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Chemical reactions are fundamental processes that drive various aspects of everyday life, from the production of energy to the creation of pharmaceuticals. In order to optimize these reactions, it is crucial to assess their efficiency. Efficiency represents the ratio between the actual output of a reaction and the maximum possible output. By determining the efficiency, scientists and engineers can evaluate the extent to which a chemical reaction is successful in converting reactants into desired products. This knowledge enables the identification of areas for improvement, ultimately leading to the design of more sustainable and cost-effective chemical processes. In this article, we will explore the different methods and calculations utilized to quantify the efficiency of a chemical reaction and understand its implications for practical applications.
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In chemistry, the theoretical amount is the maximum amount of product a chemical reaction can produce based on the chemical equation. In fact, most reactions are incomplete. If you do the experiment, you will get less product called the actual quantity . You can calculate the reaction yield using the following formula: %efficiency = (actual quantity/theoretical quantity) x 100 . A 90% reaction efficiency means that the reaction is 90% productive, and 10% of the materials have been wasted (they either don’t react, or the product isn’t recovered at all).
Steps
Find all reactants
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- For example, oxygen and glucose react with each other to form carbon dioxide and water: 6O2+OLD6HtwelfthO6{displaystyle 6O_{2}+C_{6}H_{12}O_{6}} → 6OLDO2+6H2O{displaystyle 6CO_{2}+6H_{2}O}
Each side has 6 carbon atoms (C), 12 hydrogen atoms (H) and 18 oxygen atoms (O). The equation is balanced. - Read this guide if the problem requires balancing an equation.
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- For example, an oxygen molecule ( O2{displaystyle O_{2}} ) has two oxygen atoms.
- The mp mass of oxygen is 16 g/mp. (You can find a more precise value on the periodic table.)
- 2 oxygen atoms x 16 g/mp = 32 g/mp, molecular mass mp O2{displaystyle O_{2}} .
- For example another reactant, glucose ( OLD6HtwelfthO6{displaystyle C_{6}H_{12}O_{6}} ) has a mass mp of (6 C atoms x 12 g C/mp) + (12 H atoms x 1 g H/mp) + (6 O atoms x 16 g O/mp) = 180 g/mp.
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- For example, let’s say you have 40 grams of oxygen and 25 grams of glucose.
- 40 g O2{displaystyle O_{2}} / (32 g/mp) = 1.25 mp of oxygen.
- 25g OLD6HtwelfthO6{displaystyle C_{6}H_{12}O_{6}} / (180 g/mp) = about 0.139 mp of glucose.
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- You initially have 1.25 mp of oxygen and 0.139 mp of glucose. The ratio of oxygen molecules to glucose molecules is 1.25/0.139 = 9.0. That means you have 9 oxygen molecules for every glucose molecule.
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- The left side of the equation is 6O2+OLD6HtwelfthO6{displaystyle 6O_{2}+C_{6}H_{12}O_{6}} . The coefficients indicate that there are 6 oxygen molecules and 1 glucose molecule. So the ideal ratio of this reaction is 6 oxygen/1 glucose = 6.0.
- Remember to arrange the reactants in the correct order for the two ratios. If you use the oxygen/glucose ratio for the theoretical reaction and the glucose/oxygen ratio for the actual quantity, the latter result will be wrong.
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- If the actual ratio is greater than the ideal ratio then you have more of the above reactants than necessary. The lower reactant in the ratio is the complete reactant.
- If the actual ratio is less than the ideal ratio then you don’t have enough of the above reactant, so it is an empty reactant.
- In the above example, the actual ratio of oxygen/glucose (9.0) is greater than the ideal ratio (6.0) so the bottom reactant, glucose, is the empty reactant.
Calculate theoretical quantity
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- Continuing with the example above, you are analyzing the reaction 6O2+OLD6HtwelfthO6{displaystyle 6O_{2}+C_{6}H_{12}O_{6}} → 6OLDO2+6H2O{displaystyle 6CO_{2}+6H_{2}O} . The right side has two products, carbon dioxide and water. Calculate the amount of carbon dioxide, OLDO2{displaystyle CO_{2}} .
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- In the example above, you know glucose is the end reactant so you would start with 0.139 mp of glucose.
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- The equilibrium equation is 6O2+OLD6HtwelfthO6{displaystyle 6O_{2}+C_{6}H_{12}O_{6}} → 6OLDO2+6H2O{displaystyle 6CO_{2}+6H_{2}O} . There are six desired product molecules, carbon dioxide ( OLDO2{displaystyle CO_{2}} ). One molecule of reactant is all glucose ( OLD6HtwelfthO6{displaystyle C_{6}H_{12}O_{6}} ).
