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Analyzing chemical equations

Module by: Sunil Kumar Singh. E-mail the author

Application of mole concept requires a balanced chemical equation. The different constituents of the reaction – reactants and products – bear a simple whole number proportion same as the proportion of the coefficients associated with constituents. According to mole concept, the molar mass of constituents participates in this proportion. For a generic consideration as given :

x A + y B → A x B y x A + y B → A x B y

Here, 1 mole of compound ( A x B y A x B y ) involves x mole of A and y mole of B. Using symbols :

x moles of A ≡ y moles of B ≡ 1 mole of A x B y x moles of A ≡ y moles of B ≡ 1 mole of A x B y

The point to emphasize here is that this is a relation, which is connected by "equivalence sign (≡)" - not by "equal to (=)" sign. We know that 2 moles of hydrogen reacts with 1 mole of oxygen to form 2 moles of water. Clearly, we can not equate like 2=1=2. We need to apply unitary method to interpret this relation of equivalence. We say that since x moles of A react with y moles of B. Hence, 1 mole of A reacts with y/x moles of "B". Similarly, 1 mole of B reacts with x/y moles of "A". Once we know the correspondence for 1 mole, we can find correspondence for any other value of participating moles of either A or B.

Mass of participating entities in a reaction

Mole concept is used to calculate mass of individual constituent of a chemical reaction. The proportion of molar mass is converted to determine proportion of mass in which entities are involved in a reaction. The symbolic mass relation for the chemical reaction as given above is :

x M A gm of A ≡ y M B gm of B ≡ M A x B y gm of A x B y x M A gm of A ≡ y M B gm of B ≡ M A x B y gm of A x B y

We apply unitary method on the mass relation related with equivalent sign (≡) to determine mass of different entities of the reaction.

Example 1

Problem : Calculate mass of lime (CaO) that can be prepared by heating 500 kg of 90 % pure limestone ( C a C O 3 C a C O 3 .

Solution : Purity of CaCO3 is 90 %. Hence,

Mass of C a C O 3 = 0.9 X 500 = 450 k g Mass of C a C O 3 = 0.9 X 500 = 450 k g

The chemical reaction involved here is :

C a C O 3 → C a O + C O 2 C a C O 3 → C a O + C O 2

Applying mole concept :

⇒ 1 mole of C a C O 3 ≡ 1 mole of CaO ⇒ 1 mole of C a C O 3 ≡ 1 mole of CaO

⇒ (40 + 12 + 3X 16) gm of C a C O 3 ≡ (40 + 16) gm of CaO ⇒ (40 + 12 + 3X 16) gm of C a C O 3 ≡ (40 + 16) gm of CaO

⇒ 100 gm of C a C O 3 ≡ 56 gm of CaO ⇒ 100 gm of C a C O 3 ≡ 56 gm of CaO

⇒ 100 kg of C a C O 3 ≡ 56 kg of CaO ⇒ 100 kg of C a C O 3 ≡ 56 kg of CaO

Applying unitary method :

⇒ mass of CaO produced = 56 X 450 100 = 252 k g ⇒ mass of CaO produced = 56 X 450 100 = 252 k g

Example 2

Problem : Igniting M n O 2 M n O 2 converts it quantitatively to M n 3 O 4 M n 3 O 4 . A sample of pyrolusite contains 80% M n O 2 M n O 2 , 15 % S i O 2 S i O 2 and 5 % water. The sample is ignited in air to constant weight. What is the percentage of manganese in the ignited sample ? ( A M n = 55 A M n = 55 )

Solution : The sample contains three components. Since this question involves percentage, we shall consider a sample of 100 gm. Water component weighing 5 gm evaporates on ignition. S i O 2 S i O 2 weighing 15 gm does not change. On the other hand, 80 gm of M n O 2 M n O 2 converts as :

3 M n O 2 → M n 3 O 4 + O 2 3 M n O 2 → M n 3 O 4 + O 2

Applying mole concept,

3 moles of M n O 2 ≡ 1 mole of M n 3 O 4 3 moles of M n O 2 ≡ 1 mole of M n 3 O 4

3 X 55 + 2 X 16 gm of M n O 2 ≡ 1 X 3 X 55 + 4 X 16 gm of M n 3 O 4 3 X 55 + 2 X 16 gm of M n O 2 ≡ 1 X 3 X 55 + 4 X 16 gm of M n 3 O 4

