Percent Yield Calculations

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Percent Yield Calculations Percent yield calculations are one of the most commonly used calculations in organic chemistry lab. Unfortunately the calculations are also one of the most common areas where students make mistakes. Understanding how to routinely and accurately carry out percent yield calculations will bolster understanding of the chemistry behind the calculation and simultaneously help improve grades. Percent Yield calculations can be broken down into 4 simple steps. 1. Determine the moles (not grams, not mls) of each of the reactants. 2. Calculate how many moles of product can form based on the moles of each reactant. The stoichiometry of the equation must be used for this calculation. The reactant which will produce the smallest amount of the product is the limiting reagent. The limiting reagent is the starting material which will be depleted first. The amount of product formed from the limiting reagent is the theoretical maimum amount of product which can form. 3. Convert the theoretical maimum amount of product which can form in moles (from #2) into grams. 4. Determine the percent yield of the reaction by using knowledge of the actual grams of product formed and #3 above. Many times steps one, two and three are done together in a single step using dimensional analysis. At the beginning it may be easier to do each step individually, but with eperience conducting all three steps in tandem will become more efficient and simple. Three practice problems will be shown using both individual steps and tandem steps. Reminders * Density is used to convert from volume to mass. * Molecular weight is used to convert from mass to moles. * Reaction Stoichiometry is used to convert from moles of reactants to moles product. * If a solution as opposed to a neat liquid is used, then the volume (in liters) of solution used can be multiplied by the molarity (M = moles/l) of the solution to determine moles. * The theoretical amount of product formed must be calculated using moles of limiting reagent (not grams, not mls). * Keep track of all units. Often when a student is unsure how to proceed, a close eamination of the units will direct the correct net multiplication or division. * Be very aware of the stoichiometry of the reaction. * All data provided may not have to be used to determine answer.

Practice problem 1. One material measured as a solid ( one material as a neat liquid (ml). 2.897 g of maleic anhydride (MA) is reacted with 2.56 ml of cycloheadiene (CHD) to produce 3.979 grams of the bicyclic product. What is the limiting reagent and what is the percent yield? name cycloheadiene (CHD) maleic anhydride (MA) 5,6-anhydride [2.2.2]bicyclo-2- octene density (g/ml) 0.8405 1.314 mol. Wt. 80.15 98.06 178.20 (g/mol) volume (ml) 2.56 - - mass( - 2.897 3.979 Tandem steps 1,2&3. Steps one, two and three can be carried out at the same time. This calculation must be carried out for all reactants. For maleic anhydride 2.897 _ g _ MA( 1_ mole _ MA( mol) 1_ mol _ Pr oduct 178.20 _ grams 98.06grams _ MA( 1_ mol _ MA 1_ mol _ Pr _ product oduct 5.265 _ grams _ Pr oduct Step 1 Step 2 Step 3 For cycloheadiene 0.845gCHD 1_ moles _ CHD 1_ moles _ product 178.20grams _ product 2.45 _ ml _ CHD 4.78 _ g _ Pr 1_ ml _ CHD 80.15 _ g _ CHD 1_ mole _ CHD 1_ mole _ product oduct Step 1 Step 2 Step 3 Since 4.78 is smaller than 5.265, cycloheadiene is the limiting reagent and 4.78 g is the theoretical maimum amount of product which can form. Individual steps Step 1. Alternatively steps one, two and three may be done individually. Determine the moles of all reagents. -If the reagent is in grams then just use the molecular weight to convert to moles. -Keep track of all units to ensure the correct multiplication is taking place. 2 2.897 _ g _ MA 1_ mole _ MA 2.95410 moles _ MaleicAnhy dride 98.06 _ grams _ MA.56 _ ml _ CHD 0.8405 _ g _ CHD 1_ mole _ CHD 2.6810 moles _ Cycloheadiene 1_ ml _ CHD 80.15 _ g _ CHD 2 2 The reaction stoichiometry must be used. This reaction has a 1:1 stoichiometry. Determine the

limiting reagent..95410 moles _ MA 1_ moles _ product 2.95410 1_ mole _ MA 2 ( fromma) 2.6810 moles _ CHD 1_ moles _ product 2.6810 ( fromchd) 1_ mole _ CHD The reactant which will product the smallest amount of product is the limiting reagent. Since 2.6810-2 is smaller than 2.95410-2, cycloheadiene is the limiting reagent in this problem. Step 3. Using the moles of product which form from the limiting reagent, determine the grams of product that can form. This amount is the theoretical maimum of product. 2.6810 moles _ Pr oduct 178.20 _ grams _ product 1_ mole _ product 4.78 _ grams _ Pr oduct Both methods give the same answer. Initially students may want to carry out the steps individually, but the tandem method is more efficient. 100% % yield _ Pr oduct Theoretical _ Maimum _ Mass _ Pr oduct ( 3.979 _ g _ Pr oduct 100% 83.2 _ % _ yield _ Pr oduct 4.78 _ g _ product Practice Problem 2. One material measured as a solution (M) one as a mass (. 4.24 ml of 2-butanol is reacted with 10.2 ml of 6.0 M HCl to form 2.8798 grams of 2-chlorobutane. What is the limiting reagent and what is the percent yield. name 2-butanol Hydrochloric Acid 2-chlorobutane Water (BuOH) (6.0 M) density 0.81 1.314 1.00 (g/ml) mol. Wt. 74.1 36.5 92.6 18.00 (g/mol) volume 4.24 10.2 - (ml) mass( - - 2.8798 Tandem steps 1,2 and 3 4.24 _ ml _ BuOH 0.810 _ g _ BuOH 1_ ml _ BuOH 1_ mole _ BuOH 74.1_ g _ BuOH 1_ mol _ product 92.6 _ g _ Pr od 1_ mol _ BuOH 1_ mol _ Pr od 4.28grams Pr oduct 10.2 _ ml _ HCl 1L _ HCl 6.0 _ mol _ HCl 1_ mol _ product 92.6 _ g _ Pr od 5.7 _ grams Pr oduct 1000 _ ml _ HCl 1_ L _ HCl 1_ mol _ HCl 1_ mol _ Pr od Since 4.28 is less than 5.7, 2-butanol is the limiting reagent and the maimum theoretical amount of product which can form is 4.28 g.

