STOICHIOMETRY

           In our day to day life, it’s child’s play for us to ‘count’ things or ‘weigh’ things. For example, you can easily count and tell me that there are 4 bananas or there are 6 apples kept on the table. Shopkeepers can easily weigh things on a weighing balance and tell us ‘this packet of rice weighs 2 pounds’ or ‘this packet of tea weighs 500 g’. But things change when you start dealing with atoms and molecules. You cannot simply count or weigh atoms or molecules, as you can do with the common items that you encounter in your daily life. Here it gets a little tricky and something termed as ‘Stoichiometry’ kicks in.
How do you define stoichiometry?
Stoichiometry is the branch of chemistry that deals with the relationship between the relative quantities of substances taking part in a chemical reaction
         Let’s write a general chemical reaction where there are two reactants, A and B that react together to form two products, C and D respectively
A + B → C + D
The concept of stoichiometry will help us answer questions like
1. if ‘x’ grams of A is present, then how many grams of C or D or both will be produced
2. if ‘y’ grams of B is present, then how many grams of C or D or both will be produced
3. if ‘x’ grams of A is present and ‘y’ grams of B is present, which is the limiting reactant and which is the reactant present in excess
4. if ‘x’ grams of A react with ‘y’ grams of B, how much of B will remain unreacted (or unconsumed) after the reaction is over
5. how many grams of A will produce ‘z’ grams of C
The concept of stoichiometry can be best understood by solving real problems involving chemical reactions, reactants and products
Just two things to remember before we get started:
1. the first step in any stoichiometric problem is to always ensure that the chemical reaction you are dealing with is balanced,
2. clarity of the concept of a ‘mole’ and the relationship between ‘amount (grams)’ and ‘moles’.
‘Mole concept’ in brief
               Let’s write a simple balanced chemical reaction, where hydrogen gas reacts with oxygen gas to produce water.
\redD{2}2start color redD, 2, end color redDHHH_22start subscript, 2, end subscript + OOO_22start subscript, 2, end subscript → \redD{2}2start color redD, 2, end color redDHHH_22start subscript, 2, end subscriptOOO
              In the above reaction the prefixes (shown in red) refer to the moles of the molecules involved in the chemical reaction. These are also referred to as coefficients. Here, two moles of hydrogen gas are reacting with one mole of oxygen gas to produce two moles of water. ‘Mole’ simply refers to an enormously big number which is 6.023 X 10^{23}23start superscript, 23, end superscript.
Let’s see how a particular element is represented in a periodic table. Let’s pick chlorine (Cl).

Problem 1
If 138.6 g of KClO_33start subscript, 3, end subscript is heated and decomposes completely, how many grams of oxygen gas would be produced?
KClO3→KCl+O2

Step 1: Ensure that the reaction is balanced
In this case the reaction is not balanced. After balancing, the equation looks like
2KClO3→2KCl+3O2

Step 2: Convert grams of KClO_33start subscript, 3, end subscript to moles
For this we first need to calculate the molar mass of KClO_33start subscript, 3, end subscript = 1 (atomic mass of K) + 1 (atomic mass of Cl) + 3 (atomic mass of O) = 1 (39.098) + 1 (35.453) + 3 (15.999) = 39.098 + 35.453 + 47.997 = 122.548 g
i.e. 122.548 g of KClO_33start subscript, 3, end subscript = 1 mole of KClO_33start subscript, 3, end subscript
therefore, 1 g of KClO_33start subscript, 3, end subscript = (1 mole/ 122.548 g) x 1 g
so, 138.6 g of KClO_33start subscript, 3, end subscript = (1 mole/ 122.548 g) x 138.6 g = 1.130 moles

Step 3: Use mole ratios (from the balanced chemical reaction) to calculate the moles of the relevant reactant or product
In this example, we need to calculate the mass of O_22start subscript, 2, end subscript produced. So we first need to calculate the moles of O_22start subscript, 2, end subscript that would be produced from 1.130 moles of KClO_33start subscript, 3, end subscript
               Looking at the balanced chemical reaction, we can infer that 2 moles of KClO_33start subscript, 3, end subscript produce 3 moles of O_22start subscript, 2, end subscript
So, 1.130 moles of KClO_33start subscript, 3, end subscript will produce [(3 moles of O_22start subscript, 2, end subscript/ 2 moles of KClO_33start subscript, 3, end subscript) x 1.130 moles of KClO_33start subscript, 3, end subscript] = 1.695 moles of O_22start subscript, 2, end subscript.
Step 4: Convert moles back to ‘grams’ using molar mass of O_22start subscript, 2, end subscript
             Molar mass of O_22start subscript, 2, end subscript = 2 (atomic mass of O) = 2 (15.999) = 31.998 g i.e. 1 mole of O_22start subscript, 2, end subscript = 31.998 g of O_22start subscript, 2, end subscript therefore, 1.695 moles of O_22start subscript, 2, end subscript = 31.998 g/mol x 1.695 mol = 54.236 g
Answer: If 138.6 g of KClO_33start subscript, 3, end subscript is heated and decomposes completely, 54.236 g of oxygen gas will be produced

Problem 2
How many grams of iodine must react to give 4.63 grams of ferric iodide?
2Fe+3I2→2FeI3

Step 1: Ensure that the reaction is balanced
The above reaction is balanced!!!