261 gm of M n O 2 ≡ 229 gm of M n 3 O 4 261 gm of M n O 2 ≡ 229 gm of M n 3 O 4

Since sample is ignited in air to constant weight, it means that all of Mn O 2 Mn O 2 in the sample is converted. Using unitary method, we determine mass of converted mass of M n 3 O 4 M n 3 O 4 for 80 gm of Mn O 2 Mn O 2 :

Mass of M n 3 O 4 on conversion = 229 261 X 80 = 70.2 g m Mass of M n 3 O 4 on conversion = 229 261 X 80 = 70.2 g m

We are required to find the percentage of Mn in the ignited sample. Thus, we need to determine the mass of the ignited sample. The ignited sample contains 70.2 gm of M n 3 O 4 M n 3 O 4 and 15 gm of S i O 2 S i O 2 . Total mass of ignited sample is 70.2+15 = 85.4 gm. On the other hand, amount of Mn in M n 3 O 4 M n 3 O 4 is calculated from its molecular constitution :

229 gm of M n 3 O 4 ≡ 3 X 55 gm of Mn ≡ 165 gm of Mn 229 gm of M n 3 O 4 ≡ 3 X 55 gm of Mn ≡ 165 gm of Mn

Amount of Mn in 70.2 gm of M n 3 O 4 = 165 229 X 70.2 = 0.72 X 70.2 = 50.54 g m Amount of Mn in 70.2 gm of M n 3 O 4 = 165 229 X 70.2 = 0.72 X 70.2 = 50.54 g m

Clearly, 85.4 gm of ignited sample contains 50.54 gm of Mn. Hence,

Percentage amount of manganese in the ignited sample = 50.54 85.4 X 100 = 59.2 % Percentage amount of manganese in the ignited sample = 50.54 85.4 X 100 = 59.2 %

Reaction involving gas

In analyzing chemical reaction involving gas, we make the assumption that gases are ideal gases. In general, there are three different situations in which we may use gas volume in chemical analysis :

  • Gas volumes are given at STP.
  • Gas volumes are given at other temperature and pressure condition.
  • Only gas volumes are involved in calculation

In certain situation, reaction involving gas enables us to use gas volumes itself (not the moles) to analyze the reaction. We can extend the concept of molar proportions to volume proportions directly. Such is the case, when only gas volumes are involved in the calculation.

Gas volumes are given at STP

For analyzing gas volumes at STP, we make use of Avogadro’s hypothesis. A volume of 22.4 litres of ideal gas contains 1 mole of gas. The number of moles present in a volume “V” at STP is :

n = V 22.4 n = V 22.4

Example 3

Problem : Determine the amount of concentrated H 2 S O 4 H 2 S O 4 required to neutralize 20 litres of ammonia gas at STP.

Solution : The chemical reaction involved is :

2 N H 3 + H 2 S O 4 → N H 4 2 S O 4 2 N H 3 + H 2 S O 4 → N H 4 2 S O 4

Applying mole concept :

2 moles of N H 3 ≡ 1 mole of H 2 S O 4 2 moles of N H 3 ≡ 1 mole of H 2 S O 4

2 X 22.4 litres of N H 3 ≡ 98 gm of H 2 S O 4 2 X 22.4 litres of N H 3 ≡ 98 gm of H 2 S O 4

Amount of H 2 S O 4 for 20 litres of N H 3 = 98 44.8 X 20 = 43.8 g m Amount of H 2 S O 4 for 20 litres of N H 3 = 98 44.8 X 20 = 43.8 g m

Gas volumes are given at other temperature and pressure condition

Under this condition, we first need to convert gas volumes to volumes at STP. We make use of ideal gas law,

P 1 V 1 T 1 = P 2 V 2 T 2 P 1 V 1 T 1 = P 2 V 2 T 2

We may specify one of the suffix like “1” to represent the given condition and suffix “2” to represent the STP condition.

Example 4

Problem : Determine the amount of magnesium required to liberate 900 cc of hydrogen from a solution of HCl at 27°C and 740 mm of Hg. A M g = 24 A M g = 24 .

