Stoichiometry & Stoichiometric Calculations

What is Stoichiometry?

The word stoichiometry comes from the Greek words stoicheion (element) and metron (measure). It deals with the quantitative relationships between reactants and products in a chemical reaction.

In simple terms, stoichiometry answers questions like:

• How much product can I make from a given amount of reactant?
• How much reactant do I need to produce a desired amount of product?
• If I have fixed amounts of two reactants, which one runs out first?

The foundation of stoichiometry is the balanced chemical equation, which tells us the mole ratios of all substances involved.

Key Point: Stoichiometry uses balanced equations to calculate the quantities of reactants and products. The coefficients in a balanced equation represent mole ratios.

Balanced Chemical Equations — The Language of Stoichiometry

A balanced chemical equation satisfies the Law of Conservation of Mass — the same number of atoms of each element appear on both sides.

Information from a Balanced Equation

Consider: $$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(g)$$

This equation tells us:

Reading	Reactants	Products
Molecules	1 molecule $\text{CH}_4$ + 2 molecules $\text{O}_2$	1 molecule $\text{CO}_2$ + 2 molecules $\text{H}_2\text{O}$
Moles	1 mol $\text{CH}_4$ + 2 mol $\text{O}_2$	1 mol $\text{CO}_2$ + 2 mol $\text{H}_2\text{O}$
Mass	16 g + 64 g = 80 g	44 g + 36 g = 80 g
Volume (STP)	22.4 L + 44.8 L	22.4 L + 44.8 L

Notice: Total mass of reactants (80 g) = Total mass of products (80 g) ✓

The coefficients (1, 2, 1, 2) give us the mole ratio, which is the key to all stoichiometric calculations.

[Board Important] Always balance the equation FIRST before attempting any stoichiometric calculation. An unbalanced equation will give wrong answers.

Key Point: Coefficients in a balanced equation represent mole ratios, which can be converted to mass ratios (using molar masses) or volume ratios (for gases at same T, P).

Stoichiometric Calculations — The General Method

Almost every stoichiometry problem follows this pattern:

Given quantity → Moles → Use mole ratio → Moles of desired substance → Convert to desired unit

Step-by-Step Method

1. Write and balance the chemical equation
2. Convert the given quantity (mass, volume, or particles) to moles
3. Use the mole ratio from the balanced equation to find moles of the desired substance
4. Convert moles of the desired substance to the required unit (mass, volume, or particles)

Example

How many grams of $\text{CO}_2$ are produced by burning 48 g of methane ($\text{CH}_4$)?

$$\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}$$

1. Moles of $\text{CH}_4$ = $\frac{48}{16}$ = 3 mol
2. From the equation: 1 mol $\text{CH}_4$ produces 1 mol $\text{CO}_2$
3. So 3 mol $\text{CH}_4$ produces 3 mol $\text{CO}_2$
4. Mass of $\text{CO}_2$ = $3 \times 44$ = 132 g

[JEE Tip] The mole ratio from balanced equations is the bridge between any two substances in the reaction. Always identify the mole ratio first.

Key Point: The universal stoichiometry method: Given → Moles (÷ molar mass) → Mole ratio → Moles of target → Answer (× molar mass or × $N_A$ etc.).

Limiting Reagent (Limiting Reactant)

In most real-world situations, reactants are NOT present in the exact stoichiometric ratio. One reactant will be completely consumed first — this is the limiting reagent (or limiting reactant). The other reactant(s) are said to be in excess.

The limiting reagent determines the maximum amount of product that can be formed.

How to Identify the Limiting Reagent

Method 1 (Mole Ratio Comparison):

1. Calculate moles of each reactant
2. Divide each by its stoichiometric coefficient
3. The reactant with the smallest value is the limiting reagent

Method 2 (Product Calculation):

1. Calculate the amount of product formed from each reactant separately (assuming the other is in excess)
2. The reactant that gives the least product is the limiting reagent

Classic Example: Ammonia Synthesis

$$\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$$

If we start with 50 kg of $\text{N}_2$ and 10 kg of $\text{H}_2$:

• Moles of $\text{N}_2$ = $\frac{50000}{28}$ = 1785.7 mol
• Moles of $\text{H}_2$ = $\frac{10000}{2}$ = 5000 mol

Required ratio: $\text{N}_2 : \text{H}_2$ = 1 : 3

For 1785.7 mol $\text{N}_2$, we need $1785.7 \times 3$ = 5357.1 mol $\text{H}_2$. But we only have 5000 mol $\text{H}_2$.

