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Yield and Atom Economy

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Yield and Atom Economy Revision

Yield and Atom Economy

In the chemical industry it is useful to know how efficient and effective the process used are at producing the desired products. With millions of pounds or dollars of materials being used, it is important to make sure that there is a little waste as possible. To make this possible, chemists consider two factors:

1. The percentage yield of a reaction. A measure of how effective an industrial process is in producing a product.

2. The atom economy of a reaction. A measure of how efficiently a reaction uses the atoms of reactant supplied. 

Percentage Yield

Percentage yield is a measure of how effective an industrial process is in producing a desired product. The percentage yield of a reaction is defined as the actual yield of product as a percentage of the theoretically possible yield in a reaction:

\text{Percentage Yield}=\frac{\text{Mass of Product}}{\text{Maximum Theoretical Mass}} \times{100}

The yield of a reaction is the mass of the products formed.

Though mass is always conserved in a reaction, the percentage yield of a product is never 100\%. The mass of the product actually made is typically less than what could theoretically be produced for a few reasons.

  • Product may be lost during separation from the reaction mixture. If the product is crystalized (the process in which a dissolved substance solidifies from a solution) some of it may not solidify and so remain in the reaction mixture. This would be lost upon filtration. In some cases, some liquid product may be lost to evaporation if the reaction is undertaken at high temperatures. Product may also be lost when transferred between containers. All of these things will lead to a lower percentage yield.


  • In many reactions there will be unaccounted for or unavoidable side reactions. In these cases the reactants or products may be able to react with solvents or impurities in the reaction mixture. If there are side reactions with the reactants then there will be fewer reactant atoms available to form desired product. If the product its self is able to react, then some of its mass will be lost in these reactions.


  • It is also possible that the reactants don’t completely react. This is particularly common when one or more reactants are solid. It is also a problem when working with reversible reactions. In these cases, for example in the Haber Process, the forward desired reaction takes place in parallel to the reverse reaction. This means that some amount of reactant will end up not forming product, lowering the percentage yield.

Processes that have a low percentage yield are less favorable as they will consume more reactant than is theoretically necessary. This is unstainable and unprofitable. 

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Calculating the Maximum Theoretical Product

To calculate the percentage yield of a reaction, we need to know how much of the desired product it is theoretically possible to form. To do this we need to know two things:

1. The moles of reactants used in the reaction.

2. The balanced equation of the reaction. The balanced equation of the reaction will tell us ratio of moles of reactant to moles of product.

From here, we can calculate how many moles of product a given amount of reactant should produce, and hence its mass. 

Take, for example, the reaction of magnesium hydroxide(\text{Mg(OH)}_2) and sulfuric acid (\text{H}_2\text{SO}_4) to produce magnesium sulfate (\text{MgSO}_4):

\text{Mg(OH)}_2 + \text{H}_2\text{SO}_4 \rarr \text{MgSO}_4 + 2\text{H}_2\text{O}

This reaction gives us a ratio of magnesium hydroxide to magnesium sulfate as 1:1. This tells us that the maximum amount of  moles theoretically possible is equal to the moles of magnesium hydroxide added. We can therefore work out the theatrically possible mass by using the below equation.

\text{Mass of MgSO}_4 = \text{Moles of MgSO}_4 \times \text{M}_r \text{ MgSO}_4

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Atom Economy

Atom economy is a measure of the efficiency of reaction. The atom economy of a reaction is defined as the total relative formula mass of desired products as percentage of the total relative formula mass of the reactants in a reaction:

\text{Atom Economy} =\frac{\text{Total M}_r \text{ Desired Products}}{\text{Total M}_r \text{ Reactants}} \times 100

Atom economy measures the efficiency of a reaction by reveling how much reactant will be used in forming a desired product. A higher atom economy means we have a less wasteful and more environmentally friendly reaction. The atom economy assumes a 100\% yield for all the products involved. As such, the atom economy of a reaction tells us more about how efficiently reactants are used as opposed to how sucsedully they have reacted. 

Atom economy will be affected by the amount of side product created in a reaction. For example, if we want to produce sodium chloride, there are two ways to go about it. In the first method, we react bicarbonate of soda \left(\text{NaHCO}_3\right) with hydrochloric acid in the following reaction:

\text{NaHCO}_3 + \text{HCl} \rarr \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2

In the second, we react sodium hydroxide (NaOH) with HCl instead of NaHCO3:

\text{NaOH} + \text{HCl} \rarr \text{NaCl} + \text{H}_2\text{O} 

If we wanted to produce sodium chloride in the greenest way possible, we would use the second method. This reaction has an atom economy of about 76\%. The first reaction is much less efficient, with an atom economy of about 48\%

In the first reaction, we have a needlessly heavy reactant that then produces two unhelpful side products (carbon dioxide and water). In the second reaction, the reactants are much lighter and only one unwanted side product is produced.

In some cases, a reaction may have an atom economy of 100%. Take the forwards reaction of the formation of ammonia in the Haber Process:

\text{N}_2 + \text{H}_2 \rarr 2\text{NH}_3

In this reaction there is only one product, ammonia. This means that (assuming a 100\% yield) all of the reactant mass is converted into desired product.