Solution : The chemical reaction involved is :

M g + 2 H C l → M g C l 2 + H 2 M g + 2 H C l → M g C l 2 + H 2

Applying mole concept :

1 mole of Mg ≡ 1 mole of H 2 1 mole of Mg ≡ 1 mole of H 2

24 gm of Mg ≡ 22400 cc of H 2 at STP 24 gm of Mg ≡ 22400 cc of H 2 at STP

Using ideal gas law,

⇒ V 2 = P 1 V 1 T 2 P 2 T 1 = 740 X 900 X 273 760 X 300 = 797 c c ⇒ V 2 = P 1 V 1 T 2 P 2 T 1 = 740 X 900 X 273 760 X 300 = 797 c c

Using this data to mole relation,

Amount of Mg = 24 X 797 22400 = 0.85 g m Amount of Mg = 24 X 797 22400 = 0.85 g m

Only gas volumes are involved in calculation

In this case, we are required to consider volumes of gases only. What it means that there is no solid or liquid elements involved in calculation. The reaction may involve solid or liquid components but they are not involved in calculation. In such situation, we can connect volumes of gas components of the reaction directly. Consider the reaction :

H 2 + C l 2 → 2 H C l H 2 + C l 2 → 2 H C l

x litres of H 2 ≡ x litres of C l 2 ≡ 2x litres of H C l x litres of H 2 ≡ x litres of C l 2 ≡ 2x litres of H C l

This direct relation is deducted from mole concept under two assumptions (i) all gases are ideal and (ii) temperature and pressure conditions are same for measuring all gases. Under these assumptions, equal numbers of moles of ideal gases occupy same volume. Hence, we can correlate volumes directly as above.

Example 5

Problem : A 60 cc mixture of N 2 O N 2 O and NO is mixed with excess H 2 H 2 and the resulting mixture of gas is exploded. The resulting mixture contains 38 cc of N 2 N 2 . Determine the volume composition of the original mixture.

Solution : The chemical reactions involved are :

N 2 O + H 2 → N 2 + H 2 O N 2 O + H 2 → N 2 + H 2 O

2 N O + 2 H 2 → N 2 + 2 H 2 O 2 N O + 2 H 2 → N 2 + 2 H 2 O

Let volume of N 2 O N 2 O be x in the original mixture. Hence, volume of NO in the sample is 60 – x. Now, the corresponding mole relations are :

x cc of N 2 O ≡ x cc of N 2 x cc of N 2 O ≡ x cc of N 2

(60−x) cc of N 2 O ≡ 60 − x 2 cc of N 2 (60−x) cc of N 2 O ≡ 60 − x 2 cc of N 2

According to question,

x + 60 − x 2 = 38 x + 60 − x 2 = 38

x + 60 = 76 x + 60 = 76

⇒ Volume of N 2 O = x = 16 c c ⇒ Volume of N 2 O = x = 16 c c

⇒ Volume of NO = 60 − x = 60 − 16 = 44 c c ⇒ Volume of NO = 60 − x = 60 − 16 = 44 c c

Limiting reactant

The participating mass of the reactants may not be exactly same as required by balanced chemical reaction. The reactant which limits the progress of reaction is called limiting reactant (reagent). If gA and gB be the mass of two reactants, then moles present are :

n A = g A M A n A = g A M A

n B = g B M B n B = g B M B

According to mole concept :

x moles of A ≡ y moles of B x moles of A ≡ y moles of B

nA moles of A ≡ y n A x moles of B nA moles of A ≡ y n A x moles of B

For reaction to complete as the limiting case,

y n A x = n B y n A x = n B

y n A = x n B y n A = x n B

For A to be limiting,

n B > y n A x n B > y n A x

y n A < x n B y n A < x n B

For B to be limiting,

n B < y n A x n B < y n A x

x n B < y n A x n B < y n A

y n A > x n B y n A > x n B

Example 6

Problem : Equal amounts of iron and sulphur are heated together to form FeS. What fraction of iron or Sulphur are converted into FeS. Here, A F e = 56 A F e = 56 and A S = 32 A S = 32 .

