1.6  Stoichiometry

Stoichiometry is very important in chemistry. It is the section involved with calculations of the amounts of reactants and products. It helps us determine how much reactants will react to give a specific amount of products. Using stoichiometry can help chemists identify ways to increase efficiency of a reaction by reducing the waste of reactants and raw materials.

Before proceeding, lets try to understand how to name compounds and write chemical formulae for them. Also make sure you know the symbols of most of the elements by now, as this will make the chapter a lot easier.

Chemical Formula & Equations

Naming ionic compounds are easy. You first the take name of the metal as it is. Then you slightly change the the name of the non-metal by adding -ide. For example chlorine----chloride, oxygen----oxide, sulfur----sulfide, bromine----bromide, etc.

So NaCl is sodium chloride. LiF would be lithium fluoride. NaBr would be sodium bromide. CaCl2 would be calcium chloride. So if you see calcium sulfide you should know that it is CaS.

Simple covalent molecules are also easy to name. However it gets a bit confusing when naming larger molecules since you have to consider the number of atoms of each element present in the molecules. Some of the molecules also have specific names that are used, for example H2O is water. But with practice it gets easier so lets take a step at a time. The non-metal with the lowest group number is named first and its name is unchanged. The other non-metal name is slightly changed by adding -ide.


So HCl will be hydrogen chloride. Also when there are two or more atoms of the same element present in the molecule, the prefix di, tri, tetra, etc are used. For example CO2 is called carbon dioxide. PCl3 is called phosphorous trichloride. So sulfur hexafluoride should be SF. 


As stated earlier some molecules have specific names that are used. For example CH4 is is not called carbon tetrachloride but instead it is simply referred to as methane. 

In chemical reactions, the number of molecules or atoms of 1 reactant is not always the same as the other reactant. That is to say, 1 molecule or 1 atom of one reactant does not necessarily react with only 1 molecule or 1 atom of the other.This is because of the valence electrons.


For example, Na and Cl. Sodium has one valence electron and Cl has 7 valence electrons. Therefore, chlorine only needs 1 more to get a full shell. So 1 atom of Na will only react with 1 atom of Cl. In this case the ratio of reactants is 1:1.

Now let's take another example, Ca and Cl. In this calcium has 2 valence electrons to give away but Cl needs only one. So 1 atom of Ca will react with 2 atoms of Cl. The ratio is reactants is 2:1.

The number of molecules or atoms each reactant uses to react with the other reactant must be shown in the chemical equation. These numbers are called coefficients. They are written in front of the chemical formula in the chemical equation. These coefficients balance the amount of reactants and products so that the conservation of mass is met. Conservation of mass states that the mass of reactants must be equal to the mass of products since mass can neither be created nor destroyed.

Not a coefficient

When writing chemical equations it is very important to balance the equation before doing any calculations. If not the answers obtained will be wrong and the amount of reactants used will not match the yeild of product. Let's practice balancing some equations.

First look at the reaction between carbon and a oxygen molecule. 1 carbon atom reacts with 1 molecule to form 1 molecule of CO2 as shown below. If you look at the number of atoms in the reactants and number of atoms in the products they should match. 1 atom of carbon in the reactant side and product side. 2 atoms of oxygen in the reactant side and product side. Therefore, this equation is already balanced.


Next look at the reaction for water. 1 molecule of hydrogen should react with 1 molecule of oxygen to give a molecule of water as shown below. But is this correct?


No, its not. If you look closely you'll see the number of atoms of each molecule of the left hand side does not match the right hand side. There are 2 atoms of oxygen on the reactants side but only one atom on the product side. Therefore, this equation needs balancing. We could use fractions to balance or use whole numbers to balance it. For IGCSE I think it would be best to use whole numbers instead of fractions so that the you don't mistakenly split atoms.


As you can see in each case the equations are balanced. In the first equation you can see that 1/2 of an oxygen molecule is 1 atom which now matches the right hand side of the equation. Or in the second equation, to get 2 oxygen atoms on the right hand side you multiply the water molecule by 2. If you multiply it by 2 you get 2 oxygen atoms but you also get 4 hydrogen atoms now which doesn't match the left hand side. Therefore, hydrogen molecule in the left hand side will have to be multiplied by 2 to finally balance the equation.

Look at one final example below. This reaction shows methane reacting with oxygen to form carbon dioxide and water. This is a combustion reaction where methane is burned to make energy. After it's balanced the number of atoms on the left hand side match the right hand side.


