ grams of S
⇒ 0.4813 grams of BaSO4 will contain $0.4813 \times \frac{32}{233}$ grams of S
2. Original mass of the organic compound was 0.157 grams.
• So the percentage of S in this compound = $\frac{0.4813 \times \frac{32}{233}}{0.157}\times 100$ = 42.10%
Estimation of phosphorus can be done by two methods.
• Method 1 can be written in 4 steps:
1. A known mass (m) of an organic compound is heated with fuming nitric acid.
2. Phosphorus in the organic compound gets oxidized to phosphoric acid
3. Ammonia and ammonium molybdate are added.
• The phosphoric acid gets precipitated as ammonium phosphomolybdate ((NH4)3PO4.12MoO3).
• The precipitate is weighed and the mass mA is noted.
4. Using this mA, the mass of phosphorus and it’s percentage can be calculated.
• It can be written in 2 steps:
(i) Molar mass of (NH4)3PO4.12MoO3 = 1877 grams
• So 1877 grams of (NH4)3PO4.12MoO3 will contain 31 grams of S
(Molar mass of P atom is 31 grams)
⇒ 1 gram of (NH4)3PO4.12MoO3 will contain $\frac{31}{1877}$ grams of P
⇒ mA grams of (NH4)3PO4.12MoO3 will contain $m_A \times \frac{31}{1877}$ grams of P
(ii) Original mass of the organic compound was m grams.
• So the percentage of P in this compound = $\frac{m_A \times \frac{31}{1877}}{m}\times 100=\frac{31 \times m_A \times 100}{1877 \times m}$
• Method 2 can be written in 4 steps. First two steps are the same.
1. A known mass (m) of an organic compound is heated with fuming nitric acid.
2. Phosphorus in the organic compound gets oxidized to phosphoric acid
3. Magnesia mixture is added.
• The phosphoric acid gets precipitated as MgNH4PO4.
4. This precipitate is ignited. We get: Mg2P2O7
• This is weighed and the mass mA is noted.
5. Using this mA, the mass of phosphorus and it’s percentage can be calculated.
• It can be written in 2 steps:
(i) Molar mass of Mg2P2O7 = 222 grams
• So 222 grams of Mg2P2O7 will contain 62 grams of S
(Molar mass of P atom is 31 grams)
⇒ 1 gram of Mg2P2O7 will contain $\frac{62}{222}$ grams of P
⇒ mA grams of Mg2P2O7 will contain $m_A \times \frac{62}{222}$ grams of P
(ii) Original mass of the organic compound was m grams.
• So
the percentage of P in this compound = $\frac{m_A \times
\frac{62}{222}}{m}\times 100=\frac{62 \times m_A \times 100}{222
\times m}$
Now we will see the estimation of oxygen.
• First we will see the indirect method. It can be written in 4 steps:
1. Consider the following percentages:
♦ Percentage of C = PC %
♦ Percentage of H = PH %
♦ Percentage of N = PN %
♦ Percentage of X = PX %
♦ Percentage of S = PS %
♦ Percentage of P = PP %
♦ Percentage of O = PO %
2. Suppose that, the given organic compound contains C, H, N and O.
• Then we can write: (PC + PH + PN + PO) = 100
• From this we get: PO = [100 - (PC + PH + PN)]
3. Suppose that, the given organic compound contains C, H, S and O.
• Then we can write: (PC + PH + PS + PO) = 100
• From this we get: PO = [100 - (PC + PH + PS)]
4. Usually we follow this method to find the percentage of O.
• That is., we add the percentages of all other elements in the given compound.
• Then we subtract that sum from 100.
Now we will see the direct method. It can be written in 11 steps:
1. A known mass (m) of an organic compound is heated in a stream of nitrogen gas.
2. The organic compound gets decomposed and we get a gaseous mixture. Oxygen is contained in this mixture.
3. This gaseous mixture is passed over red hot coke.
• All the oxygen in the mixture will be converted into CO
• The equation is: 2C + O2 ⟶ 2CO
4. This mixture is then passed through warm iodine pentoxide (I2O5). All the CO will get oxidized to CO2.
• The equation is: I2O5 + 5CO ⟶ I2 + 5CO2
5. We see that:
♦ In the equation in (3), CO is on the right side.
♦ In the equation in (4), CO is on the left side.
• Let us make the coefficients of CO equal.
6. We can multiply (3) by ‘5’, which is the coefficient of CO in (4)
We get: 10C + 5O2 ⟶ 10CO
7. We can multiply (4) by ‘2’, which is the coefficient of CO in (3)
We get: 2I2O5 + 10CO ⟶ 2I2 + 10CO2
8. Adding (6) and (7), we get:
10C + 5O2 + 2I2O5 + 10CO ⟶ 10CO + 2I2 + 10CO2
• Canceling 10 CO on either sides, we get:
10C + 5O2 + 2I2O5 ⟶ 2I2 + 10CO2
9. In this equation, the I2O5 was externally added. It remains as such.
• So we can write:
Five moles of O2 give ten moles of CO2.
• This is same as:
One mole O2 gives moles of CO2.
10. Thus, by using the mass of CO2 produced, we can find the mass of O2 in the original organic compound.
• It can be explained in 5 steps:
(i) Let the mass of CO2 produced be mC grams
(ii) Molar mass of CO2 = 44 grams.
• So number of moles of CO2 produced = $\frac{m_C}{44}$
(iii) Availability of one mole of CO2 means that, 0.5 moles of O2 is present.
• So the availability of $\frac{m_C}{44}$ moles of CO2 means that, $0.5 \times \frac{m_C}{44}$ moles of O2 is present.
(iv) One mole of O2 has a mass of 32 grams.
• So $0.5 \times \frac{m_C}{44}$ moles will have a mass of $0.5 \times \frac{m_C}{44} \times 32=\frac{32 \times m_C}{88}$ grams
(v) If m is the mass of the original organic compound, the percentage of O2 will be given by: $\frac{\frac{32 \times m_C}{88}}{m} \times 100=\frac{32 \times m_C \times 100}{88 \times 100}$
11. Note that, there is I2 in the final equation in (8).
• We see that, five moles of O2 give two moles of I2.
• So by using the mass of I2 also, we can find the mass of O2 in the original organic compound.
• The link below gives the folder containing additional solved examples on this chapter.
• Parts 3 and 4 are related to purification, qualitative analysis and quantitative analysis
We have completed the discussions in this chapter. In the next chapter we will see hydrocarbons.
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