Stoichiometry Formula
Quantitative relationships between reactants and products in chemical reactions
Core Stoichiometric Relationships
Mole Ratio from Balanced Equation
aA + bB → cC + dD
n(A)/a = n(B)/b = n(C)/c = n(D)/d
Mass to Mole Conversion
n = m / M
moles = mass / molar mass
General Conversion Pathway
mass A → moles A → moles B → mass B
(÷M_A) → (×ratio) → (×M_B)
Step-by-Step Process
Balance the Chemical Equation
Ensure equal atoms on both sides
Convert Given Mass to Moles
Use molar mass: n = m/M
Apply Mole Ratio
Use coefficients from balanced equation
Convert Back to Mass (if needed)
Multiply by molar mass: m = n × M
Worked Examples
Example 1: Mass-to-Mass Calculation
Problem: How many grams of H₂O form from 10.0 g H₂?
2H₂ + O₂ → 2H₂O
Solution:
Step 1: Convert H₂ to moles
n(H₂) = 10.0 g / 2.02 g/mol = 4.95 mol
Step 2: Apply mole ratio (2:2 = 1:1)
n(H₂O) = 4.95 mol × (2/2) = 4.95 mol
Step 3: Convert to mass
m(H₂O) = 4.95 mol × 18.02 g/mol
m(H₂O) = 89.2 g
Example 2: Finding Reactant Mass Needed
Problem: How much N₂ needed to produce 34.0 g NH₃?
N₂ + 3H₂ → 2NH₃
Solution:
n(NH₃) = 34.0 / 17.03 = 2.00 mol
n(N₂) = 2.00 × (1/2) = 1.00 mol
m(N₂) = 1.00 × 28.02
m(N₂) = 28.0 g
Example 3: Combustion Reaction
Problem: How much O₂ needed to burn 8.00 g CH₄?
CH₄ + 2O₂ → CO₂ + 2H₂O
Solution:
n(CH₄) = 8.00 / 16.04 = 0.499 mol
n(O₂) = 0.499 × (2/1) = 0.998 mol
m(O₂) = 0.998 × 32.00
m(O₂) = 31.9 g
Common Mistakes
Unbalanced Equations
Must balance equation first - mole ratios only work with balanced equations
Wrong Molar Mass
Must account for subscripts: H₂O is 18.02 g/mol, not 17.01
Inverted Mole Ratio
2H₂ + O₂ → 2H₂O: ratio H₂:O₂ is 2:1, not 1:2
Limiting Reagent Scenarios
When both reactants given, identify limiting reagent first