When writing chemical equations it is useful the indicate the state of each of the reactants and products. (s) is used to show the reactant or product is solid. (l) shows it's liquid. (g) shows it's in the gas phase. This is shown below:


Now let's move onto calculations.

Amount of Substance

We briefly looked at how to quantify the mass of an atom in the 'Atoms and Elements' section. Atomic mass units are the units used for the mass of an atom. The mass of a proton is 1.667x10^(-24)g which is 0.000000000000000000000001667 g. This is so small and having to do calculations using such minuscule numbers can be time consuming and increase the chances of errors in calculations. But by saying that 1 amu = 1.667x10^(-24)g and using these units instead of grams makes our life a lot easier. 

1 amu = 1.667x10^(-24)g. 1 amu = mass of 1 proton = mass of 1 neutron. The mass of electron is neglected because its only 1/1836 the mass of a proton which means it is incredibly small even compared to a proton.

Ok, so we know what the mass of a proton and neutron is but how do scientists make the connection of 1 amu = 1.667x10^(-24) g .


Well, measuring the mass of all the different types of atoms of each element is not exactly practical because they are minuscule. Comparing them like this is also not very efficient. Therefore, scientists had to set a standard atom. They had to choose one atom that will be used as a reference to compare masses and assign masses to other atoms relatively. The atom that they eventually chose was the carbon-12 atom. It has 6 protons and 6 neutrons (nucleon number 12). Its atomic mass was set to 12 amu which means that each proton and each neutron is 1 amu. Therefore, the mass of all other atoms of all the other elements will be based or calculated of of the standard mass of the carbon-12 atom. So 1 amu is 1/12 th the mass of a carbon-12 atom. 

So to calculate the mass of an atom of an element relative to carbon-12  you can use the following equation:

Atomic mass = (No. of protons + No. of neutrons) x (1/12) x 12
Since you are assigning masses relative to the carbon-12 atoms there are no units.

Hydrogen has only 1 proton; therefore its mass is 1 x (1/12) x 12 = 1

Boron has 5 protons and 5 neutrons; therefore its mass is (5+5) x (1/12) x 12 = 10

But if you look at the periodic table you will see that boron's atomic mass is 10.8 rather than 10. This is because boron has isotopes. It has a lot of  isotopes but the most stable ones are B-10 and B-11. B-10 has 5 protons / 5 neutrons. B11 has 5 protons / 6 neutrons. Each of these have different atomic masses because they have a different number of neutrons. Unfortunately, we can't just choose one of these atoms and take that as the atomic mass of boron because a sample of boron will contain all these isotopes. Therefore, the average mass of all these isotopes will have to be calculated and the answer used as the atomic mass of hydrogen. This average mass is known as the relative atomic mass or A .It has no unitsThe masses shown in the periodic table are the relative atomic masses. 

Let's calculate the relative atomic mass of boron. It is given that for every 1000 atoms 199 are B-10 and 801 are B-11.

We can use the following equation:

Ar = (% abundance of isotope 1 x nucleon number of isotope 1) + (% abundance of isotope 2 x nucleon number of isotope 2) + ..........


Aof boron = (199/1000 x 10) + (801/1000 x 11) = 10.8

So now we know how to calculate and determine the relative atomic mass of  atoms. But what about molecules? How would we calculate the relative molecular mass (Mr)? This is very easy. All you have to do is sum up the relative atomic masses of all the atoms in the molecule.

For example the Mr of CO2 is :

Ar of carbon + Ar of oxygen x 2 = 12 + 16 x 2 = 44

It's pretty straight forward. Relative molecular mass is also known as relative formula mass when dealing with ionic compounds.

Another relation chemists use to make calculations easier is the Avogadro's constant. It states that 12 g of carbon-12 has a 6.022 x 10^23 atoms. This value is  named after the Italian scientist Amedeo Avogadro. Scientists came up with a new unit to define the amount of a substance using the Avogadro's constant. The unit is known as mole and is abbreviated as mol.


So 1 mole of any substance has 6.022 x 10^23 units of that substance.


For example 1 mole of Silver has 6.022 x 10^23 atoms of silver.

1 mole of water (H2O) has 6.022 x 10^23 molecules of water. That is 6.022 x 10^23 atoms of hydrogen and (2 x 6.022 x 10^23) atoms of oxygen.


The reason this makes calculations easier is that the mass of  1 mole of an atom or molecule is equal to the value of Ar or Min grams. For example 1 mole of boron has a mass of 10.8 g. Why?