Step 2: Convert grams of FeI_33start subscript, 3, end subscript to moles
For this we first need to calculate the molar mass of FeI_33start subscript, 3, end subscript = 1 (atomic mass of Fe) + 3 (atomic mass of I) = 1 (55.845) + 3 (126.904) = 55.845 + 380.712 = 436.557 g
i.e. 436.557 g of FeI_33start subscript, 3, end subscript = 1 mole of FeI_33start subscript, 3, end subscript
therefore, 1 g of FeI_33start subscript, 3, end subscript = (1 mole/ 436.557 g) x 1 g
so, 4.63 g of FeI_33start subscript, 3, end subscript = (1 mole/ 436.557 g) x 4.63 g = 0.0106 moles

Step 3: Use mole ratios (from the balanced chemical reaction) to calculate the moles of the relevant reactant or product
In this example, we need to calculate the mass of I_22start subscript, 2, end subscript that must react. So we first need to calculate the moles of I_22start subscript, 2, end subscript that would be required to produce 0.0106 moles of FeI_33start subscript, 3, end subscript
Looking at the balanced chemical reaction, we can infer that 3 moles of I_22start subscript, 2, end subscriptare required to produce 2 moles of FeI_33start subscript, 3, end subscript
             So, 0.0106 moles of FeI_33start subscript, 3, end subscript will require [(3 moles of I_22start subscript, 2, end subscript / 2 moles of FeI_33start subscript, 3, end subscript) x 0.0106 moles of FeI_33start subscript, 3, end subscript] = 0.0159 moles of I_22start subscript, 2, end subscript

Step 4: Convert moles back to ‘grams’ using molar mass of I_22start subscript, 2, end subscript
Molar mass of I_22start subscript, 2, end subscript = 2 (atomic mass of I) = 2 (126.904) = 253.808 g
i.e. 1 mole of I_22start subscript, 2, end subscript = 253.808 g of I_22start subscript, 2, end subscript
therefore, 0.0159 moles of I_22start subscript, 2, end subscript = (253.808 x 0.0159) g = 4.035 g
Answer: 3.807 grams of iodine must react to give 4.035 grams of ferric iodide

Problem 3
How many grams of H_22start subscript, 2, end subscriptO will be produced when you burn 25 grams of methane?
CH4+O2→CO2+H2O

Step 1: Ensure that the reaction is balanced
In this case the reaction is not balanced. After balancing, the equation looks like
CH4+2O2→CO2+2H2O

Step 2: Convert grams of CH_44start subscript, 4, end subscript to moles
For this we first need to calculate the molar mass of CH_44start subscript, 4, end subscript = 1 (atomic mass of C) + 4 (atomic mass of H) = 1 (12.011) + 4 (1.008) = 12.011 + 4.032 = 16.043 g
i.e. 16.043 g of CH_44start subscript, 4, end subscript = 1 mole of CH_44start subscript, 4, end subscript
therefore, 1 g of CH_44start subscript, 4, end subscript = (1 mole/ 16.043 g) x 1 g
so, 25 g of CH_44start subscript, 4, end subscript = (1 mole/ 16.043 g) x 25 g = 1.558 moles

