Stoichiometry

The quantitative relationship among reactants and products is called stoichiometry. The term stoichiometry is derived from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). On this subject, you often are required to calculate quantities of reactants or products.

Stoichiometry calculations are based on the fact that atoms are conserved. They cannot be destroyed or created. Numbers and kinds of atoms before and after the reactions are always the same. This is the basic law of nature.

From the atomic and molecular point of view, the stoichiometry in a chemical reaction is very simple. However, atoms of different elements and molecules of different substances have different weights. Thus, simple stoichiometry at the atomic level appears to be complicated when amounts (quantities) are measured in units of g, kg, L or mL. When quantities in moles are used, the relationships (or ratios) are really simple. For example, one mole of oxygen reacts with two moles of hydrogen,
2 H2 + O2 -> 2 H2O
or one mole of hydrogen reacts with half a mole of oxygen,
H2 + ½ O2 -> H2O
one mole of carbon reacts with one mole of oxygen.
C + O2 -> CO2

This is a major and important topic that you have to master. In order to accomplish this, you have to be able to do several things. First, you have to be able to convert amounts of substances between mass units of g (or kg) to moles and vice versa. Then, you have to understand chemical reactions (changes). In this case, you not only know what are the reactants and products, you can write a balanced equation to explain the reaction. Sometimes you may be told what the reactions are.

There are many chemical reactions, but they can be divided into a few types as a summary.

In a chemical reaction, not all reactants are necessarily consumed. One of the reactants may be in excess and the other may be limited. The reactant that is completely consumed is called limiting reactant, whereas unreacted reactants are called excess reactants.

Amounts of substances produced are called yields. The amounts calculated according to stoichiometry are called theoretical yields whereas the actual amounts are called actual yields. The actual yields are often expressed in percentage, and they are often called percent yields.

Problems  

1.    Mole-Mole Problems
Problem:  How many moles of HCl are needed to react with 0.87 moles of Al?

        Step 1: Balance The Equation & Calculate the Ratios


            2Al:6HCl (1:3)        2Al:2AlCl₃ (1:1)        2Al:3H₂ (1:1.5)
        Step 2:  Find the Moles of the Given             0.87 moles of aluminum are reacted with hydrochloric acid
          Step 3: Calculate the moles using the ratios            moles HCl = 0.87molAl x  3molHCl/1molAl = 2.6 mol HCl
           

2.    Mass-Mass Problems (Strategy: Mass g Mole g Mole g Mass)
Problem:  How many grams of Al can be created decomposing 9.8g of  Al₂O₃?

        Step 1: Balance The Equation & Calculate the Ratios  

            2Al₂O₃:4Al (1:2)        2Al₂O₃:3O₂ (1:1.5)

Step 2:  Find the Mass of the Given

            9.8g Al₂O₃ are decomposed

Step 3: Calculate the moles of the given (mol/g)

            9.8g Al₂O₃ x (1mol Al₂O₃/102g Al₂O₃) = 0.096 mol Al₂O₃
       

Step 4: Calculate the moles using the ratios

        0.096 mol Al₂O₃ x (2 mol Al/1 mol Al₂O₃) = 0.19 mol Al
      
        Step 5: Calculate the mass using the new moles
        0.19 mol Al x (27g Al/1 mol Al) = 5.1g Al
       

3.    Mass-Volume Problems (Strategy: Mass g Mole g Mole g Volume)
        Problem:  How many liters of H2 are created from the reaction of 20.0g K?

        Step 1: Balance The Equation & Calculate the Ratios



            2K:2H₂O (1:1)    2K:2KOH (1:1)    2K:1H₂ (2:1)

Step 2:  Find the Mass of the Given

            20.0g K are used in the reaction

Step 3: Calculate the moles of the given (mol/g)

            20.0g K x (1 mol K / 39g K) = 0.513 mol K
           

Step 4: Calculate the moles using the ratios

            0.51 mol K x (1mol H₂ /2mol K) = 0.266mol H₂
           
        Step 5: Calculate the volume using the new moles
        0.266 mol H₂ x (22.4L H₂ /1mol H₂) = 5.75L H₂
       

4.    Volume-Volume Problems
        Problem:  How many liters of SO₂ will be produced from 26.9L O₂?

Step 1: Balance The Equation & Calculate the Ratios

            2O₂:1S₂ (2:1)    2O₂:2SO₂ (1:1)

Step 2: Find the volume of the given

            26.9L O₂

Step 3: Calculate the moles of the given

            26.9L O₂ x (1 mol O₂ / 22.4L) = 1.20 mol O₂
       

Step 4: Calculate the moles using the ratios

            1.20 mol O₂ x (1mol SO₂ /1mol  O₂) = 1.20 mol SO₂
       
        Step 5: Calculate the volume using the new moles
        1.20 mol O₂ x (1mol SO₂ /1mol  O₂) x (22.4L /1mol) = 26.9L SO₂