Balancing numbers give the reacting ratio in moles

Definition

Stoichiometry

The ratio of the amounts in moles of the reactants and products in a reaction, shown by the balancing numbers in its balanced equation.

In $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ , two moles of magnesium react with one mole of oxygen to make two moles of magnesium oxide.
This mole ratio, $2 : 1 : 2$ , is the stoichiometry of the reaction.
The reacting masses come from multiplying each amount in moles by its relative formula mass.
1. Two moles of magnesium have a mass of $\times 24.3 = 48.6\,\text{g}$ , and they react with $g32.0\,\text{g}$ of oxygen to make $g80.6\,\text{g}$ of magnesium oxide.
The mass ratio, $48.6 : 32.0 : 80.6$ , is therefore not the same as the mole ratio.
The mole ratio holds at any scale, so $mol0.10\,\text{mol}$ of magnesium reacts with $mol0.050\,\text{mol}$ of oxygen.

Common Mistake

Do not compare reacting masses directly with the balancing numbers, because the numbers count moles and a mole of each substance has a different mass.

Reacting masses can be turned into balancing numbers

Convert the mass of each reactant and product into moles by dividing by its $M_r$ .
Divide every amount by the smallest amount to get a simple ratio.
1. If a value is a half, such as $1.5$ , multiply every value by $2$ to reach whole numbers.
Use the whole-number ratio as the balancing numbers in the equation.
Check that the equation also balances atom by atom, which confirms that the formulae and the ratio fit together.

Example

Deducing an equation from reacting masses

$g4.6\,\text{g}$ of sodium reacts with $g1.6\,\text{g}$ of oxygen, $O2\text{O}_2$ , to form $g6.2\,\text{g}$ of sodium oxide, $Na2O\text{Na}_2\text{O}$ ; deduce the balanced equation ( $A_r$ : Na = 23.0, O = 16.0).
Moles of $Na=4.623.0=0.20\text{Na} = \dfrac{4.6}{23.0} = 0.20$ , moles of $O2=1.632.0=0.050\text{O}_2 = \dfrac{1.6}{32.0} = 0.050$ and moles of $Na2O=6.262.0=0.10\text{Na}_2\text{O} = \dfrac{6.2}{62.0} = 0.10$ .
Divide each amount by the smallest, $0.050$ , to give the ratio $4 : 1 : 2$ .
The balanced equation is $4Na+O2→2Na2O4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O}$ .

Calculating reacting masses using the mole ratio

Write the balanced equation and pick out the ratio between the substance you know and the one you want.
Convert the known mass into moles using $moles=massMr\text{moles} = \dfrac{\text{mass}}{M_r}$ .
Use the mole ratio to find the moles of the substance you want.
Convert those moles into a mass using $mass=moles×Mr\text{mass} = \text{moles} \times M_r$ .
The same chain works in reverse to find the mass of reactant needed for a given mass of product.
1. To make $g9.53\,\text{g}$ of magnesium chloride, $M_r = 95.3$ , by $Mg+2HCl→MgCl2+H2\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$ , you need $mol\dfrac{9.53}{95.3} = 0.100\,\text{mol}$ of $MgCl2\text{MgCl}_2$ and so $mol0.100\,\text{mol}$ of magnesium, which is $0.100 \times 24.3 = 2.43\,\text{g}$ .

Example

Calculate the mass of iron made from $g80.0\,\text{g}$ of iron oxide in the reaction $Fe2O3+3CO→2Fe+3CO2\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ ( $A_r$ : Fe = 55.8, O = 16.0).
$Mr(Fe2O3)=(2×55.8)+(3×16.0)=159.6M_r(\text{Fe}_2\text{O}_3) = (2 \times 55.8) + (3 \times 16.0) = 159.6$ .
Moles of $Fe2O3=80.0159.6=0.50125\text{Fe}_2\text{O}_3 = \dfrac{80.0}{159.6} = 0.50125$ .
The ratio of $Fe2O3\text{Fe}_2\text{O}_3$ to $Fe\text{Fe}$ is $1 : 2$ , so moles of $Fe=2×0.50125=1.0025\text{Fe} = 2 \times 0.50125 = 1.0025$ .
Mass of iron $\times 55.8 = 55.9\,\text{g}$ to three significant figures.

The limiting reactant is used up first

Definition

Limiting reactant

The reactant that is used up first, which stops the reaction and sets the maximum amount of product.

Definition

Excess reactant

A reactant present in more than the amount needed, so some is left over when the reaction stops.

A reaction stops when one reactant runs out, even if some of the others remain.
Any reactant left over at the end was present in excess.
The amount of product depends on the limiting reactant, so doubling the limiting reactant doubles the mass of product.
1. Adding more of a reactant that is already in excess makes no more product.
Chemists often use a cheap reactant in excess, so that all of a more expensive reactant is used up.
When marble chips are left in the flask after the fizzing stops, the hydrochloric acid was the limiting reactant.

Example

A sandwich shop with 20 slices of bread and 6 portions of filling can make only 6 sandwiches, so the filling is limiting and 8 slices of bread are left in excess.

Find the limiting reactant before calculating product mass

Convert each reactant mass into moles.
Divide each amount by its balancing number in the equation.
1. The reactant with the smaller result is the limiting reactant.
Use the moles of the limiting reactant, never the excess one, to calculate the mass of product.
The amount of excess reactant left over is the amount supplied minus the amount that reacts.

Example

Limiting reactant and mass of iron sulfide

$g5.58\,\text{g}$ of iron is heated with $g4.00\,\text{g}$ of sulfur, $Fe+S→FeS\text{Fe} + \text{S} \rightarrow \text{FeS}$ ; find the limiting reactant and the mass of iron sulfide formed ( $A_r$ : Fe = 55.8, S = 32.1).
Moles of $Fe=5.5855.8=0.100\text{Fe} = \dfrac{5.58}{55.8} = 0.100$ and moles of $S=4.0032.1=0.125\text{S} = \dfrac{4.00}{32.1} = 0.125$ .
The ratio is $1 : 1$ , so iron has the smaller amount and is the limiting reactant.
Moles of $FeS=0.100\text{FeS} = 0.100$ , and $Mr(FeS)=55.8+32.1=87.9M_r(\text{FeS}) = 55.8 + 32.1 = 87.9$ .
Mass of iron sulfide $\times 87.9 = 8.79\,\text{g}$ .

Self review

What do the balancing numbers in $N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ tell you?
How can you deduce the balancing numbers of an equation from measured reacting masses?
What mass of calcium oxide forms when $g50.0\,\text{g}$ of calcium carbonate decomposes by $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ ( $A_r$ : Ca = 40.1, C = 12.0, O = 16.0)?
Why does a reaction stop when the limiting reactant is used up?
Why does adding more of an excess reactant not increase the mass of product?

Balancing numbers give the reacting ratio in moles

Definition

Stoichiometry

The ratio of the amounts in moles of the reactants and products in a reaction, shown by the balancing numbers in its balanced equation.

In $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ , two moles of magnesium react with one mole of oxygen to make two moles of magnesium oxide.
This mole ratio, $2 : 1 : 2$ , is the stoichiometry of the reaction.
The reacting masses come from multiplying each amount in moles by its relative formula mass.
1. Two moles of magnesium have a mass of $\times 24.3 = 48.6\,\text{g}$ , and they react with $g32.0\,\text{g}$ of oxygen to make $g80.6\,\text{g}$ of magnesium oxide.
The mass ratio, $48.6 : 32.0 : 80.6$ , is therefore not the same as the mole ratio.
The mole ratio holds at any scale, so $mol0.10\,\text{mol}$ of magnesium reacts with $mol0.050\,\text{mol}$ of oxygen.