Step 3: Use mole ratios (from balanced chemical reaction) to calculate the moles of the relevant reactant or product
In this example, we need to calculate the mass of H_22start subscript, 2, end subscriptO produced. So we first need to calculate the moles of H_22start subscript, 2, end subscriptO that would be produced from 1.558 moles of CH_44start subscript, 4, end subscript
Looking at the balanced chemical reaction, we can infer that 1 mole of CH_44start subscript, 4, end subscriptproduces 2 moles of H_22start subscript, 2, end subscriptO
So, 1.558 moles of CH_44start subscript, 4, end subscript will produce [(2 moles of H_22start subscript, 2, end subscriptO / 1 mole of CH_44start subscript, 4, end subscript ) x 1.558 moles of CH_44start subscript, 4, end subscript] = 3.116 moles of H_22start subscript, 2, end subscriptO
Step 4: Convert moles back to ‘grams’ using molar mass of H_22start subscript, 2, end subscriptO
Molar mass of H_22start subscript, 2, end subscriptO = 2 (atomic mass of H) + 1 (atomic mass of O) = 2 (1.008) + 1 (15.999) = 18.015 g
i.e. 1 mole of H_22start subscript, 2, end subscriptO = 18.015 g of H_22start subscript, 2, end subscriptO
therefore, 3.116 moles of H_22start subscript, 2, end subscriptO = (18.015 x 3.116) g = 56.134 g
Answer: 56.134 g of H_22start subscript, 2, end subscriptO will be produced when we burn 25 grams of methane
After going through the above problems, we should now have a fair idea of how to calculate the amount of a reactant used or the amount of a product formed in a chemical reaction from a given amount of any one of the reactants or products.
Concept of ‘limiting reactant’ and ‘excess reactant’
              Limiting Reactant is defined as the reactant in a chemical reaction that limits the amount of product that is formed. The reaction will stop when all of the limiting reactant is consumed.
Excess Reactant is defined as the reactant in a chemical reaction that is in excess and remains unconsumed when a reaction stops because the limiting reactant has been completely consumed.
          Let’s understand this concept with an example. Suppose there are 4 kids and 9 pairs of flip-flops. Each kid will wear a pair of flip-flops, leaving behind 5 pairs of flip-flops.
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            No matter how many pairs of flip-flops there are, only 4 pairs of flip-flops can be worn because there are only 4 kids. Likewise in chemical reaction, when the limiting reagent (kid) is used up, the reaction will stop and the unconsumed excess reagent (pairs of flip-flops) will remain in the reaction mixture.
Now the key task is to be able to identify which reactant is the limiting reactant in a given chemical reaction.
          The easiest way to identify which reactant is the ‘limiting reactant’ and which one is the ‘excess reagent’ is by taking the number of moles of each reactant and dividing it by its molar coefficient, as indicated in the balanced chemical reaction. The substance that has the lowest value is the limiting reagent.
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             As illustrated above, in an ideal world 36 popcorns (product) should pop out of 36 seeds (reactant). In this case, ‘36’ is referred to as the ‘theoretical yield’. But actually, only 27 popcorns pop. Here, ‘27’ is referred to as the ‘actual yield. The percent yield of this popcorn making process would be defined as ( \dfrac {actual yield}{theoretical yield}theoreticalyieldactualyieldstart fraction, a, c, t, u, a, l, y, i, e, l, d, divided by, t, h, e, o, r, e, t, i, c, a, l, y, i, e, l, d, end fraction x 100) = 27/ 36 x 100 = 75%. In other words, the popcorn making process was only 75% efficient.
             The above analogy can be applied to any chemical reaction. As we have learned in the first part of this tutorial, the amount of product formed can be theoretically calculated from the given amount of the limiting reagent and the mole ratios of the reactants and products. This is called the theoretical yield of a reaction.
              However, the amount of product actually formed in a reaction is usually less than the theoretical yield and is referred to as the actual yield. This is because reactions often have "side reactions" that also compete for reactants and produce undesired products. To evaluate the efficiency of the reaction, chemists compare the actual and theoretical yields by calculating the percent yield of a reaction:
Percent yield of a reaction = \dfrac {actual yield}{theoretical yield}theoreticalyieldactualyieldstart fraction, a, c, t, u, a, l, y, i, e, l, d, divided by, t, h, e, o, r, e, t, i, c, a, l, y, i, e, l, d, end fraction x 100
Let’s attempt a problem now.

Problem 5
Let’s take the example of preparation of nitrobenzene (C_66start subscript, 6, end subscriptH_55start subscript, 5, end subscriptNO_22start subscript, 2, end subscript), starting with 35.15 g of benzene (C_66start subscript, 6, end subscriptH_66start subscript, 6, end subscript) and an excess of nitric acid (HNO_33start subscript, 3, end subscript). The balanced chemical reaction for this is
C_6H_6 + HNO_3 \rightarrow C_6H_5NO_2 + H_2OC6H6+HNO3→C6H5NO2+H2OC, start subscript, 6, end subscript, H, start subscript, 6, end subscript, plus, H, N, O, start subscript, 3, end subscript, right arrow, C, start subscript, 6, end subscript, H, start subscript, 5, end subscript, N, O, start subscript, 2, end subscript, plus, H, start subscript, 2, end subscript, O
Let’s first calculate the theoretical yield of C_66start subscript, 6, end subscriptH_55start subscript, 5, end subscriptNO_22start subscript, 2, end subscript

Step 1: Convert grams of benzene to moles
Molar mass of C_66start subscript, 6, end subscriptH_66start subscript, 6, end subscript = 6 (atomic mass of C) + 6 (atomic mass of H) = 6 (12.011) + 6 (1.008) = 72.066 + 6.048 = 78.114 g
i.e. 78.114 g of C_66start subscript, 6, end subscriptH_66start subscript, 6, end subscript = 1 mole of C_66start subscript, 6, end subscriptH_66start subscript, 6, end subscript
therefore, 1 g of C_66start subscript, 6, end subscriptH_66start subscript, 6, end subscript = (\dfrac {mole} {78.114 g}78.114gmolestart fraction, m, o, l, e, divided by, 78, point, 114, g, end fraction) x 1 g
so, 35.15 g of C_66start subscript, 6, end subscriptH_66start subscript, 6, end subscript = (\dfrac {1 mole}{ 78.114 g}78.114g1molestart fraction, 1, m, o, l, e, divided by, 78, point, 114, g, end fraction) x 35.15 g = 0.449 moles