 # Calculating Moles and Mass in Chemical Reactions Using Mole Ratios (stoichoimetric ratios) Chemistry Tutorial

## Key Concepts

• A balanced chemical equation can tell us:

⚛ The ratio of the number of molecules of each type reacting and produced.

⚛ The ratio of the moles of each reactant and product.

• The ratio of moles of each reactant and product in a reaction is known as the mole ratio (or stoichiometric ratio)
• The mole ratio (stoichiometric ratio) can be used to calculate the mass of reactants and products in a chemical reaction.

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## Mole Ratio (stoichiometric ratio)

The mole ratio is the stoichiometric ratio of reactants and products and is the ratio of the stoichiometric coefficients for reactants and products found in the balanced chemical equation.

 For the reaction aA + bB → cC + dD mol ratio for A : B : C : D is a : b : c : d

### Examples of Mole Ratios (stoichiometric ratios)

1.  in the reaction 2Mg(s) + O2(g) → 2MgO(s) the mole ratio of Mg : O2 : MgO is 2 : 1 : 2

That is, the complete reaction requires twice as many moles of magnesium as there are moles of oxygen.

n moles of oxygen gas will react with (2 × n) moles of magnesium to produce (2 × n) moles of magnesium oxide.

2.  in the reaction 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O the mole ratio of Al(OH)3 : H2SO4 : Al2(SO4)3 : H2O is 2 : 3 : 1 : 6

For each mole of Al2(SO4)3 produced, twice as many moles of Al(OH)3 are required to react with three times as many moles of H2SO4.

n moles of Al2(SO4)3 are produced when (2 × n) moles of Al(OH)3 react with (3 × n) moles of H2SO4. (6 × n) moles of H2O will also be produced.

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## Chemical Reactions and Moles of Reactants and Products

For the balanced chemical reaction shown below:

2Mg(s) + O2(g) → 2MgO(s)

the mole ratio of Mg : O2 : MgO is 2:1:2

That is, it requires 2 moles of magnesium and 1 mole of oxygen to produce 2 moles of magnesium oxide.

If only 1 mole of magnesium was present, it would require 1 ÷ 2 = ½ mole of oxygen gas to produce 2 ÷ 2 = 1 mole magnesium oxide.

If 10 moles of magnesium were present, it would require (1 ÷ 2) × 10 = 5 moles of oxygen gas to produce (2 ÷ 2) × 10 = 10 moles of magnesium oxide.

For n moles of magnesium :

• (1 ÷ 2) × n = ½n moles of oxygen gas are required
• (2 ÷ 2) × n = n moles of magnesium oxide are produced

The table below shows the moles of MgO produced when various amounts of Mg in moles react with the stoichiometric ratio of O2:

reaction 2Mg + O2 → 2MgO
mole ratio Mg
2 :
O2
1 :
MgO
2
example 1 0.50 mol 0.25 mol 0.50 mol
example 2 1.00 mol 0.50 mol 1.00 mol
example 3 1.50 mol 0.75 mol 1.50 mol
example 4 2.00 mol 1.00 mol 2.00 mol

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## Chemical Reactions and Masses of Reactants and Products

It is possible to calculate the mass of each reactant and product using the mole ratio (stoichiometric ratio) from the balanced chemical equation and the mathematical equation moles = mass ÷ molar mass

For the balanced chemical equation shown below:

2Mg(s) + O2(g) → 2MgO(s)

Given a mass of m grams of magnesium:

• mass O2 = moles(O2) × molar mass(O2)
(a) Calculate moles Mg = mass(Mg) ÷ molar mass(Mg)

moles(Mg) = m ÷ 24.31

(b) Use the balanced chemical equation to determine the mole ratio O2:Mg

moles(O2) : moles(Mg)   is   1 : 2

(c) Use the mole ratio to calculate moles O2

moles(O2) = (1 ÷ 2) × moles(Mg)

substitute value for moles(Mg)

moles(O2) = ½ × (m ÷ 24.31)

(d) Calculate mass of O2

mass(O2) = moles(O2) × molar mass(O2)

= [½ × (m ÷ 24.31)] × [2 × 16.00]

= [½ × (m ÷ 24.31)] × [32.00]

• mass MgO = moles(MgO) × molar mass(MgO)
(a) Calculate moles Mg = mass(Mg) ÷ molar mass(Mg)

moles(Mg) = m ÷ 24.31

(b) Use the balanced equation to determine the mole ratio MgO:Mg

moles(MgO) : moles(Mg)   is   2:2   which is the same as   1:1

(c) Use the mole ratio to calculate moles MgO

moles(MgO) = 1 × moles(Mg)

substitute in the value for moles(Mg)

moles(MgO) = 1 × (m ÷ 24.31)

(d) Calculate mass MgO

mass(MgO) = moles(MgO) × molar mass(MgO)

= [1 × (m ÷ 24.31)] × [24.31 + 16.00]

= [1 × m ÷ 24.31] × 40.31

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## Worked Example of Using Mole Ratio to Calculate Mass of Reactant or Product

The Question: 12.2 g of magnesium metal (Mg(s)) reacts completely with oxygen gas (O2(g)) to produce magnesium oxide (MgO(s)).

Calculate the mass of oxygen consumed during the reaction and the mass of magnesium oxide produced.

(1) Write the balanced chemical equation for the chemical reaction:

2Mg(s) + O2(g) → 2MgO(s)

(2) Determine the mole ratio (stoichiometric ratio) from the equation, Mg : O2 : MgO

moles(Mg) : moles(O2) : moles(MgO)   is   2:1:2

(3) Use the mole ratios to calculate the mass of O2 consumed and MgO produced as shown below:

• mass O2 = moles(O2) × molar mass(O2)

(a) Calculate moles(Mg) = mass(Mg) ÷ molar mass(Mg)

moles(Mg) = 12.2 ÷ 24.31 = 0.50 mol

(b) Use the balanced chemical equation to determine the mole ratio O2:Mg

moles(O2) : moles(Mg)   is   1:2

(c) Use the mole ratio to calculate moles O2

moles(O2) = (1 ÷ 2) × moles(Mg)

substitute in the value for moles of Mg

moles(O2) = ½ × 0.50 = 0.25 mol

(d) Calculate mass O2

mass(O2) = moles(O2) × molar mass(O2)

mass(O2) = 0.25 × (2 × 16.00) = 0.25 × 32.00

mass(O2) = 8.03 g

• mass MgO = moles(MgO) × molar mass(MgO)

(a) Calculate moles Mg

moles(Mg) = mass(Mg) ÷ molar mass(Mg)

moles(Mg) = 12.2 ÷ 24.31 = 0.50 mol

(b) Use the balanced equation to determine the mole ratio MgO:Mg

moles(MgO) : moles(Mg)   is   2:2   which is the same as   1:1

(c) Use the mole ratio to calculate moles MgO

moles(MgO) = 1 × moles(Mg)

substitute in the value for moles of Mg

moles(MgO) = 1 × 0.50 = 0.50 mol

(d) Calculate mass MgO

mass(MgO) = moles(MgO) × molar mass(MgO)

mass(MgO) = (1 × 0.50) × (24.31 + 16.00) = (1 × 0.50) × 40.31

mass(MgO) = 20.16 g