Chapter 12.23 - Estimation of Halogens, Sulphur, Phosphorus and Oxygen

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 m_AgX is noted.
4. Using this m_AgX, 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%

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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 m_B is noted.
4. Using this m_B, 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%

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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 m_A is noted.
4. Using this m_A, 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
⇒ m_A 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 m_A is noted.
5. Using this m_A, 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
⇒ m_A 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 = $\frac{m_A \times \frac{62}{222}}{m}\times 100=\frac{62 \times m_A \times 100}{222 \times m}$

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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 = P_C %
    ♦ Percentage of H = P_H %
    ♦ Percentage of N = P_N %
    ♦ Percentage of X = P_X %
    ♦ Percentage of S = P_S %
    ♦ Percentage of P = P_P %
    ♦ Percentage of O = P_O %
2. Suppose that, the given organic compound contains C, H, N and O.
• Then we can write: (P_C + P_H + P_N + P_O) = 100
• From this we get: P_O = [100 - (P_C + P_H + P_N)]
3. Suppose that, the given organic compound contains C, H, S and O.
• Then we can write: (P_C + P_H + P_S + P_O) = 100
• From this we get: P_O = [100 - (P_C + P_H + P_S)]
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.

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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 m_C 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.

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• 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

Additional solved examples

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We have completed the discussions in this chapter. In the next chapter we will see hydrocarbons.

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Chapter 6.5 - Enthalpy Change of Reaction

In the previous section, we saw how 'enthalpy changes' can be determined using calorimetry. In this section, we will see 'enthalpy change of reaction'.