Reactions with a high atom economy are desirable as they are greener and more efficient. Reactions with low atom economy will consume unnecessary resources and produce more waste. They are also less profitable as more money has to be spent on starting materials.

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Choosing a Reaction

When setting up industrial scale chemical reactions, it is important to choose the right ones for the products you want to produce. Many factors will need to be considered. Chief amongst these choices are:

  • Atom Economy: it is both greener and more cost effective to use reaction path ways that have a high atom economy. Reducing waste and improving efficiency creates more profitable and more sustainable industries.


  • Usefulness of By-Products: Many chemical reactions will produce by-products. In fact, it is often the case that there are no available reactions that produce no by-products. In these cases, reactions are chosen for the usefulness of these by-products. In some cases, the by-products of a reaction may be useful as the starting materials in other industries, or as useful chemicals in themselves.


  • The Rate of the Reaction: The rate of the reaction (the speed with which it progresses) is important in finding a viable reaction pathway. It may be that one method of forming a product has a very high atom economy and relatively cheap conditions but an impractically slow rate of reaction. Reactions must have practical rates to be useful in industry; a reaction with an atom economy and percentage yield of 100\% be useless if it took thousands years to happen.


When designing industrial processes these considerations will all need to be balanced along with other considerations like reaction conditions and environmental impact. 

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Example 1: Calculating Percentage Yield

In an industrial process, a factory produces \textcolor{#00bfa8}{120\text{ million}\text{ g} (1.20\times 10^8)} of ammonia from a theoretically possible yield of \textcolor{#f21cc2}{158\text{ million}\text{ g} (1.58\times10^8)}. Calculate the percentage yield of this process:

[1 mark]

\text{Percentage Yield} = \frac{\text{Mass of Ammonia Produced}}{\text{Mass of Ammonia Theoretically Possible}} \times 100\\ \, \\=\frac{\textcolor{#00bfa8}{1.20\times10^8}}{\textcolor{#f21cc2}{1.58\times10^8}} \times 100\\ \, \\= \textcolor{#008d65}{76\%}

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Example 2: Calculating Maximum Yield 

Another industrial process has a percentage yield of \textcolor{#00bfa8}{68\%}. If the process produces \textcolor{#f21cc2}{780\text{ tonnes}} of product, calculate the maxium theoretical yeild:

[1 mark]

\text{Percentage Yield} =\frac{\text{Mass of Product Produced}}{\text{Mass of Product Theoretically Possible}} \times 100


\text{Mass of Product}\text{Theoretically Possible} =\frac{\text{Mass of Product Produced} \times 100}{\text{Percentage Yield}}\\ \, \\ =\frac{\textcolor{#f21cc2}{780} \times 100}{\textcolor{#00bfa8}{68}}\\ \, \\=\textcolor{#008d65}{1147\text{ tonnes}}

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Example 3: Calculating Percentage Mass from an Equation

Calculate the percentage yield when \textcolor{#00bfa8}{11.5\text{ g}} of magnesium sulfate (\text{MgSO}_4\text{, }\text{M}_r =\textcolor{#f21cc2}{120}) is produced from \textcolor{#327399}{7.4\text{ g}} of magnesium hydroxide(\text{Mg(OH)}_2\text{, }\text{M}_r =\textcolor{#a233ff}{58}):

[3 marks]

Step 1: Calculate the maximum theoretically possible moles of magnesium sulfate that we can produce. From the equation, we know that the moles of magnesium sulfate that can be produced in theory is equal to the moles of magnesium hydroxide used in the reaction. 

\text{Mg(OH)}_2 + \text{H}_2\text{SO}_4 \rarr \text{MgSO}_4 + 2\text{H}_2\text{O}

\text{Moles of MgSO}_4=\text{Moles of Mg(OH)}_2=\frac{\text{Mass of Mg(OH)}_2}{\text{M}_r \text{Mg(OH)}_2}\\ \, \\=\frac{\textcolor{#327399}{7.4}}{\textcolor{#a233ff}{58}}\\ \, \\=\textcolor{#008d65}{0.13\text{ mol}}

Step 2: Calculate theoretical mass.

\text{Theoretical Mass MgSO}_4=\text{Moles MgSO}_4 \times \text{M}_r \text{ MgSO}_4\\ \, \\= 0.13 \times \textcolor{#f21cc2}{120}\\ \, \\=\textcolor{#008d65}{15.6\text{ g}}

Step 3: calculate the percentage yield of this reaction:

\text{Percentage Yield}=\frac{\text{Mass MgSO}_4\text{ Produced}}{\text{Theoretically Possible Mass MgSO}_4} \times 100


\text{Percentage Yield}=\frac{\textcolor{#00bfa8}{11.5}}{15.6} \times 100 = \textcolor{#008d65}{74\%}

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Example 4: Calculating Atom Economy

In an industrial process, a factory produces a product with an \text{M}_r equal to \textcolor{#00bfa8}{198} for reactants with a total \text{M}_r equal to \textcolor{#f21cc2}{375}. Calculate the atom economy of this process:

