Empirical Formulae

Empirical and Molecular Formulae

What is an empirical formula?

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

What does an empirical formula tell you?

The simplest ratio in which the atoms of each element are present in a compound.

Does an empirical formula always show the actual number of atoms in a molecule?

No. It only shows the simplest whole-number ratio.

What is the difference between a molecular formula and an empirical formula?

A molecular formula shows the actual number of each type of atom in a molecule, whereas an empirical formula shows their simplest whole-number ratio.

What is the empirical formula of H₂O?

H₂O, because the 2:1 ratio cannot be simplified.

What is the empirical formula of glucose, C₆H₁₂O₆?

CH₂O.

What are the main steps for calculating an empirical formula from masses?

1. Divide the mass of each element by its relative atomic mass (Aᵣ). 2. Divide all answers by the smallest value. 3. Convert the resulting ratio to the simplest whole numbers. 4. Use these numbers in the empirical formula.

What are the main steps for calculating an empirical formula from percentage composition?

Divide each percentage by the element's relative atomic mass, divide all answers by the smallest value, then convert to the simplest whole-number ratio.

Why do you divide the mass of each element by its relative atomic mass?

To calculate the relative number of moles of each element.

After dividing each mass by Aᵣ, what should you do next?

Divide all the answers by the smallest answer.

Why do you divide all the values by the smallest value?

To obtain the simplest ratio of the elements.

What should you do if the resulting ratio contains simple fractions?

Multiply all values by a suitable number to obtain whole numbers.

Why might calculated ratios not be exact whole numbers?

Because of experimental error.

How many significant figures should generally be kept during empirical formula calculations?

At least two significant figures to avoid inappropriate rounding.

How should a ratio of 1 : 1.5 be converted to whole numbers?

Multiply both values by 2 to give 2 : 3.

How should a ratio of 1 : 1.33 be converted to whole numbers?

Multiply both values by 3 to give approximately 3 : 4.

How should a ratio of 1 : 1.25 be converted to whole numbers?

Multiply both values by 4 to give 4 : 5.

How should a ratio of 1 : 1.2 be converted to whole numbers?

Multiply both values by 5 to give 5 : 6.

Why should you not automatically round a value such as 1.9 to 2 too early?

Premature rounding can produce the wrong empirical formula.

In the copper oxide experiment, how can copper oxide be converted into copper?

Heat copper oxide in a stream of hydrogen gas or natural gas.

What happens to the oxygen in copper oxide during the experiment?

It reacts with hydrogen to form water or steam.

What colour change occurs as copper oxide is reduced to copper?

The solid gradually changes to orange-brown.

Why is excess gas burned off at the end of the tube?

For safety reasons.

Why is the copper heated again after cooling and weighing?

To check whether its mass changes and confirm that the reaction is complete.

What does heating to constant mass indicate?

The reaction is complete when further heating causes no change in mass.

If the mass of copper oxide is 4.28 g and the mass of copper is 3.43 g, what is the mass of oxygen?

4.28 − 3.43 = 0.85 g.

For 3.43 g Cu and 0.85 g O, what is the amount ratio before simplification?

Cu: 3.43 ÷ 63.5 = 0.0540; O: 0.85 ÷ 16.0 = 0.0531.

What is the empirical formula obtained from 3.43 g Cu and 0.85 g O?

CuO, because the ratio is approximately 1 : 1.

How do you calculate an empirical formula when percentage composition is given?

Treat each percentage as a mass, divide each by its Aᵣ, then find the simplest whole-number ratio.

A compound contains 38.4% C, 4.8% H and 56.8% Cl. What values are obtained after dividing by Aᵣ?

C = 38.4 ÷ 12.0 = 3.2; H = 4.8 ÷ 1.0 = 4.8; Cl = 56.8 ÷ 35.5 = 1.6.

What is the simplest ratio for 3.2 : 4.8 : 1.6?

2 : 3 : 1.

What is the empirical formula of a compound containing 38.4% C, 4.8% H and 56.8% Cl?

C₂H₃Cl.

How do you calculate the percentage of an element when its percentage is not provided?

Subtract the total percentage of the other elements from 100%.

A compound contains 29.1% Na and 40.5% S, with the remainder oxygen. What percentage is oxygen?

100 − (29.1 + 40.5) = 30.4%.

For 29.1% Na, 40.5% S and 30.4% O, what values are obtained after dividing by Aᵣ?

Na = 29.1 ÷ 23.0 = 1.27; S = 40.5 ÷ 32.1 = 1.26; O = 30.4 ÷ 16.0 = 1.90.

What ratio is obtained after dividing 1.27 : 1.26 : 1.90 by the smallest value?

Approximately 1 : 1 : 1.5.

How is the ratio 1 : 1 : 1.5 converted to whole numbers?

Multiply everything by 2 to give 2 : 2 : 3.

What is the empirical formula for a compound containing 29.1% Na, 40.5% S and 30.4% O?

Na₂S₂O₃.

What is combustion analysis used for?

To determine the empirical formula of an organic compound by analysing the products formed when it is completely burned.

Which elements are commonly present in organic compounds analysed by combustion?

Carbon and hydrogen, or carbon, hydrogen and oxygen.

What products are measured when an organic compound containing C, H and O is completely burned?

Carbon dioxide and water.

Where does the carbon in the carbon dioxide produced during combustion come from?

The carbon in the original organic compound.

Where does the hydrogen in the water produced during combustion come from?

The hydrogen in the original organic compound.

How can the mass of carbon be calculated from the mass of CO₂?

Mass of C = mass of CO₂ × (12.0 ÷ 44.0).

How can the mass of hydrogen be calculated from the mass of H₂O?

Mass of H = mass of H₂O × (2.0 ÷ 18.0).

Why is 12/44 used to find the mass of carbon in CO₂?

Carbon contributes 12.0 of the total relative molecular mass of 44.0.

Why is 2/18 used to find the mass of hydrogen in H₂O?

The two hydrogen atoms contribute 2.0 of the total relative molecular mass of 18.0.

A 1.87 g organic compound produces 2.65 g CO₂. What mass of carbon was in the compound?

2.65 × (12.0 ÷ 44.0) = 0.723 g.

A 1.87 g organic compound produces 1.63 g H₂O. What mass of hydrogen was in the compound?

1.63 × (2.0 ÷ 18.0) = 0.181 g.

If a 1.87 g compound contains 0.723 g C and 0.181 g H, what mass of oxygen does it contain?

1.87 − 0.723 − 0.181 = 0.966 g.

How do you find the mass of oxygen in a combustion-analysis question?

Subtract the masses of carbon and hydrogen from the original mass of the compound.

For masses C = 0.723 g, H = 0.181 g and O = 0.966 g, what are the mole values?

C = 0.723 ÷ 12.0 = 0.0603; H = 0.181 ÷ 1.0 = 0.181; O = 0.966 ÷ 16.0 = 0.0604.

What is the simplest ratio of 0.0603 : 0.181 : 0.0604?

Approximately 1 : 3 : 1.

What is the empirical formula from the combustion-analysis example?

CH₃O.

When calculating an empirical formula, should you divide by atomic number or relative atomic mass?

Relative atomic mass.

For oxygen in empirical formula calculations, what value should you divide by?

16.0.

Why should you not divide oxygen by 8 or 32 in an empirical formula calculation?

8 is its atomic number and 32 is the relative molecular mass of O₂; empirical formula calculations require the relative atomic mass, 16.0.