What is percent yield?
The percent yield section is very simple mathematically. However, the components of the math equation for this section are often not clearly labeled. On top of that, percent yield will quite frequently be mixed in with other concepts like stoichiometry. In light of this, I will try to show a clear definition of the percent yield equation and give examples that are mixed with other concepts. The equation is below.
Percent Yield = | Actual | * 100 |
Theoretical |
The hardest part about this equation is to decide what number is your actual and what number is your theoretical. The best guide I have found in chemistry is that the theoretical number almost always comes from a conversion calculation. The actual number usually is accompanied by some kind of phrase like “at the end of the reaction” or “when the reaction is complete”. The theoretical is what you were supposed to get if everything went perfect in the process. However, like life nothing ever goes perfect in chemistry. The actual is the amount you do get when you take into account any mistakes that were made. The percent yield has no units. Which means the actual and theoretical yield have to have the same units when you divide them.
Examples: Give the part of the percent yield calculation missing.
If your actual yield is 17g and your theoretical yield is 43g then what is your percent yield?
Answer: 39.5 %
From the reaction you were supposed to get 21g of CO_{2} but instead you got 15g of CO_{2}. What was your percent yield?
Answer: 71.4%
If your percent yield of O_{2} from the reaction below was 86% then what is your actual yield if you started with 63g of H_{2}O?
2 H_{2}O_{(l)} —> 2 H_{2(g)} + O_{2(g)}
Answer: 49.9 g O_{2} (hint you have to convert the 63g of H_{2}O to get the theoretical)
VIDEO Percent Yield Demonstrated Example 1: Your calculations indicated that you would produce 3.7g of LiOH but instead you produced 5.2g of LiOH. What is your percent yield?
Step 1:
What information does the question supply us with?
Answer:
Actual = 5.2 g LiOH
Theoretical = 3.7 g LiOH
Step 2:
What section of the formula does the question ask?
Answer: percent yield = ?
Step 3:
What is the formula?
Percent Yield = | Actual | * 100 |
Theoretical |
Step 4:
Fill in the formula with the information above
Percent Yield = | 5.2 g LiOH | * 100 |
3.7 g LiOH |
Step 5:
How do you do the calculations?
Answer: (5.2 * 100) / (3.7) = 140
140 % = | 5.2 g LiOH | * 100 |
3.7 g LiOH |
Step 6:
COMPLETE ANSWER: Percent yield = 140 %
VIDEO Percent Yield Demonstrated Example 2: If you had a percent yield of 93% and at the end of the chemical reaction found you weighed out 8.4g. What was your theoretical yield?
Step 1:
What information does the question supply us with?
Answer:
percent yield = 93%
Actual = 8.4 g
Step 2:
What section of the formula does the question ask?
Answer: theoretical yield = ?
Step 3:
What is the formula?
Percent Yield = | Actual | * 100 |
Theoretical |
Step 4:
Fill in the formula with the information above
93 = | 8.4 g | * 100 |
Theoretical |
Step 5:
How do you solve for volume?
Answer: First multiply both sides by theoretical (red)
93 * Theoretical = | 8.4 g | * 100 * Theoretical |
Theoretical |
Step 6:
Cross out Theoretical on the right side
93 * Theoretical = | 8.4 g | * 100 * Theoretical |
Theoretical |
Step 7: Simplify
93 * Theoretical = | 8.4 g | * 100 |
1 |
Step 8:
Now divide both sides by 93 (red)
93 * Theoretical = | 8.4 g | * 100 |
93 | 93 |
Step 9:
Cross out 93 on the left side
93 * Theoretical = | 8.4 g | * 100 |
93 | 93 |
Step 10: Simplfy
Theoretical = | 8.4 g | * 100 |
93 |
Step 11:
How do I do the calculations?
Answer: (8.4 * 100) / (93) = 7.8
7.8 g = | 8.4 g | * 100 |
93 |
Step 12:
COMPLETE ANSWER: 7.8 g
VIDEO Percent Yield Demonstrated Example 3: After the reaction you were left with 9g of NH_{3}. What was your percent yield if you started with 4.2g of H_{2}. You will need the periodic table for this problem.
N_{2(g)} + 3 H_{2(g)} —-> 2 NH_{3(g)}
Step 1:
What information does the question supply us with?
Answer:
Actual = 9 g NH_{3}
Theoretical = 4.2 g H_{2}
Step 2:
What section of the formula does the question ask?