So  from Avogadro's constant we know that 1 mol of boron has 6.022 x 10^23 atoms. Each atom has a nucleon number (proton + neutron) of 10.8. Therefore 1 mol of boron has (6.022 x 10^23) x 10.8 = 6.5 x 10^24 protons and neutrons combined.


We know that 1 amu = mass of 1 proton = mass of 1 neutron. This means that mass of 1 mol of boron in amu is (6.5 x 10^24) x 1 amu = 6.5 x 10^24 amu.


We also know that 1 amu = 1.667 x 10^-24 g. Therefore, mass in grams of 1 mol of boron = (6.5 x 10^24) x (1.667 x 10^-24)  = 10.8 g. This is called the molar mass.

It is easy to mix up molar mass and relative molecular mass since they have the same value. However, it is important to note that they are different. Mr, relative molecular mass is the mass relative to carbon-12 and therefore has no units. Whereas molar mass is the mass in grams per mole of substance. 

Similarly, 1 mole of CO2, which has a Mr = 12 + 16 x 2 = 44, has a mass of 44g.


Therefore, the mass in grams of 1 mol of the substance depends on the compound. We can convert this concept into the equation shown below.

n = m/M where

n = Number of moles (mol)

m = mass of sample (g)  

M = molar Mass (g/mol)

Molar mass is the mass in grams of 1 mole of the sample.

This equation can be used to calculate the number of moles in a given sample if you know its mass and vice versa.

This formula is very important and I suggest you memorize it because you will use it through out your chemistry education as the basis for most calculations. 

So now if we take the molecule CO2, we know that for every mole of carbon, there are 2 moles of oxygen. So if we take 2 moles of CO2 there are 2 moles of C and 4 moles of O.

Let's assume we have analyzed an unknown compound and discovered that it has 0.2 moles of carbon and 0.2 moles of hydrogen. This is a ratio of 1 : 1; therefore we can conclude that the formula of the compound should be CH. But is this formula correct?  It is correct if we want a formula that shows the simplest ratio of atoms in a molecule. This formula is called the empirical formula. 

But what if the sample that was analyzed had 0.1 moles of the unknown compound. Then to get 0.2 moles of carbon and 0.2 moles of hydrogen from 0.1 moles of the compound the formula would have to be C2H2.  This is the actual formula showing the actual number of atoms or unit ions in the compound.  It is called the molecular formula. So CH is the empirical formula of the compound and C2H2 is the molecular formula.

The molecular formula can be the same as or a multiple of the empirical formula. If we can determine the relative molecular mass (Mr) of the compound we can calculate the molecular formula from the empirical formula. Lets work out an example:

Assume a sample contains 0.24 g of magnesium, 0.32 g of sulfur and 2.56 g of oxygen. Assume it has an actual Mr of 120 g/mol. Lets determine its empirical formula and then calculate the molecular formula. The easiest way to do these calculations is to draw up a table similar to the one shown below. It's best to do them like this until you get enough practice.

emp calc.png

Let me explain the layout of the table. So the Ar row shows the relative atomic masses for magnesium, sulfur and oxygen. 
The mass row shows the masses for all the elements. These values were given in the problem.

The n row shows how to calculate the number of moles of each element.  The equation used was the one shown above.

The ratio is calculated by dividing the number of moles of each element by the smallest number of moles among all the elements. In this case the smallest value is 0.01 mols.

Mr based on the empirical formula was calculated by using the ratio calculated for the empirical formula.

This value will then have to be divided by the actual Mr value of the compound to calculate the factor.

Then the ratio we calculated for the empirical formula will be multiplied by the calculated factor.

This is will give you the actual ratio for the molecular formula. In this case the empirical formula and molecular formula are the same.  

Lets do one more example. Assume a sample has 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its relative molecular mass is 180 g/mol. In this case the values are in percent not in grams. But since it is percent by mass you can assume there is 100 g of the compound and use the percent values as the mass. The calculations are shown in the table below.

emp calc2.png

In this example the number of atoms of each element in the empirical formula increased by factor of 6 to give the molecular formula. This is the molecular formula of a glucose molecule.

Chemical reactions are not a 100% efficient. That is to say a 100 % of the reactants are not converted to product. Some of it lost and wasted in the process.The product obtained is not a 100% pure either. It may contain unreacted reactants and other impurities. Chemists and chemical engineers experiment by changing the reaction conditions to achieve a compromise where a maximum amount of product with maximum purity is made from the reactants. This saves chemical companies a lot of money because the least amount of reactants is wasted. Calculating % yield and % purity is very important.

