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Percent Yield Lab Report

1158 Words5 Pages

This report presents how to optimize experiment and equipment to obtain highest percent yield possible. Percent yield is the percent ratio of actual yield to the theoretical yield. % yield= (Actual yield)/(Theoretical yield)×100% Where the actual yield is the amount of product actually produced in a chemical reaction in grams, theoretical yield is the amount of product obtained from the stoichiometry or balanced equation using the limiting reactant to determine the product. Units for both actual and theoretical yield need to be the same (moles of grams) (”Percent Yield Definition and Formula”, 2018). This report discusses the study of how the chemical composition and the mass of specific substances may result in higher percent yields. This report presents a chemical reaction between copper (ll) sulphate pentahydrate and magnesium metal to obtain a base percent yield. Where the purpose of the experiment will be to obtain a percent yield higher than 71 %. If 0.1g Magnesium reacts with 0.7g Copper ll Sulfate pentahydrate, then a percent yield higher than 71 % will be obtained because Magnesium is higher on the reactivity scale than aluminum as well Magnesium has an electronegativity of 1.3 …show more content…

As copper (II) chloride has a higher hydrogen bond donor and acceptor count than copper (II) chloride, as copper (II) chloride has 0 both hydrogen bond donor and acceptor count where copper (II) sulphate has 5 donor count and 9 acceptor count that shows which solution is stronger. As well Copper (II) sulphate has more complexity of 62.2 while copper (II) chloride has a complexity of 2.8 which indicates which substances are better than the other thus this hypothesis is accepted as copper (II) sulphate is better mixture than copper (II chloride. (”Cupric chloride. (n.d.). Retrieved from” ) (“Cupric sulphate. (n.d.). Retrieved

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