Molar Ratio Calculators

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A molar ratio is the ratio of moles of one substance to moles of another in a chemical reaction or formula, derived from the balanced chemical equation. It is the cornerstone of stoichiometric calculations — converting between amounts of reactants and products in a reaction. For a balanced equation aA + bB → cC + dD, the molar ratios are a:b:c:d. Molar ratios are used to calculate how much product can be made from a given amount of reactant, to identify the limiting reagent, and to determine theoretical yield in synthesis reactions.

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Molar Ratio from Balanced Equations

In the reaction: N₂ + 3H₂ → 2NH₃

Molar ratios: 1 mol N₂ : 3 mol H₂ : 2 mol NH₃. To produce 4 mol NH₃: need 2 mol N₂ and 6 mol H₂. These ratios are always in the proportions given by the balanced equation stoichiometric coefficients.

Molar Ratio Calculations

Step-by-step method:

  1. Write and balance the equation
  2. Convert grams to moles: n = mass / molar mass
  3. Use molar ratio to convert moles of A to moles of B
  4. Convert moles of B back to grams if needed: mass = n × molar mass

Example: how many grams of H₂O form from 5.0 g H₂? n(H₂) = 5.0/2.016 = 2.48 mol. H₂ + ½O₂ → H₂O: ratio 1:1. n(H₂O) = 2.48 mol. Mass H₂O = 2.48 × 18.015 = 44.7 g.

Limiting Reagent

The limiting reagent is the reactant that runs out first and determines the maximum yield. Find the moles of each reactant, divide by its stoichiometric coefficient — the smallest quotient identifies the limiting reagent. Theoretical yield is calculated from the limiting reagent using the molar ratio to the product.

Glossary

Molar Ratio
The ratio of moles of one substance to another in a balanced chemical equation; derived from stoichiometric coefficients; used to convert between moles of reactants and products.
Limiting Reagent
The reactant that is completely consumed first in a reaction, limiting the amount of product formed; identified by dividing moles of each reactant by its stoichiometric coefficient.
Theoretical Yield
The maximum mass of product that can form from the limiting reagent with 100% conversion; calculated from moles of limiting reagent × molar ratio to product × molar mass of product.

Frequently Asked Questions

A molar ratio is the ratio of moles of substances in a balanced chemical equation. For N₂ + 3H₂ → 2NH₃: the molar ratio of N₂ to H₂ is 1:3; H₂ to NH₃ is 3:2. Use molar ratios to convert between moles of different substances in a reaction. Example: given 6 mol H₂, how much NH₃ forms? Ratio H₂:NH₃ = 3:2, so NH₃ = 6 × (2/3) = 4 mol. This conversion is the core calculation step in all stoichiometry problems.

Step 1: Convert all reactant amounts to moles. Step 2: Divide each by its stoichiometric coefficient. Step 3: The reactant with the smallest quotient is limiting. Example: 3.0 mol N₂ and 6.0 mol H₂ react via N₂ + 3H₂ → 2NH₃: N₂: 3.0/1 = 3.0; H₂: 6.0/3 = 2.0. H₂ is limiting (smaller quotient). Theoretical yield NH₃ = 6.0 mol H₂ × (2 mol NH₃/3 mol H₂) = 4.0 mol NH₃. N₂ remaining = 3.0 − (6.0/3) = 3.0 − 2.0 = 1.0 mol N₂ in excess.

Molar ratios are based on counting atoms and molecules (moles); mass ratios are based on weighing. For H₂O (MW = 18.015 g/mol) from H₂ (MW = 2.016) + ½O₂ (MW = 32.00): molar ratio H₂:H₂O = 1:1 (one mole of each). Mass ratio H₂:H₂O = 2.016:18.015 = 0.112:1 (very different from the molar ratio). Chemical reactions always follow molar (not mass) ratios because atoms and molecules react in fixed number ratios. Converting moles to mass (or vice versa) using molar mass is how you go between counting and weighing.

Theoretical yield is the maximum amount of product that could form if the limiting reagent is completely consumed with no side reactions. Steps: (1) Identify limiting reagent. (2) Convert limiting reagent moles to product moles using molar ratio from balanced equation. (3) Convert product moles to grams. Example: 5.0 g Fe₂O₃ (MW = 159.7) reacts with excess Al: Fe₂O₃ + 2Al → Al₂O₃ + 2Fe. n(Fe₂O₃) = 5.0/159.7 = 0.0313 mol. n(Fe) = 0.0313 × 2 = 0.0626 mol. Theoretical yield Fe = 0.0626 × 55.85 = 3.50 g.