Welcome to science intervention! Today we are revising how to calculate empirical formula DO NOW: Work out the relative masses of these elements and compounds. The first one has been done for you 1) 2) 3) 4) 5) 6) 7) 8) H2O ( 1 + 1 + 16= 18) Mg CaCO3 H2SO4 CaO 0 SO2 02 Today you will need your calculators Today we are learning how to calculate empirical formula Answers 1) H2O ( 1 + 1 + 16= 18) 2)Mg = 24 3)CaCO3= 100 4)H2SO4= 98 5)CaO= 56 6)0= 16 7)SO2 = 64 8)02 = 32 Today we are learning 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. Today we are learning 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 C Step 2: Underneath each element, write down the masses/percentages in the question 27.3 72.7 27.3 12 = 2.3 72.7 16 =4.5 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… 2.3 = 1 2.3 So the formula is CO2! O 4.5= 2 2.3 Don’t forget to write the formula! Today we are learning how to calculate empirical formula Past exam question 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) Hints: 1. Write the elements! 2. Write the masses/percentages under each element 3. Divide the number’s by the atomic masses of those elements 4. Divide by the smaller of the two numbers Write the formula! Today we are learning 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 Answer: Pb2Cl2CO Hints: 1. Write the elements! 2. Write the masses/percentages under each element 3. Divide the number’s by the atomic masses of those elements 4. Divide by the smaller of the two numbers Write the formula! Today we are learning 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. Answer: C2H6O! Hints: 1. Write the elements! 2. Write the masses/percentages under each element 3. Divide the number’s by the atomic masses of those elements 4. Divide by the smaller of the two numbers Write the formula! Today we are learning how to calculate empirical formula Past exam question- Read the hint! Calculate the simplest (empirical) formula of this substance: Hint: If the 70 % of iron (Fe) and 30 % of oxygen (O) ratio is .5 (eg 2.5), don’t round up! Double both Relative atomic masses: O = 16; Fe = 56. numbers Eg 2.5: 3 Answer: Fe2O3! becomes 5: 6 Today we are learning how to calculate empirical formula Lets look at a trickier one… Calculate the simplest (empirical) formula of this substance: 70 % of iron (Fe) and 30 % of oxygen (O) Relative atomic masses: O = 16; Fe = 56. Answer: Fe2O3!