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Percent Yield

Percent Yield.

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Percent Yield

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  1. Percent Yield • In theory, everything is perfect. People do what is right, even if it hurts them, simply because that is what is right. If a person works hard toward a goal, one day he will achieve it. If I buy expensive, energy efficient light bulbs that are supposed to last for 5 years, I will only have to buy light bulbs once every 5 years. • Unfortunately, we do not live in a theoretical world. In our world, things go wrong. And as we’ve seen with many other things, as goes the world, goes chemistry. • Therefore, our perfectly balanced chemical equations that say if we put in so many grams/moles of X we will get back so many grams/moles of Y are victims of our imperfect world also. In the real world, we will only get a percentage of the Y out that the chemical equation predicted for us.

  2. Percent Yield…continued • Chemical Equation produces →Theoretical (perfect) Yield • Real World produces → Actual (imperfect) Yield • The actual yield can never be greater than the theoretical yield. • The actual Yield can never be more than 100% of the theoretical yield. • The discrepancy is not due to an incorrect equation. It is due to imperfections in the process such as: • Reactants or products leak out, especially when they are gases. • The reactants are not 100% pure. • Some product is lost when it is purified. • There are also many chemical reasons, including: • The products decompose back into reactants (as with the ammonia process). • The products react to form different substances. • Some of the reactants react in ways other than the one shown in the equation. • These are called side reactions. • The reaction occurs very slowly. This is especially true of reactions involving organic substances.

  3. Steps to Solve Percent Yield Problems • Identify the Limiting Reactant in the reaction. • Calculate the Theoretical Yield of the substance in question. • Identify what you know and what you do not know and write it down. • Convert any masses given to moles. • Mass of substance (g) = Number of Moles Molar Mass of substance (g/mol) • Determine the Mole Ratio needed. • Coefficient of unknown substance in BALANCED chemical equation Coefficient of Limiting Reactant in BALANCED chemical equation • # mol unknown molecule # mol Limiting Reactant • # mol comes from BALANCED chemical equation NOT the problem itself. • Use the mole ratio to determine the amount of product the known reactant will produce. • Known Moles of Limiting Reactant X Mole Ratio = Moles of Unknown • Convert moles of product to grams. • # of Moles Calculated X Total Molar Mass (g/mol) = Theoretical Mass of substance (g) • Determine % Yield • Mass of Actual Yield (g) X 100 = % Yield Mass of Theoretical Yield (g)

  4. We Percent Yield Practice 1. Dichlorine monoxide, Cl2O, is sometimes used as a powerful chlorinating agent in research. It can be produced by passing chlorine gas over heated mercury (II) oxide according to the following equation: HgO + Cl2 → HgCl2 + Cl2O What is the % yield, if the quantity of reactants is sufficient to produce 0.86g of Cl2O but only 0.71g is obtained?

  5. You Percent Yield Practice …continued 2. Acetylene, C2H2, can be used as an industrial starting material for the production of many organic compounds. Sometimes, it is first brominated to form 1,1,2,2-tetrabromoethane, CHBr2CHBr2, which can then be reacted in many different ways to make other substances. The equation for the bromination of acetylene follows: C2H2 + 2Br2 → CHBr2CHBr2 If 72.0g of C2H2 reacts with excess bromine and 729g of the product is recovered, what is the % yield of the reaction?

  6. Percent Error Notes Percent error is a calculation of how efficient your lab technique is. Just like % Yield, it compares your Theoretical yield to your Actual yield. However, the focus is on what was not produced instead of what was produced. Steps to Solve % error Problems: • Identify your Actual and Theoretical Yields • Mass of Theoretical Yield(g) - Mass of Actual Yield (g)X 100 = Mass of Theoretical Yield (g)

  7. We Percent Error Practice 1. Dichlorine monoxide, Cl2O, is sometimes used as a powerful chlorinating agent in research. It can be produced by passing chlorine gas over heated mercury (II) oxide according to the following equation: HgO + Cl2 → HgCl2 + Cl2O What is the % yield, if the quantity of reactants is sufficient to produce 0.86g of Cl2O but only 0.71g is obtained?

  8. You Percent Error Practice …continued 2. Acetylene, C2H2, can be used as an industrial starting material for the production of many organic compounds. Sometimes, it is first brominated to form 1,1,2,2-tetrabromoethane, CHBr2CHBr2, which can then be reacted in many different ways to make other substances. The equation for the bromination of acetylene follows: C2H2 + 2Br2 → CHBr2CHBr2 If 72.0g of C2H2 reacts with excess bromine and 729g of the product is recovered, what is the % yield of the reaction?

  9. Homework • Modern Chemistry Book page 321 questions 22-30 • Modern Chemistry Book page 318 Practice Questions 1 and 2 • Whoosh Bottle Questions • Al Leftovers Lab Part II

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