Determine the maximum amount of product that can be formed from a chemical reaction using stoichiometric ratios. Enter reactant mass, molar masses, and stoichiometric coefficients.
Theoretical yield is the maximum amount of product a chemical reaction can produce if it runs to completion with zero loss. It's a stoichiometric ceiling, not a real-world result — but knowing the ceiling is what lets you measure how well a reaction actually performs. Without a theoretical yield, you can't compute percent yield, can't plan how much starting material to use, and can't compare reactions across different scales or conditions.
The math is straightforward: convert reactant mass to moles using its molar mass, scale by the reaction's stoichiometric ratio (the coefficients in the balanced equation), then convert moles of product back to mass using the product's molar mass. The trickiest part for new students isn't the math but identifying the limiting reagent — when you have multiple reactants, only the one that runs out first sets the theoretical yield. The others are "in excess" and don't constrain how much product forms.
This calculator handles the standard "one limiting reagent" case. Enter the limiting reagent's mass and molar mass, the product's molar mass, and the stoichiometric coefficients from the balanced equation. The output is the theoretical yield in grams — the maximum product mass if the reaction went perfectly. Combine with the percent-yield calculator after the experiment to grade your actual performance.
**Scenario:** Burn 10.0 g of methane (CH₄, 16.04 g/mol) completely: CH₄ + 2 O₂ → CO₂ + 2 H₂O. Find theoretical CO₂ (44.01 g/mol). **Calculation:** Moles CH₄ = 10.0 / 16.04 = 0.6235 mol. Moles CO₂ = 0.6235 × (1/1) = 0.6235 mol. Mass CO₂ = 0.6235 × 44.01 = 27.44 g theoretical CO₂. **Result:** 10 g methane combusts completely to 27.44 g CO₂ (and ~22.5 g water). A typical residential gas stove burning 1 kg methane per day releases ~2.7 kg CO₂.
**Scenario:** Industrial lime kiln: 1 ton (1000 kg) of CaCO₃ (100.09 g/mol). Reaction: CaCO₃ → CaO + CO₂. Theoretical CaO (56.08 g/mol)? **Calculation:** Moles CaCO₃ = 1,000,000 g / 100.09 = 9991 mol. Moles CaO = 9991 × (1/1) = 9991 mol. Mass CaO = 9991 × 56.08 = 560,400 g = 560.4 kg. **Result:** 1 ton limestone gives 560 kg lime theoretically. Industrial yields ~95%, so actual delivery is ~530 kg. The other 440 kg "lost" is CO₂ released — a major source of industrial emissions from cement and lime production.
**Scenario:** Grignard reaction: 2.5 g RMgBr (175 g/mol) + 1.2 g aldehyde (148 g/mol) → alcohol product (200 g/mol). Stoichiometry 1:1:1. **Calculation:** Moles RMgBr = 2.5/175 = 0.01429 mol. Moles aldehyde = 1.2/148 = 0.00811 mol. Aldehyde has fewer moles → limiting. Theoretical product = 0.00811 × 200 = 1.62 g. **Result:** Theoretical yield is 1.62 g (limited by aldehyde). If Grignard typical yield is 70%, expect ~1.13 g actual. The excess Grignard (0.62 g, half unreacted) is consumed in workup or unwanted side reactions. Sometimes adding excess Grignard improves yield; sometimes it causes more side products — it depends on the specific substrate.
**Calculate theoretical yield to:**
- **Set up experiments correctly**: ensure you have enough starting material to produce the target product mass. - **Choose the right scale**: if a paper reports 65% yield at 5 g scale, plan starting material accordingly when scaling up. - **Compute percent yield**: actual / theoretical × 100, the standard metric for reaction efficiency. - **Identify the limiting reagent**: which reactant runs out first determines maximum product. - **Plan reagent ordering**: convert "I need 50 g of product" backward into "I need 75 g of starting material" via known yields. - **Compare reaction conditions**: same theoretical yield, different actual yields = different efficiency. - **Industrial process optimization**: knowing the theoretical ceiling sets the target for improvement.
**Steps for stoichiometric problem-solving:**
1. **Write the balanced equation.** No way around this. 2. **Identify limiting reagent** if multiple reactants are given. 3. **Convert reactant mass to moles** using molar mass. 4. **Apply stoichiometric ratio** (product coefficient / limiting reagent coefficient). 5. **Convert product moles to mass** using product molar mass. 6. **Sanity check**: does the answer make sense? Theoretical yield should never exceed the combined mass of reactants minus byproducts.
**Common reaction patterns:**
- **1:1 stoichiometry**: product_coeff/reactant_coeff = 1. Mass scales linearly through molar mass ratio. - **2:1 (two products from one reactant, like 2 H + O → 2 H₂O)**: product moles = 2 × reactant moles. - **N₂ + 3 H₂ → 2 NH₃**: ratio (2/3) for NH₃ from H₂, (2/1) for NH₃ from N₂. - **Combustion**: CₐHᵦ + (a + b/4) O₂ → a CO₂ + (b/2) H₂O. Coefficients depend on the hydrocarbon.
**Theoretical yield in industrial chemistry:**
- **Haber process** (NH₃): theoretical yield is 100% per pass, but equilibrium limits to ~15%; unconverted reactants are recycled. The plant's "real-world" yield approaches 100% overall through recycling. - **Ostwald process** (HNO₃): catalytic oxidation of NH₃; theoretical 100%, real-world 95-97% per pass. - **Contact process** (H₂SO₄): theoretical 100%, real-world >99% via SO₂/SO₃ recycling. - **Petroleum cracking**: theoretical varies with feedstock; real-world yields constrained by side reactions and thermodynamics.
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e.g. NaCl = 58.44
e.g. Na = 22.99
Theoretical Yield
3.934 g
Moles Product
0.17112 mol
Mole Ratio
1.00
| Parameter | Value |
|---|---|
| Reactant Mass | 10.0000 g |
| Reactant Molar Mass | 58.4400 g/mol |
| Moles of Reactant | 0.171116 mol |
| Mole Ratio (product/reactant) | 1/1 = 1.0000 |
| Moles of Product | 0.171116 mol |
| Product Molar Mass | 22.9900 g/mol |
| Theoretical Yield | 3.9339 g |
| Theoretical Yield (mg) | 3933.95 mg |