# Percentage yield and atom economy

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##### Intros
###### Lessons
1. Why do chemists do chemical reactions?
2. Practical problems in chemical reactions.
3. Atom economy and percentage yield.
4. Worked example: Calculate atom economy.
5. Worked example: Calculate percentage yield.
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##### Examples
###### Lessons
1. Find the atom economy and percentage yield of chemical reactions.
Water can be produced by reaction of hydrogen and oxygen gas according to this equation:

2 H$_2 +$O$_2 \,$$\,$2 H$_2$O

1. What is the atom economy of this reaction?
2. If the reaction was performed and 150 g was the theoretical yield of water and the actual yield was only 92 g, what is the percentage yield?
2. Find the atom economy and percentage yield of chemical reactions.
The reaction to make iron metal by reduction uses carbon. CO$_2$ is an unwanted side-product. The reaction is as follows:

2 Fe$_2$O$_3 +$3 C$\,$$\,$3 CO$_2 +$4 Fe

1. What is the atom economy of this reaction?
2. 280 g of Fe$_2$O$_3$ was used in this reaction and 104 g of Fe metal was collected. Calculate the theoretical yield and then the percentage yield of this reaction.
3. This reaction was done again and a percentage yield of 90% was achieved this time. If 250 g of Fe was collected this time, how much Fe$_2$O$_3$ was used up?
3. Compare the viability of two reactions by finding the atom economy and percentage yield.
Magnesium reacts with hydrochloric acid as shown in the following equation:

Mg$+$2 HCl$\,$$\,$MgCl$_2 +$H$_2$

1. If this reaction was performed to make hydrogen gas, what would be the atom economy?
2. 55 g of Mg metal was used in this reaction with HCl in excess. Only 4.21 g of H$_2$ gas was produced from this reaction. What is the percentage yield of this reaction?
3. Hydrogen gas could also be made by electrolysis of water, in the following reaction:
2 H$_2$O→2 H$_2 +$O$_2$

Find the atom economy of this reaction. Is it a better or worse way to make hydrogen than reacting magnesium with hydrochloric acid?
###### Topic Notes
In this lesson, we will learn:
• The meaning of atom economy, percentage yield and the difference between the terms.
• How to calculate atom economy and percentage yield from example chemical reactions.
• To explain the importance of atom economy as a chemist when planning and running chemical processes.

Notes:

• In Moles, excess and limiting reagents, we learned that in chemical reactions, knowing your excess and limiting reagents is important practical information. When your limiting reagent runs out, any amounts in excess have nothing to react with, so you will stop making your products.

• Percentage yield and atom economy are two other practical considerations when doing chemical reactions. They are both related to the amount of useful product generated in a reaction, compared to undesirable side products or unreacted starting material.

• The atom economy of a reaction is the percentage of atomic mass of useful products in a reaction. It is calculated by:

• $Atom \, Economy \,$ (%) = $\frac{Atomic \, mass \, of \, useful \, products } {Total \, atomic \, mass \, of \, produtcs } \, * \,$ 100

A high atom economy means most of what the process makes is useful! In the same way, a low atom economy tells you that a reaction is mostly producing unwanted side products.
The atom economy is used to show the efficiency of a reaction: does it make a lot of waste products that will require storage and disposal, or is most of the product valuable to us?
• Reactions with a single product have a 100% atom economy because the only chemical being produced is the desired product.
• Atom economy is used as an application of the conservation of mass: no atoms are created or destroyed in chemical reactions, they are only re-arranged by breaking and forming substances. We are using the amount of atomic mass as a measure of the reaction efficiency.

• The percentage yield of a reaction is the mass of products formed as a percentage of how much could have been formed given the mass of reactants used. The equation to calculate percentage yield is:

• $Percentage \, yield \, = \, \frac{actual \, yield } {theoretical \, yield} \, * \,$100

Where:
• Actual yield is the yield of product obtained in the experiment (in g or moles).
• Theoretical yield is the yield of product based on the limiting reagent, i.e. the calculations done in Moles, excess and limiting reagents.

A low percentage yield means that not much of the reactants you used has become products. A high percentage yield therefore means that a lot of the reactant chemicals you used successfully reacted to make the products.

• Worked example: Calculate the atom economy of a reaction.
The reaction below shows the production of iron metal by reacting iron oxide (Fe2O3) with carbon.

• 2Fe2O3 (s) + 3C (s) $\,$$\,$ 4Fe (s)+ 3CO2 (g)

This reaction is intended to produce iron metal with CO2 as an unwanted side product. What is the atom economy of this reaction?

To begin finding atom economy, we need to know what is desirable in the products. We already know that Fe is the desired product and CO2 is undesirable side product.
Now we need to find the atomic mass of both of these – how much of the product, in terms of atomic mass, is valuable?

%$\, Atom \, economy \, = \,$ $\large \frac{4 \, * \, (55.8 \, g \, mol^{-1}\, Fe )} { (4 \, * \, (55.8 \, g \, mol^{-1} \, Fe ) ) \, + \, (3 \, * \, (44 \, g \, mol \, CO_{2})) } \, * \,$ 100 = 62.8 %

• Worked example: Calculate the percentage yield of a reaction.
The reaction below shows the production of iron metal by reacting iron oxide with carbon:

• 2Fe2O3 (s) + 3C (s) $\,$$\,$ 4Fe (s) + 3CO2 (g)

In one run of this reaction, the limiting reagent is Fe2O3. 750g of Fe2O3 was reacted in this run, and an experimental yield of 460 g of Fe metal was obtained.

What is the theoretical and percentage yield of this run of the reaction?

To begin, we need to find the theoretical yield. If we only have 750g of Fe2O3 and we are told this is the limiting reagent then we know this is going to run out and this amount dictates how much Fe can possibly be made.

You can use conversion factors to find the theoretical yield of Fe metal product from the amount of Fe2O3 reactant.

750$g \, Fe_{2}O_{3} \, * \,$ $\large \frac{1 \, mol \, Fe_{2}O_{3} } {159.6g \, Fe_{2}O_{3} } \, * \, \frac{4\, mol \, Fe } {2 \, mol \, Fe_{2}O_{3} } \, * \, \frac{55.8 \, g \, Fe } {1 \, mol \, Fe } \,$ = 524.4 $\, g \, Fe$

This 524.4g Fe tells us the theoretical yield of the reaction. It is the maximum possible amount of product we can get in this reaction, because we only have 750g of the Fe2O3 limiting reagent. In terms of percentage yield, this is your 100% mark.

Now, we have the 100% mark, and we have an experimental yield of 460g of Fe. This is our actual “score”. Let’s convert 460/524.4 into a percentage.

$Percentage \, yield \, = \,$ $\large \frac{460 \, g \, Fe } {524.4 \, g \, Fe} \, * \,$100 = 87.7%

This is the percentage yield of the reaction.

• Know the difference between yield and atom economy!
• Atom economy is about how wasteful the reaction is. If your reaction has low atom economy, it will always be wasteful; most of the product made is simply not valuable to you. This is a chemical problem, not a practical one.
The unwanted products are a waste of money/resources and an environmental problem because waste has to be stored or disposed of safely.
• Yield is about how much product was successfully made. If your yield is low, this is probably a practical problem, such as not enough time to react or your conditions may need to be changed.