1. Stoichiometry © D Scott; CHS

2. Stoichiometry Consider the chemical equation: 4NH 3 + 5O 2  6H 2 O + 4NO There are several numbers involved. What do they all mean? “stochio” = Greek for element “metry” = measurement Stoichiometry is about measuring the amounts of elements and compounds involved in a reaction. © D Scott; CHS

3. Stoichiometry Recall that Chemical formulas represent numbers of atoms 4NH 3 + 5O 2  6H 2 O + 4NO NH 3 1 nitrogen and 3 hydrogen atoms O2O2 2 oxygen atoms H2OH2O 2 hydrogen atoms and 1 oxygen atom NO 1 nitrogen atom and 1 oxygen atom © D Scott; CHS

4. Stoichiometry Recall that Chemical formulas are balanced with coefficients 4NH 3 + 5O 2  6H 2 O + 4NO 4 X NH 3 = 4 nitrogen + 12 hydrogen 5 X O 2 = 10 oxygen 6 X H 2 O= 12 hydrogen + 6 oxygen 4 X NO= 4 nitrogen + 4 oxygen © D Scott; CHS

5. Stoichiometry With Stoichiometry we find out that 4 : 5 : 6 : 4 do more than just multiply atoms. 4NH 3 + 5O 2  6H 2 O + 4NO © D Scott; CHS

6. Stoichiometry 4NH 3 + 5O 2  6H 2 O + 4NO 4 : 5 : 6 : 4 Can mean either: 4 molecules of NH 3 react with 5 molecules of O 2 to produce 6 molecules of H 2 O and 4 molecules of NO OR 4 moles of NH 3 react with 5 moles of O 2 to produce 6 moles of H 2 O and 4 moles of NO © D Scott; CHS 4 : 5 : 6 : 4 is what we call a mole ratio.

7. 4NH 3 + 5O 2  6H 2 O + 4NO How many moles of H 2 O are produced if 2.00 moles of O 2 are used? Stoichiometry Question (1) 2.00 mol O 2 2.40 mol H 2 O = Notice that a correctly balanced equation is essential to get the right answer 6 mol H 2 O 5 mol O 2 © D Scott; CHS

8. 4 mol NO 6 mol H 2 O 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many moles of NO are produced in the reaction if 15 mol of H 2 O are also produced? Stoichiometry Question (2) 15 mol H 2 O 10. mol NO = © D Scott; CHS

10. 4 NH 3 + 5 O 2  6 H 2 O + 4 NO MOLE RATIO (coefficients)

11. Stoichiometry Recall that Chemical formulas have molar masses: 4NH 3 + 5O 2  6H 2 O + 4NO NH 3 17 g/mol O2O2 32 g/mol H2OH2O18 g/mol NO30 g/mol © D Scott; CHS

12. 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of H 2 O are produced if 2.2 mol of NH 3 are combined with excess oxygen? Stoichiometry Question (3) © D Scott; CHS MOLE RATIO (coefficients) (H 2 O) (NH 3 )

13. 18.02 g H 2 O 1 mol H 2 O 6 mol H 2 O 4 mol NH 3 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of H 2 O are produced if 2.2 mol of NH 3 are combined with excess oxygen? Stoichiometry Question (3) 2.2 mol NH 3 59 g H 2 O = © D Scott; CHS

14. 5 mol O 2 6 mol H 2 O 32 g O 2 1 mol O 2 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of O 2 are required to produce 0.3 mol of H 2 O? Stoichiometry Question (4) 0.3 mol H 2 O 8 g O 2 = © D Scott; CHS

15. 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of NO are produced if 12 g of O 2 is combined with excess ammonia? Stoichiometry Question (5) © D Scott; CHS MOLE RATIO (coefficients) (NO) (O 2 )

16. 4 NH 3 + 5 O 2  6 H 2 O + 4 NO How many grams of NO are produced if 12 g of O 2 is combined with excess ammonia? 4 mol NO 5 mol O 2 x Stoichiometry Question (5) 12 g O 2 9.0 g NO = 30.01 g NO 1 mol NO x 1 mol O 2 32 g O 2 x © D Scott; CHS

17. Moving along the stoichiometry path We always use the same type of information to make the jumps between steps: grams (x)  moles (x)  moles (y)  grams (y) Molar mass of x Molar mass of y Mole ratio from balanced equation © D Scott; CHS

18. Many stoichiometry problems follow a pattern: grams(x)  moles(x)  moles(y)  grams(y) Converting grams to grams We can start anywhere along this path depending on the question we want to answer Notice that we cannot directly convert from grams of one compound to grams of another. Instead we have to go through moles. © D Scott; CHS

19. Have we learned it yet? Try these on your own - 4 NH 3 + 5 O 2  6 H 2 O + 4 NO a) How many moles of H 2 O can be made using 1.6 mol NH 3 ? b) what mass of NH 3 is needed to make 0.75 mol NO? c) how many grams of NO can be made from 47 g of NH 3 ? © D Scott; CHS

20. 4 NH 3 + 5 O 2  6 H 2 O + 4 NO a) b) c) Answers 6 mol H 2 O 4 mol NH 3 x 1.6 mol NH 3 2.4 mol H 2 O = 4 mol NH 3 4 mol NO x 0.75 mol NO 13 g NH 3 = 17.04 g NH 3 1 mol NH 3 x 4 mol NO 4 mol NH 3 x 47 g NH 3 83 g NO = 30.01 g NO 1 mol NO x 1 mol NH 3 17.04 g NH 3 x © D Scott; CHS