Chemical Stoichiometry refers to the quantitative study of the reactants and products involved in a chemical reaction. The word “ stoichiometry” is derived from the Greek word “stoikhein” meaning element and “metron” meaning measure.
The term Stoichiometry was first coined or discovered by a German chemist named Jeremias Richter. Even though this tongue-twisting word can sound complicated and big but it is a simple idea. In this lesson, we will learn about what it means and discuss the different aspects of this concept.
What Is Stoichiometry?
In simple words, we can define,
Stoichiometry as the calculation of products and reactants in a chemical reaction. It is basically concerned with numbers.
Stoichiometry is an important concept in chemistry that helps us use balanced chemical equations to calculate amounts of reactants and products. Here, we make use of ratios from the balanced equation. In general, all the reactions that take place are dependent on one main factor, how much substance is present.
Stoichiometry helps us determine how much substance is needed or is present. Things that can be measured are;
- Reactants’ and Products mass
- Molecular weight
- Chemical equations
Stoichiometric coefficient or stoichiometric number is the number of molecules that participate in the reaction. If you look at any balanced reaction you will notice that there are an equal number of elements on both sides of the equation. The stoichiometric coefficient is basically the number present in front of atoms, molecules or ions.
Stoichiometric coefficients can be fractions as well as whole numbers. In essence, the coefficients help us to establish the mole ratio between reactants and products.
Balanced Reactions and Mole Ratios
Atoms and molecules are extremely small in size and their numbers in a very small amount of a substance are very large. Therefore, to represent atoms and molecules in bulk, a mole concept was introduced. One mole of any substance contains 6.022 x 1023 numbers of that substance. This number is also known as Avogadro’s number.
The mass of one mole of a substance in grams is called molar mass. The molar mass of one mole of a substance is numerically equal to the atomic/molecular formula mass.
Let us take one example of a balanced chemical equation.
3Fe(s) + 4H2O(l) ⇾ Fe3O4 (s)+ 4H2 (g)
The quantitative information drawn from this balanced chemical equation is
- 3 mole of Fe reacts with 4 moles of H2O to yield one mole of Fe3O4 and 4 moles of H2.
- 168g ( 56×3) of Fe reacts with 72g( 18×4) of H20 to yield 231g of Fe3O4 and 8g of H2 gas.
If the reactants and products are in gaseous form, then the molar volume is taken into consideration. One mole of any gas occupies 22.4 Litres.
CH4(g) + 2O2(g)⇾ CO2(g)+ 2H20 (g)
In the above reaction, 22.4 Litres of CH4 reacts with 44.8 (2 x 22.4) litres of 02 to yield 22.4 Litres of CO2 and 44.8 litres of H2O
In a chemical reaction, it is possible that one of the reactants is present in excess amount. Some of these excess reactants will, therefore, be left over when the reaction is complete; the reaction stops immediately as soon as one of the reactants is totally consumed.
The substance that is totally consumed in a reaction is called the limiting reagent.
Let us take one example of a chemical reaction to understand limiting reagent concept.
N2 + 3H2 ➝ 2NH3
Suppose we have one mole of N2 reacting with one mole of H2. But from the balanced chemical equation, one mole of N2 requires three moles of H2.So, the limiting reagent in this reaction is H2.
Stoichiometry in Chemical Analysis
Stoichiometric calculations which follow a quantitative analysis methodology is often used by chemists to determine the concentrations of substances present in a sample. There are basically two main types of analysis. We will discuss them below.
1. Gravimetric Analysis
In analytical chemistry, the gravimetric analysis describes the quantitative determination of analyte based on the mass of the solid. The gravimetric analysis gives the most accurate results out of all other analytical analysis as the weight of a substance can be measured with great accuracy compared to other fundamental quantities.
Gravimetric analysis can be generalized into the following types.
- Precipitation gravimetry – It involves isolation of ion in solution by a precipitation reaction, filtering, washing the precipitate free of contaminants, and finally weighing the precipitate and determining its mass by difference.
- Volatilization gravimetry – volatilization gravimetry involves separating components of a mixture by heating or chemically decomposing the sample.
- Electrogravimetry – It involves the electrochemical reduction of metal ions at the cathode and simultaneous deposition of ions on the cathode. The cathode is weighed before and after electrolysis and the weight, difference corresponds to the mass of analyte initially present in the sample.
2. Volumetric Analysis
The volumetric analysis involves the quantitative measurement of substance in terms of volume.
Principle: In volumetric analysis, a known volume (V1) of the substance whose concentration (N1) is known is reacted with the unknown volume (V2) of the solution of the substance whose concentration(N2) is to be calculated. The volume, V1 is noted at the endpoint of the reaction. The concentration N2 is calculated using the following equation.
N1x V1 = N2 x V2
The endpoint of such reaction is indicated by a change in colour or precipitation etc.
Terms involved in volumetric analysis;
- Titration – The process of finding out of the volume of solution required to react completely with the volume of another solution is called titration.
- Titrant – The solution of known strength is called titrant.
- Titrate – The solution whose concentration is to be estimated.
- Indicator – Indicators are reagents which change their colour when the reaction is complete.
Stoichiometry Problems With Solutions
1. Calculate the mass of sodium hydroxide required to make 500ml of 0.10 M solution.
The molar mass of NaOH = 40g
Volume of NaOH= 500ml = 0.5 L
Molarity = 0.10M
Molarity = moles / volume in litres
⇒ weight of NaOH = molarity x molar mass of NaOH x volume
= 0.10 x 40 x 0.5
2. How much volume of 11M HCl has to be diluted with water to prepare 3 M 400 ml HCl?
M1 = 11M
M2 = 3M
V1 = ?
Now, M1 x V1= M2 x V2
⇒V1= (3×400)/ 11
= 109 ml
3. How many carbon atoms are present in 0.5 moles of oxalic acid (C2H2O4)?
1 mole of oxalic acid = 6.022 x 10 23 number of oxalic acid
0.5 mole of oxalic acid = 6.022 x 10 23 x 0.5 number of oxalic acid
Since there are 2 carbon per oxalic acid,
Therefore number of carbon atoms in 0.5 mole of oxalic acid = 6.022 x 10 23 x 0.5 x 2
= 6.022 x 10 23
4. 0.5216g of a solid mixture containing Na2SO4 is dissolved in water and treated with an excess of BaCl2, resulting in the precipitation of 0.6168g of BaSO4. What percentage of the mixture was BaS04?
Na2SO4 + BaCl2 → BaSO4 + 2NaCl
233g of BaSO4 is obtained from 142g of Na2SO4
So, 0.6168g of BaSO4 is obtained from = (142×0.6168) / 233
Since the mass of solid mixture is 0.5216g
Therefore, the percentage of BaSO4 is solid mixture = (0.37/0.5216) x 100
5. A solution containing 5g of KOH and Ca(OH)2 is neutralized by an acid. If it consumes 0.3g equivalents of the acid, Calculate the composition of the solution.
Let mass of KOH present in mixture = x
Mass of Ca(OH)2 = (5-x)g
Eq. mass of KOH = 56; Eq. mass of Ca(OH)2 = 74/2
Gram equivalent of KOH + Gram equivalent of Ca(OH)2 = gram equivalent of acid
+ = 0.1
⇒ x = 3.83g
Mass of KOH in the sample = 3.83g
Percentage of KOH = (3.83/5) x 100
Percentage of Ca(OH)2 = x100