Relative atomic mass

Definition

Relative atomic mass

The weighted mean mass of the atoms of an element, taking into account the abundance of each isotope, compared with one-twelfth of the mass of a carbon-12 atom, given the symbol $A_r$ .

A single atom has a mass far too small to measure on a balance, so chemists compare the masses of atoms with one another instead.
The standard for this comparison is the carbon-12 atom, which is given a mass of exactly $12$ on the scale.
One-twelfth of the mass of a carbon-12 atom is the unit that every other atom is compared against.
Most elements are a mixture of isotopes with different masses, so the relative atomic mass, $A_r$ , is an average that allows for how common each isotope is.
This averaging is why some relative atomic masses are not whole numbers, such as $Ar(Cl)=35.5A_r(\text{Cl}) = 35.5$ .
$A_r$ has no units, because it is a comparison of one mass with another.
The relative atomic mass of each element is printed in its box on the Periodic Table you are given in the exam.
1. Read the box carefully, because the proton number is printed there too and is easy to mistake for $A_r$ .

Relative molecular and formula mass

Definition

Relative molecular mass

The sum of the relative atomic masses of all the atoms shown in the formula of a molecule, given the symbol $M_r$ .

Definition

Relative formula mass

The sum of the relative atomic masses of all the atoms shown in the formula of an ionic compound, given the symbol $M_r$ .

The relative molecular mass, $M_r$ , is used for substances made of molecules, such as water and carbon dioxide.
The relative formula mass, also given the symbol $M_r$ , is used for ionic compounds, because they are made of ions rather than molecules.
Both are found the same way, by adding up the $A_r$ values of every atom shown in the formula.
Multiply the $A_r$ of each element by the subscript written after its symbol.
When a formula contains a bracket, multiply every atom inside the bracket by the number outside it.
Like $A_r$ , the value of $M_r$ has no units.

Example

Relative molecular mass

These examples use $A_r$ values of $H=1\text{H} = 1$ , $C=12\text{C} = 12$ and $O=16\text{O} = 16$ .
Water, $H2O\text{H}_2\text{O}$ , contains two hydrogen atoms and one oxygen atom.

M_r(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18

Carbon dioxide, $CO2\text{CO}_2$ , contains one carbon atom and two oxygen atoms.

M_r(\text{CO}_2) = 12 + (2 \times 16) = 44

Example

Relative formula mass

These examples use $A_r$ values of $H=1\text{H} = 1$ , $N=14\text{N} = 14$ , $O=16\text{O} = 16$ , $Al=27\text{Al} = 27$ , $S=32\text{S} = 32$ and $Ca=40\text{Ca} = 40$ .
Calcium hydroxide, $Ca(OH)2\text{Ca(OH)}_2$ , contains one calcium atom and two $OH\text{OH}$ groups.

M_r(\text{Ca(OH)}_2) = 40 + 2 \times (16 + 1) = 74

Ammonium sulfate, $(NH4)2SO4(\text{NH}_4)_2\text{SO}_4$ , contains two $NH4\text{NH}_4$ groups, one sulfur atom and four oxygen atoms.

M_r((\text{NH}_4)_2\text{SO}_4) = 2 \times (14 + 4) + 32 + (4 \times 16) = 132

Aluminium sulfate, $Al2(SO4)3\text{Al}_2(\text{SO}_4)_3$ , contains two aluminium atoms and three $SO4\text{SO}_4$ groups.

M_r(\text{Al}_2(\text{SO}_4)_3) = (2 \times 27) + 3 \times (32 + 64) = 342

Common Mistake

Do not multiply only the first atom inside a bracket, because the number outside a bracket applies to every atom inside it.
Do not give $A_r$ or $M_r$ a unit such as grams, because both are relative values.
Do not call the $M_r$ of an ionic compound such as $NaCl\text{NaCl}$ a relative molecular mass, because ionic compounds are not made of molecules.

Reacting masses

Definition

Conservation of mass

The principle that atoms are neither created nor destroyed in a chemical reaction, so the total mass of the products equals the total mass of the reactants.

A balanced equation shows the fixed ratio in which formulae react, so it also fixes the ratio of their masses.
To turn an equation into a mass ratio, multiply the $M_{r}$ of each substance by the coefficient in front of it.
1. For $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ , the mass ratio is $100 : 56 : 44$ .
2. This means $g100\ \text{g}$ of calcium carbonate breaks down to give $g56\ \text{g}$ of calcium oxide and $g44\ \text{g}$ of carbon dioxide.
The total mass on the left of an equation always equals the total mass on the right, because atoms are only rearranged.
1. In the example above, $56 + 44 = 100$ , so mass is conserved.
A mass ratio works in any mass unit, so the same ratio gives $56$ tonnes of calcium oxide from $100$ tonnes of calcium carbonate.
When a reaction in an open container gives off a gas, the mass of the container and its contents falls, because the gas escapes into the air.

