Empirical formulae from masses

Definition

Empirical formula

A formula that shows the simplest whole-number ratio of the atoms or ions of each element in a compound.

A formula shows a ratio of atoms, but analysing a compound gives the masses of the elements in it.
Dividing each mass by the $A_r$ of that element turns the masses into amounts in moles, and moles of atoms can be compared directly.
Find an empirical formula by following these steps:
1. Write down the mass of each element in the compound.
2. Divide each mass by the $A_r$ of that element to find its amount in moles.
3. Divide every amount by the smallest amount.
4. If any value is not close to a whole number, multiply all the values by the same small whole number.
5. Write the empirical formula using the whole numbers as subscripts.
When the data are percentages by mass, treat each percentage as the mass in grams of that element in a $g100\ \text{g}$ sample.
When the mass of one element is not given, find it by subtracting the masses of the other elements from the total mass of the compound.
Some ratios must be scaled up rather than rounded.
1. A value close to $1.5$ means every value is multiplied by $2$ .
2. A value close to $1.33$ or $1.67$ means every value is multiplied by $3$ .
3. A value close to $1.25$ or $1.75$ means every value is multiplied by $4$ .
Round to the nearest whole number only when a value is very close to it, such as $1.98$ or $3.02$ , because small differences come from rounding in the data.

Example

Empirical formula from masses

A compound contains $g11.2\ \text{g}$ of iron and $g4.80\ \text{g}$ of oxygen.

\text{Fe}: \dfrac{11.2}{56} = 0.200\ \text{mol} \qquad \text{O}: \dfrac{4.80}{16} = 0.300\ \text{mol}

Dividing both amounts by $0.200$ gives the ratio $1 : 1.5$ .
Multiplying both values by $2$ gives the whole-number ratio $2 : 3$ .

\text{Fe}_2\text{O}_3

Example

Empirical formula from percentages

A hydrocarbon contains $85.7%85.7\%$ carbon and $14.3%14.3\%$ hydrogen by mass, so a $g100\ \text{g}$ sample contains $g85.7\ \text{g}$ of carbon and $g14.3\ \text{g}$ of hydrogen.

\text{C}: \dfrac{85.7}{12} = 7.14\ \text{mol} \qquad \text{H}: \dfrac{14.3}{1} = 14.3\ \text{mol}

Dividing both amounts by $7.14$ gives the ratio $1 : 2.00$ .

\text{CH}_2

Practical

Finding the formula of magnesium oxide

Aim: to find the empirical formula of magnesium oxide by measuring the mass of magnesium that burns and the mass of oxide it forms.
Apparatus: a crucible with a lid, a pipeclay triangle, a tripod, a Bunsen burner, a heat-resistant mat, crucible tongs, a balance reading to $g0.01\ \text{g}$ , about $cm15\ \text{cm}$ of magnesium ribbon, emery paper and eye protection.
Method:
- Rub the magnesium ribbon with emery paper until it is shiny, which removes the dull layer of oxide from its surface.
- Weigh the empty crucible with its lid and record the mass.
- Coil the ribbon loosely, place it in the crucible, and weigh the crucible, lid and magnesium together.
- Stand the crucible in a pipeclay triangle on a tripod, put the lid on, and heat it strongly with a roaring blue flame.
- Once the magnesium starts to glow, use tongs to lift the lid slightly every minute or so, which lets oxygen in while keeping the white smoke of magnesium oxide inside.
- Keep heating until the magnesium no longer glows when the lid is lifted, which shows that it has finished reacting.
- Let the crucible cool on the heat-resistant mat with the lid on, then weigh the crucible, lid and contents.
- Heat, cool and reweigh again until two readings agree, which shows that all the magnesium has reacted.
Results:
- The mass of the empty crucible and lid is $g24.62\ \text{g}$ .
- The mass of the crucible, lid and magnesium is $g24.86\ \text{g}$ , so the magnesium has a mass of $g0.24\ \text{g}$ .
- The mass of the crucible, lid and magnesium oxide is $g25.02\ \text{g}$ , so the oxide has a mass of $g0.40\ \text{g}$ .
- The oxygen that combined has a mass of $0.16\ \text{g}$ .
- The shiny grey ribbon has turned into a white powder.
Calculation:
- The amount of magnesium is $mol\frac{0.24}{24} = 0.010\ \text{mol}$ and the amount of oxygen is $mol\frac{0.16}{16} = 0.010\ \text{mol}$ .
- The ratio $0.010 : 0.010$ simplifies to $1 : 1$ , so the empirical formula is $MgO\text{MgO}$ .
Watch out:
- Lifting the lid too far lets white magnesium oxide smoke escape, which makes the final mass too low.
- Stopping the heating too early leaves some magnesium unreacted, which makes the mass of oxygen too low.
Safety: wear eye protection, avoid looking straight at the burning magnesium because its light is very bright, and handle the hot crucible only with tongs.

Molecular formulae from empirical formulae

Definition

Molecular formula

A formula that shows the actual number of atoms of each different element in one molecule of a substance.

The molecular formula is always a whole-number multiple of the empirical formula.
To find the molecular formula, you need the empirical formula and the $M_r$ of the compound.
Work through these steps:
1. Add up the $A_r$ values in the empirical formula to find the empirical formula mass.
2. Divide the $M_r$ of the compound by the empirical formula mass.
3. Multiply every subscript in the empirical formula by this whole number.
If the division does not give a whole number, there is an error in an earlier step.

Example

Molecular formula of glucose

A compound has the empirical formula $CH2O\text{CH}_2\text{O}$ and an $M_r$ of $180$ .

\text{empirical formula mass} = 12 + (2 \times 1) + 16 = 30

\dfrac{180}{30} = 6

Multiplying every subscript by $6$ gives the molecular formula $C6H12O6\text{C}_6\text{H}_{12}\text{O}_6$ .

Example

Molecular formula of a hydrocarbon

The hydrocarbon with the empirical formula $CH2\text{CH}_2$ has an $M_r$ of $56$ .

\text{empirical formula mass} = 12 + (2 \times 1) = 14

\dfrac{56}{14} = 4

Multiplying every subscript by $4$ gives the molecular formula $C4H8\text{C}_4\text{H}_8$ .

Percentage composition by mass

Definition

Percentage composition by mass

The percentage of the total mass of a compound that is made up by one particular element.

The percentage by mass of an element in a compound is found using $element×ArMr×100\text{percentage by mass} = \dfrac{\text{number of atoms of the element} \times A_r}{M_r} \times 100$ .
Count every atom of the element, including any atoms inside brackets.
The percentages of all the elements in a compound add up to $100%100\%$ , which gives a useful check.
Percentage composition lets you compare compounds, such as which of two compounds supplies more nitrogen per kilogram.

Example

Percentage of nitrogen in two compounds

Calcium nitrate, $Ca(NO3)2\text{Ca(NO}_3\text{)}_2$ , has $Mr=40+2×(14+48)=164M_r = 40 + 2 \times (14 + 48) = 164$ and contains two nitrogen atoms.

\%\ \text{N} = \dfrac{2 \times 14}{164} \times 100 = 17.1\%

Ammonium nitrate, $NH4NO3\text{NH}_4\text{NO}_3$ , has $M_r = 14 + 4 + 14 + 48 = 80$ and also contains two nitrogen atoms.

\%\ \text{N} = \dfrac{2 \times 14}{80} \times 100 = 35.0\%

Percentage yield

Definition

Theoretical yield

The maximum mass of product that could form from the reactants, calculated from the balanced equation.

Definition

Actual yield

The mass of product that is actually obtained at the end of a reaction.

Definition

Percentage yield

The actual yield of a product expressed as a percentage of its theoretical yield.

The theoretical yield is calculated from the balanced equation, assuming that all of the limiting reactant turns into product.
The actual yield is the mass of product that is really collected at the end.
The percentage yield is $yield×100\text{percentage yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100$ .
The actual yield is almost always smaller than the theoretical yield.
1. The reaction may not go to completion, especially if it is reversible.
2. Some product is lost when it is filtered, poured from one container to another or purified.
3. Some of the reactants may form other, unwanted products.
A percentage yield above $100%100\%$ is impossible, and it usually means the product is still wet or contains impurities.

