Why do we do stoichiometry? If we want to do any quantitative chemistry (that is, involving numbers and stuff), we need to know how much of a reactant a reaction uses and how much of a product is formed. Stoichiometry is the foundation of all mathematical calculations regarding chemistry and chemical equations.
Step 1: Balance equation and find out mole ratio
The first step to solving stoichiometry is to balance the equation which is given in the question. In order to balance the equation, one must first look at all the elements present in the equation, and balance using the following principles below.
Principles of Balancing Equations
- There must be the same number of elements on each side. If there are 2 oxygen atoms in the reactants, there must be 2 oxygen atoms in the products.
- For example: $N_2 (g) +3H_2 (g) \rightarrow 2NH_3 (g)$
- There are equal number of atoms of each element on both sides of the equation
- The charge on both sides on any equation must be the same.
- For example: $Cu^{2+} (aq) + Zn (s) \rightarrow Cu (s) + Zn^{2+}$
- Here, both sides have an overall charge of 2+
- There should not be any of the same species on both sides
Now the next step is to find out the mole ratio. What is the mole ratio? It is a chemical ratio of any two (or more) species within a chemical equation. For example, using one of the equations above,
$N_2 (g) +3H_2 (g) \rightarrow 2NH_3 (g)$ It takes 1 nitrogen gas molecule to form 3 ammonia molecules. 1 nitrogen gas molecule reacts with 3 hydrogen gas molecules and so on.
Step 2: Convert the quantities you have into moles
You may be given quantities in different ways depending on the examiner’s mood. Sometimes they’ll give you the mass, sometimes they’ll give you the volume (especially of a gas), the concentration of a solution, or if they’re feeling nice you’ll be given the number of moles in which case you can happily skip this step.
If the quantity is given is given in:
Concentration and volume
Multiply concentration and volume. If the concentration is given in $mol \ dm^{-3)$, then you will get the number of moles. If the concentration is given in $g dm^{-3}$, you will get the mass.
Watch the units though!
Mass
Divide the mass by molar mass. (Since the molar mass is the mass per mole of a substance) The molar mass is calculated by the sum of the atomic masses of all the atoms in the molecule. For example, the molar mass of $CO_2$ is 12.01 (Ar of carbon)+ 2x16.00 (Ar of oxygen) = 42.01
Gas Volume
To convert gas volume into moles, the total volume must be divided by the molar gas constant for the given condition. Put simply, the molar gas constant is calculated from the ideal gas law $PV=nRT$, and it is the volume which one mole of any gas occupies under standard conditions. There are two standard conditions, one for standard pressure and temperature (STP) and standard atmospheric temperature and pressure (SATP). STP conditions are 273K (0$^{\circ}$C) and 100kPa and the constant is 22.7 $dm^3$ (L). SATP conditions are 298 (25$^{\circ}$C) and 100 kPa and the constant is 24.8 $dm^3$ (L). The value of 22.7 $dm^3$ is given in the IB data booklet, and 24.8 $dm^3$ is usually memorised. It can also be derived from the ideal gas law.
Step 3: If there are quantities of more than reagent, find the limiting reagent
If you have the moles of only one reactant, you can skip this step :)
The limiting reagent is the one that will be used up first in a reaction. Hence, the limiting reagent determines how much of the reaction will occur. (assuming the reaction goes to completion) For example, a question gives that 1 mole of carbon is burnt in 2 moles of oxygen gas. Assuming complete combustion, the chemical equation is:
$C(s)+O_2(g) \rightarrow CO_2(g)$
Let’s calculate the required amount of oxygen gas using the amount of carbon. Given that the mole ratio of carbon and oxygen gas is 1:1, 1 mole of carbon will require only 1 mole of oxygen for combustion. Hence, there will be excess oxygen and carbon is the limiting reagent.
Step 4: Using the mole ratio, find the number of moles of the species required
Once the number of moles which has been used up or produced in the reaction has been calculated, this can then be converted into the moles of the species required. This is done by using the mole ratio. Using the reaction of the production of ammonia,
$N_2 (g) +3H_2 (g) \rightarrow 2NH_3 (g)$
Let’s say 4 moles of $NH_3$ is produced and the question asks for the number of moles of hydrogen gas required. For every 2 moles of ammonia, 3 moles of hydrogen gas is required. Hence, 4 moles of ammonia will require 6 moles of hydrogen gas.
Step 5: Convert the moles of the species into the quantity required
This is similar to step 3, but in reverse. For mass and gas volume, you can multiply it by the molar mass of the substance or 22.7 respectively (assuming STP). For concentration, simply divide the number of moles by the volume in $dm^3$.
Worked example
2.431g of magnesium was added to 100ml of 3$mol \ dm^{-3}$ HCl. Determine the volume of hydrogen gas liberated and find the concentration of the $Cl^-$ in the resultant solution. Assume that the reaction occurs at STP and at constant pressure.
Answer:
The balanced chemical equation is:
$Mg(s)+ 2HCl(aq) \rightarrow H_2 (g) + MgCl_2 (aq)$
Since the Ar of Magnesium is 24.31g $mol^{-1}$, to calculate the number of moles of Magnesium,
Moles of magnesium added $=\frac{2.431}{24.31}=0.1$ moles
To calculate the moles of HCl, multiply the volume (in $dm^3$) by the concentration
$100ml = \frac{100}{1000}dm^3 = 0.1 dm^3$
Actual Moles of HCl present in solution $0.1\times3=0.3$ moles
Since the mole ratio of Magnesium to HCl is 1:2, 0.1 moles of magnesium would require 0.2 moles of HCl to react fully. Since there are 0.3 moles of HCl present in solution, HCl is in excess and hence Magnesium is the limiting reagent
The mole ratio of magnesium to hydrogen gas is 1:1. Hence, the number of moles of hydrogen gas produced is 0.1mol.
To convert it into volume,
$Volume(H_2)=0.1\times22.7$
$= 2.27dm^3$
Hence, the amount of hydrogen gas produced is 2.27$dm^3$
Extension - assumes knowledge of ionic equations
The ionic equation is:
$Mg(s)+ 2H^+(aq) \rightarrow H_2 (g) + Mg^{2+} (aq)$
Since chloride is a spectator ion and does not participate in the reaction, its concentration does not change. Hence, the concentration of chloride does not change. Hence, $[Cl^-]=3\ mol \ dm^{-3}$
Author: lmaozedong
