Chapter 12.22 - Quantitative Analysis

In the previous section, we saw qualitative analysis. In this section, we will see quantitative analysis of organic compounds.

• We have seen the applications of percentage composition, empirical formula and molecular formula in the first chapter. (Section 1.2).
• There we saw solved examples involving percentage composition. We were directly given the percentage composition in the questions.
• It was mentioned that, chemical tests are required to find the percentage composition of a substance. In this section, we will see some of those chemical tests.

Carbon and Hydrogen

• First we will see the method to find the amount of carbon present in a substance.
• Along with that, we will see the method for hydrogen also. Because both carbon and hydrogen are calculated using a single test. It can be explained in 12 steps:
1. A known mass (m) of the given organic compound is burnt in the presence of excess oxygen and copper(II) oxide (CuO).
• Note the apparatus shown in the text book. We see an inlet for supplying 'pure dry oxygen'.
• The oxygen must be pure so that only O₂ molecules are present in the supply. If other molecules are present, they will also enter into the reactions, causing errors.
• The oxygen must be dry also. If water content is present, those H₂O molecules will also enter into reactions causing errors.
• The CuO is placed as pallets in the apparatus. It is a strong oxidizing agent.
2. The carbon in the given organic compound will be oxidized to CO₂.
• If there are x number of C atoms in the organic compound, the same x number of O₂ molecules will be required to convert all those carbon atoms to CO₂ molecules.
• The equation is: x C + x O₂ ⟶ x CO₂
3. The hydrogen in the given organic compound will be oxidized to H₂O.
• If there are y number of H atoms in the organic compound, (y/4) number of O₂ molecules will be required to convert all those H atoms to H₂O molecules.
• The equation is: y H + y/4 O₂ ⟶ (y/2) H₂O
4. Now we know the importance of supplying excess oxygen. It will ensure that all C and H atoms will be oxidized. If enough oxygen is not available, some of the C and H atoms will remain as such.
5. All the C atoms have become CO₂. Also all the H atoms have become H₂O.
• The H₂O molecules will be in gaseous form because of the heat.
• Since we supply excess oxygen, some unreacted O₂ molecules will also be present.
• So what we have now, is a gaseous mixture of CO₂, H₂O and O₂.
6. This gaseous mixture is made to pass through a U-tube containing anhydrous calcium chloride (CaCl₂).
• All the H₂O molecules will be absorbed by the anhydrous CaCl₂.
7. So the mixture now contains CO₂ and O₂ molecules only. This mixture is made to pass through another U-tube containing concentrated solution of potassium hydroxide (KOH).
• All the CO₂ molecules will be absorbed by the KOH solution.
8. So the mixture now contains O₂ molecules only. Those molecules are expelled through the outlet.
9. Consider the U-tube mentioned in step 6. It’s mass is measured before attaching it to the apparatus. We will denote this mass as m_H,i
   ♦ ‘H’ indicates that, this U-tube is used to measure the mass of hydrogen.
   ♦ ‘i’ indicates that, it is the initial mass.
• This is the total mass of the u-tube and the anhydrous CaCl₂.
• When the experiment is complete, we detach it from the apparatus and measure the mass again. We will denote this mass as m_H,f
   ♦ ‘f’ indicates that, it is the final mass.
• This is the total mass of the u-tube, the anhydrous CaCl₂ and the newly produced H₂O molecules.
• Thus we get:
Mass (m_H2O) of the newly produced H₂O molecules = (m_H,f - m_H,i)
10. Consider the U-tube mentioned in step 7. It’s mass is also measured before attaching it to the apparatus. We will denote this mass as m_C,i
   ♦ ‘C’ indicates that, this U-tube is used to measure the mass of carbon.
   ♦ ‘i’ indicates that, it is the initial mass.
• This is the total mass of the u-tube and the concentrated KOH solution.
• When the experiment is complete, we detach it from the apparatus and measure the mass again. We will denote this mass as m_C,f
   ♦ ‘f’ indicates that, it is the final mass.
• This is the total mass of the u-tube, the concentrated KOH solution and the newly produced CO₂ molecules.
• Thus we get:
Mass (m_CO2) of the newly produced CO₂ molecules = (m_C,f - m_C,i)
11. Now we can begin the calculations. First we will see the calculations for H. It can be written in 4 steps:
(i) We know that 18 grams of H₂O will contain 2 grams  of H.
• So 1 gram of H₂O will contain $\frac{2}{18}$ grams of H.
(ii) In our present case, we have m_H2O grams of H₂O.
• So there will be $\rm{m_{H2O} \times \frac{2}{18}}$ grams of H.
(iii) Recall that, we took 'm' grams of the organic compound for the experiment.
So we can write:
m grams of the given organic compound contains ($\rm{m_{H2O} \times \frac{2}{18}}$) grams of H
(iv) Thus the percentage of H in the organic compound
= $\rm{\frac{m_{H2O} \times \frac{2}{18}}{m}\times 100}$
⇒ Percentage of H in the organic compound = $\rm{\frac{2 \times m_{H2O} \times 100}{18 \times m}}$
12. Next we will see the calculations for C. It can be written in 4 steps:
(i) We know that 44 grams of CO₂ will contain 12 grams  of C.
• So 1 gram of CO₂ will contain $\frac{12}{44}$ grams of H.
(ii) In our present case, we have m_CO2 grams of CO₂.
• So there will be $\rm{m_{CO2} \times \frac{12}{44}}$ grams of C.
(iii) Recall that, we took 'm' grams of the organic compound for the experiment.
So we can write:
m grams of the given organic compound contains ($\rm{m_{CO2} \times \frac{12}{44}}$) grams of C
(iv) Thus the percentage of C in the organic compound
= $\rm{\frac{m_{CO2} \times \frac{12}{44}}{m}\times 100}$
⇒ Percentage of C in the organic compound = $\rm{\frac{12 \times m_{CO2} \times 100}{44 \times m}}$

