Mole Concept & Stoichiometry – Study Notes
Manipal MET – Chemistry
1. Simple Explanation
| Term | What it means | How we use it |
|---|---|---|
| Mole (mol) | A counting unit that tells us how many atoms / molecules / ions are present. | Think of it as a “mole‑bucket” that always holds the same number of particles. |
| Avogadro’s Number | (6.022 \times 10^{23}) entities per mole | The conversion factor between the microscopic (atoms/molecules) and macroscopic (grams) world. |
| Molar Mass | Mass of 1 mol of a substance (g mol⁻¹) | Found in the periodic table (g/mol). |
| Stoichiometry | The quantitative part of a chemical reaction – how many moles of each species relate to each other. | Obtained from the balanced chemical equation. |
| Limiting Reactant | The reagent that is completely consumed first, thereby limiting product formation. | Determines the actual amount of product you can make. |
| Percent Yield | Ratio of the actual yield to the theoretical yield, expressed as a percentage. | (\displaystyle %\text{Yield}= \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100%) |
2. Key Points
| Concept | Key Formula / Idea | Important Notes |
|---|---|---|
| Mole–mass relationship | (n = \frac{m}{M}) (n = moles, m = mass in g, M = molar mass) | Works for any pure substance. |
| Mole–mole relationship (stoichiometry) | Use the coefficients from a balanced equation. Example: (2,\text{H}_2 + \text{O}_2 \to 2,\text{H}_2\text{O}) → 2 mol H₂ reacts with 1 mol O₂. | Coefficients are mole ratios. |
| Molar volume (ideal gas) | (V_m = 22.4;\text{L mol}^{-1}) at STP (0 °C, 1 atm) | Use (V_m = 24.0;\text{L mol}^{-1}) at 25 °C, 1 atm (approx). |
| Molarity (M) | (M = \frac{n_{\text{solute}}}{V_{\text{solution}}}) (mol L⁻¹) | Common in solution chemistry. |
| Limiting reagent calculation | 1. Convert masses to moles. 2. Divide each mole value by its coefficient. 3. The smallest quotient is the limiting reagent. | The other reagent is called the excess reagent. |
| Theoretical yield | Calculate using the limiting reagent and stoichiometry. | Gives the maximum possible product. |
| Percent yield | (\displaystyle %\text{Yield}= \frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100%) | Indicates efficiency of a reaction. |
| Conversion factors | Use (1;\text{mol} = 6.022\times10^{23}) entities and (1,\text{mol} = 22.4,\text{L}) (ideal gas at STP). | Helpful in quick mental calculations. |
3. Worked Example
Problem
A student reacts 10 g of hydrogen gas ((\text{H}_2)) with excess oxygen to produce water.
- What is the theoretical mass of water produced?
- If the student actually isolated 9.0 g of water, what is the percent yield?
Step‑by‑Step Solution
| Step | Action | Calculation | Result |
|---|---|---|---|
| 1. Balance the reaction | (\displaystyle 2,\text{H}_2 + \text{O}_2 \to 2,\text{H}_2\text{O}) | ||
| 2. Find molar masses | (M_{\text{H}2}=2.016;\text{g mol}^{-1}) (M{\text{H}_2\text{O}}=18.015;\text{g mol}^{-1}) |
||
| 3. Convert mass of (\text{H}_2) to moles | (n_{\text{H}_2}= \frac{10.0;\text{g}}{2.016;\text{g mol}^{-1}} = 4.96;\text{mol}) | 4.96 mol | |
| 4. Determine limiting reagent | Reaction coefficient: 2 mol H₂ → 2 mol H₂O. Moles of water that can form: (n_{\text{H}_2\text{O}} = 4.96;\text{mol}) (same number because coeffs are 2:2) |
4.96 mol water | |
| 5. Convert moles of water to mass | (m_{\text{H}2\text{O}} = n{\text{H}2\text{O}}\times M{\text{H}_2\text{O}} = 4.96,\text{mol}\times18.015,\text{g mol}^{-1}) | (= 89.4;\text{g}) | 89.4 g (theoretical) |
| 6. Calculate percent yield | (\displaystyle %\text{Yield}= \frac{9.0;\text{g}}{89.4;\text{g}}\times100 = 10.1%) | 10.1 % |
Answer
- Theoretical mass of water = **89.