Yield

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Why Calculations Do Not Always Match Observations

Rememer: The principle of conservation of mass states that mass in a closed system is neither created nor destroyed by chemical reactions or physical transformations.

Even though no atoms are gained or lost in a chemical reaction, it is not always possible to obtain the calculated amount of a product because:

The reaction may not go to completion because it is reversible

Some of the product may be lost when it is separated from the reaction mixture

Some of the reactants may react in different ways to those which we expected

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Non-enclosed systems

Gases

One of the important properties of gases is that they diffuse into all available space. This is because their molecules have a lot of kinetic energy, so they have a lot of translational freedom.

Non-enclosed system: A system where the gas is free to escape

Example: An open beaker or flask

Reaction that produces gas

There may be an apparent decrease in mass between reactants and products if this reaction is carried out in a non-enclosed system.

This is because some of the reactants have been turned into gas and have escaped the reaction vessel.

If we could weigh the gas produced, we would find that no mass is lost.

Reactions that involve heating of solids

There may be an apparent increase in mass between reactants and products if this reaction is carried out in a non-enclosed system.

This apparent gain is because some of the solid reactants have reacted with gases to form solid products.

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Yield

The actual yield of a reaction is the actual mass of product obtained (Measured by weighing, not calculating!).

The theoretical yield is the mass of product we would obtain according to the balanced equation.

Percentage Yield = \(100 \% \times \frac{ \text{Actual Yield}}{\text{Theoretical Yield}}\)

There are some reasons why the mass of a reaction may be less than the maximum possible:

Reaction is reversible and may not be complete

Side reactions lead to unwanted products

Product may need to be purified resulting in loss of mass

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Example 1 - Theoretical Yield

Example 1:

Copper(II) carbonate decomposes to obtain copper(II) oxide. What is the theoretical yield of copper (II) oxide obtainable from 5.78g of copper(II) carbonate?

Answer:

The reaction takes place according to the equation \[CuCO_3 \rightarrow CuO+CO_2\]

When considering the theoretical yield, we imagine how much product we would have if all of the copper carbonate reacted completely to form copper oxide and carbon dioxide only. That would mean that one mole of copper carbonate woudl produce one mole of copper oxide and one mole of carbon dioxide.

The molecular mass of copper carbonate is 123.5 \(g \text{ } mol^{-1}.\)

Find the number of moles of copper carbonate: \[Moles = \frac{Mass}{\text{Molecular mass}}\]\[Moles = \frac{5.78}{123.5} = 0.0468mol\]

So, in theory, we should obtain 0.0468mol of copper oxide and 0.0468mol of carbon dioxide from our reaction.

The molecular mass of copper oxide is 79.5g.

Therefore, the theoretical yield of copper(II) oxide is: \[Mass = Moles \times \text{Molecular mass}\]\[\text{Mass of copper oxide} = 0.0468 \times 79.5 = 2.97g\]

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Example 2 - Theoretical Yield

Example:

Magnesium phosphate can be prepared from magnesium by reacting it with phosphoric acid. The equation for the reaction is: \[3Mg + H_3PO_4 \rightarrow Mg_3(PO_4)_2 +3H_2\]

What is the theoretical yield of magnesium phosphate obtainable from 5.62g of magnesium?

Answer:

The ratio of magnesium to magnesium phosphate is 1:3. This means that 3 moles of magnesium should, in theory, produce one mole of magnesium phosphate.

The atomic mass of magnesium is \(24.3 g \text{ }mol^{-1}\).\\

The molecular mass of magnesium phosphate is \(262.9g \text{ } mol^{-1}\).

Calculate the number of moles of magnesium: \[Moles = \frac{Mass}{\text{Molecular mass}}\]\[ \text{Moles of magnesium} = \frac{5.62}{24.3} = 0.231mol\]

Therefore, in theory we should have \(\frac{0.231}{3}=0.0770mol \) of magnesium phosphate. \[Mass= Moles \times \text{Molecular Mass}\]\[\text{Mass of magnesium phosphate =}0.770 \times 262.9 = 20.2g\]

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Example 3 - Percentage Yield

Example:

A manufacturer uses carbon monoxide and hydrogen to obtain methanol according to the following equation: \[CO+2H_2 \rightarrow CH_3OH\]

He obtains 4.07 tonnes of methanol starting from 4.32 tonnes of carbon monoxide. What is the percentage yield?

Answer:

The equation shows that one mole of carbon monoxide should produce one mole of methanol. The molecular masses are as follows:

Carbon monoxide: \(28g \text{ }mol^{-1}\)

Methanol: \(32 g \text{ }mol^{-1}\)

First find the theoretical yield. \[Moles = \frac{Mass}{\text{Molecular mass}}\] \[ \therefore \text{Moles of carbon monoxide} = \frac{4.32 \times 10^6}{28}=1.54 \times 10^5 mol\]Therefore, in theory we should aso have \(1.54 \times 10^5 mol\) of methanol. \\

Next calculate the actual yield: \[ \therefore \text{Moles of methanol} = \frac{4.07 \times 10^6}{32}=1.27 \times 10^5 mol\]

The percentage yield can be found by \[\text{Percentage Yield} = 100 \% \times \frac{ \text{Actual Yield}}{\text{Theoretical Yield}}\]\[=100 \% \times \frac{1.27 \times 10^5}{1.54 \times 10^5} = 82.4 \% \]

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Atom economy

Factors to take into consideration when assessing the suitability of an industrial process:

Percentage yield

Scarcity of non-renewable raw materials

Cost of raw materials

Quantity of energy needed

Atom economy

\[\text{Atom economy =} \frac{ \text{Molar mass of the desired product}}{ \text{Sum of the molar masses of all the products}} \times 100\%\]

If there is only one product, then the atom economy will always be 100%.

Reaction types and atom economy:

Addition reaction have 100% atom economy

Elimination and substitution reactions have lower atom economies

Multistep reactions have lower atom economies

A low atom economy can be offset if the manufacturer can sell some of the waste products.

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Example 4 - Atom Economy

Example:

Ethanol can be manufactured by the hydration of ethene. What is the atom economy of this process?

Answer:

The reaction takes place according to the equation:

\[C_2H_4+ H_2O \rightarrow C_2H_5OH \]

This reaction has only one product, therefore its atom economy must be 100%.

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Example 5 - Atom Economy

Example:

Ethene can be manufactured by the dehydration of ethanol. What is the atom economy of this process?

Answer:

The reaction takes place according to the following equation:

\[C_2H_5OH \rightarrow C_2H_4 + H_2O\]

The desired product is ethene - its molar mass is \( 28g \text{ }mol^{-1}\)

The sum of the molar masses of all the products is \( 46g \text{ }mol^{-1}\)

\[\text{Atom economy =} \frac{ \text{Molar mass of the desired product}}{ \text{Sum of the molar masses of all the products}} \times 100\%\]

\[ = \frac{28}{46} \times 100\% = 60.87%\]