Example 1 — Direct Percent Yield
Theoretical product = 8.00 g; actual isolated = 7.20 g.
Percent Yield Calculator computes theoretical and actual yields, limiting reagent, efficiency percentage, step-by-step stoichiometry, with clear, LaTeX equations online.
%Yield = (Actual / Theoretical) × 100 =
(7.5 / 8.0) × 100 =
93.75%
A Percent Yield Calculator helps chemists and students evaluate the efficiency of a chemical reaction. Laboratory reactions often experience losses due to side reactions, incomplete conversion, transfer, or purification steps. The calculator compares the actual yield obtained experimentally with the theoretical yield predicted from stoichiometry and the limiting reagent. It provides step-by-step guidance: identifying the limiting reagent, converting masses to moles, calculating theoretical product mass, and computing percent yield. Equation previews and unit checks ensure accuracy and ease of verification.
Enter reactant identities (optional), molar masses, and starting amounts (mass or moles). Using the balanced chemical equation
the tool determines the limiting reagent by comparing available mole ratios to required stoichiometric ratios. The theoretical product moles follow from the limiting reagent, then multiplying by the product's molar mass gives the theoretical mass. Input the experimentally recovered product mass to compute percent yield. Optional outputs include percent error and overall yield for multistep syntheses. All formulas are displayed in LaTeX for reports or lab notebooks.Moles from mass:
Limiting reagent for }aA + bB \to cC\text{:}
Theoretical mass of product:
Percent yield:
Percent error (optional):
Overall multistep yield:
Theoretical product = 8.00 g; actual isolated = 7.20 g.
Reaction: \( A + 2B \to C \), \( M_A = 50, M_B = 20, M_C = 90\,\text{g/mol} \).
Starting amounts: \( m_A = 10\,\text{g} \Rightarrow n_A = 0.20\,\text{mol} \), \( m_B = 12\,\text{g} \Rightarrow n_B = 0.60\,\text{mol} \).
Required B = 0.40 mol (available 0.60), so A is limiting.
\[ n_{C,\text{theor}} = 0.20\,\text{mol} \Rightarrow m_{C,\text{theor}} = 18.0\,\text{g} \]
If \( m_{C,\text{actual}} = 15.3\,\text{g} \), then
\[ \%\text{Yield} = \frac{15.3}{18.0} \times 100\% = 85\% \]
Step yields: 80%, 75%, 90%
\[ \%\text{Overall} = 0.80 \times 0.75 \times 0.90 \times 100\% = 54\% \]
It is the ratio of actual product obtained to the theoretical maximum, expressed as a percentage.
Losses occur due to side reactions, incomplete conversion, transfer, purification, or product instability.
Apparent yields above 100% usually result from impurities or residual solvent; purification reduces it.
Yes. Correct stoichiometric coefficients are essential to identify the limiting reagent and compute theoretical yield.
Include the correct molar mass (including bound water or solvent) or convert consistently to anhydrous mass.
Convert volumes to moles using \( n = C \times V \) with consistent units before applying stoichiometry.
Report according to the least precise measurement, typically 2–3 significant figures in introductory labs.
Impure reactants reduce effective moles. Adjust \( n \) by the purity fraction to avoid overestimating theoretical yield.
Percent yield measures the efficiency of isolation, while atom economy measures theoretical resource utilization in reaction design.
Multiply step yields as fractions and convert to a percentage:
It can push conversion toward completion but cannot exceed the theoretical maximum set by the limiting reagent.
Product may remain dissolved or be lost during transfer/filtration; optimized technique and solvent choice minimize losses.