Individual steps Step 1. Determine the moles of all reagents. -If the reagent is in grams then just use the molecular weight to convert to moles. - If the reagent is in volume with a set molartiy, then use volume and molarity to determine moles. Remember Molarity has units of mole/liter Moles of 2-butanol.24 _ ml _ 2 BuOH 0.810 _ g _ BuOH 1_ mole _ BuOH 1_ ml _ BuOH 74.1_ g _ BuOH Moles of HCl 4 2 4.6310 10.2 _ ml _ HCl 1_ L _ HCl 6.0 _ mole _ HCl 6.110 moles _ HCl 1000 _ ml HCl 1_ L _ HCl moles _ 2 Bu tan ol The reaction stoichiometry must be used. This reaction has a 1:1 stoichiometry. Determine the limiting reagent. 4.6310 moles _ BuOH 1_ moles _ product 4.6310 1_ mole _ BuOH 6.110 moles _ HCl 1_ moles _ product 6.110 1_ mole _ HCl The reactant which will product the smallest amount of product is the limiting reagent. Since 4.6310-2 is smaller than 6.110-2, 2-butanol is the limiting reagent. Step 3. Using the moles of the limiting reagent,determine the theoretical maimum grams of product that can form. Keep track of units to avoid errors. 4.6310 moles _ Pr oduct 92.6 _ grams _ product 1_ mole _ product 4.28 _ grams _ product This same answer was obtained when the steps were carried out sequentially. 100% % yield _ Pr oduct Theoretical _ Maimum _ Mass _ Pr oduct ( 2.8798 _ g _ Pr oduct 100% 67.3 _ % _ yield _ 2 ChloroBu tan e 4.28 _ g _ product Practice Problem 3. Non 1 :1 stoichiometry. Both materials measured as a neat liquid (ml). 7.3 ml of methylbenzoate is reacted with 10.0 ml of bromobenzene in an ecess of magnesium shavings and an acid work-up to produce triphenylmethanol. What is the limiting reagent, the theoretical maimum amount of product which can form, and the percent yield? (An ecess of magnesium is used.)

name methyl benzoate Bromobenzene triphenylmethanol (MB) (BrB) density (g/ml) 1.09 1.50 mol. Wt. (g/mol) 136.1 157.02 260.3 volume (ml) 7.3 10.0 - mass( - - 10.655 Tandem steps 1,2 and 3 7.3 _ ml _ MB 1.09 _ g _ MB 1_ mole _ MB 1_ mol _ Pr od 260.5 _ g _ Pr od 15 _ g _ Pr oduct 1_ ml _ MB 136.1_ g _ MB 1_ mol _ MB 1_ mol _ prod 10.0 _ ml _ BrB 1.50 _ g _ BrB 1_ mol _ BrBe 1_ mol _ Pr od 260.5 _ g _ Pr od 12.5 _ g _ Pr oduct 1.00 _ ml _ BrB 157.02 _ g _ BrB 2 _ mol _ BrB 1_ mol _ Pr od Because 12.5 is smaller than 15, bromobenzene is the limiting reagent and the theoretical maimum amount of product which can form is 12.5 grams. IndividuallyStep 1. Determine the moles of all reagents. -If the stoichiometry is not 1:1 pay close attention when calculating the limiting reagent. -Keep vigilant concerning units. -It is not necessary to calculate this for magnesium since it was used in ecess. 7.3 _ ml _ MeBenzoate 1.09 _ g _ MeBenzoate 1_ mole _ MeBenzoate 5.810 moles _ Methyl _ Benzoate 1_ ml _ MeBenzoate 136.1_ g _ MeBenzoate 10.0 _ ml _ Br Benzene 1.50 _ g _ Br Benzene 1_ mol _ Br Benzene 9.5510 moles _ BromoBenzene 1.00 _ ml _ Br Benzene 157.02 _ g _ BrBenzene The reaction stoichiometry must be used. Determine Limiting Reagent. This reaction does NOT have a 1:1 stoichiometry in regard to all reactants. Eamine the reaction. 5.810 moles _ MB 1_ moles _ product 5.810 1_ mole _ MB 9.55 moles _ BrB 1_ moles _ product 4.7810 2 _ mole _ BrB 4.78 10-2 is smaller than 5.8 10-2, therefore Bromobenzene is the limiting reagent. Note that the limiting reagent in this case is the reagent which is present in the larger volume. Step 3. Using the moles of the limiting reagent, determine the theoretical maimum grams of product that can form. Keep track of units to avoid errors. 4.7810 moles _ MethylBenzoate 1_ moles _ Triphenylmethanol 260.5 _ g _ Triphenylmethanol 12.5 _ grams _ product 1_ mole _ MethylBenzoate 1_ mole _ Triphenylmethanol 10.655 _ g _ Pr oduct 100% 100% 85.2 _ % _ yield Theoretical _ Maimum _ Mass _ Pr oduct ( 12.5 _ g _ product

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