Mass-mass stoichiometry calculations determine the amount of product formed or reactant consumed in a chemical reaction, given the mass of another reactant or product. These calculations rely on understanding mole ratios from balanced chemical equations and molar masses.
Understanding the Process
The core principle lies in converting mass to moles, using mole ratios from balanced equations to relate different substances, and then converting moles back to mass. This three-step process is summarized by the equation:
moles A x (mole ratio of B/A) x molar mass of B = mass of B
Where:
- moles A: Moles of the substance whose mass is known.
- mole ratio of B/A: The ratio of moles of substance B (the desired substance) to moles of substance A (the known substance), derived from the balanced chemical equation.
- molar mass of B: The mass of one mole of substance B (grams/mol).
- mass of B: The mass of substance B calculated.
Step-by-Step Guide
- Balance the chemical equation: Ensure the equation accurately reflects the reaction's stoichiometry. This provides the crucial mole ratios.
- Convert mass to moles: Use the molar mass of substance A to convert its given mass into moles. (Moles = mass / molar mass)
- Use the mole ratio: Apply the mole ratio from the balanced equation to find the moles of substance B.
- Convert moles to mass: Use the molar mass of substance B to convert its moles into mass. (Mass = moles x molar mass)
Example
Let's say we want to find the mass of water (H₂O) produced from the complete combustion of 10 grams of methane (CH₄):
CH₄ + 2O₂ → CO₂ + 2H₂O
- Molar masses: CH₄ = 16 g/mol; H₂O = 18 g/mol
- Moles of CH₄: 10 g / 16 g/mol = 0.625 mol
- Mole ratio: From the equation, 1 mol CH₄ produces 2 mol H₂O.
- Moles of H₂O: 0.625 mol CH₄ x (2 mol H₂O / 1 mol CH₄) = 1.25 mol H₂O
- Mass of H₂O: 1.25 mol x 18 g/mol = 22.5 g
Therefore, 22.5 grams of water are produced.
Beyond Simple Reactions
The fundamental principles remain the same, even for more complex reactions involving multiple steps or limiting reactants. Identifying the limiting reactant is crucial for accurate mass-mass calculations in such scenarios. Remember to always start with a balanced chemical equation.