Common Mistake

Do not compare reacting masses directly with the balancing numbers, because the numbers count moles and a mole of each substance has a different mass.

Reacting masses can be turned into balancing numbers

Convert the mass of each reactant and product into moles by dividing by its $M_r$ .
Divide every amount by the smallest amount to get a simple ratio.
1. If a value is a half, such as $1.5$ , multiply every value by $2$ to reach whole numbers.
Use the whole-number ratio as the balancing numbers in the equation.
Check that the equation also balances atom by atom, which confirms that the formulae and the ratio fit together.

Example

Deducing an equation from reacting masses

$g4.6\,\text{g}$ of sodium reacts with $g1.6\,\text{g}$ of oxygen, $O2\text{O}_2$ , to form $g6.2\,\text{g}$ of sodium oxide, $Na2O\text{Na}_2\text{O}$ ; deduce the balanced equation ( $A_r$ : Na = 23.0, O = 16.0).
Moles of $Na=4.623.0=0.20\text{Na} = \dfrac{4.6}{23.0} = 0.20$ , moles of $O2=1.632.0=0.050\text{O}_2 = \dfrac{1.6}{32.0} = 0.050$ and moles of $Na2O=6.262.0=0.10\text{Na}_2\text{O} = \dfrac{6.2}{62.0} = 0.10$ .
Divide each amount by the smallest, $0.050$ , to give the ratio $4 : 1 : 2$ .
The balanced equation is $4Na+O2→2Na2O4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O}$ .

Calculating reacting masses using the mole ratio

Write the balanced equation and pick out the ratio between the substance you know and the one you want.
Convert the known mass into moles using $moles=massMr\text{moles} = \dfrac{\text{mass}}{M_r}$ .
Use the mole ratio to find the moles of the substance you want.
Convert those moles into a mass using $mass=moles×Mr\text{mass} = \text{moles} \times M_r$ .
The same chain works in reverse to find the mass of reactant needed for a given mass of product.
1. To make $g9.53\,\text{g}$ of magnesium chloride, $M_r = 95.3$ , by $Mg+2HCl→MgCl2+H2\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$ , you need $mol\dfrac{9.53}{95.3} = 0.100\,\text{mol}$ of $MgCl2\text{MgCl}_2$ and so $mol0.100\,\text{mol}$ of magnesium, which is $0.100 \times 24.3 = 2.43\,\text{g}$ .

Example

Calculate the mass of iron made from $g80.0\,\text{g}$ of iron oxide in the reaction $Fe2O3+3CO→2Fe+3CO2\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$ ( $A_r$ : Fe = 55.8, O = 16.0).
$Mr(Fe2O3)=(2×55.8)+(3×16.0)=159.6M_r(\text{Fe}_2\text{O}_3) = (2 \times 55.8) + (3 \times 16.0) = 159.6$ .
Moles of $Fe2O3=80.0159.6=0.50125\text{Fe}_2\text{O}_3 = \dfrac{80.0}{159.6} = 0.50125$ .
The ratio of $Fe2O3\text{Fe}_2\text{O}_3$ to $Fe\text{Fe}$ is $1 : 2$ , so moles of $Fe=2×0.50125=1.0025\text{Fe} = 2 \times 0.50125 = 1.0025$ .
Mass of iron $\times 55.8 = 55.9\,\text{g}$ to three significant figures.

The limiting reactant is used up first

Definition

Limiting reactant

The reactant that is used up first, which stops the reaction and sets the maximum amount of product.

Definition

Excess reactant

A reactant present in more than the amount needed, so some is left over when the reaction stops.

A reaction stops when one reactant runs out, even if some of the others remain.
Any reactant left over at the end was present in excess.
The amount of product depends on the limiting reactant, so doubling the limiting reactant doubles the mass of product.
1. Adding more of a reactant that is already in excess makes no more product.
Chemists often use a cheap reactant in excess, so that all of a more expensive reactant is used up.
When marble chips are left in the flask after the fizzing stops, the hydrochloric acid was the limiting reactant.