Yield is the amount of product obtained from the reactants in a chemical reaction. % yeild shows the amount of product that was actually formed relative to the hypothetical amount of product that would have been formed had the reaction been a 100% efficient. The formula is straightforward and is shown below:  


% purity is the amount of pure product relative to the total amount of impure product obtained. The equation is shown below:



Lets recap avagadro's constant for a second. Avagodro's constant is 6.022 x 10^23. This is the number of atoms that 1 mol any type of substance contains. 

But when it comes to calculations concerning gases, Avagadro's law is applied instead. This law states that at a fixed temperature and pressure, equal volumes of gases contain the same amount of molecules. For example 1 liter of oxygen gas has the same amount of molecules that 1 liter of carbon dioxide will have at 30 C and a pressure of 1 atm. 

To make comparisons between calculations easier we can set a standard value for this gas law. We can assume that the volume of 1 mol of any gas at standard room temperature and pressure is 24 dm^3. Standard room temperature is 20 C and standard pressure is 1 atm. 1 dm^3 = 1000 cm^3 = 1 liter. This standard volume is called the molar volume.

We can come up with an equation to define this law. If at standard conditions, 1 mol of gas occupies 24 liters then 2 mols should occupy 48 liters and so on. We can express it in the form shown below:

mol gas.png

Note that molar volume is 24dm^3 only in standard conditions.

When using equations make sure the units are uniform or else your answers will be wrong. Let's do a calculation. 

When carbon reacts with oxygen , carbon dioxide is formed. The chemical equation is shown below:


Assume 5 g of carbon react. How much gas is produced under standard conditions?


The steps are shown below.

First find the number of moles in 5 g of carbon. Then by looking at the chemical equation above you can determine the ratio of carbon to carbon dioxide. Using this ratio you can determine the number of moles of carbon dioxide formed. Finally using Avogadro's law you can calculate the volume at standard conditions.



Let's move onto solutions. When dealing with solutions the important thing is figuring out how much solute has been dissolved in a unit of the solution.


Molar concentration of the solution is the amount of solute in moles that has been dissolved in 1 liter or 1 dm^3 of solution. Its units are (mol/dm^3) or (mol/litre). These units are called molarity and is denoted by M. So a 1 M solution is a solution where 1 mol of solute is dissolved in 1 liter of solution. Or 2 M solution is a solution where 2 mols of solute is dissolved in 1 liter solution.


Difference between concentration and molar concentration or molarity are the units. Concentration can be measured in g/L, Kg/L, mol/l and other units. But molarity is exclusively mol/L. The equation is shown below:


Finally, let's look at titration. Titration is a method used to determine the concentration of an unknown solution.


First a solution, with known concentration is selected to react with the unknown solution. This is called the titrant. The unknown solution is called the analyte. The analyte is laced with an indicator that will change color once all of the unknown solution has reacted with the titrant. The titrant is gradually added until the color changes. Once the color changes the flow of titrant is stopped and volume of titrant used is noted. Using this volume and the concentration of the known solution, the concentration of the unknown solution can be calculated.

Lets look at a neutralization reaction which is one type of the reactions used in titration.

chemix (6).png

A neutralization reaction involves an acid and alkali. Neutralization as the name suggests is a reaction where an acid reacts with an alkali to become neutral.


A specific volume of alkali along with some drops of indicator can be placed in the flask. The indicator will change color when the solution becomes acidic


The burette can be filled up with the acid. Then the acid is added drop by drop while the flask is swirled to make sure the drops of acids are distributed in the solution . Each drop of acid will react with the alkali to become neutral. Eventually all the alkali will react with the acid and neutralize. The moment the alkali is all used up and the next drop of acid makes the solution acidic, the indicator will change color. The flow is immediately stopped and the volume of acid used up is noted. If we know the concentration of either the acid or alkali, the unknown concentration of the solution can be calculated. It is important to note that pure acid or pure alkali is not used because this can be dangerous. Instead aqueous solutions are used.  We will talk about acids and alkali in detail later on. So for now just focus on how to do the concentration calculations similar to the ones shown below.

Let's do a calculation. Assume 25 cm^3 of a 1M solution of sodium hydroxide (alkali) reacts with 30 cm^3 hydrochloric acid. Let's determine the concentration of the acid. 

First we need to write the chemical equation so that we can figure how many moles of acid reacted with the alkali. The equation is shown below:


Calculation is explained below: