Adding relative atomic masses gives relative formula mass

Definition

Relative atomic mass

The average mass of an atom of an element compared with one-twelfth of the mass of a carbon-12 atom.

Definition

Relative formula mass

The mass of one formula unit of a substance compared with one-twelfth of the mass of a carbon-12 atom, given the symbol Mr.

Each element's relative atomic mass, $A_r$ , is given on the Periodic Table.
To find $M_r$ , multiply each $A_r$ by the number of atoms of that element in the formula and add the results.
A small number after a symbol applies only to that atom, so $H2O\text{H}_2\text{O}$ contains two hydrogen atoms and one oxygen atom.
A number after a bracket multiplies everything inside it, so $Ca(OH)2\text{Ca(OH)}_2$ contains one calcium, two oxygen and two hydrogen atoms.
Relative formula masses have no units, because they compare masses with the carbon-12 standard.

Example

Calculate the relative formula mass of calcium hydroxide, $Ca(OH)2\text{Ca(OH)}_2$ ( $A_r$ : Ca = 40.1, O = 16.0, H = 1.0).
Count the atoms: $1$ Ca, $2$ O and $2$ H.
Multiply and add: $Mr=40.1+(2×16.0)+(2×1.0)=74.1M_r = 40.1 + (2 \times 16.0) + (2 \times 1.0) = 74.1$ .

Relative formula masses balance on both sides of equations

A large number in front of a formula multiplies the whole formula, so $2H2O2\text{H}_2\text{O}$ means two water molecules.
To find the mass of a species in an equation, multiply its $M_r$ by this balancing number.
In $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ $2 Mg + O_{2} \to 2 MgO$ , the reactants total $(2×24.3)+32.0=80.6(2 \times 24.3) + 32.0 = 80.6$ $(2 \times 24.3) + 32.0 = 80.6$ .
1. The product also totals $\times (24.3 + 16.0) = 80.6$ , because no atoms are gained or lost.
If the two totals differ, the equation has not been balanced correctly.

Common Mistake

Do not count oxygen once for $O2\text{O}_2$ , because an oxygen molecule contains two atoms, giving $M_r = 32.0$ .
Do not round $A_r$ values such as Mg = 24.3 to whole numbers unless the values you are given are whole numbers.

The empirical formula is the simplest whole-number ratio

Definition

Empirical formula

The simplest whole-number ratio of atoms of each element in a compound.

Definition

Molecular formula

The actual number of atoms of each element in one molecule of a compound.

To find an empirical formula from a molecular formula, divide the numbers of atoms by the largest number that goes into all of them.
1. Ethane, $C2H6\text{C}_2\text{H}_6$ , divides by $2$ to give the empirical formula $CH3\text{CH}_3$ .
2. Glucose, $C6H12O6\text{C}_6\text{H}_{12}\text{O}_6$ , divides by $6$ to give $CH2O\text{CH}_2\text{O}$ .
When the numbers share no common factor, as in $H2O\text{H}_2\text{O}$ or $CO2\text{CO}_2$ , the molecular formula is already the empirical formula.
Ethene, $C2H4\text{C}_2\text{H}_4$ , and propene, $C3H6\text{C}_3\text{H}_6$ , share the empirical formula $CH2\text{CH}_2$ , so an empirical formula alone may not identify a compound.

Note

Ionic compounds such as sodium chloride are giant lattices with no separate molecules, so their formula, $NaCl\text{NaCl}$ , is always an empirical formula.

Reading empirical formulae from models and diagrams

In a ball and stick model or displayed formula, count the atoms of each element first.
Then write the counts as a ratio and simplify it to the smallest whole numbers.
For a giant structure diagram, such as part of a sodium chloride lattice, work out the ratio in which the ions repeat.
Going the other way, the empirical formula $CH2\text{CH}_2$ $CH_{2}$ tells you that a matching model must contain twice as many hydrogen atoms as carbon atoms.
1. A model of ethene with $2$ carbon and $4$ hydrogen atoms fits this ratio, while one with $2$ carbon and $6$ hydrogen atoms does not.

Example

A diagram of a molecule shows $4$ carbon atoms, $8$ hydrogen atoms and $2$ oxygen atoms; find its empirical formula.
Write the ratio of atoms: C : H : O = $4 : 8 : 2$ .
Divide each number by $2$ , the largest number that goes into all three: $2 : 4 : 1$ .
Write the empirical formula: $C2H4O\text{C}_2\text{H}_4\text{O}$ .

Self review

How do you calculate the relative formula mass of a compound?
What is the relative formula mass of $MgCl2\text{MgCl}_2$ ( $A_r$ : Mg = 24.3, Cl = 35.5)?
What does the number $3$ in $3CO23\text{CO}_2$ tell you?
What is the empirical formula of $C4H10\text{C}_4\text{H}_{10}$ ?
Why is the formula of sodium chloride always an empirical formula?

Adding relative atomic masses gives relative formula mass

Definition

Relative atomic mass

The average mass of an atom of an element compared with one-twelfth of the mass of a carbon-12 atom.

Definition

Relative formula mass

The mass of one formula unit of a substance compared with one-twelfth of the mass of a carbon-12 atom, given the symbol Mr.

Each element's relative atomic mass, $A_r$ , is given on the Periodic Table.
To find $M_r$ , multiply each $A_r$ by the number of atoms of that element in the formula and add the results.
A small number after a symbol applies only to that atom, so $H2O\text{H}_2\text{O}$ contains two hydrogen atoms and one oxygen atom.
A number after a bracket multiplies everything inside it, so $Ca(OH)2\text{Ca(OH)}_2$ contains one calcium, two oxygen and two hydrogen atoms.
Relative formula masses have no units, because they compare masses with the carbon-12 standard.

Example

Calculate the relative formula mass of calcium hydroxide, $Ca(OH)2\text{Ca(OH)}_2$ ( $A_r$ : Ca = 40.1, O = 16.0, H = 1.0).
Count the atoms: $1$ Ca, $2$ O and $2$ H.
Multiply and add: $Mr=40.1+(2×16.0)+(2×1.0)=74.1M_r = 40.1 + (2 \times 16.0) + (2 \times 1.0) = 74.1$ .

Relative formula masses balance on both sides of equations

A large number in front of a formula multiplies the whole formula, so $2H2O2\text{H}_2\text{O}$ means two water molecules.
To find the mass of a species in an equation, multiply its $M_r$ by this balancing number.
In $2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ $2 Mg + O_{2} \to 2 MgO$ , the reactants total $(2×24.3)+32.0=80.6(2 \times 24.3) + 32.0 = 80.6$ $(2 \times 24.3) + 32.0 = 80.6$ .
1. The product also totals $\times (24.3 + 16.0) = 80.6$ , because no atoms are gained or lost.
If the two totals differ, the equation has not been balanced correctly.

Common Mistake

Do not count oxygen once for $O2\text{O}_2$ , because an oxygen molecule contains two atoms, giving $M_r = 32.0$ .
Do not round $A_r$ values such as Mg = 24.3 to whole numbers unless the values you are given are whole numbers.

The empirical formula is the simplest whole-number ratio

Definition

Empirical formula

The simplest whole-number ratio of atoms of each element in a compound.

Definition

Molecular formula

The actual number of atoms of each element in one molecule of a compound.

To find an empirical formula from a molecular formula, divide the numbers of atoms by the largest number that goes into all of them.
1. Ethane, $C2H6\text{C}_2\text{H}_6$ , divides by $2$ to give the empirical formula $CH3\text{CH}_3$ .
2. Glucose, $C6H12O6\text{C}_6\text{H}_{12}\text{O}_6$ , divides by $6$ to give $CH2O\text{CH}_2\text{O}$ .
When the numbers share no common factor, as in $H2O\text{H}_2\text{O}$ or $CO2\text{CO}_2$ , the molecular formula is already the empirical formula.
Ethene, $C2H4\text{C}_2\text{H}_4$ , and propene, $C3H6\text{C}_3\text{H}_6$ , share the empirical formula $CH2\text{CH}_2$ , so an empirical formula alone may not identify a compound.

Note

Ionic compounds such as sodium chloride are giant lattices with no separate molecules, so their formula, $NaCl\text{NaCl}$ , is always an empirical formula.

Reading empirical formulae from models and diagrams

In a ball and stick model or displayed formula, count the atoms of each element first.
Then write the counts as a ratio and simplify it to the smallest whole numbers.
For a giant structure diagram, such as part of a sodium chloride lattice, work out the ratio in which the ions repeat.
Going the other way, the empirical formula $CH2\text{CH}_2$ $CH_{2}$ tells you that a matching model must contain twice as many hydrogen atoms as carbon atoms.
1. A model of ethene with $2$ carbon and $4$ hydrogen atoms fits this ratio, while one with $2$ carbon and $6$ hydrogen atoms does not.

Example

A diagram of a molecule shows $4$ carbon atoms, $8$ hydrogen atoms and $2$ oxygen atoms; find its empirical formula.
Write the ratio of atoms: C : H : O = $4 : 8 : 2$ .
Divide each number by $2$ , the largest number that goes into all three: $2 : 4 : 1$ .
Write the empirical formula: $C2H4O\text{C}_2\text{H}_4\text{O}$ .

Self review

How do you calculate the relative formula mass of a compound?
What is the relative formula mass of $MgCl2\text{MgCl}_2$ ( $A_r$ : Mg = 24.3, Cl = 35.5)?
What does the number $3$ in $3CO23\text{CO}_2$ tell you?
What is the empirical formula of $C4H10\text{C}_4\text{H}_{10}$ ?
Why is the formula of sodium chloride always an empirical formula?

2.1.2 Relative formula mass and empirical formula

Adding relative atomic masses gives relative formula mass

Relative formula masses balance on both sides of equations

The empirical formula is the simplest whole-number ratio

Reading empirical formulae from models and diagrams

Review 2.1.2 Relative formula mass and empirical formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.1.2 Relative formula mass and empirical formula

Adding relative atomic masses gives relative formula mass

Relative formula masses balance on both sides of equations

The empirical formula is the simplest whole-number ratio

Reading empirical formulae from models and diagrams

Review 2.1.2 Relative formula mass and empirical formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.1.2 Relative formula mass and empirical formula

Adding relative atomic masses gives relative formula mass

Relative formula masses balance on both sides of equations

The empirical formula is the simplest whole-number ratio

Reading empirical formulae from models and diagrams

Review 2.1.2 Relative formula mass and empirical formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.1.2 Relative formula mass and empirical formula Revision Guide

2.1.2 Relative formula mass and empirical formula

Adding relative atomic masses gives relative formula mass

Relative formula masses balance on both sides of equations

The empirical formula is the simplest whole-number ratio

Reading empirical formulae from models and diagrams

Review 2.1.2 Relative formula mass and empirical formula by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

2.1.2 Relative formula mass and empirical formula Revision Guide