$\text{H}_2$ is the limiting reagent (we don't have enough of it).

Product formed: 5000 mol $\text{H}_2$ produces $\frac{2}{3} \times 5000$ = 3333.3 mol $\text{NH}_3$ Mass = $3333.3 \times 17$ = 56666.7 g ≈ 56.67 kg

[JEE Important] Limiting reagent problems are extremely common in JEE. Always check which reactant limits the reaction.

Key Point: The limiting reagent is completely consumed and determines the amount of product. Excess reagent is left over after the reaction.

Concentration of Solutions

Many reactions occur in solutions rather than between pure substances. We need ways to express how much solute is dissolved in a solution. The most common concentration terms are:

1. Mass Percentage (w/w)

$$\text{Mass \%} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$$

Example: A 10% NaCl solution means 10 g of NaCl in 100 g of solution (not 100 g of solvent!).

2. Mole Fraction ($x$)

For a two-component solution (solute A in solvent B): $$x_A = \frac{n_A}{n_A + n_B} \quad \text{and} \quad x_B = \frac{n_B}{n_A + n_B}$$

where $n_A$ and $n_B$ are the number of moles.

Important: $x_A + x_B = 1$ always.

3. Molarity ($M$)

$$M = \frac{\text{Moles of solute}}{\text{Volume of solution in litres}}$$

Unit: mol/L or mol L$^{-1}$ (also written as M)

Example: 1 M NaOH means 1 mole of NaOH (= 40 g) dissolved in enough water to make 1 L of solution.

[NEET Important] Molarity changes with temperature because volume changes with temperature. This is an important conceptual distinction.

4. Molality ($m$)

$$m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}}$$

Unit: mol/kg or mol kg$^{-1}$ (also written as $m$)

[JEE Tip] Molality does NOT change with temperature because it depends on mass (not volume). This makes it more reliable for precise work than molarity.

Key Point: Molarity = moles/volume (changes with T), Molality = moles/mass of solvent (independent of T). Mole fraction has no units and sum = 1.

Dilution Formula

When you add water (or solvent) to a solution, the volume increases but the amount of solute (in moles) stays the same. This gives us the dilution formula:

$$M_1 V_1 = M_2 V_2$$

where:

• $M_1$ = initial molarity, $V_1$ = initial volume
• $M_2$ = final molarity, $V_2$ = final volume

Why This Works

Moles of solute = $M \times V$ (in litres)

Since moles don't change during dilution: $M_1 V_1 = M_2 V_2$.

Example

You have 200 mL of 2 M HCl. You dilute it to 500 mL. What is the new molarity?

$$M_2 = \frac{M_1 V_1}{V_2} = \frac{2 \times 200}{500} = 0.8 \text{ M}$$

[Board Important] The dilution formula $M_1V_1 = M_2V_2$ is one of the most commonly used formulas in solution chemistry.

Key Point: On dilution, moles of solute remain constant: $M_1V_1 = M_2V_2$. Adding solvent decreases concentration but doesn't change the amount of solute.

Solved Examples

Example 1: Mass-to-Mass Stoichiometry

How many grams of oxygen are required to completely burn 100 g of propane ($\text{C}_3\text{H}_8$)?

$$\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$$

Solution:

1. Moles of $\text{C}_3\text{H}_8$ = $\frac{100}{44}$ = 2.273 mol
2. Mole ratio: 1 mol $\text{C}_3\text{H}_8$ requires 5 mol $\text{O}_2$
3. Moles of $\text{O}_2$ needed = $2.273 \times 5$ = 11.364 mol
4. Mass of $\text{O}_2$ = $11.364 \times 32$ = 363.6 g

Final Answer: 363.6 g of oxygen is required.