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Nitrogen

There are two methods for the estimation of nitrogen: Dumas method and Kjeldahl's method.
First we will see the Dumas method. It can be written in 10 steps:
1. A known mass (m) of the given organic compound is heated with copper(II) oxide (CuO) in an atmosphere of carbon dioxide.
• This reaction produces free nitrogen, carbon dioxide and water.
2. The mixture of gases thus produced may contain some traces of nitrogen oxides.
• We need to extract pure nitrogen from those oxides also.
• For that, the gaseous mixture is passed over a heated copper gauze. The oxides will be reduced to nitrogen.
3. The final gas mixture is passed through aqueous solution of KOH.
• The CO₂ will be absorbed by KOH. Pure nitrogen gets collected on top of the KOH solution.
• The KOH is kept in a graduated tube. So we can directly measure the volume (V₁) of the nitrogen.
4. So we obtained the volume of the newly produced nitrogen. Next we want it's pressure also. This can be written in 5 steps:
(i) Consider the apparatus shown in the text book. The nitrogen is collected in the graduated tube. We see a flask to the right side of the graduated tube. The flask contains aqueous KOH solution.
(ii) The graduated tube is connected to the flask through a tube. We see that, the level of KOH in the flask is same as that in the graduated tube. The KOH in the flask is open to the atmosphere. So the gas collected above KOH in the graduated tube will also be experiencing the same atmospheric pressure.
(iii) The gas collected above the KOH solution will be a mixture of nitrogen and water vapor. This is because, the KOH solution is an aqueous solution.
(iv) So the total pressure experienced by the gaseous mixture will be the sum of two items:
   ♦ Pressure experienced by pure nitrogen.
   ♦ Pressure experienced by water vapor (aqueous tension).
We can write: P_Total = P_Nitrogen + Aqueous tension
(see the notes related to Eq.5.5 in section 5.4) 
(v) We want the pressure experienced by the pure nitrogen.
• So we must subtract the aqueous tension from total pressure.
• In (ii), we saw that, total pressure is same as atmospheric pressure.
• So we can write:
Pressure experienced by nitrogen (P₁) = Atmospheric pressure - Aqueous tension
5. The temperature (T₁) of the nitrogen in the graduated tube will be same as the room temperature.
• Thus we obtained the pressure (P₁), volume (V₁) and temperature (T₁) of the nitrogen in the graduated tube.
6. We want the mass of this nitrogen. For that we use the familiar equation:
$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$
Where
   ♦ P₁ = Atmospheric pressure - Aqueous tension
   ♦ V₁ = Volume read out from the graduated tube
   ♦ T₁ = Room temperature at which experiment is carried out
   ♦ P₂ = pressure experienced by the same mass of nitrogen when it is at STP.
         ✰ STP is standard temperature and pressure.
         ✰ So at STP, it will be experiencing 760 mm pressure.
   ♦ V₂ = Volume occupied by the same mass of nitrogen when it is at STP.
   ♦ T₂ = Temperature experienced by the same mass of nitrogen when it is at STP.
         ✰ STP is standard temperature and pressure.
         ✰ So at STP, it will be experiencing 273 K temperature.
7. Now from the equation in (6), We get:
$\frac{P_1 V_1}{T_1}=\frac{760 \times V_2}{273}$
⇒ $V_2=\frac{P_1 V_1 \times 273}{760 \times T_1}$
8. Now we can find the mass of the nitrogen in the graduated tube. It can be written in 3 steps:
(i) We know that, 28 grams of nitrogen will occupy a volume of 22400 mL at STP.
(ii) So 1 mL of nitrogen at STP will have a mass of $\frac{28}{22400}$ grams.
(iii) So V₂ mL of nitrogen at STP will have a mass of $\frac{28 \times V_2}{22400}$ grams
9. We can write:
Mass of the nitrogen in the graduated tube
= Mass of nitrogen in m grams of the given organic compound = $\frac{28 \times V_2}{22400}$ grams
10. Thus the percentage of N in the organic compound
= $\rm{\frac{\frac{28 \times V_2}{22400}}{m}\times 100}$