Calculating masses by simple proportion

Once you know the mass ratio, you can find any reacting mass by simple proportion.
Work through these steps in order:
1. Write the balanced equation.
2. Work out the $M_r$ of the two substances in the question.
3. Multiply each $M_r$ by its coefficient to give the mass ratio.
4. Find the scale factor by dividing the mass you are given by its value in the ratio.
5. Multiply the other value in the ratio by the scale factor.
The same method works in either direction, so it can find the mass of product formed or the mass of reactant needed.
Keep every mass in the same unit throughout, whether that is grams, kilograms or tonnes.

Example

Calcium oxide from limestone

\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2

$Mr(CaCO3)=40+12+(3×16)=100M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100$ and $Mr(CaO)=40+16=56M_r(\text{CaO}) = 40 + 16 = 56$ .
The mass ratio is $CaCO3:CaO=100:56\text{CaCO}_3 : \text{CaO} = 100:56$ .
Heating $g250\ \text{g}$ of calcium carbonate gives a scale factor of $250100=2.5\frac{250}{100} = 2.5$ .

\text{mass of CaO} = 56 \times 2.5 = 140\ \text{g}

Example

Mass of reactant needed

2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}

The mass ratio of $Mg:MgO\text{Mg}:\text{MgO}$ is $\times 24):(2 \times 40) = 48:80$ .
Making $g20\ \text{g}$ of magnesium oxide gives a scale factor of $2080=0.25\frac{20}{80} = 0.25$ .

\text{mass of Mg} = 48 \times 0.25 = 12\ \text{g}

The oxygen used is $8\ \text{g}$ , which agrees with $32 \times 0.25 = 8\ \text{g}$ from the ratio.

Example

Masses in tonnes

\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2

$Mr(Fe2O3)=(2×56)+(3×16)=160M_r(\text{Fe}_2\text{O}_3) = (2 \times 56) + (3 \times 16) = 160$ , and two iron atoms have a mass of $\times 56 = 112$ .
The mass ratio is $Fe2O3:Fe=160:112\text{Fe}_2\text{O}_3 : \text{Fe} = 160:112$ .
Starting from $800$ tonnes of iron(III) oxide gives a scale factor of $800160=5\frac{800}{160} = 5$ .

\text{mass of Fe} = 112 \times 5 = 560\ \text{tonnes}

Exam technique

Reacting mass calculations

Show the $M_r$ values and the mass ratio, because each correct step can earn credit even if the final answer is wrong.
Include the coefficients from the equation, because leaving them out gives a ratio that is wrong by a whole-number factor.
Give the answer in the same unit as the mass in the question, because a missing unit usually costs the final mark.
Check that the answer is sensible, for example that calcium oxide made from calcium carbonate has a smaller mass than the calcium carbonate.

Self review

Which atom is the standard used to compare the masses of atoms?
Why is the $A_r$ of chlorine not a whole number?
What is the $M_r$ of $Ca(OH)2\text{Ca(OH)}_2$ , using $A_r$ values of $Ca=40\text{Ca} = 40$ , $O=16\text{O} = 16$ and $H=1\text{H} = 1$ ?
Why is the term relative formula mass used for sodium chloride?
What mass of calcium oxide forms when $g50\ \text{g}$ of calcium carbonate is heated?

Relative atomic mass

Definition

Relative atomic mass

The weighted mean mass of the atoms of an element, taking into account the abundance of each isotope, compared with one-twelfth of the mass of a carbon-12 atom, given the symbol $A_r$ .

A single atom has a mass far too small to measure on a balance, so chemists compare the masses of atoms with one another instead.
The standard for this comparison is the carbon-12 atom, which is given a mass of exactly $12$ on the scale.
One-twelfth of the mass of a carbon-12 atom is the unit that every other atom is compared against.
Most elements are a mixture of isotopes with different masses, so the relative atomic mass, $A_r$ , is an average that allows for how common each isotope is.
This averaging is why some relative atomic masses are not whole numbers, such as $Ar(Cl)=35.5A_r(\text{Cl}) = 35.5$ .
$A_r$ has no units, because it is a comparison of one mass with another.
The relative atomic mass of each element is printed in its box on the Periodic Table you are given in the exam.
1. Read the box carefully, because the proton number is printed there too and is easy to mistake for $A_r$ .