Solution : The chemical reaction involved in the conversion is :

F e + S → F e S F e + S → F e S

1 mole of Fe ≡ 1 moles of S 1 mole of Fe ≡ 1 moles of S

Here, x=y=1. Now, equal amounts are used. Let “g” gm of each takes part in the reaction. The numbers of moles corresponding to “g” gm are :

n F e = g 56 n F e = g 56

and

y n F e = g 56 y n F e = g 56

Similarly,

x n S = g 32 x n S = g 32

Here, y n F e < x n S y n F e < x n S . It means that iron is in short supply and is the limiting reactant. Clearly, all iron is converted into FeS. Hence, fraction of iron used is 1. On the other hand, the moles of Sulphur converted are x 56 x 56 only as reaction is limited by iron.

⇒ Mass of Sulphur converted in the reaction = x 56 X M S = 32 x 56 ⇒ Mass of Sulphur converted in the reaction = x 56 X M S = 32 x 56

⇒ Fraction of suphur converted = 32 x / 56 x = 32 56 = 0.571 ⇒ Fraction of suphur converted = 32 x / 56 x = 32 56 = 0.571

Multiple reactions

Some chemical reaction under analysis involves simultaneous or concurrent reactions. Consider the example in which a mixture of KCl and KI are treated with excess of silver nitrate ( A g N O 3 A g N O 3 ) :

K C l + A g N O 3 → K N O 3 + A g C l K C l + A g N O 3 → K N O 3 + A g C l

K I + A g N O 3 → K N O 3 + A g I K I + A g N O 3 → K N O 3 + A g I

We apply mole concept to two concurrent reactions as :

1 mole of KCl ≡ 1 mole of A g N O 3 ≡ 1 mole of K N O 3 ≡ 1 mole of AgCl 1 mole of KCl ≡ 1 mole of A g N O 3 ≡ 1 mole of K N O 3 ≡ 1 mole of AgCl

and

1 mole of KI ≡ 1 mole of A g N O 3 ≡ 1 mole of K N O 3 ≡ 1 mole of AgI 1 mole of KI ≡ 1 mole of A g N O 3 ≡ 1 mole of K N O 3 ≡ 1 mole of AgI

Example 7

Problem : 2.2 g of a mixture of KCl and KI yields 3.8 gm of AgCl and AgI, when treated with excess silver nitrate ( A g N O 3 A g N O 3 ). Find the mass of KCl and KI in the original mixture.

Solution : Since the mixture is treated with excess silver nitrate, both chloride and iodide are completely consumed, forming silver chloride and iodide. Let the initial mixture contains x and y grams of KCl and KI respectively. Then, according to question :

x + y = 2.2 x + y = 2.2

This is first of two linear equations, which will be used to determine x and y. Now, chemical reactions are :

K C l + A g N O 3 → K N O 3 + A g C l K C l + A g N O 3 → K N O 3 + A g C l

K I + A g N O 3 → K N O 3 + A g I K I + A g N O 3 → K N O 3 + A g I

The amount of AgCl and AgI produced is 3.8 gm. Applying mole concept to first equation,

1 mole of KCl ≡ 1 mole of AgCl 1 mole of KCl ≡ 1 mole of AgCl

x 39 + 35.5 moles of KCl ≡ x 39 + 35.5 moles of AgCl ≡ x 74.5 moles of AgCl x 39 + 35.5 moles of KCl ≡ x 39 + 35.5 moles of AgCl ≡ x 74.5 moles of AgCl

Therefore, mass of AgCl is :

x 74.5 moles of AgCl = x 74.5 X 108 + 35.5 g m = 143.5 x 74.5 g m x 74.5 moles of AgCl = x 74.5 X 108 + 35.5 g m = 143.5 x 74.5 g m

Similarly, applying mole concept to second equation,

1 mole of KI ≡ 1 mole of AgI 1 mole of KI ≡ 1 mole of AgI

x 39 + 127 moles of KCl ≡ y 39 + 127 moles of AgI ≡ y 166 moles of AgI x 39 + 127 moles of KCl ≡ y 39 + 127 moles of AgI ≡ y 166 moles of AgI

Therefore, mass of AgI is :

y 166 moles of AgI = y 166 X 108 + 127 g m = 235 y 166 g m y 166 moles of AgI = y 166 X 108 + 127 g m = 235 y 166 g m

