Rendimento teórico

Theoretical yield: stoichiometric maximum

Theoretical yield is the most product a reaction could ever give you. Imagine a perfect world where every molecule of the limiting reagent reacts, nothing gets lost, and no side reactions skim off material. Reaching the number takes four steps. First you find the limiting reagent. Then you convert its mass to moles with n = m/MM. The coefficient ratio from the balanced equation tells you how many moles of product that makes, and you finish by multiplying by the product's molar mass. Suppose 5.85 g NaCl (MM 58.5) reacts with excess AgNO₃ to form AgCl. That works out to 0.1 mol NaCl; with a 1:1 ratio you get 0.1 mol AgCl; and 0.1 · 143.3 = 14.33 g of theoretical product. Whatever you actually recover from the flask comes in under that figure, and actual/theoretical · 100 gives your percent yield.

Applications

Chemists turn to it when planning a synthesis. A pharmaceutical team will use it to size a batch, working out how much API comes out of N kg of intermediate. You also see it throughout chemistry coursework and in ENEM and vestibular exam problems, while process engineers treat it as a yield benchmark during cost analyses.

FAQ

Why is actual yield always lower? A handful of things chip away at it. Side reactions burn off reagents. Some product disappears during filtration or transfer. Equilibrium reactions stop short of full conversion, and impurities pull the purity down. For organic synthesis, ending up somewhere between 60 and 80% is perfectly normal.

What if the reagent isn't limiting? Base the calculation only on the moles of the limiting reagent. Whatever is in excess just sits there unreacted and adds nothing to the product. Not sure which one is limiting? Run that tool first.

Does the equation need to be balanced? Yes. Everything rides on the coefficient ratio, and that ratio only comes from a balanced equation. Leave it unbalanced and your mole ratios are off, an error that carries straight through to a wrong theoretical mass.