Relative molecular and formula mass

Definition

Relative molecular mass

The sum of the relative atomic masses of all the atoms shown in the formula of a molecule, given the symbol $M_r$ .

Definition

Relative formula mass

The sum of the relative atomic masses of all the atoms shown in the formula of an ionic compound, given the symbol $M_r$ .

The relative molecular mass, $M_r$ , is used for substances made of molecules, such as water and carbon dioxide.
The relative formula mass, also given the symbol $M_r$ , is used for ionic compounds, because they are made of ions rather than molecules.
Both are found the same way, by adding up the $A_r$ values of every atom shown in the formula.
Multiply the $A_r$ of each element by the subscript written after its symbol.
When a formula contains a bracket, multiply every atom inside the bracket by the number outside it.
Like $A_r$ , the value of $M_r$ has no units.

Example

Relative molecular mass

These examples use $A_r$ values of $H=1\text{H} = 1$ , $C=12\text{C} = 12$ and $O=16\text{O} = 16$ .
Water, $H2O\text{H}_2\text{O}$ , contains two hydrogen atoms and one oxygen atom.

M_r(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18

Carbon dioxide, $CO2\text{CO}_2$ , contains one carbon atom and two oxygen atoms.

M_r(\text{CO}_2) = 12 + (2 \times 16) = 44

Example

Relative formula mass

These examples use $A_r$ values of $H=1\text{H} = 1$ , $N=14\text{N} = 14$ , $O=16\text{O} = 16$ , $Al=27\text{Al} = 27$ , $S=32\text{S} = 32$ and $Ca=40\text{Ca} = 40$ .
Calcium hydroxide, $Ca(OH)2\text{Ca(OH)}_2$ , contains one calcium atom and two $OH\text{OH}$ groups.

M_r(\text{Ca(OH)}_2) = 40 + 2 \times (16 + 1) = 74

Ammonium sulfate, $(NH4)2SO4(\text{NH}_4)_2\text{SO}_4$ , contains two $NH4\text{NH}_4$ groups, one sulfur atom and four oxygen atoms.

M_r((\text{NH}_4)_2\text{SO}_4) = 2 \times (14 + 4) + 32 + (4 \times 16) = 132

Aluminium sulfate, $Al2(SO4)3\text{Al}_2(\text{SO}_4)_3$ , contains two aluminium atoms and three $SO4\text{SO}_4$ groups.

M_r(\text{Al}_2(\text{SO}_4)_3) = (2 \times 27) + 3 \times (32 + 64) = 342

Common Mistake

Do not multiply only the first atom inside a bracket, because the number outside a bracket applies to every atom inside it.
Do not give $A_r$ or $M_r$ a unit such as grams, because both are relative values.
Do not call the $M_r$ of an ionic compound such as $NaCl\text{NaCl}$ a relative molecular mass, because ionic compounds are not made of molecules.

Reacting masses

Definition

Conservation of mass

The principle that atoms are neither created nor destroyed in a chemical reaction, so the total mass of the products equals the total mass of the reactants.

A balanced equation shows the fixed ratio in which formulae react, so it also fixes the ratio of their masses.
To turn an equation into a mass ratio, multiply the $M_{r}$ of each substance by the coefficient in front of it.
1. For $CaCO3→CaO+CO2\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ , the mass ratio is $100 : 56 : 44$ .
2. This means $g100\ \text{g}$ of calcium carbonate breaks down to give $g56\ \text{g}$ of calcium oxide and $g44\ \text{g}$ of carbon dioxide.
The total mass on the left of an equation always equals the total mass on the right, because atoms are only rearranged.
1. In the example above, $56 + 44 = 100$ , so mass is conserved.
A mass ratio works in any mass unit, so the same ratio gives $56$ tonnes of calcium oxide from $100$ tonnes of calcium carbonate.
When a reaction in an open container gives off a gas, the mass of the container and its contents falls, because the gas escapes into the air.

Calculating masses by simple proportion

Once you know the mass ratio, you can find any reacting mass by simple proportion.
Work through these steps in order:
1. Write the balanced equation.
2. Work out the $M_r$ of the two substances in the question.
3. Multiply each $M_r$ by its coefficient to give the mass ratio.
4. Find the scale factor by dividing the mass you are given by its value in the ratio.
5. Multiply the other value in the ratio by the scale factor.
The same method works in either direction, so it can find the mass of product formed or the mass of reactant needed.
Keep every mass in the same unit throughout, whether that is grams, kilograms or tonnes.