Enthalpy change of reaction can be explained in 13 steps. While writing those steps, we will see enthalpy change of formation also:
1. Consider the equation of a simple reaction: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
• This is a balanced equation. We know that, a balanced equation will give the number of moles of each of the reactants and products involved in the reaction.
• For example, in this reaction, one mol zinc reacts with two mol HCl to give one mol ZnCl₂ and one mol H₂
2. We can write the general form of a balanced equation:
a₁R₁ + a₂R₂ + a₃R₃ + . . . →  b₁P₁ + b₂P₂ + b₃P₃ + . . .   
• R₁, R₂, R₃, . . . are the reactants
    ♦ a₁, a₂, a₃, . . . are the ‘number of mol’ of each of those reactants.
• P₁, P₂, P₃, . . . are the products
    ♦ b₁, b₂, b₃, . . . are the ‘number of mol’ of each of those products.
3. Now we apply the concept of enthalpy to each reactant and product:
• Let ΔH_f[R₁] be the enthalpy change when one mole of R₁ is formed.
    ♦ The subscript 'f' denotes 'formation'.
• This enthalpy change must be determined only after carefully considering two aspects:
(i) R₁ can be formed in many ways.
For example:
• H₂O can be formed by the combination of H₂ and O₂
    ♦ 2H₂ + O₂ → 2H₂O
• H₂O can be formed as one of the products in a reaction
    ♦ HCl + NaOH → NaCl + H₂O
• We must consider only that reaction in which R₁ is formed from it's elements.
• So in the case of H₂O, we must consider only: 2H₂ + O₂ → 2H₂O
(ii) While carrying out the reaction between H₂ and O₂, different labs may use different temperatures and pressures. This will give different ΔH_f values.
• In order to avoid such a confusion, scientists have given a set of rules:
    ♦ The elements from which R₁ is formed must be
          ✰ Under a pressure of 1 bar
          ✰ At a temperature of 298.15 K
          ✰ At the most stable state
          ✰ For example, oxygen is most stable when it exist as O₂ molecules, not as individual O atoms.
4. When the above two aspects are obeyed, the ΔH_f value obtained is written as: ΔH_^⊖f
• The superscript '⊖' indicates that, it is the standard value.
• ΔH_^⊖f is known as the standard enthalpy change of formation.
5. It is interesting to note that, ΔH_^⊖f values for elements is zero.
• For example: ΔH_^⊖f[O₂] = 0,  ΔH_^⊖f[C] = 0   etc.,
    ♦ This is because:
          ✰ O₂ is formed from O₂. There is no enthalpy change.
          ✰ C is formed from C. There is no enthalpy change.
• We can look up the ΔH_^⊖f values of most compounds from the data book or text book.
6. Now consider the general reaction mentioned in (2)
• Let us write down the ΔH_^⊖f values of the reactants R₁, R₂, R₃, . . .
    ♦ They can be written as: ΔH_^⊖f[R₁], ΔH_^⊖f[R₂], ΔH_^⊖f[R₃], . . .
• Let us write down the ΔH_^⊖f values of the products P₁, P₂, P₃, . . .
    ♦ They can be written as: ΔH_^⊖f[P₁], ΔH_^⊖f[P₂], ΔH_^⊖f[P₃], . . .
7. The above ΔH_^⊖f values that we obtain from the data book, are related to one mole.
• For example, the ΔH_^⊖f value of Al₂O₃ is -1675.7 kJ mol^_-1
    ♦ That means, when one mol Al₂O₃ is formed from Al and O₂, the enthalpy change is -1675.7 kJ
• But in our present case, we have:
    ♦ a₁ mol of reactant R₁
    ♦ a₂ mol of reactant R₂ . . . so on
• So we must multiply the values by the corresponding mol number.
• Thus the step (6) can be modified as:
    ♦ ΔH_^⊖f values of the reactants: a₁ΔH_^⊖f[R₁], a₂ΔH_^⊖f[R₂], a₃ΔH_^⊖f[R₃], . . .
    ♦ ΔH_^⊖f values of the products: b₁ΔH_^⊖f[P₁], b₂ΔH_^⊖f[P₂], b₃ΔH_^⊖f[P₃], . . .
8. Now we find two sums:
(i) Sum for the reactants:
$\mathbf\small{\rm{\sum\limits_{i}{a_i \Delta {H^\circleddash}_f[R_i]}}}$ = a₁ΔH_^⊖f[R₁] + a₂ΔH_^⊖f[R₂] + a₃ΔH_^⊖f[R₃] . . .
(ii) Sum for the products:
$\mathbf\small{\rm{\sum\limits_{i}{b_i \Delta {H^\circleddash}_f[P_i]}}}$ = b₁ΔH_^⊖f[P₁] + b₂ΔH_^⊖f[P₂] + b₃ΔH_^⊖f[P₃] . . .
9. Next we subtract the first sum from the second sum.
• The result is: standard enthalpy change of reaction.
    ♦ It is denoted as: ΔH_^⊖r
10. So for the general reaction mentioned in (2), we can write:
Eq.6.9: $\mathbf\small{\rm{\Delta {H^\circleddash }_r=\sum\limits_{i}{b_i \Delta {H^\circleddash}_f[P_i]}-\sum\limits_{i}{a_i \Delta {H^\circleddash}_f[R_i]}}}$
11. Using this equation, let us find the ΔH_^⊖r of the reaction mentioned in (1), which is:
Zn (s) + 2HCl (aq) → ZnCl₂ (s) + H₂ (g)
It can be done in 2 steps:
(i) From the data book (values can be obtained online also) , we have:
    ♦ ΔH_^⊖f[Zn(s)] = 0 kJ mol^_-1
    ♦ ΔH_^⊖f[HCl(aq)] = -167.16 kJ mol^_-1
    ♦ ΔH_^⊖f[ZnCl₂(aq)] = -488.2 kJ mol^_-1
    ♦ ΔH_^⊖f[H₂(g)] = 0 kJ mol^_-1
(ii) Applying Eq.6.9, we get:
ΔH_^⊖r = [-415.1 -(2 × -167.16)] = -153.88 kJ mol^_-1
12. Now we will see the significance of the sign (+ve or -ve) of the ΔH value.
• It can be written in 4 steps:
(i) We have: U_B = U_A + Q_P  – W
(Here we asume that, the system absorbs Q_P. So it is given a positive sign)
⇒ U_B = U_A + Q_P – P(V_B – V_A)
⇒ (U_B + PV_B) – (U_A + PV_A) = Q_P
⇒ ΔH = H_B – H_A = Q_P
(ii) We see that, if H_B is greater than H_A, ΔH will be positive.
• Q_P will also be positive
    ♦ That means our assumption is correct.
    ♦ That means, the system absorbs heat.
(iii) So we can conclude that:
A positive ΔH indicates that the system absorbs heat. It is an endothermic reaction.
(iv) We can write the converse also:
A negative ΔH indicates that the system releases heat. It is an exothermic reaction.
13. We have seen how ΔH_^⊖r is calculated for the reaction between Zn and HCl.
• Let us see another example. This time we want the ΔH_^⊖r of the following reaction:
CaCO₃(s) → CaO(s) + CO₂(g)
• This is the decomposition reaction of calcium carbonate. The answer can be written in 3 steps:
(i) From the data book, we have:
    ♦ ΔH_^⊖f[CaCO₃(s)] = -1206.92 kJ mol^_-1
    ♦ ΔH_^⊖f[CaO(aq)] = -635.09 kJ mol^_-1
    ♦ ΔH_^⊖f[ZnCl₂(aq)] = -393.51 kJ mol^_-1
(ii) Applying Eq.6.9, we get:
ΔH_^⊖r = [-635.09 - 393.51 - (-1206.92)] = 178.32 kJ mol^_-1
• We get a positive value. That means, we have to supply 178.32 kJ mol^_-1 for the reaction to take place.
(iii) From the balanced equation, it is clear that, one mol CaCO₃ is undergoing decomposition.
• So we can write: 178.32 kJ is required for the decomposition of one mol CaCO₃
• If we know the mass (in grams) of CaCO₃ at the beginning of the reaction, we can calculate the number of moles present in that mass.
• Using that ‘number of moles’, the heat energy required can be calculated.
• This is the advantage of knowing the ΔH_^⊖r value of a reaction.