= $\rm{\frac{28 \times V_2 \times 100}{22400 \times m}}$

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Let us see a solved example:

Solved example 12.21
In Dumas method for estimation of nitrogen, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300 K temperature and 715 mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300 K = 15 mm)
Solution:
1. First we apply the equation
$\frac{P_1 V_1}{T_1}=\frac{P_2 V_2}{T_2}$
Where
   ♦ P₁ = Atmospheric pressure - Aqueous tension
   ♦ V₁ = Volume read out from the graduated tube
   ♦ T₁ = Room temperature at which experiment is carried out
   ♦ P₂ = pressure experienced by the same mass of nitrogen when it is at STP.
         ✰ STP is standard temperature and pressure.
         ✰ So at STP, it will be experiencing 760 mm pressure.
   ♦ V₂ = Volume occupied by the same mass of nitrogen when it is at STP.
   ♦ T₂ = Temperature experienced by the same mass of nitrogen when it is at STP.
         ✰ STP is standard temperature and pressure.
         ✰ So at STP, it will be experiencing 273 K temperature.
• Given that, atmospheric pressure is 715 mm and aqueous tension is 15 mm
   ♦ So P₁ = (715 - 15) = 700 mm
• Given that, volume read out from the graduated tube = 50 mL
• Given that temperature at which the experiment is carried out is 300 K. So T₁ = 300 K.
• Substituting the known values, we get:
$\frac{700 \,\rm{(mm)} \times  50 \,\rm{(mL)}}{300 \,\rm{(K)}}=\frac{760 \, \rm{(mm)} \times V_2}{273 \,\rm{(K)}}$
• Thus we get: V₂ = 41.9 mL
2. Now we apply the equation:
Percentage of N in the organic compound = $\rm{\frac{28 \times V_2 \times 100}{22400 \times m}}$
• Substituting the known values, we get:
Percentage of N in the organic compound = $\rm{\frac{28 \, \rm{(g)} \times 41.9 \, \rm{(mL)} \times 100}{22400 \, \rm{(mL)} \times 0.3 \, \rm{(g)}}}$ = 17.46%

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Now we will see the Kjeldahl's method. It can be written in 8 steps:
1. A known mass (m) of the given organic compound is heated with concentrated sulfuric acid.
• Nitrogen in the organic compound will get converted into ammonium sulphate ((NH₄)₂SO₄). The equation is:
Organic compound + H₂SO₄ ⟶ (NH₄)₂SO₄
2. The resulting acid mixture containing ammonium sulphate is transferred to another flask. It is heated with excess NaOH.
• All the nitrogen will be converted to ammonia. The equation is:
(NH₄)₂SO₄ + 2NaOH ⟶ Na₂SO₄ + 2NH₃ + 2H₂O
3. The ammonia gas is made to pass through an excess of known volume of sulfuric acid.
    ♦ The volume of this sulfuric acid is known.
    ♦ The molarity of this sulfuric acid is also known.
    ♦ Thus we will get the number of moles of H₂SO₄ taken in this step.
4. All the ammonia will be converted to ammonium sulphate. The reaction is:
2NH₃ + H₂SO₄ ⟶ (NH₄)₂SO₄
5. The remaining sulfuric acid is titrated against NaOH solution of known molarity.
• Thus we will get the number of moles of remaining sulfuric acid.
(see fig.8.7 in section 8.14)
6. We subtract this 'remaining number' from the number initially taken in step (3)
• Thus we get the number of moles of H₂SO₄ used up in the reaction in (4)
7. In the equation in (4), we see that, two moles of NH₃ reacts with one mole of H₂SO₄.
• That means, the number of moles of NH₃ produced in (2) is twice the number that we obtain in (6)
• Thus we get the number of moles of NH₃ produced in (2)
8. From this number of moles of NH₃, we can find the mass of nitrogen.
• This mass is the same mass of nitrogen present in the m grams of original organic compound. So we can calculate the percentage of nitrogen.
• The following solved example will make the process clear. 

