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page 100. mass. relative atomic or formula mass. 2 x (1+14+16+16+16) = 126g. 0.5 x (40+12+16+16+16) = 50g. 5.4 x (24+16+16+1+1)) = 113g. The number of protons and neutrons the atom contains. 11:33 PM. page 100. no. of moles = mass of chemical molar mass.

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  1. page 100 mass relative atomic or formula mass 2 x (1+14+16+16+16) = 126g 0.5 x (40+12+16+16+16) = 50g 5.4 x (24+16+16+1+1)) = 113g The number of protons and neutrons the atom contains. 11:33 PM

  2. page 100 no. of moles = mass of chemical molar mass 37 = 37 = 0.5 moles (40+32+2) 74 mass 112 = 112 = 1.75 moles (32+32) 64 200 = 200 = 2.5 moles (63+16) 79 11:33 PM

  3. no. of moles = mass of chemical molar mass mass of chemical = no. of moles x molar mass page 100 mass of chemical = 0.75 x (24+16) = 30g mass of chemical = 0.025 x (207+71) = 6.95g mass of chemical = 2 x (28+8+32+64) = 2 x 132 = 264g 11:33 PM

  4. 08/11/2014 • Empirical Formulae 1 • L.O. • I can calculate the empirical formula of a substance 11:33 PM

  5. What is the Empirical Formula? SEE WORKSHEET The Empirical formula is the simplest ratio of the atoms within the formula of a compound 1. What is the empirical formula of glucose, C6H12O6? 1. What is the empirical formula of methanoic acid, HCOOH? CH2O2 CH2O 1. What is the empirical formula of ethanoic acid, CH3COOH? CH2O

  6. What is the Empirical Formula? SEE WORKSHEET The Empirical formula is the simplest ratio of the atoms within the formula of a compound The empirical formula of the compound is:- CO2 Step 4: element present in the smallest amount is:- Step 1: mass of each element 36g 96g Step 2: Ar of each element Step 5: divide by the smallest:- 1. 132g of a compound contains 36g of carbon and 96g of oxygen. Calculate it’s empirical formula? Step 3: convert to moles of each element 3 = 1 6 = 2 3 3 12 16 36 = 3 96 = 6 12 16 CARBON

  7. What is the Empirical Formula? SEE WORKSHEET The Empirical formula is the simplest ratio of the atoms within the formula of a compound Step 4: element present in the smallest amount is:- CARBON Step 1: mass of each element in 100g of the compound 56.5g 8.7g 34.8g Step 5: divide by the smallest:- 1.45 = 2 0.725 = 1 2.175 = 3 0.725 0.725 0.725 The empirical formula of the compound is:- K2CO3 Step 2: Ar of each element 39g 12g 16g Step 3: convert to moles of each element 56.5 = 1.45 8.7 = 0.725 34.8 = 2.175 39 12 16 2. Calculate the empirical formula of a compound containing 56.5% of potassium, 8.7% of carbon and 34.8% of oxygen?

  8. Now try the examples on the question sheet

  9. Complete the homework sheet (C5-L2e)

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