According to question,

143.5 x 74.5 + 235 y 166 = 3.8 143.5 x 74.5 + 235 y 166 = 3.8

In order to render coefficient of first term of the equation equal to 1 and hence simplify the equation, we multiply the equation by 74.5 143.5 74.5 143.5 ,

x + 74.5 X 235 y 143.5 X 166 = 3.8 X 74.5 143.5 x + 74.5 X 235 y 143.5 X 166 = 3.8 X 74.5 143.5

⇒ x + 0.73 y = 1.97 ⇒ x + 0.73 y = 1.97

Thus, we have two linear equations and two unknowns. Substituting for y in terms of x from first linear equation derived earlier, we have :

⇒ x + 0.73 2.2 − x = 1.97 ⇒ x + 0.73 2.2 − x = 1.97

⇒ x + 1.61 − 0.73 x = 1.97 ⇒ x + 1.61 − 0.73 x = 1.97

⇒ 0.27 x = 0.36 ⇒ 0.27 x = 0.36

⇒ x = 1.34 g m ⇒ x = 1.34 g m

⇒ y = 2.2 − x = 2.2 − 1.34 = 0.86 g m ⇒ y = 2.2 − x = 2.2 − 1.34 = 0.86 g m

Chain reaction

A chemical process involves multiple reactions in which product of one reaction becomes reactant in the second reaction. Consider example of formation of sulphuric acid from pyrite( F e S 2 F e S 2 ) :

4 F e S 2 + 11 O 2 → 2 F e 2 O 3 + 8 S O 2 4 F e S 2 + 11 O 2 → 2 F e 2 O 3 + 8 S O 2

2 S O 2 + O 2 → 2 S O 3 2 S O 2 + O 2 → 2 S O 3

S O 3 + H 2 O → H 2 S O 4 S O 3 + H 2 O → H 2 S O 4

Here, first reaction supplies S O 2 S O 2 required as reactant in the second reaction. We, therefore, need to multiply second reaction 4 times so that S O 2 S O 2 molecules are balanced. Similarly, third reaction should be multiplied by 8 to match 8 S O 3 8 S O 3 molecules produced in the second reaction. Thus, balanced chain reaction is :

4 F e S 2 + 11 O 2 → 2 F e 2 O 3 + 8 S O 2 4 F e S 2 + 11 O 2 → 2 F e 2 O 3 + 8 S O 2

8 S O 2 + 4 O 2 → 8 S O 3 8 S O 2 + 4 O 2 → 8 S O 3

8 S O 3 + 8 H 2 O → 8 H 2 S O 4 8 S O 3 + 8 H 2 O → 8 H 2 S O 4

From balanced chain equations, we conclude that :

⇒ 4 moles of F e S 2 ≡ 8 moles of H 2 S O 4 ⇒ 4 moles of F e S 2 ≡ 8 moles of H 2 S O 4

⇒ 1 mole of F e S 2 ≡ 2 moles of H 2 S O 4 ⇒ 1 mole of F e S 2 ≡ 2 moles of H 2 S O 4

Alternatively, we can argue that one molecule of F e S 2 F e S 2 contains 2 atoms of Sulphur(S). On the other hand, one molecule of H 2 S O 4 H 2 S O 4 contains 2 atoms of Sulphur(S). Further, there is no loss or gain of Sulphur in chain reactions. Hence, to conserve mass of suphur,

⇒ 1 mole of F e S 2 ≡ 2 moles of H 2 S O 4 ⇒ 1 mole of F e S 2 ≡ 2 moles of H 2 S O 4

Example 8

Problem : How many grams of sulphuric acid can be obtained from 100 gm of pyrite?

Solution : Here,

⇒ 1 mole of F e S 2 ≡ 2 moles of H 2 S O 4 ⇒ 1 mole of F e S 2 ≡ 2 moles of H 2 S O 4

⇒ (56+2X32) gm of F e S 2 ≡ 2X(2X1+32+4X16) moles of H 2 S O 4 ⇒ (56+2X32) gm of F e S 2 ≡ 2X(2X1+32+4X16) moles of H 2 S O 4

⇒ 120 gm of F e S 2 ≡ 196 gm of H 2 S O 4 ⇒ 120 gm of F e S 2 ≡ 196 gm of H 2 S O 4

Applying unitary method,

⇒ mass of H 2 S O 4 = 192 120 X 100 = 160 g m ⇒ mass of H 2 S O 4 = 192 120 X 100 = 160 g m

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