1. Entry Task: Feb 5 th - 6 th Block #1 Write the Question How many formula units are in 50.0 grams of Iron II oxide? You have 5 minutes!

2. Agenda: Discuss Particles Moles Grams Self Check on Concepts Discuss Self-Check Notes on Percent Composition HW: Percent Composition ws

3. I can… Calculate particles  moles  grams Explain what is meant by the percent composition of a compound.

5. Discuss Particles  Moles  Grams Avervision for answers

6. Clear off your Desk!! Calculators Periodic Tables Needed

8. Discuss Self Check Avervision for answers

10. Percent Composition is the percent of each element in a compound. Percent composition is a step in determining the ratio in elements in a compound/molecule.

11. Example: Magnesium oxide- MgO Its molar mass is 24.305 g for Mg + 15.999 g for O, = 40.304 g By just looking at this compound, It LOOKS like it 50/50 of magnesium oxide. But according to its molar mass, it is not 50/50. 24.305g (Magnesium mass) 40.304 grams (mass of MgO) Magnesium = 0.60 X 100 = 60% Oxygen 15.999 g (oxygen mass) 40.304 grams (mass of MgO) = 0.40 X 100 = 40%

12. Fe 2 O 3 Fe = 55.845 x 2 = 111.69 g O = 15.999x 3 = 47.997g 159.687 g  159.687 g 0.6994 X 100 = 69.9% 0.3005 X 100 = 30.1% Iron III oxide  159.687 g

13. Barium hydroxide Ba(OH) 2 O= 15.999 g X 2 = 31.998 g Ba= 137.33 g X 1 = 137.33 g H= 1.0079 g X 2 = 2.0158 g 171.3438 g  171.3438 g = 0.801 X 100 = 80.1% = 0.187 X 100 = 18.7% = 0.0117 X 100 = 1.2%

14. Lead (II) nitrate Pb(NO 3 ) 2 Pb= 207.2 g X 1 = 207.2 g N= 14.007 g X 2 = 28.014 g O= 15.999 g X 6 = 95.994 g  331.208 g = 0.626 x100 = 62.6% 331.208 g  331.208 g = 0.0845 x100 = 8.46%  331.208 g = 0.289 x100 = 28.9%

15. Ammonium sulfate (NH 4 ) 2 SO 4 N= 14.007 g X 2 = 28.014 g H= 1.0079 g X 8 = 8.06 g S= 30.974 g X 1 = 30.974 g 1 mole = 131.044 grams O= 15.999 g X 4 = 63.996 g  131.044 X 100 = 21.4 %  131.044 X 100 = 6.2%  131.044 X 100 = 23.6 %  131.044 X 100 = 48.8.4 % add up % = 100%

16. Hydronium nitride (H 3 O) 3 N H= 1.0079 g X 9 = 9.071 g N= 14.007 g X 1 = 14.007 g O= 15.999 g X 3 = 47.997 g 71.1 grams  71.1 grams x 100= = 12.7 %  71.1 grams x 100= = 67.6 %  71.1 grams x 100= = 19.7 %

17. Aluminum cyanide Al(CN) 3 Al= 26.98 g X 1 = 26.98 g N= 14.007 g X 3 = 42.021 g C= 12.0107 g X 3 = 36.032 g 105 grams  105 grams x 100= = 25.6 %  105 grams x 100= = 34.3 %  105 grams x 100= = 40.0 %