---

Example 2: Mass-to-Volume Stoichiometry

What volume of $\text{CO}_2$ (at STP) is produced when 10 g of calcium carbonate is completely decomposed?

$$\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$$

Solution:

1. Moles of $\text{CaCO}_3$ = $\frac{10}{100}$ = 0.1 mol
2. Mole ratio: 1 mol $\text{CaCO}_3$ → 1 mol $\text{CO}_2$
3. Moles of $\text{CO}_2$ = 0.1 mol
4. Volume at STP = $0.1 \times 22.4$ = 2.24 L

Final Answer: 2.24 L of $\text{CO}_2$ at STP.

Example 3: Limiting Reagent Problem

50 kg of $\text{N}_2$ and 10 kg of $\text{H}_2$ react to form $\text{NH}_3$. Calculate the mass of $\text{NH}_3$ produced. Identify the limiting reagent and find the excess reagent remaining.

$$\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$$

Solution:

1. Moles: $\text{N}_2$ = $\frac{50000}{28}$ = 1785.7 mol; $\text{H}_2$ = $\frac{10000}{2}$ = 5000 mol

2. Check limiting reagent: For 1785.7 mol $\text{N}_2$, we need $1785.7 \times 3$ = 5357.1 mol $\text{H}_2$. We have only 5000 mol. $\text{H}_2$ is limiting.

3. $\text{NH}_3$ produced (from $\text{H}_2$): $$\text{mol NH}_3 = \frac{2}{3} \times 5000 = 3333.3 \text{ mol}$$ $$\text{Mass} = 3333.3 \times 17 = 56666.7 \text{ g} \approx 56.67 \text{ kg}$$

4. $\text{N}_2$ consumed: $\frac{1}{3} \times 5000$ = 1666.7 mol = $1666.7 \times 28$ = 46666.7 g ≈ 46.67 kg

5. $\text{N}_2$ remaining: 50 − 46.67 = 3.33 kg excess

Final Answer: 56.67 kg of $\text{NH}_3$ is produced. $\text{H}_2$ is the limiting reagent. 3.33 kg of $\text{N}_2$ remains unused.

---

Example 4: Limiting Reagent — Method 2

10 g of hydrogen reacts with 32 g of oxygen. Which is the limiting reagent? How much water is formed?

$$2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$$

Solution:

1. Moles: $\text{H}_2$ = 10/2 = 5 mol; $\text{O}_2$ = 32/32 = 1 mol
2. From equation: 2 mol $\text{H}_2$ needs 1 mol $\text{O}_2$
3. For 5 mol $\text{H}_2$, we need 2.5 mol $\text{O}_2$. We have only 1 mol. $\text{O}_2$ is limiting.
4. From 1 mol $\text{O}_2$: $\text{H}_2\text{O}$ = 2 mol = $2 \times 18$ = 36 g
5. $\text{H}_2$ consumed: 2 mol. Excess $\text{H}_2$: 5 − 2 = 3 mol = 6 g

Final Answer: 36 g of water is formed. $\text{O}_2$ is limiting. 6 g of $\text{H}_2$ is in excess.

Example 5: Molarity Calculation

Calculate the molarity of a solution prepared by dissolving 5.85 g of NaCl in 250 mL of solution. (Molar mass of NaCl = 58.5 g/mol)

Solution:

1. Moles of NaCl = $\frac{5.85}{58.5}$ = 0.1 mol
2. Volume = 250 mL = 0.25 L
3. Molarity = $\frac{0.1}{0.25}$ = 0.4 M

Final Answer: Molarity = 0.4 M or 0.4 mol/L.

---

Example 6: Molality Calculation

Calculate the molality of a solution containing 20 g of NaOH in 500 g of water. (Molar mass of NaOH = 40 g/mol)

Solution:

1. Moles of NaOH = $\frac{20}{40}$ = 0.5 mol
2. Mass of solvent = 500 g = 0.5 kg
3. Molality = $\frac{0.5}{0.5}$ = 1 mol/kg = 1 m

Final Answer: Molality = 1 m.

