# CHSE Odisha Chemistry – Stoichiometry

# Stoichiometry

This word is derived from 2 Greek words **Stoicheion** and **Metron**. **“Stoicheion”** means element and **“Metron”** means to measure

Stoichiometry calculations deal with the mass or volumes of products and reactants. Or you can say, Stoichiometry is the calculation of reactants and products in chemical reactions.

#### Definition

Stoichiometry is the study of the quantitative relationships or ratios between two or more substances undergoing a physical or chemical change (chemical reaction).

It is a branch of chemistry that deals with the application of two major laws of chemical reaction.

1. **Conservation of mass principle**: This is founded on the law of conservation of mass where the total mass of the reactants is always equals to the total mass of the products.

2. **The laws of definite proportions**: The relations among quantities of reactants and products typically form a ratio of positive integers.

### What are the Stoichiometric Calculations?

Calculations based on chemical equations are known as stoichiometric calculations. Following factors may be taken in to account while doing these calculations.

- Molecules
- Molecular Mass
- Mole Concept
- Gram Molar Volume etc.

**Example**: Let’s take the balanced equation of the combustion of Ethane (C2H6) and understand it from various perspectives like the Molar concept, Molecular mass & Molar volume.

**2C**_{2}H_{6(g)} + 7O_{2(g)} = 4CO_{2(g)} + 6H_{2}O_{(g)}

_{2}H

_{6(g)}+ 7O

_{2(g)}= 4CO

_{2(g)}+ 6H

_{2}O

_{(g)}

1. **According to Molar Concept**: 2 moles of Ethane reacts with 7 moles of oxygen to produce 4 moles of Carbon-dioxide and 6 moles of Water vapor.

2. **According to Molecular Mass**: 60 grams of Ethane can react with 224 grams of Oxygen to produce 176 grams of Carbon-dioxide and 108 grams of water vapor.

3. **According to the Volume(Avogadro’s Concept ie 1 mole of a gas occupies 22.4L of volume at STP)**: It means 44.8L of Ethane can react with 156.8L of Oxygen to produce 89.6L of Carbon-dioxide and 134.4L of water vapor.

From this example we know that mole, molecular mass and volume all are co-related.

We can classify the calculations based on chemical reactions into 3 types.

- Calculations based on mass to mass relationships
- Calculations based on mass to volume relationships
- Calculations based on volume to volume relationships

**Calculations based on mass to mass relationships**

Following steps should kept in mind while doing the calculations

- First of all we need a balanced equation.
- Write down the molecular masses along with their respective coefficients multiplied with molecular masses.
- Finally establish the mass to mass relationship and find out the mass of the required species by unitary method.

**Example:** Calculate the mass of Oxygen required for completely burning 30g of Ethane to produce CO_{2}. Also, calculate the mass of H_{2}O produced during the reaction.

**Solution: **Balanced equation for combustion reaction of ethane is given below.

**2C**_{2}H_{6(g)} + 7O_{2(g)} = 4CO_{2(g)} + 6H_{2}O_{(g)}

_{2}H

_{6(g)}+ 7O

_{2(g)}= 4CO

_{2(g)}+ 6H

_{2}O

_{(g)}

Now let us find the molecular masses of the compounds involved in the reaction taking care of the respective co-efficient.

2(12×2+1×6) + 7(16×2) = 4(12+16×2) + 6(1×2+16)

=> 2(30) + 7(32) = 4(44) + 6(18)

=> 60 + 224 = 176 + 108

From above we know that,

To burn 60g of ethane we need 224g of oxygen.

To burn 30g of ethane we need (224/60) x 30 = 112 gram of oxygen.

Similarly we also know, 60g of ethane can produce 108g of water vapor in this reaction.

So 30g of ethane can produce (108/60) x 30 = 54 gram of water vapor.

**Let’s understand the same example in a different way.**

Suppose we allow 60 grams of Ethane to make reaction with 112 grams of oxygen. What will happen?

From the previous example we get 112 grams of oxygen is required to burn 30grams of ethane. Here we are trying to burn 60 grams of ethane with 112gms of oxygen. So 30 grams of ethane will remain unburned. Here we can say Oxygen is acting as **Limiting reagent**.

### What is Limiting Reagent?

In a chemical reaction, the reactant present in the lesser amount limits the amount of product formed. So we can call them as Limiting reagents.

Limiting reagents plays the major role while making Stoichiometry calculations.

