Grade 10 → Stoichiometry and Chemical Calculations ↓
Stoichiometric calculations and chemical yield
Stoichiometry is a branch of chemistry that deals with the relative amounts of reactants and products in a chemical reaction. It involves calculations that estimate the amounts of substances consumed and produced in a reaction. One of the main parts of stoichiometry is understanding how stoichiometric calculations are performed and how these relate to chemical yield, which is the amount of product obtained in a chemical reaction.
Basic concepts of stoichiometry
Stoichiometry is based on the conservation of mass, where the total mass of the reactants is equal to the total mass of the products. The key is to add the reactants and products quantitatively using a balanced chemical equation. This includes:
- Understanding the mole concept and Avogadro's number.
- Using molar mass to convert between grams and moles.
- Using the coefficients in balanced equations to determine mole ratios.
Example 1: Balancing a chemical equation
Consider the combustion reaction of propane:
C 3 H 8 + O 2 → CO 2 + H 2 O
First, balance the equation:
- Balance the carbon atoms:
C 3 H 8 + O 2 → 3 CO 2 + H 2 O - Balance the hydrogen atoms:
C 3 H 8 + O 2 → 3CO 2 + 4H 2 O - Balance the oxygen atoms:
C 3 H 8 + 5 O 2 → 3CO 2 + 4H 2 O
The balanced equation is:
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O
Stoichiometric calculations
Stoichiometric calculations usually involve determining the mass of one substance in a chemical equation given the mass of another substance. Here's how to do it:
Very basic stoichiometry steps
- Convert a given mass of a substance into moles using its molar mass.
- Use the stoichiometric coefficients (mole ratios) from the balanced equation to convert moles of one substance to moles of another.
- Convert moles back to grams if necessary.
Example 2: Simple stoichiometric calculation
Suppose you burn 11.0 g of propane (C 3 H 8 ). Find the mass of carbon dioxide produced.
1. Calculate the moles of C 3 H 8:
Molar mass of C 3 H 8 = 44.1 g/mol Moles of C 3 H 8 = 11.0 g / 44.1 g/mol = 0.249 moles
2. Use the balanced equation to find the moles of CO2:
C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O 0.249 mol C 3 H 8 × (3 mol CO 2 / 1 mol C 3 H 8 ) = 0.747 mol CO 2
3. Convert moles of CO2 to grams:
Molar mass of CO2 = 44.0 g/mol Mass of CO2 = 0.747 mol × 44.0 g/mol = 32.868 g
The mass of CO2 produced is about 32.87 grams.
Chemical byproducts
Chemical yield is the amount of product obtained in a chemical reaction. It is divided into two types:
- Theoretical yield: The amount of product predicted by stoichiometry, based on a balanced chemical equation where the reaction proceeds completely to completion.
- Actual yield: The amount of product actually obtained from a reaction.
The percent yield is then calculated to see how efficient the reaction was.
Percentage yield = (actual yield / theoretical yield) × 100
Example 3: Calculating chemical yield
Consider a reaction where the theoretical yield of a product is 20.0 g, but only 15.0 g of product is obtained experimentally.
Calculate the percentage yield:
Percent yield = (15.0 g / 20.0 g) × 100 = 75%
The percentage yield is 75%, which shows the efficiency of the process.
Visualization of stoichiometry through reactions
Let's imagine a simple reaction:
Water is formed by the reaction of hydrogen and oxygen:
2H 2 + O 2 → 2H 2 O
Visualize breaking and making bonds:
The shape of each molecule, represented by a rectangular box, changes as bonds are broken and water is formed.
Conclusion
Understanding stoichiometric calculations and chemical yields are essential for predicting the outcomes of chemical reactions. Mastering these fundamental concepts is important for anyone studying chemistry.
By practicing with different chemical equations and scenarios, one can gain confidence in performing these calculations and understanding the relationship between reactants and products in any reaction. This will help in making accurate predictions about quantities in chemical experiments and industrial processes.
We hope this detailed explanation provides a solid understanding of stoichiometric calculations and chemical yield. Remember to take the time to practice these concepts with different chemical equations to further broaden your understanding.