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• It is clear that, the ΔH_^⊖f values of individual reactants and products play an important role in the ΔH_^⊖r value of the overall reaction.
• In step (3), we have seen two important aspects about ΔH_^⊖f
    ♦ Now we will see them in some more detail
• It can be written in 5 steps:
1.We have seen that, the ΔH_^⊖f values can be looked up from the data book.
• Those values in the data book are determined by scientists using precision instruments in the lab.
• Those experiments are carried out under standard conditions (1 bar pressure and 298.15 K)
    ♦ So whenever the experiments are repeated in different labs, the same ΔH_^⊖f values will be obtained.
2. The experiments are chosen in such a way that the required compound is formed from the constituent elements only.
• This point can be explained using an example:
(i) The following reaction → CaCO₃ as the product:
CaO(s) + CO₂(g) → CaCO₃(s)
    ♦ ΔH for the reaction is: -178.3 kJ mol_^-1
(iii) CaCO₃ is the only product. Also, only one mol CaCO₃ is formed.
• So it appears that ΔH_^⊖f of CaCO₃ is -178.3 kJ mol_^-1
• But it is not the acceptable value because, in this reaction, CaCO₃ is formed from other compounds.
• For the ΔH to be acceptable as ΔH_^⊖f, there must be only Ca, C and O (in their pure forms) in the left side of the equation in (i)
3. Consider the reaction given below:
H₂(g) + Br₂(l) → 2HBr(g)
• The ΔH for this reaction is -72.8 kJ mol^-1
• Can we take -72.8 kJ mol^-1 as the ΔH_^⊖f value of HBr?
   ♦ On the left side only H₂ and Br₂ is present.
   ♦ On the right side, only HBr is present.
   ♦ So at a first glance, it appears that, -72.8 kJ mol-1 is indeed the ΔH_^⊖f value of HBr.
• But remember that, value is related to the formation of one mole of a compound.
   ♦ In our present case, two moles of HBr is formed
   ♦ So 72.8 kJ is the energy released when two moles of HBr is formed.
   ♦ Thus, 72.8 kJ mol^-1 is not acceptable.
• However, if we divide the equation through out out by 2, we will get:
¹⁄₂H₂(g) + ¹⁄₂Br₂(l) → HBr(g)
   ♦ This time, only one mol HBr is formed. So the energy released will be half of 72.8 kJ
   ♦ That means, when one mol HBr is formed, the energy released is (¹⁄₂ × 72.8) = 36.4
   ♦ So we can write: ΔH_^⊖f of HBr is -36.4 kJ mol^-1
4. Let us compare ΔH_^⊖f and ΔH_^⊖r
• We  know that, both are 'enthalpy changes' taking place during reactions.
• But ΔH_^⊖f is a special case of ΔH_^⊖r
   ♦ Because, for an enthalpy change to be acceptable as ΔH_^⊖f, the three rules mentioned above must be satisfied.
• The three rules can be summarized as follows:
(i) standard conditions should be adopted.
(ii) On the left side of the reaction equation, there must be the constituent elements (in standard form) only.
• On the right side, there must be only one compound
(iii) On the right side, there must be only one mol of the compound.
5. The reader must try and become convinced that:
    ♦ All ΔH_^⊖f values are ΔH_^⊖r values
    ♦ But all ΔH_^⊖r values are not ΔH_^⊖f values

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In the next section we will see thermochemical equations

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