[1 mark]

\text{Atom Economy} =\frac{\text{Total M}_r\text{ Desired Products}}{\text{Total M}_r\text{ Reactants}} \times 100\\ \, \\=\frac{\textcolor{#00bfa8}{198}}{\textcolor{#f21cc2}{375}} \times 100\\ \, \\=\textcolor{#008d65}{53\%}

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Example 5: Calculating Mr from Atom Economy

Another industrial process has an atom economy of \textcolor{#00bfa8}{68\%}. If the reactants have a total \text{M}_r equal to \textcolor{#f21cc2}{598}, calculate the total \text{M}_r of the desired products:

[1 mark]

\text{Atom Economy} =\frac{\text{Total M}_r\text{ Desired Products}}{\text{Total M}_r\text{ Reactants}} \times 100


\text{Total M}_r\text{ Desired Products}=\frac{\text{Atom Economy}}{100} \times \text{Total M}_r\text{ Reactants}\\ \, \\=\frac{\textcolor{#00bfa8}{68}}{100} \times \textcolor{#f21cc2}{598}\\ \, \\=\textcolor{#008d65}{407}
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Example 6: Calculating Atom Economy from an Equation

Iron (Fe) can be extracted from its ore in the following reaction:

3\text{CO} + \text{Fe}_2\text{O}_3 \rarr 3\text{CO}_2 + \text{Fe}

Calculate the atom economy of this reaction:

[3 marks]

Step 1: calculate them total \text{M}_r of the products.

\text{M}_r =  2\times\text{A}_r\text{Fe}+3\times\text{A}_r\text{C} + 6\times\text{A}_r\text{O}\\ \, \\= 3(12)+2(56)+6(16) \\ \, \\=\textcolor{#00bfa8}{244}

Step 2: calculate the formula mass of the desired product:

\text{M}_r\text{Fe} =\textcolor{#f21cc2}{56}

Step 3: calculate the atom economy:

\text{Atom Economy} =\frac{\textcolor{#f21cc2}{56}}{\textcolor{#00bfa8}{244}} \times 100 =\textcolor{#008d65}{23\%}

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Yield and Atom Economy Example Questions

  • Atom economy is a measure of the efficiency of a reaction. Percentage yield is a measure of the effectiveness of a reaction.
  • Atom economy is calculated using the relative molecular masses of reactants and products. Percentage yield is calculated using masses of a product and its theoretically possible mass.
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\text{Atom Economy}=\frac{\text{Total M}_r\text{ Desired Products}}{\text{Total M}_r\text{ Reactants}}\\ \, \\ =\frac{57}{130} \times 100\\ \, \\=\underline{44\%}

(One mark for correct formula. One mark for correct answer.)

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\text{Mass of Product Theoretically Possible}=\frac{\text{Mass of Product Producted}}{\text{Percentage Yeild}}\times 100\\ \, \\ =\frac{1268 \times 100}{72}\\ \, \\=\underline{1761\text{ kg}}

(One mark for correct formula. One mark for correct answer.)

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Industrial processes need to use reactions with practical rates of reaction. If a reaction took thousands of years to happen, it would be unusable in industry. Reactions should be chosen that have quick enough rates to be useable in industry.

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Calculation should be broken down into steps:

Step 1 (1 mark): Calculation of \text{Cu}_2\text{O} to \text{Cu}.

  • In the equation 2 moles of \text{Cu}_2\text{O} react to form 6 moles of copper. This gives a ratio of 2:6
  • Simplified, this ratio is 1:3

Step 2 (2 marks): conversion of tonnes to grams.

1\text{ tonne}=1000\text{ kg}


1\text{ kg}=1000\text{ g}


6\text{ tonnes}=6 \times 1000\times 1000 =\underline{6.00\times10^6\text{ g}}


7.87\text{ tonnes}=7.87 \times 1000\times 1000 =\underline{7.87\times10^6\text{ g}}

Step 3 (1 mark): Calculation of the Moles of copper oxide.

\begin{aligned}\text{Moles}&=\frac{\text{Mass}}{\text{M}_r}\\ &=\frac{6.00\times10^6}{144}\\ &=\underline{4.20 \times 10^4\text{ mol}}\end{aligned}

Step 4 (1 mark): Calculation of the moles of copper.



\begin{aligned}\text{Moles Cu} &= 3 \times \text{Moles Cu}_2\text{S}\\ & = 3\times 4.20\times 10^4 \\ &=\underline{1.26\times 10^5\text{ mol}}\end{aligned}

Step 5 (1 mark): Calculation of theoretical mass of copper.

\text{Theoretical Mass Cu}=\text{Moles Cu} \times \text{M}_r \text{ Cu}\\ \, \\=1.25 \times 10^5 \times 63.5 \\ \, \\ =\underline{8.00 \times 10^6\text{ g}}

Step 6 (1 mark): Calculation of percentage yield.

\text{Percentage Yield}=\frac{7.78 \times 10^6}{8.00\times10^6}=\underline{98\%}


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Yield and Atom Economy Worksheet and Example Questions

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Yield and atom economy of chemical reactions. Questions

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