2.00 ECS 1.00 IonsMass Finding the Empirical Formula of a Compound © Chemworks

2.00 ECS 1.00 IonsMass The Empirical Formula is just the simplest formula of a compound. For example the empirical formula for water is H 2 O It shows that hydrogen and oxygen are in the ratio of 2 : 1

2.00 ECS 1.00 IonsMass However not all chemicals have their formulae in such a simple ratio. For example the chemical formula for hydrogen peroxide is H 2 O 2 It shows that hydrogen and oxygen are in the ratio of 2 : 2 The ratio could be simplified to 1 : 1 So the empirical formula is HO, not H 2 O 2

2.00 ECS 1.00 IonsMass The same thing happens with dinitrogen tetroxide. The chemical formula for dinitrogen tetroxide is N 2 O 4 It shows that the nitrogen and oxygen are in the ratio of 2 : 4 The ratio could be simplified to 1 : 2 So the empirical formula is NO 2, not N 2 O 4

2.00 ECS 1.00 IonsMass Example 1: A compound contains 24g carbon and 8g hydrogen. What is its empirical formula? Let’s try another example … Elements C H Mass 24g 8g g.f.mass 12 1 moles ratio 2 8 simplest ratio 1 4 So the empirical formula is CH 4

2.00 ECS 1.00 IonsMass Example 2: A compound is 27.3% carbon and 72.7% oxygen. What is its empirical formula? We seem to have a problem as there are no masses, only percentages. Elements C O Mass But if we assume that we had 100g of the compound it would contain 27.3g of carbon and 72.7g of oxygen. 27.3g 72.7g

2.00 ECS 1.00 IonsMass Example 2: A compound is 27.3% carbon and 72.7% oxygen. What is its empirical formula? This does not look like a simple ratio, but let’s divide both numbers by the smallest number. Elements C O Mass 27.3g 72.7g g.f.mass moles 27.3g 72.7g ratio

2.00 ECS 1.00 IonsMass Example 2: A compound is 27.3% carbon and 72.7% oxygen. What is its empirical formula? Elements C O Mass 27.3g 72.7g g.f.mass moles 27.3g 72.7g ratio ratio 1 2 So the formula for the compound is CO 2

2.00 ECS 1.00 IonsMass Now try the following questions:

2.00 ECS 1.00 IonsMass Calculate the empirical formula for: a compound that is 40% sulphur and 60% oxygen Elements S O Mass 40g 60g g.f.mass moles ratio ratio 1 3 The empirical formula is SO 3

2.00 ECS 1.00 IonsMass Calculate the empirical formula for: a compound that is 80% copper and 20% oxygen Elements Cu O Mass 80g 20g g.f.mass moles ratio ratio 1 1 The empirical formula is CuO

2.00 ECS 1.00 IonsMass Calculate the empirical formula for a compound that is 2% hydrogen, 32.7% sulphur and 65.3% oxygen Elements H S O Mass 2g 32.7g 65.3% g.f.mass moles ratio ratio The empirical formula is H 2 SO 4

2.00 ECS 1.00 IonsMass Calculate the empirical formula for a compound that is 43.4% sodium, 11.3% carbon and 45.3% oxygen Elements Na C O Mass 43.4g 11.3g 45.3% g.f.mass moles ratio ratio The empirical formula is Na 2 CO 3 moles

2.00 ECS 1.00 IonsMass 16g of a copper oxide compound is reduced to 12.8g copper. What is the empirical formula of the copper oxide compound? Elements Cu O Mass 12.8g 3.2g g.f.mass moles ratio ratio 2 1 The empirical formula is Cu 2 O moles

2.00 ECS 1.00 IonsMass Well done!