Example

Percentage yield of calcium oxide

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

Heating $g10.0\ \text{g}$ of calcium carbonate uses $mol\frac{10.0}{100} = 0.100\ \text{mol}$ .
The mole ratio is $1 : 1$ , so the theoretical yield of calcium oxide is $0.100 \times 56 = 5.60\ \text{g}$ .
The actual yield collected is $g4.76\ \text{g}$ .

\text{percentage yield} = \dfrac{4.76}{5.60} \times 100 = 85.0\%

Percentage purity

Definition

Percentage purity

The mass of the pure substance in a sample expressed as a percentage of the total mass of the sample.

Many substances, such as medicines, foods and metal ores, contain impurities that are not the substance you want.
The percentage purity is $sample×100\text{percentage purity} = \dfrac{\text{mass of pure substance}}{\text{total mass of sample}} \times 100$ .
The mass of pure substance is often found from a reaction, using the mass or volume of a product and the balanced equation.
Purity matters most in medicines and foods, because even a small amount of a harmful impurity can make a product unsafe.

Example

Purity of a limestone sample

\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}

A $g5.00\ \text{g}$ sample of limestone reacts with excess acid and gives $dm31.08\ \text{dm}^3$ of carbon dioxide at r.t.p.
The amount of carbon dioxide is $mol\frac{1.08}{24} = 0.0450\ \text{mol}$ , so the sample contained $mol0.0450\ \text{mol}$ of calcium carbonate.
The mass of pure calcium carbonate is $0.0450 \times 100 = 4.50\ \text{g}$ .

\text{percentage purity} = \dfrac{4.50}{5.00} \times 100 = 90.0\%

Common Mistake

Do not divide masses by proton numbers, because an empirical formula calculation always uses $A_r$ values.
Do not round a ratio such as $1 : 1.5$ to $1 : 2$ , because it must be doubled to give $2 : 3$ .
Do not put the theoretical yield on the top of the percentage yield fraction.
Do not put the mass of the impurity on the top of the percentage purity fraction.

Exam technique

Formula and percentage calculations

Set out empirical formula working in columns, with a row for mass, a row for moles and a row for the ratio, so every step is clear to the examiner.
Keep at least three significant figures in mole values until the final ratio, because early rounding can hide a ratio such as $1 : 1.33$ .
Check that any percentage yield or percentage purity you calculate is below $100%100\%$ .
When asked for a molecular formula, show the empirical formula mass and the multiplier as well as the final formula.

Self review

Why must the mass of each element be divided by its $A_r$ when finding an empirical formula?
What should you do when a mole ratio comes out as $1 : 1.5$ ?
How do you find a molecular formula from an empirical formula and an $M_r$ ?
What is the percentage by mass of nitrogen in $NH4NO3\text{NH}_4\text{NO}_3$ ?
Which yield goes on the bottom of the percentage yield fraction?

Empirical formulae from masses

Definition

Empirical formula

A formula that shows the simplest whole-number ratio of the atoms or ions of each element in a compound.

A formula shows a ratio of atoms, but analysing a compound gives the masses of the elements in it.
Dividing each mass by the $A_r$ of that element turns the masses into amounts in moles, and moles of atoms can be compared directly.
Find an empirical formula by following these steps:
1. Write down the mass of each element in the compound.
2. Divide each mass by the $A_r$ of that element to find its amount in moles.
3. Divide every amount by the smallest amount.
4. If any value is not close to a whole number, multiply all the values by the same small whole number.
5. Write the empirical formula using the whole numbers as subscripts.
When the data are percentages by mass, treat each percentage as the mass in grams of that element in a $g100\ \text{g}$ sample.
When the mass of one element is not given, find it by subtracting the masses of the other elements from the total mass of the compound.
Some ratios must be scaled up rather than rounded.
1. A value close to $1.5$ means every value is multiplied by $2$ .
2. A value close to $1.33$ or $1.67$ means every value is multiplied by $3$ .
3. A value close to $1.25$ or $1.75$ means every value is multiplied by $4$ .
Round to the nearest whole number only when a value is very close to it, such as $1.98$ or $3.02$ , because small differences come from rounding in the data.

Example

Empirical formula from masses

A compound contains $g11.2\ \text{g}$ of iron and $g4.80\ \text{g}$ of oxygen.

\text{Fe}: \dfrac{11.2}{56} = 0.200\ \text{mol} \qquad \text{O}: \dfrac{4.80}{16} = 0.300\ \text{mol}

Dividing both amounts by $0.200$ gives the ratio $1 : 1.5$ .
Multiplying both values by $2$ gives the whole-number ratio $2 : 3$ .

\text{Fe}_2\text{O}_3

Example

Empirical formula from percentages

A hydrocarbon contains $85.7%85.7\%$ carbon and $14.3%14.3\%$ hydrogen by mass, so a $g100\ \text{g}$ sample contains $g85.7\ \text{g}$ of carbon and $g14.3\ \text{g}$ of hydrogen.

\text{C}: \dfrac{85.7}{12} = 7.14\ \text{mol} \qquad \text{H}: \dfrac{14.3}{1} = 14.3\ \text{mol}

Dividing both amounts by $7.14$ gives the ratio $1 : 2.00$ .

\text{CH}_2

Practical

Finding the formula of magnesium oxide

Aim: to find the empirical formula of magnesium oxide by measuring the mass of magnesium that burns and the mass of oxide it forms.
Apparatus: a crucible with a lid, a pipeclay triangle, a tripod, a Bunsen burner, a heat-resistant mat, crucible tongs, a balance reading to $g0.01\ \text{g}$ , about $cm15\ \text{cm}$ of magnesium ribbon, emery paper and eye protection.
Method:
- Rub the magnesium ribbon with emery paper until it is shiny, which removes the dull layer of oxide from its surface.
- Weigh the empty crucible with its lid and record the mass.
- Coil the ribbon loosely, place it in the crucible, and weigh the crucible, lid and magnesium together.
- Stand the crucible in a pipeclay triangle on a tripod, put the lid on, and heat it strongly with a roaring blue flame.
- Once the magnesium starts to glow, use tongs to lift the lid slightly every minute or so, which lets oxygen in while keeping the white smoke of magnesium oxide inside.
- Keep heating until the magnesium no longer glows when the lid is lifted, which shows that it has finished reacting.
- Let the crucible cool on the heat-resistant mat with the lid on, then weigh the crucible, lid and contents.
- Heat, cool and reweigh again until two readings agree, which shows that all the magnesium has reacted.
Results:
- The mass of the empty crucible and lid is $g24.62\ \text{g}$ .
- The mass of the crucible, lid and magnesium is $g24.86\ \text{g}$ , so the magnesium has a mass of $g0.24\ \text{g}$ .
- The mass of the crucible, lid and magnesium oxide is $g25.02\ \text{g}$ , so the oxide has a mass of $g0.40\ \text{g}$ .
- The oxygen that combined has a mass of $0.16\ \text{g}$ .
- The shiny grey ribbon has turned into a white powder.
Calculation:
- The amount of magnesium is $mol\frac{0.24}{24} = 0.010\ \text{mol}$ and the amount of oxygen is $mol\frac{0.16}{16} = 0.010\ \text{mol}$ .
- The ratio $0.010 : 0.010$ simplifies to $1 : 1$ , so the empirical formula is $MgO\text{MgO}$ .
Watch out:
- Lifting the lid too far lets white magnesium oxide smoke escape, which makes the final mass too low.
- Stopping the heating too early leaves some magnesium unreacted, which makes the mass of oxygen too low.
Safety: wear eye protection, avoid looking straight at the burning magnesium because its light is very bright, and handle the hot crucible only with tongs.

Molecular formulae from empirical formulae

Definition

Molecular formula

A formula that shows the actual number of atoms of each different element in one molecule of a substance.

The molecular formula is always a whole-number multiple of the empirical formula.
To find the molecular formula, you need the empirical formula and the $M_r$ of the compound.
Work through these steps:
1. Add up the $A_r$ values in the empirical formula to find the empirical formula mass.
2. Divide the $M_r$ of the compound by the empirical formula mass.
3. Multiply every subscript in the empirical formula by this whole number.
If the division does not give a whole number, there is an error in an earlier step.

Example

Molecular formula of glucose

A compound has the empirical formula $CH2O\text{CH}_2\text{O}$ and an $M_r$ of $180$ .

\text{empirical formula mass} = 12 + (2 \times 1) + 16 = 30

\dfrac{180}{30} = 6

Multiplying every subscript by $6$ gives the molecular formula $C6H12O6\text{C}_6\text{H}_{12}\text{O}_6$ .

Example

Molecular formula of a hydrocarbon

The hydrocarbon with the empirical formula $CH2\text{CH}_2$ has an $M_r$ of $56$ .

\text{empirical formula mass} = 12 + (2 \times 1) = 14

\dfrac{56}{14} = 4

Multiplying every subscript by $4$ gives the molecular formula $C4H8\text{C}_4\text{H}_8$ .

Percentage composition by mass

Definition

Percentage composition by mass

The percentage of the total mass of a compound that is made up by one particular element.

The percentage by mass of an element in a compound is found using $element×ArMr×100\text{percentage by mass} = \dfrac{\text{number of atoms of the element} \times A_r}{M_r} \times 100$ .
Count every atom of the element, including any atoms inside brackets.
The percentages of all the elements in a compound add up to $100%100\%$ , which gives a useful check.
Percentage composition lets you compare compounds, such as which of two compounds supplies more nitrogen per kilogram.

Example

Percentage of nitrogen in two compounds

Calcium nitrate, $Ca(NO3)2\text{Ca(NO}_3\text{)}_2$ , has $Mr=40+2×(14+48)=164M_r = 40 + 2 \times (14 + 48) = 164$ and contains two nitrogen atoms.

\%\ \text{N} = \dfrac{2 \times 14}{164} \times 100 = 17.1\%

Ammonium nitrate, $NH4NO3\text{NH}_4\text{NO}_3$ , has $M_r = 14 + 4 + 14 + 48 = 80$ and also contains two nitrogen atoms.

\%\ \text{N} = \dfrac{2 \times 14}{80} \times 100 = 35.0\%

Percentage yield

Definition

Theoretical yield

The maximum mass of product that could form from the reactants, calculated from the balanced equation.

Definition

Actual yield

The mass of product that is actually obtained at the end of a reaction.

Definition

Percentage yield

The actual yield of a product expressed as a percentage of its theoretical yield.

The theoretical yield is calculated from the balanced equation, assuming that all of the limiting reactant turns into product.
The actual yield is the mass of product that is really collected at the end.
The percentage yield is $yield×100\text{percentage yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100$ .
The actual yield is almost always smaller than the theoretical yield.
1. The reaction may not go to completion, especially if it is reversible.
2. Some product is lost when it is filtered, poured from one container to another or purified.
3. Some of the reactants may form other, unwanted products.
A percentage yield above $100%100\%$ is impossible, and it usually means the product is still wet or contains impurities.

Example

Percentage yield of calcium oxide

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

Heating $g10.0\ \text{g}$ of calcium carbonate uses $mol\frac{10.0}{100} = 0.100\ \text{mol}$ .
The mole ratio is $1 : 1$ , so the theoretical yield of calcium oxide is $0.100 \times 56 = 5.60\ \text{g}$ .
The actual yield collected is $g4.76\ \text{g}$ .

\text{percentage yield} = \dfrac{4.76}{5.60} \times 100 = 85.0\%

Percentage purity

Definition

Percentage purity

The mass of the pure substance in a sample expressed as a percentage of the total mass of the sample.

Many substances, such as medicines, foods and metal ores, contain impurities that are not the substance you want.
The percentage purity is $sample×100\text{percentage purity} = \dfrac{\text{mass of pure substance}}{\text{total mass of sample}} \times 100$ .
The mass of pure substance is often found from a reaction, using the mass or volume of a product and the balanced equation.
Purity matters most in medicines and foods, because even a small amount of a harmful impurity can make a product unsafe.

Example

Purity of a limestone sample

\text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)}

A $g5.00\ \text{g}$ sample of limestone reacts with excess acid and gives $dm31.08\ \text{dm}^3$ of carbon dioxide at r.t.p.
The amount of carbon dioxide is $mol\frac{1.08}{24} = 0.0450\ \text{mol}$ , so the sample contained $mol0.0450\ \text{mol}$ of calcium carbonate.
The mass of pure calcium carbonate is $0.0450 \times 100 = 4.50\ \text{g}$ .

\text{percentage purity} = \dfrac{4.50}{5.00} \times 100 = 90.0\%

Common Mistake

Do not divide masses by proton numbers, because an empirical formula calculation always uses $A_r$ values.
Do not round a ratio such as $1 : 1.5$ to $1 : 2$ , because it must be doubled to give $2 : 3$ .
Do not put the theoretical yield on the top of the percentage yield fraction.
Do not put the mass of the impurity on the top of the percentage purity fraction.

Exam technique

Formula and percentage calculations

Set out empirical formula working in columns, with a row for mass, a row for moles and a row for the ratio, so every step is clear to the examiner.
Keep at least three significant figures in mole values until the final ratio, because early rounding can hide a ratio such as $1 : 1.33$ .
Check that any percentage yield or percentage purity you calculate is below $100%100\%$ .
When asked for a molecular formula, show the empirical formula mass and the multiplier as well as the final formula.

Self review

Why must the mass of each element be divided by its $A_r$ when finding an empirical formula?
What should you do when a mole ratio comes out as $1 : 1.5$ ?
How do you find a molecular formula from an empirical formula and an $M_r$ ?
What is the percentage by mass of nitrogen in $NH4NO3\text{NH}_4\text{NO}_3$ ?
Which yield goes on the bottom of the percentage yield fraction?

3.3.3 Empirical formulae, yield and purity calculations

Empirical formulae from masses

Empirical formula from masses

Empirical formula from percentages

Finding the formula of magnesium oxide

Molecular formulae from empirical formulae

Molecular formula of glucose

Molecular formula of a hydrocarbon

Percentage composition by mass

Percentage of nitrogen in two compounds

Percentage yield

Percentage yield of calcium oxide

Percentage purity

Purity of a limestone sample

Formula and percentage calculations

Review 3.3.3 Empirical formulae, yield and purity calculations by teaching Genie

3.3.3 Empirical formulae, yield and purity calculations

Empirical formulae from masses

Empirical formula from masses

Empirical formula from percentages

Finding the formula of magnesium oxide

Molecular formulae from empirical formulae

Molecular formula of glucose

Molecular formula of a hydrocarbon

Percentage composition by mass

Percentage of nitrogen in two compounds

Percentage yield

Percentage yield of calcium oxide

Percentage purity

Purity of a limestone sample

Formula and percentage calculations

Review 3.3.3 Empirical formulae, yield and purity calculations by teaching Genie

3.3.3 Empirical formulae, yield and purity calculations

Empirical formulae from masses

Empirical formula from masses

Empirical formula from percentages

Finding the formula of magnesium oxide

Molecular formulae from empirical formulae

Molecular formula of glucose

Molecular formula of a hydrocarbon

Percentage composition by mass

Percentage of nitrogen in two compounds

Percentage yield

Percentage yield of calcium oxide

Percentage purity

Purity of a limestone sample

Formula and percentage calculations

Review 3.3.3 Empirical formulae, yield and purity calculations by teaching Genie

3.3.3 Empirical formulae, yield and purity calculations Revision Guide

3.3.3 Empirical formulae, yield and purity calculations

Empirical formulae from masses

Empirical formula from masses

Empirical formula from percentages

Finding the formula of magnesium oxide

Molecular formulae from empirical formulae

Molecular formula of glucose

Molecular formula of a hydrocarbon

Percentage composition by mass

Percentage of nitrogen in two compounds

Percentage yield

Percentage yield of calcium oxide

Percentage purity

Purity of a limestone sample

Formula and percentage calculations

Review 3.3.3 Empirical formulae, yield and purity calculations by teaching Genie

3.3.3 Empirical formulae, yield and purity calculations Revision Guide