0.1 Analyzing chemical equations (Page 15/3)

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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 :

$x A + y B \to A_{x} B_{y}$

Here, 1 mole of compound ( $A_{x} B_{y}$ ) involves x mole of A and y mole of B. Using symbols :

$x moles of A \equiv y moles of B \equiv 1 mole of A_{x} B_{y}$

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 :

$x M_{A} gm of A \equiv y M_{B} gm of B \equiv M_{A_{x} B_{y}} gm of A_{x} B_{y}$

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

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

Solution : Purity of CaCO3 is 90 %. Hence,

$Mass of C a C O_{3} = 0.9 X 500 = 450 k g$

The chemical reaction involved here is :

$C a C O_{3} \to C a O + C O_{2}$

Applying mole concept :

$\Rightarrow 1 mole of C a C O_{3} \equiv 1 mole of CaO$

$\Rightarrow (40 + 12 + 3X 16) gm of C a C O_{3} \equiv (40 + 16) gm of CaO$

$\Rightarrow 100 gm of C a C O_{3} \equiv 56 gm of CaO$

$\Rightarrow 100 kg of C a C O_{3} \equiv 56 kg of CaO$

Applying unitary method :

$\Rightarrow mass of CaO produced = \frac{56 X 450}{100} = 252 k g$

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

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. $S i O_{2}$ weighing 15 gm does not change. On the other hand, 80 gm of $M n O_{2}$ converts as :

$3 M n O_{2} \to M n_{3} O_{4} + O_{2}$

Applying mole concept,

$3 moles of M n O_{2} \equiv 1 mole of M n_{3} O_{4}$

$3 X (55 + 2 X 16) gm of M n O_{2} \equiv 1 X (3 X 55 + 4 X 16) gm of M n_{3} O_{4}$

$261 gm of M n O_{2} \equiv 229 gm of M n_{3} O_{4}$

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

$Mass of M n_{3} O_{4} on conversion = \frac{229}{261} X 80 = 70.2 g m$

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 $M n_{3} O_{4}$ and 15 gm of $S i O_{2}$ . Total mass of ignited sample is 70.2+15 = 85.4 gm. On the other hand, amount of Mn in $M n_{3} O_{4}$ is calculated from its molecular constitution :

$229 gm of M n_{3} O_{4} \equiv 3 X 55 gm of Mn \equiv 165 gm of Mn$

$Amount of Mn in 70.2 gm of M n_{3} O_{4} = \frac{165}{229} X 70.2 = 0.72 X 70.2 = 50.54 g m$

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

$Percentage amount of manganese in the ignited sample = \frac{50.54}{85.4} X 100 = 59.2 %$

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Source: OpenStax, Stoichiometry. OpenStax CNX. Jul 05, 2008 Download for free at http://cnx.org/content/col10540/1.7

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