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CIE IGCSE Chemistry Notes

3.2.3 Reacting Masses Calculation

Introduction to Reacting Masses

Reacting masses deal with the mass relationships between reactants and products in a chemical reaction. This concept is rooted in the principle of the conservation of mass and the stoichiometry of chemical reactions.

The Law of Conservation of Mass

  • Fundamental Principle: In a chemical reaction, the total mass of the reactants equals the total mass of the products.
  • Implication: The mass is neither created nor destroyed; it is conserved.

The Mole Concept

  • Definition: A mole is a unit that measures the amount of substance.
  • Significance: It links the mass of a substance to the number of particles it contains.
Illustration of the mole concept

Image courtesy of GeeksforGeeks

Steps in Calculating Reacting Masses

Calculating reacting masses requires a clear understanding of the chemical equation and the stoichiometry involved.

Step 1: Interpreting the Chemical Equation

The starting point is always a balanced chemical equation, which indicates the relative number of moles of each reactant and product involved in the reaction.

Example Equation: ( \text{2H}2 + \text{O}2 \rightarrow \text{2H}2\text{O} )

Step 2: Converting Mass to Moles

To find the amount of a substance, convert its mass to moles using the formula: ( \text{Number of moles} = \frac{\text{Mass in grams}}{\text{Molar mass}} )

Step 3: Applying Stoichiometry

Use the mole ratio from the balanced equation to find the moles of other substances involved.

Step 4: Converting Moles to Mass

Finally, convert the calculated moles back to mass using the substance's molar mass.

Detailed Examples

Example 1: Formation of Water

Objective: Calculate the mass of water produced from 10g of hydrogen reacting with oxygen.

Steps:

1. Balanced Equation: ( \text{2H}2 + \text{O}2 \rightarrow \text{2H}2\text{O} )

2. Convert Hydrogen Mass to Moles:

  • Molar mass of H₂ = 2.02 g/mol.
  • Moles of H₂ = 10 g / 2.02 g/mol = 4.95 moles.

3. Stoichiometry:

  • 2 moles of H₂ react with 1 mole of O₂ to produce 2 moles of H₂O.
  • Moles of H₂O = 4.95 moles (same as H₂ due to 1:1 ratio in product).

4. Convert Moles of Water to Mass:

  • Molar mass of H₂O = 18.02 g/mol.
  • Mass of H₂O = 4.95 moles × 18.02 g/mol = 89.2 g.

Example 2: Magnesium and Oxygen Reaction

Objective: Determine the mass of oxygen needed to react completely with 20g of magnesium.

Steps:

1. Balanced Equation: ( \text{2Mg} + \text{O}2 \rightarrow \text{2MgO} )

2. Convert Magnesium Mass to Moles:

  • Molar mass of Mg = 24.31 g/mol.
  • Moles of Mg = 20 g / 24.31 g/mol ≈ 0.82 moles.

3. Stoichiometry:

  • 2 moles of Mg react with 1 mole of O₂.
  • Moles of O₂ required = 0.82 moles / 2 = 0.41 moles.

4. Convert Moles of Oxygen to Mass:

  • Molar mass of O₂ = 32.00 g/mol.
  • Mass of O₂ = 0.41 moles × 32.00 g/mol ≈ 13.12 g.

Delving Deeper: Additional Considerations

When calculating reacting masses, several factors can affect the accuracy and complexity of the calculations:

  • Purity of Reactants: Impurities in reactants can affect the mass calculations.
  • Limiting Reactants: In some reactions, one reactant may be in excess, and the other may limit the amount of product formed.
  • Yield of Reaction: Not all reactions go to completion. The theoretical yield may differ from the actual yield.

Practice Problems

To master reacting masses calculation, practice with a variety of problems. Here are some to get started:

  1. Combustion of Propane: Calculate the mass of carbon dioxide produced when 44g of propane (C₃H₈) is burned in excess oxygen.
  2. Formation of Ammonia: What mass of nitrogen gas is needed to react completely with 3g of hydrogen to form ammonia?

Conclusion

Reacting mass calculations are fundamental in understanding the quantitative aspects of chemical reactions. By mastering these concepts, students can better predict and control the outcomes of chemical experiments. Practice is key, and applying these concepts in different scenarios will enhance understanding and proficiency.

FAQ

Avogadro's law plays a significant role in reacting mass calculations, especially when dealing with gases. Avogadro's law states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. This law is crucial for converting between volumes of gases and their corresponding moles in chemical reactions.

In reacting masses calculations involving gases, Avogadro's law allows us to directly relate the volume of a gas to the number of moles, given that the temperature and pressure are constant. For example, under standard conditions, one mole of any gas occupies 22.4 liters. If you know the volume of a gaseous reactant or product, you can use Avogadro's law to find the number of moles of that gas, and from there, apply stoichiometric principles to calculate the reacting masses of other reactants or products in the reaction.

This becomes particularly useful in scenarios where collecting and measuring the volume of a gas is more practical than weighing solid or liquid reactants or products. It also allows for the comparison and calculation of reacting masses of different gases in a reaction based on their volumes.

