Percent Yield 49g48

Theoretical and Percent Yield The 1. 2. 3. 4. 5.

general mechanism for finding percentage yield is as follows: Balance the chemical equation Find the limiting reagent Find the theoretical yield Find the actual yield Find the percentage yield

Percentage Yield =

mass of Actual Yield x 100% mass of Theoretical Yield

1. Balancing The Chemical Equation: The first step in finding theoretical and percentage yield is to balance the rel evant chemical equation. How To Balance Chemical Equations Balancing the reaction's chemical equation is the first step in calculating theo retical and percent yield. The principle of balancing an equation is to obey the law of conservation of mas s,which states simply that matter cannot be created nor destroyed, but only tran sferred to a different format. When a chemical equation is balanced, it displays the ratios in which the reacta nts react. In essence, the amount of atoms on one side must equal the amounts o f atoms on the other. That is, the amount of carbons on the reactant side must equal the number of carbon atoms on the product side, etc.

The first step in balancing any equation, is to write out the correct chemical f ormula:

For Example:

CH4 + O2 --> CO2 + H2O

Is the reaction for the combustion of methane (CH4) in excess oxygen (O2) To balance the equation, you need to find the smallest whole number coefficients so that each element is balalnced in the reaction. To solve for these coefficie nts, use a sytem of equations: For the current example: aCH4 + bO2 --> cCO2 + dH2O Multiplying each element by the coefficient gives you the following equations: Eq'n 1: a = c When the equation is balanced, there will be (1 x a) Carbon atoms from methane o n the reactants side and (1 x c) Carbon atoms from carbon dioxide on the product s side Eq'n 2:

4a = 2d When the equation is balanced, there will be (4 x a) Hydrogen atoms from methane on the reactants side and (2 x d) hydrogen atoms from water on the products sid e Eq'n 2(i): 2a = d Note that Eq'n 2 can be simplified to this format Eq'n 3: 2b = 2c + d When the equation is balalnced, there will be (2 x b) oxygen atoms from oxygen g as on the reactant side, (2 x c) oxygen atoms from carbon dioxide on the product s side and (1 x d) oxygen atoms from water on the product side. Eq'n 3(i): b = 2a Note that this formula results by subsitituting Eq'n 1 and Eq'n 2(i) into Eq'n 3 and simplifying To solve for a, b, c and d: First take the smallest whole number that satisfies Eq'n 1, which is simply a = c =1 Then, substitute that value into Eq'n 2(i): 2a = d 2(1) = d d = 2 and b = b = b =

Eqn' 3(i): 2a 2(1) 2

Your coefficients are a = 1, b = 2, c = 1 and d = 2 When writing your balanced equation, the coefficient 1 is assumed and can be omi tted, yielding the formula: CH4 + 2O2 --> CO2 + 2H2O Now, your solution: Carbon: # of reactant moles: 1 # of product moles: 1 Balanced! Hydrogen: # of reactant moles: 4 # of product moles: 4 Balanced!

Oxygen: # of reactant moles: 4 # of product moles: 4 Balanced! As all the element are balanced, the solution is correct. aAl + bCl2 --> cAlCl3 Ans: a = 2 b = 3 c = 2 aSF4 + bH2 O--> cSO2 + dHF Ans: a = 1 b = 2 c = 1 d = 4 2. Finding The Limiting Reagent: * this is the reactant which the product yield depends on, as it is not in e xcess. * To determine which reactant is the limiting reagent 1(a). Divide the mass (in grams) of the reactant by its molecula r weight (in g/mol) OR 1(b). Multiply the amount used (in mL) by its density, then divi de by its molar mass 2. Multiply the mass (your answer from steps 1(a) or 1(b)) by the number of moles of the reactant used in the reaction. Example: Limiting Reagent Finding Limiting Reagent Example: salicylic acid + acetic anhydride « acetylsalicylic acid (ASA) + acetic acid Salicylic acid: (0.211 g) / (138.1 g/mol) * 1 = 0.00153 mol Acetic anhydride: (0.480 mL) * (1.08 g/mL) / (102.1 g/mol) * 1 = 0.00508 mol Therefore, salicylic acid is the limiting reagent, since there are less moles of that than there are of acetic anhydride. 3. Theoretical Yield * this is how much product will be synthesized in ideal conditions. * To determine theoretical yield, multiply the amount of moles of the limiti ng reagent by the ratio of the limiting reagent and the synthesized product and by the molecular weight of the product. Example: Theoretical Yield Theoretical yield = 0.00153mol sal. acid *

