Stoichiometry & Reactions
Introduction
Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. By understanding stoichiometry, we can predict how much of each substance is involved in a reaction, ensuring that chemical equations are balanced and accurately represent the conservation of mass.
Basic Concepts
1. The Mole
- The mole is the standard unit of measurement in chemistry for counting atoms, molecules, or ions.
- 1 mole of any substance contains Avogadro's number of particles: \(6.022 \times 10^{23}\).
2. Molar Mass
- The molar mass of a substance is the mass of one mole of that substance, expressed in grams per mole (g/mol).
- It is equivalent to the atomic or molecular weight of the substance in atomic mass units (amu).
3. Balanced Chemical Equations
- A balanced chemical equation has the same number of atoms of each element on both sides of the equation, reflecting the law of conservation of mass.
- Example:
\[
\text{C}6\text{H}{12}\text{O}6 + 6 \text{O}2 \rightarrow 6 \text{CO}2 + 6 \text{H}2\text{O}
\]
- In this reaction, the number of carbon, hydrogen, and oxygen atoms is the same on both sides of the equation.
Stoichiometry
4. Mole-to-Mole Conversions
- Stoichiometry allows us to convert between moles of reactants and moles of products using the coefficients in a balanced chemical equation.
- Example:
- Balanced Equation: \(\text{2 H}2 + \text{O}2 \rightarrow \text{2 H}_2\text{O}\)
- Interpretation: 2 moles of \(\text{H}2\) react with 1 mole of \(\text{O}2\) to produce 2 moles of \(\text{H}_2\text{O}\).
5. Mass-to-Mass Conversions
- Use the molar mass to convert between the mass of a substance and the number of moles, and then apply stoichiometry to find the mass of another substance in the reaction.
- Example:
- Calculate the mass of \(\text{H}2\text{O}\) produced from 10 grams of \(\text{H}2\):
\[
\text{molar mass of \(\text{H}_2\)} = 2.02 \, \text{g/mol}
\]
\[
\text{moles of \(\text{H}_2\)} = \frac{10 \, \text{g}}{2.02 \, \text{g/mol}} \approx 4.95 \, \text{moles}
\]
- Use the balanced equation: \(\text{2 H}2 + \text{O}2 \rightarrow \text{2 H}_2\text{O}\)\
- Convert moles of \(\text{H}_2\text{O}\) to grams:
\[
\text{molar mass of \(\text{H}_2\text{O}\)} = 18.02 \, \text{g/mol}
\]
\[
\text{mass of \(\text{H}_2\text{O}\)} = 4.95 \, \text{moles} \times 18.02 \, \text{g/mol} \approx 89.15 \, \text{g}
\]
Balancing Chemical Equations
6. Steps to Balance a Chemical Equation
- Write the Unbalanced Equation:
- Count Atoms for Each Element:
- Balance One Element at a Time:
- Verify the Balance:
- Balanced Equation:
Applications of Stoichiometry
7. Limiting Reactants
- In a chemical reaction, the limiting reactant is the substance that is completely consumed first, limiting the amount of products formed.
- Example:
- Given 5 moles of \(\text{H}2\) and 2 moles of \(\text{O}2\), determine the limiting reactant in the reaction \(\text{2 H}2 + \text{O}2 \rightarrow \text{2 H}_2\text{O}\).
- \(\text{H}2\): \(5 \text{ moles} \times \frac{1 \text{ mole } \text{O}2}{2 \text{ moles } \text{H}2}} = 2.5 \text{ moles of } \text{O}2\) needed
- \(\text{O}_2\): 2 moles available
- \(\text{O}_2\) is the limiting reactant.
8. Percent Yield
- Percent yield is a measure of the efficiency of a reaction, calculated by comparing the actual yield to the theoretical yield.
- Formula:
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%
\]
Example Problem
9. Example Calculation
- Problem: How many grams of \(\text{CO}2\) are produced when 50 grams of \(\text{C}4\text{H}_{10}\) combust completely?
- Solution:
\[
\text{2 C}4\text{H}{10} + 13 \text{O}2 \rightarrow 8 \text{CO}2 + 10 \text{H}_2\text{O}
\]
- Find moles of \(\text{C}4\text{H}{10}\):
- Use stoichiometry to find moles of \(\text{CO}_2\):
- Convert moles of \(\text{CO}_2\) to grams:
Conclusion
Stoichiometry is an essential tool in chemistry that enables the precise calculation of reactant and product quantities in chemical reactions. By mastering mole-to-mole conversions, balancing chemical equations, and understanding the concept of limiting reactants, we can predict and control the outcomes of chemical processes.
Key Terms
| Term | Definition |
|---|---|
| Mole | A unit for measuring the amount of substance, containing \(6.022 \times 10^{23}\) particles. |
| Molar Mass | The mass of one mole of a substance, expressed in grams per mole (g/mol). |
| Balanced Equation | A chemical equation with equal numbers of each type of atom on both sides. |
| Stoichiometry | The calculation of reactants and products in chemical reactions. |
| Limiting Reactant | The reactant that is completely consumed in a reaction, limiting the amount of product formed. |
| Percent Yield | A measure of reaction efficiency: \((\text{Actual Yield}/\text{Theoretical Yield}) \times 100\%\). |
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