Starter  Work out the Ar and Mr of the following: The first one has been done for you H2O ( 1 + 1 + 16= 18) Mg CaCO3 H2SO4 CaO O SO2 02 Today you will need your calculators

Today we are revising how to calculate empirical formula Answers H2O ( 1 + 1 + 16= 18) Mg = 24 CaCO3= 100 H2SO4= 98 CaO= 56 0= 16 SO2 = 64 02 = 32

Today we are revising how to calculate empirical formula The empirical formula is the simplest ratio of the different atoms in it. For example, ethane is has the formula C2H6. Therefore the empirical formula is CH3. You can work out the empirical formula of a compound if you know the mass of each element in it. This is the reverse of finding out the % mass of a compound from its formula.

Today we are revising how to calculate empirical formula Lets look at an example question…. What is the empirical formula of a compound that contains 27.3% carbon and 72.7% oxygen by mass? Step 1: Write down the symbols for the elements in the compound Step 2: Underneath each element, write down the masses/percentages in the question Step 3: Divide each mass by the relative formula mass of the element Step 4: Divide all the answers by the smallest answer to give the ratio… C O 27.3 72.7 27.3 72.7 12 = 2.3 16 =4.5 2.3 = 1 4.5= 2 2.3 2.3 Don’t forget to write the formula! So the formula is CO2!

Today we are revising how to calculate empirical formula Past exam question Hints: Write the elements! 2. Write the masses/percentages under each element 3. Divide the number’s by the atomic masses of those elements Divide by the smaller of the two numbers Write the formula! Compound X had the following percentage composition by mass: 10.8% magnesium, 31.8% chlorine and 57.4% oxygen. Relative atomic masses: Mg = 24; Cl = 35.5; O = 16 Calculate the empirical formula of compound X. Answer: MgCl2O8! (Put the elements in the order given in the question)

Today we are revising how to calculate empirical formula Past exam question A compound is called phosgenite. Analysis of this compound shows that it contains: 76.0% lead (Pb) 13.0% chlorine (Cl) 2.2% carbon (C) 8.8% oxygen (O) Calculate the empirical formula of this compound. Relative atomic masses: Pb = 207; Cl = 35.5; C = 12 ; O = 16 Hints: Write the elements! 2. Write the masses/percentages under each element 3. Divide the number’s by the atomic masses of those elements Divide by the smaller of the two numbers Write the formula! Answer: Pb2Cl2CO

Today we are revising how to calculate empirical formula Past exam question A sample of the solvent used in one perfume contained 0.60 g of carbon, 0.15 g of hydrogen and 0.40 g of oxygen. Relative atomic masses: H = l; C = 12; O = 16. Calculate the empirical (simplest) formula of the solvent. Hints: Write the elements! 2. Write the masses/percentages under each element 3. Divide the number’s by the atomic masses of those elements Divide by the smaller of the two numbers Write the formula! Answer: C2H6O!

Today we are revising how to calculate empirical formula Calculate the simplest (empirical) formula of this substance: 70 % of iron (Fe) and 30 % of oxygen (O) Relative atomic masses: O = 16; Fe = 56. Hint: If the ratio is .5 (eg 2.5), don’t round up! Double both numbers Eg 2.5: 3 becomes 5: 6 Answer: Fe2O3!