1. What is a ratio? A ratio is a way to compare two quantities by using division Rate is a type of ratio: miles/hr A proportion shows a relationship between two ratios. I need 5 lbs of flour to make 15 loaves of bread. How many lbs of flour do I need to make 25 loaves of bread? 5 lbs = x lbs 15 loaves 25 loaves 15x = 125 x = 8.3 lbs

2. Mole Ratios A conversion factor (ratio) between #mol of any 2 substances in a balanced equation Also called: Stoichiometric Ratio Having the correct mole ratios is a vital key to solving chemistry problems! 2 Al + 3 Cl 2  2 AlCl 3 The coefficient gives the number of mols of the compound. In the above equation, what is the mol ratio between Al and Cl? 2 Al 3 Cl 2 What is the mol ratio between Al and AlCl 3 2 Al 2 AlCl 3

3. What is the mol ratio between Cl 2 and AlCl 3 3 Cl 2 2 AlCl 3

4. Practice Question Determine the mole ratio necessary to convert moles of aluminum to moles of aluminum chloride 2 Al + 3 Cl 2  2 AlCl 3 2 mol Al \ 2 mol AlCl 3 1 mol Al \ 1 mol AlCl 3

5. Steps for Solving Mol to Mol Problems Write and Balance the chemical equation Underline the given and wanted in the chemical equation Set up a mol ratio between given and wanted Set up the rest of the proportion using given and wanted Solve the wanted using the proportion

6. Mole to Mole Problems One disadvantage of burning propane (C 3 H 8 ) is that carbon dioxide (CO 2 ) is one of the products. The released carbon dioxide increases the growing concentration of CO 2 in the atmosphere. How many moles of carbon dioxide are produced when 10.0 moles of propane are burned in excess oxygen in a gas grill? C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 1.0 mol C 3 H 8 = 10.0 mol C 3 H 8 3.0 mol CO 2 x mol CO 2 X = 3.0 * 10.0 X = 30.0 mol CO 2

7. For Today: Mol to Mass Problems Mol to mol conversions Mol to Mass conversions Mass to Mass conversions

8. Mol to Mass Conversions Write and Balance the chemical equation Set up a mol ratio between given and wanted Set up the rest of the proportion using given and wanted Solve the wanted using the proportion Convert wanted in mols to grams by multiplying by molar mass of wanted

9. Mole to Mass Problems 1. Determine the mass of sodium chloride (table salt) produced when 1.25 moles of chlorine gas (Cl 2 ) reacts vigorously with sodium. Cl 2 + 2Na  2NaCl 1.00 mol Cl 2 * 1.25 mol Cl 2 2.00 mol NaCl x mol NaCl X = 2.50 mol NaCl Molar mass of NaCl = 58.44 g/mol 2.50 mol NaCl * 58.44 g/mol NaCl = 146.1 g NaCl

10. Mole to Mass Problems Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride is extracted using the following reagents: TiO 2 (s) + C (s) + 2Cl 2 (g)  TiCl 4 (s) + CO 2 If you have 1.25 mol TiO 2, what mass of Cl 2 is needed for the reaction to occur? 1 mol TiO 2 * 1.25 mol TiO 2 2 mol Cl 2 x mol Cl 2 X = 2.50 mol Cl 2 Molar mass of Cl 2 = 71.00 g/mol 71.0 x 2.50 = 177.50 g of Cl 2

11. Mole to Mass Problems Sodium chloride is decomposed into the elements sodium and chlorine by means of electrical energy. How many grams of chlorine gas can be obtained from 2.50 mol NaCl ? Chlorine is a diatomic molecule. NaCl -> Na + Cl 2 2NaCl -> 2Na + Cl 2 2 mol NaCl * 2.50 mol NaCl 1 mol Cl 2 x mol Cl 2 X = 1.25 mol Cl 2 molar mass of Cl 2 = 71.00 g/mol 1.25 * 71.00 = 88.75 g of Cl 2

12. For today: Mass to mass conversions Mol to mol conversions Mol to Mass conversions Mass to Mass conversions

13. Mass to Mass Conversions Write and Balance the chemical equation Set up a mol ratio between given and wanted Convert the given to mols by dividing by molar mass Set up the rest of the proportion using given and wanted Solve the wanted using the proportion Convert wanted to grams by multiplying by molar mass of wanted

14. Challenge Problem Balance this equation and determine how many mols of Chlorine must react to produce 12.30 g of aluminum chloride: Al + Cl 2 -> AlCl 3 2Al + 3Cl 2 -> 2AlCl 3

15. Mass to Mass Problems One in a series of reactions that inflate air bags in automobiles is the decomposition of sodium azide (NaN 3 ). 2 NaN 3 (s)  2 Na (s) + 3 N 2 (g) Determine the mass of N 2 produced if 100.0 g NaN 3 is decomposed. 2 mol NaN 3 3 mol N 2 molar mass of NaN 3 = 65.00 g/mol 100.00 g NaN 3 = 1.54 mol NaN 3 65.00 2 mol NaN 3 * 1.54 mol NaN 3 X = 2.32 mol N 2 3 mol N 2 x mol N 2 molar mass of N 2 = 28.00 g/mol 2.32 * 28.00 = 64.96 g N 2

