Stoichiometry Cheatsheet

Cheatsheet Content

### Basic Concepts - **Stoichiometry:** The quantitative relationship between reactants and products in a chemical reaction. - **Mole (mol):** The SI unit for amount of substance. $1 \text{ mol} = 6.022 \times 10^{23}$ particles (Avogadro's number). - **Molar Mass (M):** The mass of one mole of a substance, expressed in g/mol. - **Balanced Chemical Equation:** Represents a chemical reaction, showing the exact ratio of moles of reactants and products. #### Key Relationships - Moles = Mass (g) / Molar Mass (g/mol) - Moles = Number of Particles / Avogadro's Number - For gases at STP (Standard Temperature and Pressure, $0^\circ\text{C}$ and $1 \text{ atm}$): $1 \text{ mol} = 22.4 \text{ L}$ ### Stoichiometric Calculations Follow these steps for most stoichiometry problems: 1. **Balance the Chemical Equation:** Ensure the number of atoms of each element is the same on both sides. 2. **Convert Given Quantities to Moles:** Use molar mass, Avogadro's number, or gas volume at STP. 3. **Use Mole Ratios:** From the balanced equation, determine the mole ratio between the given substance and the desired substance. - Example: For $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$, the mole ratio of $\text{H}_2$ to $\text{O}_2$ is $2:1$. 4. **Convert Moles to Desired Units:** Convert the calculated moles back to mass, volume, or number of particles. #### Example: Mass-to-Mass Calculation How many grams of $\text{H}_2\text{O}$ are produced from $10.0 \text{ g}$ of $\text{H}_2$? Given reaction: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ 1. **Balanced:** Yes, already balanced. 2. **Convert $\text{H}_2$ to moles:** - Molar mass of $\text{H}_2 = 2 \times 1.008 \text{ g/mol} = 2.016 \text{ g/mol}$ - Moles $\text{H}_2 = 10.0 \text{ g} / 2.016 \text{ g/mol} = 4.96 \text{ mol}$ 3. **Use mole ratio:** From the equation, $2 \text{ mol H}_2$ produces $2 \text{ mol H}_2\text{O}$. So, the ratio is $1:1$. - Moles $\text{H}_2\text{O} = 4.96 \text{ mol H}_2 \times (2 \text{ mol H}_2\text{O} / 2 \text{ mol H}_2) = 4.96 \text{ mol H}_2\text{O}$ 4. **Convert $\text{H}_2\text{O}$ to grams:** - Molar mass of $\text{H}_2\text{O} = (2 \times 1.008) + 16.00 = 18.016 \text{ g/mol}$ - Mass $\text{H}_2\text{O} = 4.96 \text{ mol} \times 18.016 \text{ g/mol} = 89.4 \text{ g}$ ### Limiting Reactant - **Definition:** The reactant that is completely consumed in a chemical reaction. It limits the amount of product that can be formed. - **Excess Reactant:** The reactant that is left over after the reaction is complete. #### Determining the Limiting Reactant 1. **Calculate moles** for all given reactants. 2. **Divide moles of each reactant by its stoichiometric coefficient** from the balanced equation. 3. The reactant with the **smallest resulting value** is the limiting reactant. 4. Use the limiting reactant's moles to calculate the amount of product formed. #### Example: $2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$ If you start with $10.0 \text{ g H}_2$ and $10.0 \text{ g O}_2$: 1. **Moles:** - Moles $\text{H}_2 = 10.0 \text{ g} / 2.016 \text{ g/mol} = 4.96 \text{ mol}$ - Moles $\text{O}_2 = 10.0 \text{ g} / 32.00 \text{ g/mol} = 0.313 \text{ mol}$ 2. **Divide by coefficients:** - For $\text{H}_2$: $4.96 \text{ mol} / 2 = 2.48$ - For $\text{O}_2$: $0.313 \text{ mol} / 1 = 0.313$ 3. **Compare:** $0.313 ### Reaction Yields - **Theoretical Yield:** The maximum amount of product that can be formed from a given amount of reactants, calculated using stoichiometry. - **Actual Yield:** The amount of product actually obtained from a reaction in the laboratory. (Always $\le$ Theoretical Yield) - **Percent Yield:** A measure of the efficiency of a reaction. $$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$$ ### Concentration & Solutions - **Molarity (M):** Moles of solute per liter of solution. $$\text{Molarity (M)} = \frac{\text{Moles of Solute}}{\text{Volume of Solution (L)}}$$ - **Dilution Equation:** $M_1V_1 = M_2V_2$ (where $M$ is molarity and $V$ is volume) - **Stoichiometry with Solutions:** Use molarity to convert between volume of solution and moles of solute in stoichiometric calculations.