**Example**: 100g sample of CaCO_{3} is allowed to react with 75g of H_{3}PO_{4} Calculate the mass of Calcium Phosphate Ca_{3}(PO_{4})_{2} that could be produced and the mass of excess reagent that will remain unreacted.

**Solution: **Balanced equation for the said reaction is given below.

**3CaCO**_{3 }+ 2H_{3}PO_{4 }= Ca_{3}(PO_{4})_{2 }+ 3CO_{2} + 3H_{2}O

_{3 }+ 2H

_{3}PO

_{4 }= Ca

_{3}(PO

_{4})

_{2 }+ 3CO

_{2}+ 3H

_{2}O

We can solve this problem by considering 3 compounds from the above equation. These are CaCO_{3}, H_{3}PO_{4} & Ca_{3}(PO_{4})_{2}. So other compounds may be ignored.

Now let’s calculate the Molecular mass of the above compounds considering their respective co-efficient present in the balanced equation.

3(40+12+16×3) + 2(1×3+31+16×4) à 3×40+(31+16×4)x2

ie 3(40+12+48) + 2(3+31+64) => 3×40 + 2×95

ie 300 + 196 => 310

From this derivation, we know that 300g of CaCO_{3 }require 196g of H_{3}PO_{4} to produce 310g of Ca_{3}(PO_{4})_{2}

100g of CaCO_{3 }will produce (310/300) x 100 = 103.33g of Ca_{3}(PO_{4})_{2}

Again it is found that, 196g of H_{3}PO_{4} can produce 310g of Ca_{3}(PO_{4})_{2}

So 75g of H_{3}PO_{4} can produce (310/196) x 75 = 118.62g of Ca_{3}(PO_{4})_{2}

**So here CaCO _{3 }is the limiting reagent and the mass of **

**Ca**

_{3}(PO_{4})_{2}**could be produced = 103.33g**

Again we know, to produce 310g of Ca_{3}(PO_{4})_{2} 196g of H_{3}PO_{4} required**.**

Hence to produce 103.33g of Ca_{3}(PO_{4})_{2} (196/310) x 103.33 = 65.33g of H_{3}PO_{4} required**.**

Here amount of excess reagent (H_{3}PO_{4}) that remain un reacted = 75g – 65.33g = 9.67g

**Calculations based on mass to volume relationships**

**What is the relationship between mass and volume of a substance?**

1 mole or 1 gram molecular mass of a substance occupies 22.4ltrs of volume at normal temperature and pressure.

It means, **at NTP 1 mole=1 gram molecular = 22.4Ltr**

**Example:** What volume of chlorine gas at NTP will be produced when 8.7g of Manganese dioxide(MnO_{2} ) reacts with an excess of HCL?

**Solution: **Balanced equation for the said reaction is given below.

**MnO**_{2} + 4HCL_{ }= MnCl_{2 }+ Cl_{2} + 2H_{2}O

_{2}+ 4HCL

_{ }= MnCl

_{2 }+ Cl

_{2}+ 2H

_{2}O

Here 1 mole of MnO_{2} can produce 1 mole of Chlorine gas.

Molecular Mass of MnO_{2} = 55+16×2 = 87

Now we can say 87g of MnO_{2} can produce 22.4ltr of Chlorine gas

So 8.7g of MnO_{2} can produce (22.4/87) x 8.7 = 2.24 ltr of Chlorine gas

**Calculations based on volume to volume relationships**

These calculations are based on 2 major laws of chemical combination.

1. **Avogadro’s Law**: Under similar conditions of temperature and pressure, equal moles of gasses occupy equal volumes.

2. **Gay-Lussac’s Law**: Under similar conditions of temperature and pressure, gasses react in simple ratio of their volumes.

These calculations may work well for the chemical reactions having gaseous reactants and products.

**Example: **What volume of Oxygen at NTP is necessary for complete combustion of 20L Propane (C_{3}H_{8}) measured at 27^{0}C and 760mm of Hg?

**Solution: **Balanced equation for the said reaction is given below.

**C**_{3}H_{8} + 5O_{2 }= 3CO_{2} + 4H_{2}O

_{3}H

_{8}+ 5O

_{2 }= 3CO

_{2}+ 4H

_{2}O

Now let’s convert the given volume of propane to volume at NTP using Gas Equation.

But from the above equation we get, 1 mole of propane needs 5 mole of oxygen for complete combustion.

It means 22.4L of propane needs 22.4Lx5 = 112L of oxygen

So 18.2L of propane needs (112/22.4) x 18.2 = 91 Ltr of oxygen.