Example

Calcium oxide from limestone

\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2

$Mr(CaCO3)=40+12+(3×16)=100M_r(\text{CaCO}_3) = 40 + 12 + (3 \times 16) = 100$ and $Mr(CaO)=40+16=56M_r(\text{CaO}) = 40 + 16 = 56$ .
The mass ratio is $CaCO3:CaO=100:56\text{CaCO}_3 : \text{CaO} = 100:56$ .
Heating $g250\ \text{g}$ of calcium carbonate gives a scale factor of $250100=2.5\frac{250}{100} = 2.5$ .

\text{mass of CaO} = 56 \times 2.5 = 140\ \text{g}

Example

Mass of reactant needed

2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}

The mass ratio of $Mg:MgO\text{Mg}:\text{MgO}$ is $\times 24):(2 \times 40) = 48:80$ .
Making $g20\ \text{g}$ of magnesium oxide gives a scale factor of $2080=0.25\frac{20}{80} = 0.25$ .

\text{mass of Mg} = 48 \times 0.25 = 12\ \text{g}

The oxygen used is $8\ \text{g}$ , which agrees with $32 \times 0.25 = 8\ \text{g}$ from the ratio.

Example

Masses in tonnes

\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2

$Mr(Fe2O3)=(2×56)+(3×16)=160M_r(\text{Fe}_2\text{O}_3) = (2 \times 56) + (3 \times 16) = 160$ , and two iron atoms have a mass of $\times 56 = 112$ .
The mass ratio is $Fe2O3:Fe=160:112\text{Fe}_2\text{O}_3 : \text{Fe} = 160:112$ .
Starting from $800$ tonnes of iron(III) oxide gives a scale factor of $800160=5\frac{800}{160} = 5$ .

\text{mass of Fe} = 112 \times 5 = 560\ \text{tonnes}

Exam technique

Reacting mass calculations

Show the $M_r$ values and the mass ratio, because each correct step can earn credit even if the final answer is wrong.
Include the coefficients from the equation, because leaving them out gives a ratio that is wrong by a whole-number factor.
Give the answer in the same unit as the mass in the question, because a missing unit usually costs the final mark.
Check that the answer is sensible, for example that calcium oxide made from calcium carbonate has a smaller mass than the calcium carbonate.

Self review

Which atom is the standard used to compare the masses of atoms?
Why is the $A_r$ of chlorine not a whole number?
What is the $M_r$ of $Ca(OH)2\text{Ca(OH)}_2$ , using $A_r$ values of $Ca=40\text{Ca} = 40$ , $O=16\text{O} = 16$ and $H=1\text{H} = 1$ ?
Why is the term relative formula mass used for sodium chloride?
What mass of calcium oxide forms when $g50\ \text{g}$ of calcium carbonate is heated?

3.2.1 Relative masses and reacting masses

Relative atomic mass

Relative molecular and formula mass

Relative molecular mass

Relative formula mass

Reacting masses

Calculating masses by simple proportion

Calcium oxide from limestone

Mass of reactant needed

Masses in tonnes

Reacting mass calculations

Review 3.2.1 Relative masses and reacting masses by teaching Genie

3.2.1 Relative masses and reacting masses

Relative atomic mass

Relative molecular and formula mass

Relative molecular mass

Relative formula mass

Reacting masses

Calculating masses by simple proportion

Calcium oxide from limestone

Mass of reactant needed

Masses in tonnes

Reacting mass calculations

Review 3.2.1 Relative masses and reacting masses by teaching Genie

3.2.1 Relative masses and reacting masses

Relative atomic mass

Relative molecular and formula mass

Relative molecular mass

Relative formula mass

Reacting masses

Calculating masses by simple proportion

Calcium oxide from limestone

Mass of reactant needed

Masses in tonnes

Reacting mass calculations

Review 3.2.1 Relative masses and reacting masses by teaching Genie

3.2.1 Relative masses and reacting masses Revision Guide

3.2.1 Relative masses and reacting masses

Relative atomic mass

Relative molecular and formula mass

Relative molecular mass

Relative formula mass

Reacting masses

Calculating masses by simple proportion

Calcium oxide from limestone

Mass of reactant needed

Masses in tonnes

Reacting mass calculations

Review 3.2.1 Relative masses and reacting masses by teaching Genie

3.2.1 Relative masses and reacting masses Revision Guide