1. Using the MOLE

2. Percentage composition Percentage composition is the mass of individual elements in a compound expressed as a percentage of the mass of the compound. Eg The percentage of aluminium in alumina (Al 2 O 3 ) is 52.9%

3. How do we calculate Percentage Composition? We can use the molar mass of elements that make up a compound to determine it’s percentage composition.

4. How do we calculate Percentage Composition? Step 1. Find the molar mass of the compound e.g. M (Al 2 O 3 ) = (2 x 27) + (3 x 16) g mol -1 = 102 g mol -1

5. How do we calculate Percentage Composition? Step 2. Find the percentage of element in the compound e.g. % of Al in Al 2 O 3 = mass of Al in 1 mol of Al 2 O 3 mass of 1 mole of Al 2 O 3 = (2x27) x 100 102 = 52.9% There are 2 aluminium atoms in every alumina molecule therefore mass of Al is 2 x Ar(Al)

6. General Formula % by Mass = mass of element in 1 mole of compound x 100% mass of 1 mole of compound

7. Have a go! What percentage of C 7 H 14 O 2 is hydrogen? Step 1: Find molar mass of C 7 H 14 O 2 M(C 7 H 14 O 2 ) = (7x12) + (14x1) + (2x16) = 130g mol -1 Step 2: Find percentage of H in C 7 H 14 O 2 % H in C 7 H 14 O 2 = (14x1) x 100 130 = 10.8%

8. Empirical Formulas Empirical formula is the formula that gives the simplest whole number ratio (by number of moles) of each element in the compound. e.g. Water = H 2 O Therefore the whole number ratio of elements in a water compound is: H:O = 2:1

9. Empirical Formulas Empirical formulas are determined experimentally by finding the mass of each element present in a given mass of compound.

10. Measure the mass (m) of each element in the compound Calculate the amount in mole (n) of each element in the compound Calculate the simplest whole number ratio of moles of each element in a compound Empirical formula of compound

11. Example A compound of carbon and oxygen is found to contain 27.3% carbon and 72.7% oxygen by mass. Calculate the empirical formula of the compound. Step 1: Record the mass in grams of each element. If the masses are given as a percentage, assume the sample weighs 100g therefore the percentage becomes the mass in grams Mass of C: 27.3 g Mass of O: 72.7 g

12. Example A compound of carbon and oxygen is found to contain 27.3% carbon and 72.7% oxygen by mass. Calculate the empirical formula of the compound. Step 2: Calculate the amount in moles (n) of each element n(C): 27.3 = 2.27 mol 12 n(O): 72.7 = 4.54 mol 16

13. Example A compound of carbon and oxygen is found to contain 27.3% carbon and 72.7% oxygen by mass. Calculate the empirical formula of the compound. Step 3: Obtain the simplest whole number ratio: Carbon: 2.27 = 1 2.27 Oxygen: 72.7 = 2 2.27 Carbon to Oxygen ratio 1:2 Therefore Empirical Formula is CO 2

14. Have a go! 9.0g of a compound of only carbon, hydrogen and oxygen is found to contain 4.8g of oxygen and 3.6g of carbon. Calculate the empirical formula of the compound.

15. Molecular Formulas Molecular formulas give the actual number of atoms in one molecule of the compound. Therefore the molecular formula can be the same or different to the empirical formula. The molecular formula is always a whole number multiple of the empirical formula.

16. Example A compound has the empirical formula CH. The molar mass of this compound is 78g mol -1 Step 1: Find the molar mass of CH M (CH) = 12 + 1 = 13 Step 2: The number of CH units in a molecule = molar mass of compound molar mass of CH = 78g mol 13g mol = 6 Step 3: The molecular formula of the compound is therefore 6 x CH = C 6 H 6

17. Have a go! A sample was found to contain 7.2g of carbon and 1.5g of hydrogen. The molar mass of this compound was determined to be 58g mol -1 What is the molecular formula of the compound?