Percentage yield and atom economy

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Why do chemists do chemical reactions?
Practical problems in chemical reactions.
Atom economy and percentage yield.
Worked example: Calculate atom economy.
Worked example: Calculate percentage yield.

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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?
Find the atom economy and percentage yield of chemical reactions.
The reaction to make iron metal by reduction uses carbon. CO $_2$ $_{2}$ 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?
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

atom economy

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

high atom economy

show the efficiency of a reaction

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

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.

low percentage yield

high percentage yield

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.

2Fe₂O_{3 (s)} + 3C_(s)

\,

→

\,

4Fe_(s)+ 3CO_{2 (g)}

₂

What is the atom economy of this reaction?

₂

\, 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:

2Fe₂O_{3 (s)} + 3C_(s)

\,

→

\,

4Fe_(s) + 3CO_{2 (g)}

₂

₃

₂

₃

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

₂

₃

₂

₃

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

the theoretical yield of the reaction

₂

₃

Percentage \, yield \, = \,

\large \frac{460 \, g \, Fe } {524.4 \, g \, Fe} \, * \,

100 = 87.7%

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.

Percentage yield and atom economy

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Percentage yield and atom economy

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