- The ratio of carbon dioxide to glucose is 6/1 = 6. In other words, the reaction produces 6 carbon dioxide molecules from one glucose molecule.
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- With 0.139 mp glucose and the ratio of carbon dioxide to glucose is 6. So the theoretical amount of carbon dioxide is (0.139 mp glucose) x (6 mp carbon dioxide / mp glucose) = 0.834 mp carbon dioxide.
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- For example, the mp mass of CO 2 is 44 g/mp. (The mp mass of carbon is ~12 g/mp and of oxygen is ~16 g/mp, so the total is 12 + 16 + 16 = 44.)
- Take 0.834 mp CO 2 x 44 g/mp CO 2 = ~36.7 grams. So the theoretical amount of this example is 36.7 grams of CO 2 .
Calculate percentage efficiency
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- Theoretical amount is the maximum amount of product obtained by the reaction.
- Actual quantity is the actual amount of product obtained, and is measured directly on the balance.
- Percent efficiency = Actual AmountAmount of Theory∗100%{displaystyle {frac {text{Actual Amount}}{text{Theory Amount}}}*100%} . For example, the percentage efficiency is 50%, which means that the amount of product obtained is only 50% of the theoretical amount.
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- Assume the actual amount of CO2 product is 29 grams.
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- The actual amount is 29 grams while the theoretical amount is 36.7 grams. 29g36,7g=0,79{displaystyle {frac {29g}{36.7g}}=0.79} .
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- 0.79 x 100 = 79, so the percentage efficiency of this experiment is 79%. You generate 79% of the theoretical maximum CO 2 .
Advice
- Some people confuse percentage efficiency (actually obtained in total that can be obtained) with percentage error (deviation of experimental results from desired results). The exact percentage yield formula is Actual AmountAmount of Theory∗100{displaystyle {frac {text{Actual Amount}}{text{Theory Amount}}}*100} . If you subtract these two amounts, that’s the formula for percentage error.
- If you get abnormal results then unit test. If the actual amount is greater than the theoretical amount, then you have used the wrong units in the calculations. Redo the calculations and remember to double check the units of each calculation.
- If the percentage efficiency is more than 100% (and you are sure the calculations are correct) then your product is probably not pure. You need to clean the product (like drying or filtering) and reweigh.
This article is co-authored by a team of editors and trained researchers who confirm the accuracy and completeness of the article.
The wikiHow Content Management team carefully monitors the work of editors to ensure that every article is up to a high standard of quality.
This article has been viewed 129,935 times.
In chemistry, the theoretical amount is the maximum amount of product a chemical reaction can produce based on the chemical equation. In fact, most reactions are incomplete. If you do the experiment, you will get less product called the actual quantity . You can calculate the reaction yield using the following formula: %efficiency = (actual quantity/theoretical quantity) x 100 . A 90% reaction efficiency means that the reaction is 90% productive, and 10% of the materials have been wasted (they either don’t react, or the product isn’t recovered at all).
In conclusion, calculating the efficiency of a chemical reaction is an essential step in understanding the effectiveness of a reaction and determining whether it meets the desired goals. Efficiency, often measured in terms of atom economy or yield, provides valuable insights into the amount of desired product formed relative to the starting materials used.
By calculating the efficiency, chemists can identify potential areas for improvement in a reaction, such as minimizing waste or optimizing reaction conditions. This allows for the development of more sustainable and cost-effective reactions, which is crucial for advancing scientific knowledge and meeting the ever-growing demands of industries.
Various factors can influence the efficiency of a reaction, including the reaction pathway, reactant stoichiometry, and reaction conditions. Therefore, it is vital to carefully design reactions and consider all these variables to maximize the efficiency.
Calculating reaction efficiency can be done using different approaches and formulas, such as atom economy, percentage yield, or selectivity. Each method provides a unique perspective on the reaction’s efficiency, highlighting different aspects of the process.
Moreover, the efficiency calculation also aids in comparing different reactions or alternative synthetic routes, enabling scientists to choose the most efficient and sustainable methods. This not only saves resources but also mitigates the environmental impact associated with chemical processes.
In summary, the ability to calculate the efficiency of a chemical reaction is fundamental in evaluating and optimizing reactions. It empowers chemists to develop more efficient processes, minimize waste, and make informed decisions regarding the choice of reaction pathways. Pursuing high reaction efficiency is crucial for advancing sustainability in chemistry and achieving greater scientific and industrial progress.
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