Answer: percent yield = ?
Step 3:
Where do we start?
Answer: We first need to start with a conversion because our actual and theoretical units do not match. We need to convert the theoretical from g H_{2} to g NH_{3}
Step 4:
How many conversions must we do?
Answer: Look at the conversion map. We pass through 3 arrows when we go from Grams of A —> moles of A —> moles of B —> Grams of B. 3 arrows = 3 conversions
Step 5:
How do we set up the conversion?
Answer:
4.2 g H_{2} | g NH_{3} | |||
1 |
Step 6:
What is the first conversion?
Answer: molar mass (grams to mole ratio) of H_{2} found on the periodic table
Step 7:
How do I put that in?
Answer: units first, set up the units that need to cancel out (in red)
4.2 g H_{2} | mol | g NH_{3} | ||
g |
Step 8:
What is the next step?
Answer: Fill in the numbers and cross out units
4.2 g H_{2} | 1 mol | g NH_{3} | ||
2 g |
Step 9: Simplify
4.2 H_{2} | 1 mol | g NH_{3} | ||
2 |
Step 10:
What is the next conversion?
Answer: mole to mole ratio (coefficient ratio)
Step 11:
How do I set it up?
Answer: units first, set up the units that you need to cancel out (in red)
4.2 H_{2} | 1 mol | NH_{3} | g NH_{3} | |
2 | H_{2} |
Step 12:
What is the next step?
Answer: Fill in the numbers and cross out units
4.2 H_{2} | 1 mol | 2 NH_{3} | g NH_{3} | |
2 | 3 H_{2} |
Step 13: Simplify
4.2 | 1 mol | 2 NH_{3} | g NH_{3} | |
2 | 3 |
Step 14:
What is the next conversion?
Answer: molar mass (grams to mole ratio) of H_{2} found on the periodic table
Step 15:
How do I put that in?
Answer: units first, set up the units that need to cancel out (in red)
4.2 | 1 mol | 2 NH_{3} | g = | g NH_{3} |
2 | 3 | mol |
Step 16:
What is the next step?
Answer: Fill in the numbers and cross out units
4.2 | 1 mol | 2 NH_{3} | 17 g = | g NH_{3} |
2 | 3 | 1 mol |
Step 17: Simplify
4.2 | 1 | 2 NH_{3} | 17 g = | g NH_{3} |
2 | 3 | 1 |
Step 18:
How do I do the calculations?
Answer: (4.2 * 2 * 17) / (2 * 3) = 23.8
4.2 | 1 | 2 NH_{3} | 17 g = | 22.7 g NH_{3} |
2 | 3 | 1 |
Now we can compare the 23.8 g NH_{3} which is our theoretical value to our actual value of 9 g NH_{3}.
Step 19:
How do we do that?
Answer: With the percent yield formula
Percent Yield = | Actual | * 100 |
Theoretical |
Step 20:
Fill in the numbers
Percent Yield = | 9 g NH_{3} | * 100 |
23.8 g NH_{3} |
Step 21:
How do I do the calculations?
Answer: (9 * 100) / (23.8) = 37.8
37.8 % = | 9 g NH_{3} | * 100 |
23.8 g NH_{3} |
Step 20:
COMPLETE ANSWER: 37.8 %
PRACTICE PROBLEMS: Solve these percent yield problems. Don’t forget to use the periodic table and the conversion map when you need them.
Your calculations indicated that you would produce 4.8g but instead you produced 3.2g. What is your percent yield?
Answer: 66.7%
If you had a percent yield of 75% and at the end of the chemical reaction found you weighed out 36g. What was your theoretical yield?
Answer: 48 g
After the reaction you were left with 19g of O_{2}. What was your percent yield if you started with 82g of H_{2}SO_{4}.
H_{2}SO_{4(aq)} —–> H_{2(aq)} + S_{(s)} + 2 O_{2(aq)}
Answer: 35.5%
If you had a 48% yield at the end of the chemical reaction and you started with 65g of CO_{3}^{2-}. Then what is your actual yield of Fe_{2}(CO_{3})_{3}.
2 Fe^{3+}_{(aq)} + 3 CO_{3}^{2-}_{(aq)} —-> Fe_{2}(CO_{3})_{3(s)}
Answer: 50.6 g Fe_{2}(CO_{3})_{3}
Now try the Stoichiometry worksheet.