Example

A sandwich shop with 20 slices of bread and 6 portions of filling can make only 6 sandwiches, so the filling is limiting and 8 slices of bread are left in excess.

Find the limiting reactant before calculating product mass

Convert each reactant mass into moles.
Divide each amount by its balancing number in the equation.
1. The reactant with the smaller result is the limiting reactant.
Use the moles of the limiting reactant, never the excess one, to calculate the mass of product.
The amount of excess reactant left over is the amount supplied minus the amount that reacts.

Example

Limiting reactant and mass of iron sulfide

$g5.58\,\text{g}$ of iron is heated with $g4.00\,\text{g}$ of sulfur, $Fe+S→FeS\text{Fe} + \text{S} \rightarrow \text{FeS}$ ; find the limiting reactant and the mass of iron sulfide formed ( $A_r$ : Fe = 55.8, S = 32.1).
Moles of $Fe=5.5855.8=0.100\text{Fe} = \dfrac{5.58}{55.8} = 0.100$ and moles of $S=4.0032.1=0.125\text{S} = \dfrac{4.00}{32.1} = 0.125$ .
The ratio is $1 : 1$ , so iron has the smaller amount and is the limiting reactant.
Moles of $FeS=0.100\text{FeS} = 0.100$ , and $Mr(FeS)=55.8+32.1=87.9M_r(\text{FeS}) = 55.8 + 32.1 = 87.9$ .
Mass of iron sulfide $\times 87.9 = 8.79\,\text{g}$ .

Self review

What do the balancing numbers in $N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3$ tell you?
How can you deduce the balancing numbers of an equation from measured reacting masses?
What mass of calcium oxide forms when $g50.0\,\text{g}$ of calcium carbonate decomposes by $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ ( $A_r$ : Ca = 40.1, C = 12.0, O = 16.0)?
Why does a reaction stop when the limiting reactant is used up?
Why does adding more of an excess reactant not increase the mass of product?

3.1.6 Stoichiometry, limiting reactants and reacting masses

Balancing numbers give the reacting ratio in moles

Reacting masses can be turned into balancing numbers

Deducing an equation from reacting masses

Calculating reacting masses using the mole ratio

The limiting reactant is used up first

Find the limiting reactant before calculating product mass

Limiting reactant and mass of iron sulfide

Review 3.1.6 Stoichiometry, limiting reactants and reacting masses by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

3.1.6 Stoichiometry, limiting reactants and reacting masses

Balancing numbers give the reacting ratio in moles

Reacting masses can be turned into balancing numbers

Deducing an equation from reacting masses

Calculating reacting masses using the mole ratio

The limiting reactant is used up first

Find the limiting reactant before calculating product mass

Limiting reactant and mass of iron sulfide

Review 3.1.6 Stoichiometry, limiting reactants and reacting masses by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

3.1.6 Stoichiometry, limiting reactants and reacting masses

Balancing numbers give the reacting ratio in moles

Reacting masses can be turned into balancing numbers

Deducing an equation from reacting masses

Calculating reacting masses using the mole ratio

The limiting reactant is used up first

Find the limiting reactant before calculating product mass

Limiting reactant and mass of iron sulfide

Review 3.1.6 Stoichiometry, limiting reactants and reacting masses by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

3.1.6 Stoichiometry, limiting reactants and reacting masses Revision Guide

3.1.6 Stoichiometry, limiting reactants and reacting masses

Balancing numbers give the reacting ratio in moles

Reacting masses can be turned into balancing numbers

Deducing an equation from reacting masses

Calculating reacting masses using the mole ratio

The limiting reactant is used up first

Find the limiting reactant before calculating product mass

Limiting reactant and mass of iron sulfide

Review 3.1.6 Stoichiometry, limiting reactants and reacting masses by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

3.1.6 Stoichiometry, limiting reactants and reacting masses Revision Guide