Temperature and pressure are significant factors in reactions involving gases. For solid and liquid reactants and products, changes in temperature and pressure usually have negligible effects on mass. However, for gases, these factors can significantly influence the volume, and thus indirectly affect the reacting masses when these are calculated using the gas laws. It's essential to understand this, especially when working with the Ideal Gas Law (PV=nRT), where P is pressure, V is volume, T is temperature, n is the number of moles, and R is the gas constant.

For instance, if you're calculating the reacting masses of substances involved in a gaseous reaction, the volume occupied by the gases (and thus the number of moles) will change with temperature and pressure. This means that to make accurate reacting mass calculations, you may need to first convert the volumes of gaseous reactants or products to standard temperature and pressure conditions (STP) using the gas laws, and then proceed with the stoichiometric calculations.

Disproportionation reactions are a type of redox reaction where a single substance is both oxidised and reduced, forming two different products. Handling reacting mass calculations in disproportionation reactions requires careful analysis of the reaction and understanding the stoichiometry involved.

In these reactions, you must consider the balanced chemical equation to determine the mole ratio between the reactant and each of the products. Even though a single reactant breaks down to form multiple products, the law of conservation of mass still applies. Here's how to approach it:

  1. Balance the Equation: Ensure the chemical equation is balanced, including both mass and charge.
  2. Determine the Moles of Reactant: Calculate the moles of the reactant using its mass and molar mass.
  3. Apply Stoichiometry: Use the mole ratios from the balanced equation to determine the moles of each product formed.
  4. Calculate the Masses of the Products: Convert the moles of each product back to mass using their respective molar masses.

For example, in the disproportionation of hydrogen peroxide (H₂O₂), it decomposes into water (H₂O) and oxygen (O₂). The balanced equation is: 2H₂O₂ → 2H₂O + O₂. To calculate the masses of water and oxygen produced from a known mass of H₂O₂, use the stoichiometric ratios from the balanced equation after converting the mass of H₂O₂ to moles.

Yes, reacting mass calculations can be employed to determine the purity of a substance. Purity in chemistry refers to the extent to which a substance is free from impurities. In a laboratory setting, you might have a sample of a substance that is not 100% pure and want to know its purity percentage. Here's how reacting mass calculations come into play:

  1. Perform a Reaction: React the impure sample with a known quantity of another reactant in a reaction where the products and stoichiometry are known.
  2. Calculate the Theoretical Yield: Using the mass of the pure substance (assuming 100% purity) and stoichiometry, calculate the theoretical yield of the product.
  3. Measure the Actual Yield: Experimentally determine the actual yield of the product.
  4. Calculate Purity: Purity (%) = (Actual yield / Theoretical yield) × 100.

For example, if you have an impure sample of a metal that reacts with an acid to produce a salt and hydrogen gas, by measuring the amount of salt or hydrogen produced, you can back-calculate the amount of pure metal that reacted, thus determining the sample's purity.

Identifying the limiting reactant in a reacting masses calculation is crucial for accurately determining the amount of product formed. To identify the limiting reactant, follow these steps:

  1. Calculate the Moles of Each Reactant: Use the formula (number of moles = mass / molar mass) to find the moles of each reactant given their masses.
  2. Use Stoichiometry: Refer to the balanced chemical equation and compare the mole ratios of the reactants.
  3. Determine the Limiting Reactant: The reactant that produces the least amount of product (when the mole ratios are applied) is the limiting reactant. It's the substance that will be completely consumed first in the reaction.
  4. Calculate the Product Mass Using the Limiting Reactant: Once the limiting reactant is identified, use its mole quantity to calculate the mass of the products.

For example, in a reaction where 2 moles of A react with 1 mole of B to produce C, if you start with 3 moles of A and 2 moles of B, A is in excess (would produce 1.5 moles of C), while B (would produce 2 moles of C) is the limiting reactant. The reaction will stop when B is completely used up.

Practice Questions

Calculate the mass of water produced when 16g of methane (CH₄) is completely combusted in excess oxygen. (Molar Mass: CH₄ = 16 g/mol, H₂O = 18 g/mol)

Methane undergoes combustion according to the equation: CH₄ + 2O₂ → CO₂ + 2H₂O. First, calculate the moles of CH₄: moles = mass / molar mass = 16g / 16 g/mol = 1 mole. The stoichiometry shows that 1 mole of CH₄ produces 2 moles of H₂O. Therefore, 1 mole of CH₄ produces 2 moles of H₂O. Then, convert moles of water to mass: mass = moles × molar mass = 2 moles × 18 g/mol = 36g. So, 36g of water is produced.

If 10g of aluminium reacts with excess hydrochloric acid to produce aluminium chloride and hydrogen gas, what is the mass of aluminium chloride formed? (Molar Mass: Al = 27 g/mol, AlCl₃ = 133.5 g/mol)

The reaction is: 2Al + 6HCl → 2AlCl₃ + 3H₂. Calculate the moles of Al: moles = mass / molar mass = 10g / 27 g/mol ≈ 0.37 moles. The stoichiometry of Al to AlCl₃ is 2:2, so 0.37 moles of Al produces 0.37 moles of AlCl₃. Convert moles of AlCl₃ to mass: mass = moles × molar mass = 0.37 moles × 133.5 g/mol ≈ 49.4g. Therefore, the mass of aluminium chloride produced is approximately 49.4g.

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