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Why is atomic mass significant to calculating percent yield

Why is atomic mass significant to calculating percent yield
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Atomic mass is significant to calculate percent yield, since we need values of products and reactants in moles and then calculcal the yield and not by mass.

For example, let's imagine in a balanced equation one mole of A reacts with one mole of B, to give one molm of product C. Now, 5 grmas of A and 4 grams of B would have reacted to give 7 grams of product. By the mass we cannot make out yield. So first we have to know the molar mass of A and B. Then divide 5 grams with molar mass of A to get moles of A, similarly find moles of B. Whichever is less is the limiting reagent and will use as a reference to find yield.

Now find the moles of product C formed by dividing 7 grams with molar mass of C.

Now percepe yield would be moles of product, divided by moles of limiting reactant and then multiplying the resultant value with hundred.

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