1 mol ASA

*

180.2g ASA

. 1 mol sal. acid

1 mol ASA = 0.276 g ASA

Therefore, in a perfect experiment, 0.276 g of acetylsalicylic acid will be synt hesized. 4. Actual Yield * this is how much product was actually synthesized in the experiment. * Example: 0.135 g acetylsalicylic acid

Theoretical and Percent Yield This site is designed to help students calculate theoretical and percent yield. There are links at the bottom of this page for additional tutorials, such as ho w to balance equations, etc. PLEASE NOTE: LINK TO CRITIQUE IS FOUND AT THE BOTTOM OF THIS PAGE The 1. 2. 3. 4. 5.

general mechanism for finding percentage yield is as follows: Balance the chemical equation Find the limiting reagent Find the theoretical yield Find the actual yield Find the percentage yield

Percentage Yield =

mass of Actual Yield x 100% mass of Theoretical Yield

1. Balancing The Chemical Equation: The first step in finding theoretical and percentage yield is to balance the rel evant chemical equation. If you are unsure how to do this, click on the "How to Balance Chemical Equations" link below: How to Balance Chemical Equations 2. Finding The Limiting Reagent: * this is the reactant which the product yield depends on, as it is not in e xcess. * To determine which reactant is the limiting reagent 1(a). Divide the mass (in grams) of the reactant by its molecula r weight (in g/mol) OR 1(b). Multiply the amount used (in mL) by its density, then divi de by its molar mass 2. Multiply the mass (your answer from steps 1(a) or 1(b)) by the

number of moles of the reactant used in the reaction. Unsure of how to calculate molar mass or molecular weight? Click the link below: Calculating Molar Mass or Molecular Weight Example: Finding the Limiting Reagent 3. Theoretical Yield * this is how much product will be synthesized in ideal conditions. * To determine theoretical yield, multiply the amount of moles of the limiti ng reagent by the ratio of the limiting reagent and the synthesized product and by the molecular weight of the product. Example: Theoretical Yield 4. Actual Yield * this is how much product was actually synthesized in the experiment. * Example: 0.135 g acetylsalicylic acid 5. Percent Yield * The percentage yield is the ratio between the actual yield and the theoret ical yield multiplied by 100%. It indicates the percent of theoretical yield th at was obtained from the final product in an experiment. * The percentage yield can be calculated using the mass of the actual produc t obtained and the theoretical mass of the product calculated using the balanced equation of the reaction. Percentage Yield =

Mass of Actual Yield Mass of Theoretical Yield

x

100%

In the following reaction, 0.157g of p-acetaminophenol was used to react with 0. 486 g of acetic anhydride to produce acetaminophen and acetic acid. The product was purified and acetimophen was extracted. The actual mass of acetaminophen p roduced was 0.198 g. Determine the theoretical yield and the percent yield of i sopentyl acetate. p-Aminophenol cid

+

Acetic anhydride

C6H7NO

À C4H6O3

CH3COOH Find the limiting reagent

molar mass of p-aminophenol =109.1g/mol molar mass of acetic anhydride = 102.1 g/mol

Acetaminophen

+ C8H9NO2

Acetic a

moles of p-aminophenol = mass/molar mass = 0.157g/(109.1g/mol) = 0.00144 mol moles of acetic anhydride = mass/molar mass = 0.486g/(102.1g/mol) = 0.00476 mol From the balanced chemical equation we know that for every mole of p-aminophenol and acetic acid one mole of acetaminophen is produced. Since there are more moles of acetic anhydride than p-aminophenol, p-aminophenol is used up first as the reaction proceeds. We can conclude that acetic anhydri de is in excess and p-aminophenol is the limiting reagent. Therefore # moles of p-aminophenol = # of acetaminophen produced Finding the Theoretical Yield Theoretical Yield = moles of acetamiophen x molar mass of acetaminophen = 0.00144 mol x 151.2g/mol = 0.217 g

Finding the Percent Yield Percent Yield =

Actual Yield x 100% Theoretical Yield = = 91.2 %

0.198g 0.217g

x 100 %