Solved example 12.22
During a Kjeldahl's experiment to estimate nitrogen, the ammonia evolved due to the reaction in step (2) neutralized 10 mL of 1 M H₂SO₄. The mass of original organic compound taken was 0.5 gram. Find out the percentage of nitrogen.
Solution:
1. Given that, 10 mL of 1 M H₂SO₄ was neutralized.
• One L of 1 M H₂SO₄ will contain one mole of H₂SO₄
⇒ 1 mL of 1 M H₂SO₄ will contain $\frac{1}{1000}$ mole of H₂SO₄
⇒ 10 mL of 1 M H₂SO₄ will contain $\frac{10}{1000}$ mole of H₂SO₄
2. This much moles of H₂SO₄ was neutralized.
• That means, $2 \times \frac{10}{1000}=\frac{20}{1000}$ mole of NH₃ was present.
3. One mole NH₃ will contain 14 grams of nitrogen.
So $\frac{20}{1000}$ will contain $14 \times \frac{20}{1000}=\frac{280}{1000}$ grams of nitrogen.
4. So percentage of nitrogen in the 0.5 grams of the orginal organic compound
= $\frac{\frac{280}{1000}}{0.5} \times 100$

= $\frac{280 \times 100}{0.5 \times 1000}$ = 56%

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Note that, in the above example, we are directly given the number of moles of H₂SO₄ used up by the NH₃. In some cases, we will have to calculate that number, using the data related to titration with NaOH. Such an example is given below:

Solved example  12.23
A sample of 0.5 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 mL of 0.5 M H₂SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralization. Find the percentage composition of nitrogen in the compound.
Solution:
1. Given that, 60 mL of 0.5 M NaOH was required.
• One L of 0.5 M NaOH will contain 0.5 mole of NaOH
⇒ 1 mL of 0.5 M NaOH will contain $\frac{0.5}{1000}$ mole of NaOH
⇒ 60 mL of 0.5 M NaOH will contain $\frac{60 \times 0.5}{1000}$ mole of NaOH
2. This much mole of NaOH was required.
• That means $\frac{1}{2} \times \frac{60 \times 0.5}{1000}=\frac{15}{1000}$ mole of H₂SO₄ was present in the residual acid.
3. Given that, 50 mL of 0.5 M H₂SO₄ was originally taken.
• One L of 0.5 M H₂SO₄ will contain 0.5 mole of H₂SO₄
⇒ 1 mL of 0.5 M H₂SO₄ will contain $\frac{0.5}{1000}$ mole of H₂SO₄
⇒ 50 mL of 0.5 M H₂SO₄ will contain $\frac{50 \times 0.5}{1000}=\frac{25}{1000}$ mole of H₂SO₄
4. From (2) and (3), we get:
Number of moles of H₂SO₄ that reacted with NH₃
= $\frac{25}{1000} - \frac{15}{1000}= \frac{10}{1000}$
5. That means, $2 \times \frac{10}{1000}=\frac{20}{1000}$ mole of NH₃ was present.
6. One mole NH₃ will contain 14 grams of nitrogen.
• So $\frac{20}{1000}$ will contain $14 \times \frac{20}{1000}=\frac{280}{1000}$ grams of nitrogen.
7. So percentage of nitrogen in the 0.5 grams of the orginal organic compound
= $\frac{\frac{280}{1000}}{0.5} \times 100$

= $\frac{280 \times 100}{0.5 \times 1000}$ = 56%

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• In step 1 of Kjeldahl’s method, we see that, the nitrogen in the organic compound must be first converted into ammonium sulfate.
• Consider the following three cases:
    ♦ Nitrogen in the organic compound is in the nitro group.
    ♦ Nitrogen in the organic compound is in the azo group.
    ♦ Nitrogen in the organic compound is in the ring (example: pyridine).
• In these three cases, the nitrogen will not get converted into ammonium sulfate.
• So for these three cases, Kjeldahl’s method cannot be used.

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In the next section we will see estimation of halogens.

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