16. Mass to Mass Problems Sulfur dioxide (SO 2 ) reacts with oxygen (O 2 ) and water (H 2 O) in the air to form sulfuric acid (H 2 SO 4 ). Write the balanced chemical equation for this reaction. If 2.50 g SO 2 react with excess oxygen and water, how many grams of H 2 SO 4 are produced? SO 2 + O 2 + H 2 O -> H 2 SO 4 2SO 2 + O 2 + 2H 2 O -> 2H 2 SO 4 2 mol SO 2 2 mol H 2 SO 4 molar mass of SO 2 = 64.00 g/mol 2.50 g SO 2 = 0.039 mol SO 2 64.00 2 mol SO 2 * 0.039 mol SO 2 x = 0.039 mol H 2 SO 4 2 mol H 2 SO 4 x mol H 2 SO 4 molar mass of H 2 SO 4 = 98.00 g/mol 0.039 * 98.00 = 3.80 g H 2 SO 4

17. Mass to Mass Problems Titanium is a transition metal used in many alloys because it is extremely strong and lightweight. Titanium tetrachloride is extracted using the following reagents: TiO 2 (s) + C (s) + Cl 2 (g)  TiCl 4 (s) + CO 2 If you have 159.7 g of TiO 2, what mass of Cl 2 is needed for the reaction to occur? TiO 2 (s) + C (s) + 2Cl 2 (g)  TiCl 4 (s) + CO 2 1 mol TiO 2 2 mol Cl 2 Molar mass of TiO 2 = 79.90 g /mol 159.7 g TiO 2 = 2.00 mol TiO 2 79.90 1 mol TiO 2 * 2.00 mol TiO 2 x = 4.00 mol Cl 2 2 mol Cl 2 x mol Cl 2 Molar mass of Cl 2 = 71.00 g/mol 4.00 * 71.00 = 284.00 g Cl 2

18. Limiting Reactants: Think food…. Assuming unlimited bread. A PB&J sandwich consists of 1 tbsp PB and 1 tbsp Jelly. If I have a jar containing 20 tbsp PB and a jar containing 10 tbsp of jelly How many PB&J sandwiches can I make? 10 If I have a jar containing 40 tbsp PB and same amount of jelly, how many PB&J sandwiches can I make? 10

19. Limiting and Excess Reactants Limiting Reactant (Limiting reagent) A reactant that determines how much product is formed (the one that gets completely used up) Excess Reactant (Excess reagent) The portion of a reactant that remains after the chemical rxn stops (what is left over)

20. Steps to Solving LR problems Write and balance equation; ID given and wanted Calculate the molar mass of the reactants and the product Determine the mols of a reactant by dividing by molar mass of the reactant. Set up a mol ratio between the reactant and the product Cross multiply and divide to get mols of product Repeat above steps for each reactant. The reactant that gives you the smaller mass of product is the limiting reactant. Multiply by the molar mass of product to get grams of product

21. Steps to Solving Excess Probs Subtract the two moles of product values you obtained in the LR problem Make a mole ratio between the product and the excess reactant Convert to grams

22. Example The reaction between solid white phosphorus and oxygen produces solid tetraphosphorus decoxide. (P 4 O 10 ). This compound is often called diphosphorus pentoxide because its empirical formula is P 2 O 5. Determine the mass of tetraphophorus decoxide formed if 25.0 g of phosphorus (P 4 ) and 50.0 g of oxygen (O 2 ) are combined. P 4 + 5O 2  P 4 O 10

23. Example P 4 + 5O 2  P 4 O 10 25.0g P 4 x 1 mole P 4 x 1 mole P 4 O 10 = 0.2016 moles P 4 O 10 124g P 4 1 mole P 4 50.0g O 2 x 1 mole O 2 x 1 mole P 4 O 10 = 0.3125 moles P 4 O 10 32.0g O 2 5 moles O 2 Phosphorus is limiting and Oxygen is in excess! 0.2016 moles P 4 O 10 x 284g P 4 O 10 = 57.2 g P 4 O 10 1 mole P 4 O 10

24. Example How much of the excess reactant remains after the reaction stops? 0.3125 - 0.2016 = 0.1109 moles P 4 O 10 0.1109 moles P 4 O 10 x 5 moles O 2 x 32.0g O 2 = 17.7g O 2 1 mole P 4 O 10 1 mole O 2

25. Percent Yield Actual Yield Amount of product actually obtained in an experiment Theoretical Yield Amount of product that should be obtained from a balanced chemical equation Percent Yield (actual/theoretical) x 100 The closer you are to 100%, the better!

26. Percent Yield Treat these problems like a mass to mass conversion problem! Use the balanced chemical equation to get the theoretical yield. Divide the actual yield (always given in problem) by the theoretical yield and multiply by 100.

27. When K 2 CrO 4 is added to a soln with.500g AgNO 3, solid Ag 2 CrO 4 is formed. If.455g of Ag 2 CrO 4 is obtained, what is the % yield? Balanced equation K 2 CrO 4 + 2 AgNO 3  Ag 2 CrO 4 + 2KNO 3 Molar Mass of AgNO 3 = 169.9 g/mol.500 = 0.003 mol of AgNO 3 169.9 2 AgNO 3 * 0.003 AgNO 3 x =.0015 mol Ag 2 CrO 4 1 Ag 2 CrO 4 x Ag 2 CrO 4 Molar Mass of Ag 2 CrO 4 = 331.7 g/mol.0015 * 331.7 = 0.498 g of Ag 2 CrO 4 % Yield (.455/.498) x 100 = 91.4%