Note the difference from molarity: Molality uses mass of solvent (not volume of solution). If you dissolved 20 g NaOH in 500 mL of solution, the molarity would be $\frac{0.5}{0.5}$ = 1 M — same number here by coincidence, but the definitions are different.

Example 7: Mole Fraction

A solution contains 18 g of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$, M = 180) dissolved in 90 g of water. Calculate the mole fraction of glucose and water.

Solution:

1. Moles of glucose = $\frac{18}{180}$ = 0.1 mol
2. Moles of water = $\frac{90}{18}$ = 5 mol
3. Total moles = 0.1 + 5 = 5.1 mol

$$x_{\text{glucose}} = \frac{0.1}{5.1} = 0.0196$$

$$x_{\text{water}} = \frac{5}{5.1} = 0.9804$$

Check: 0.0196 + 0.9804 = 1.000 ✓

Final Answer: $x_{\text{glucose}} = 0.0196$, $x_{\text{water}} = 0.9804$.

---

Example 8: Dilution Problem

How would you prepare 500 mL of 0.1 M HCl from a stock solution of 1 M HCl?

Solution: Using $M_1V_1 = M_2V_2$:

$$1 \times V_1 = 0.1 \times 500$$ $$V_1 = 50 \text{ mL}$$

Procedure: Take 50 mL of 1 M HCl and dilute it with water to make a total volume of 500 mL.

Takeaway: The dilution formula is extremely practical — it's used daily in every chemistry laboratory.

Example 9: Mass Percent to Molarity Conversion

A solution of $\text{H}_2\text{SO}_4$ has a density of 1.84 g/mL and is 98% $\text{H}_2\text{SO}_4$ by mass. Calculate its molarity.

Solution:

1. Consider 1 L of solution:

• Mass of solution = $1000 \times 1.84$ = 1840 g
• Mass of $\text{H}_2\text{SO}_4$ = $\frac{98}{100} \times 1840$ = 1803.2 g

1. Moles of $\text{H}_2\text{SO}_4$: $$n = \frac{1803.2}{98} = 18.4 \text{ mol}$$

2. Molarity: $$M = \frac{18.4}{1} = 18.4 \text{ M}$$

Final Answer: The molarity of concentrated $\text{H}_2\text{SO}_4$ is 18.4 M.

[JEE Important] This is a very common conversion problem. The key step is considering 1 L of solution and using density to find mass.

---

Example 10: Stoichiometry with Molarity

What volume of 0.5 M $\text{H}_2\text{SO}_4$ is needed to completely react with 5.3 g of $\text{Na}_2\text{CO}_3$?

$$\text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2$$

Solution:

1. Moles of $\text{Na}_2\text{CO}_3$ = $\frac{5.3}{106}$ = 0.05 mol
2. Mole ratio: 1:1 → Moles of $\text{H}_2\text{SO}_4$ needed = 0.05 mol
3. Volume = $\frac{\text{moles}}{M} = \frac{0.05}{0.5}$ = 0.1 L = 100 mL

Final Answer: 100 mL of 0.5 M $\text{H}_2\text{SO}_4$ is needed.

Example 11: Reactions in Solution — Volumetric Problem

If 25 mL of 0.1 M NaOH is required to neutralise 10 mL of an HCl solution, what is the molarity of HCl?

$$\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}$$

Solution:

1. Moles of NaOH = $0.1 \times 0.025$ = 0.0025 mol
2. Mole ratio NaOH : HCl = 1 : 1
3. Moles of HCl = 0.0025 mol
4. Molarity of HCl = $\frac{0.0025}{0.010}$ = 0.25 M

Final Answer: The HCl solution is 0.25 M.

---

Example 12: Percentage Yield

In a reaction, 100 g of $\text{CaCO}_3$ is heated. Theoretically, 56 g of CaO should form. If only ite 48 g is obtained, what is the percentage yield?

$$\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{48}{56} \times 100 = 85.7\%$$

Final Answer: The percentage yield is 85.7%.

Takeaway: In real reactions, yields are usually less than 100% due to side reactions, incomplete reactions, or losses during processing.