Yield & Atom Economy Flashcards

All 24 cards in this deck

What is meant by the yield of a reaction?

The amount (mass) of product obtained from the reaction.

What is the formula for percentage yield?

$\%\text{ yield} = \frac{\text{mass of product actually made}}{\text{maximum theoretical mass of product}} \times 100$

Give three reasons why the actual yield of a reaction is less than the theoretical yield.

1. The reaction is reversible, so it does not go to completion.
2. Some product is lost when it is separated from the reaction mixture.
3. Some reactants react in unexpected ways (side reactions).

True or false? A percentage yield of more than 100% is possible if the reaction works especially well.

False. The maximum theoretical mass is the most that can ever be made, so percentage yield can never exceed 100%.

What are the steps to calculate the percentage yield of a product?

e.g. 4 g of product is made when the maximum theoretical mass is 5 g

1. Divide actual mass by maximum theoretical mass: $4 \div 5 = 0.8$
2. Multiply by 100: $0.8 \times 100$
3. Percentage yield $= 80\%$

What are the steps to find the mass of product actually made from the percentage yield?

e.g. percentage yield $= 80\%$, maximum theoretical mass $= 20$ g

1. Divide the percentage yield by 100: $80 \div 100 = 0.8$
2. Multiply by the maximum theoretical mass: $0.8 \times 20$
3. Mass actually produced $= 16$ g

What is meant by the maximum theoretical mass of a product?

The mass of product that would be made if all the reactant were converted, with no losses and no side reactions.

What are the steps to calculate the theoretical mass of a product from a given mass of reactant?

e.g. mass of MgO from 2.4 g of Mg in $2\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\mathrm{MgO}$ ($A_r$ Mg = 24, $M_r$ MgO = 40)

1. Moles of reactant $= 2.4 \div 24 = 0.1$ mol
2. Use the equation ratio $2:2$, so moles of MgO $= 0.1$ mol
3. Mass $= 0.1 \times 40 = 4$ g

Which piece of information from the chemical equation is needed to convert moles of reactant into moles of product?

The balancing numbers (the mole ratio of reactant to product in the balanced equation).

True or false? The theoretical mass calculated from a balanced equation allows for product lost during filtering and drying.

False. The theoretical mass assumes no losses at all; losses are why the actual yield is lower.

When a reactant is in excess, which reactant's mass is used to calculate the theoretical mass of product?

The reactant that is not in excess (the limiting reactant), because it is used up first.

What is atom economy (atom utilisation)?

A measure of the amount of the starting materials that end up as useful products.

What is the formula for percentage atom economy?

$\%\text{ atom economy} = \frac{M_r \text{ of desired product}}{\text{total } M_r \text{ of all reactants}} \times 100$

What are the steps to calculate the percentage atom economy of a reaction?

e.g. desired product CaO in $\mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}$ ($M_r$: $\mathrm{CaCO_3}=100$, CaO $=56$)

1. $M_r$ of desired product $= 56$
2. Total $M_r$ of reactants $= 100$
3. $\frac{56}{100} \times 100 = 56\%$ (round to the number of significant figures asked for)

What feature of a balanced equation shows that the atom economy of a reaction is 100%?

There is only one product, so all the reactant atoms end up in the desired product.

Give two reasons why industry uses reactions with a high atom economy.

For sustainable development and for economic reasons (a higher percentage of useful product and less waste).

True or false? The denominator in the atom economy calculation is the total $M_r$ of all the products.

False. It is the total $M_r$ of all the reactants.

Name five factors considered when choosing a reaction pathway to make a product.

Atom economy, percentage yield, rate of reaction, position of equilibrium, and usefulness of by-products.

True or false? The reaction pathway with the highest atom economy is always the one chosen in industry.

False. Yield, rate, equilibrium position and whether the by-products are useful are also considered.

Why might a pathway with a lower atom economy still be chosen?

Because its by-products are useful and can be sold or used, or because it has a higher yield or faster rate.

Why is the position of equilibrium considered when choosing a reaction pathway?

An equilibrium lying towards the products gives a higher yield of the desired product.

Why is the rate of reaction considered when choosing a reaction pathway?

A faster reaction makes the product in less time, so more can be produced and costs are lower.

True or false? A reaction with a high percentage yield must also have a high atom economy.

False. Yield measures how much of the possible product is obtained; atom economy depends only on the balanced equation.

True or false? The atom economy of a reaction can be worked out from the balanced equation alone, without knowing how much product was actually made.

True. Atom economy uses only the $M_r$ values in the balanced equation; the actual mass made affects percentage yield, not atom economy.