1. Stoichiometry Mass Calculations

2. Yesterday: Mole to Mole Ratios – relating reactants and products in a chemical equation If we have a balanced chemical equation, we can calculate the number of moles of a substance using the known ratio of reactants and products when given the number of moles of one of the reactants or products. Previous classes: Relationship between mass (g), molar mass (grams/mol) and mols of a substance  # of mols = mass / MM Molar Mass is the atomic mass of an element or formula weight of a compound (in grams/mol) 1 mole = 6.022 X 10 23 particles/atoms/molecules

3. Today: Mass to Mass Calculations If we know the reaction (from the balanced chemical equation) and the amount of one of the substances (mass, # of moles, or particles) in the reaction, we can calculate the amount of the other substances in the reaction.

4. Mass to Mass Calculations: The general strategy: Step 1: Write a balanced chemical equation Step 2: If you are given the mass or number of particles of a substance, convert it to the number of moles. Step 3: Calculate the number of moles of the required substance based on the number of moles of the given substance, using the appropriate mole ratio Step 4: Convert the number of moles of the required substances to mass or number of particles.

5. Mass (g) of given substance 2.) Convert the given mass of a substance to moles of the substance  Use the molar mass of the given substance Moles Mass (g) of the required substance 3.) Calculate the number of mols of the required substance  use mol to mol ratio from the balanced equation 4.) Convert the number of moles of the required substances to a mass or number of particles  Use the molar mass of the required substance 1.) Write out the balanced chemical equation

6. Mass to Mass Calculations for Products and Reactants Iron can be produced from iron ore, Fe 2 O 3 by reacting the ore with carbon monoxide (CO). Carbon dioxide is also produced. What mass of iron can be formed from 425g of iron ore? Step 1: Write out the balanced chemical equation Fe 2 O 3 + 3CO  2Fe + 3CO 2 m= 425gm= ? Example 1:

7. Step 2: Fill in chart with information you know Balanced Equation Fe 2 O 3 3CO 2Fe3CO 2 Mole Ratio 1323 Mass (m) 425g ? g Molar Mass (M) 159.7 g/mol Moles (n)

8. Step 3: Convert given mass into moles (n=m/M) Moles of Fe 2 O 3 = 425g 159.7g/mol = 2.66 mol of Fe 2 O 3

9. Fill in chart with information Balanced Equation Fe 2 O 3 3CO 2Fe3CO 2 Mole Ratio 1323 Mass (m) 425g ? g Molar Mass (M) 159.7 g/mol Moles (n)2.66mol

10. Mass to Mass Calculations for Products and Reactants Step 4: Calculate the number of mols of the required substance (mol to mole ratio) Fe 2 O 3 + 3CO  2Fe + 3CO 2 x mols of Fe = 2 mol of Fe 2.66 mols of Fe 2 O 3 1mol of Fe 2 O 3 2.66 mols of Fe 2 O 3 2 mol of Fe 1 mols Fe 2 O 3 or X = 5.32 mols of Fe

11. Fill in chart with information Balanced Equation Fe 2 O 3 3CO 2Fe3CO 2 Mole Ratio 1323 Mass (m) 425g ? g Molar Mass (M) 159.7 g/mol 55.85 g/mol Moles (n)2.66mol5.32 mol

12. Step 5: Convert moles of required substance to the mass mass of Fe = 5.32 mols of Fe X 55.85g/mol of Fe = 297g of Fe

13. Mass to Mass Calculations for Reactants What mass of hydrazine (N 2 H 4 ) is required to react completely with 1000g of dinitrogen tetraoxide (N 2 O 4 )? Step 1: Write out the balanced chemical equation 2N 2 H 4 + N 2 O 4  3N 2 + 4H 2 O m= ?m= 1000g Example 2:

14. Step 2: Fill in chart with information you know Balanced Equation 2N 2 H 4 N2O4N2O4 3N23N2 4H 2 O Mole Ratio 2134 Mass (m) ?g 1000g Molar Mass (M) 92 g/mol Moles (n)

15. Step 3: Convert given mass into moles (n=m/M) moles of N 2 O 4 = 1000g 92g/mol = 10.87 mol of N 2 O 4

16. Fill in chart with information you know Balanced Equation 2N 2 H 4 N2O4N2O4 3N23N2 4H 2 O Mole Ratio 2134 Mass (m) ?g 1000g Molar Mass (M) 92 g/mol Moles (n)10.87mol

17. Mass to Mass Calculations for Reactants Step 4: Calculate the number of mols of the required substance (mol to mole ratio) 2N 2 H 4 + N 2 O 4  3N 2 + 4H 2 O x mols of N 2 H 4 = 2 mols of N 2 H 4 10.87 mols of N 2 O 4 1 mol of N 2 O 4 = 21.74 mols of N 2 H 4 OR 10.87 mols of N 2 O 4 2 mol of N 2 H 4 1 mols N 2 O 4 X = 21.74 mols of N 2 H 4

18. Fill in chart with information you know Balanced Equation 2N 2 H 4 N2O4N2O4 3N23N2 4H 2 O Mole Ratio 2134 Mass (m) ?g 1000g Molar Mass (M) 32 g/mol92 g/mol Moles (n)21.74 mol10.87mol

19. Step 5: Convert moles of required substance to the mass mass of N 2 H 4 = 21.74 mols of N 2 H 4 X 32g/mol = 695.7 of N 2 H 4

20. Mass to Mass Calculations for particles/molecules Ammonia gas react with oxygen gas to produce water and nitrogen monoxide. How many molecules of oxygen are required to completely react with 23.0g of ammonia? Step 1: Write out the balanced chemical equation 4NH 3 + 5O 2  6H 2 O + 4NO m= 23.0gm= ? Step 2: Convert given mass into moles Mols of NH 3 = 23.0g of NH 3 17g/mol of NH 3 = 1.35 mols of NH 3 Example 3:

21. Mass to Mass Calculations for particles/molecules Step 3: Calculate the number of mols of the required substance (mol to mole ratio) Step 4: Convert moles of required substance to molecules x mols of O 2 = 5 mols of O 2 1.35 mols of NH 3 4 mols of NH 3 4NH 3 + 5O 2  6H 2 O + 4NO = 1.68 mols of O 2 # of molecules of O 2 = 1.68 mols of O 2 X 6.022 X 10 23 molecules of O 2 1 mol = 1.02 X 10 24 molecules

22. Practice Questions p. 244 #11,12, 14 p. 246 #15, 16, 18 p. 248 # 19, 20 p. 249 #1, 4, 6, 7