3. Conservation of Mass (again) I think it must be important? I think it must be important? Lavoisier- observed that the total mass of a substance present after a chemical reaction is the same as the total mass before the reaction. Lavoisier- observed that the total mass of a substance present after a chemical reaction is the same as the total mass before the reaction. We can conclude that atoms are not created or destroyed during a chemical reaction, they are simply rearranged. We can conclude that atoms are not created or destroyed during a chemical reaction, they are simply rearranged.

4. Chemical Equations Representations of chemical reactions. Representations of chemical reactions. Reactants to the left of the arrow Reactants to the left of the arrow Products to the right Products to the right H 2 + O 2  H 2 O H 2 + O 2  H 2 O Where did the other oxygen go??? Where did the other oxygen go???

5. We must balance!! That is the purpose of stoichiometry That is the purpose of stoichiometry Because atoms are neither created nor destroyed, a chemical equation must have equal numbers of atoms on both sides of the arrow. Because atoms are neither created nor destroyed, a chemical equation must have equal numbers of atoms on both sides of the arrow. Balance by changing coefficients Balance by changing coefficients You never change subscripts to balance. Coefficients change the amount of a substance, changing subscripts would change the identity. You never change subscripts to balance. Coefficients change the amount of a substance, changing subscripts would change the identity.

6. Balancing equations CH 4 + O 2  CO 2 + H 2 O ReactantsProducts C11 O23 H42

7. Balancing equations CH 4 + O 2  CO 2 + 2 H 2 O ReactantsProducts C11 O23 H424

8. Balancing equations CH 4 + O 2  CO 2 + 2 H 2 O ReactantsProducts C11 O23 H424 4

9. Balancing equations CH 4 + 2O 2  CO 2 + 2 H 2 O ReactantsProducts C11 O23 H424 4 4

10. Abbreviations (s)  (s)  (g)  (g)  (aq) (aq) heat heat   catalyst catalyst

11. Patterns of Reactivity We know that group 1 metals react with water to form metal hydroxides and hydrogen gas! We know that group 1 metals react with water to form metal hydroxides and hydrogen gas! Sooo…… Sooo…… 2 Na + 2 HOH  2 NaOH + H 2 2 Na + 2 HOH  2 NaOH + H 2 and.. and.. 2 K + 2HOH  2 KOH + H 2 2 K + 2HOH  2 KOH + H 2 Etc…… Etc……

12. Combustion in Air Rapid reactions that produce a flame. Rapid reactions that produce a flame. Most combustion involves oxygen from the air. Most combustion involves oxygen from the air. Usually involve burning hydrocarbons. Usually involve burning hydrocarbons. The products of burning a hydrocarbon are always water and carbon dioxide. The products of burning a hydrocarbon are always water and carbon dioxide. C 3 H 8 + 5O 2  3CO 2 + 4H 2 O C 3 H 8 + 5O 2  3CO 2 + 4H 2 O

13. Write the balanced equation: When methanol is burned in air: When methanol is burned in air: CH 3 OH + O 2  H 2 O + CO 2 CH 3 OH + O 2  H 2 O + CO 2

14. Combination & Decomposition What does the word combine mean? What does the word combine mean? What does the word decompose mean? What does the word decompose mean? If you know this, then you know what happens in a combination or decomposition reaction! If you know this, then you know what happens in a combination or decomposition reaction! In a combination reaction, two or more substances react to form one product. In a combination reaction, two or more substances react to form one product. Ex. Ca + O 2  CaO Ex. Ca + O 2  CaO

15. More on combination When reactants are nonmetal- metal, result is an ionic solid. Notice that the equation needs to be balanced. When reactants are nonmetal- metal, result is an ionic solid. Notice that the equation needs to be balanced. 2Ca + O 2  2CaO 2Ca + O 2  2CaO

16. Decomposition One substance breaks down to produce two or more different substances. One substance breaks down to produce two or more different substances. Ex. Metal Carbonates break down when heated to form their (basic) metal oxide and carbon dioxide. Ex. Metal Carbonates break down when heated to form their (basic) metal oxide and carbon dioxide. Try to write and balance the equation for the decomposition of lithium carbonate. Try to write and balance the equation for the decomposition of lithium carbonate.

17. Very Good! Li 2 CO 3  Li 2 O + CO 2 Li 2 CO 3  Li 2 O + CO 2 This works out to be already balanced! This works out to be already balanced!

18. Practice Ca(OH) 2 + H 3 PO 4  H 2 O + Ca 3 (PO 4 ) 2 Ca(OH) 2 + H 3 PO 4  H 2 O + Ca 3 (PO 4 ) 2 Cr + S 8  Cr 2 S 3 Cr + S 8  Cr 2 S 3 Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. Fe 2 O 3 (s) + Al(s)  Fe(s) + Al 2 O 3 (s) Fe 2 O 3 (s) + Al(s)  Fe(s) + Al 2 O 3 (s)

19. Meaning A balanced equation can be used to describe a reaction in molecules and atoms. A balanced equation can be used to describe a reaction in molecules and atoms. Not grams. Not grams. Chemical reactions happen molecules at a time Chemical reactions happen molecules at a time or dozens of molecules at a time or dozens of molecules at a time or moles of molecules. or moles of molecules.

20. Atomic Mass Atoms are so small, it is difficult to discuss how much they weigh in grams. Atoms are so small, it is difficult to discuss how much they weigh in grams. Use atomic mass units. Use atomic mass units. an atomic mass unit (amu) is one twelth the mass of a carbon-12 atom. an atomic mass unit (amu) is one twelth the mass of a carbon-12 atom. This gives us a basis for comparison. This gives us a basis for comparison. The decimal numbers on the table are atomic masses in amu. The decimal numbers on the table are atomic masses in amu.

21. They are not whole numbers Because they are based on averages of atoms and of isotopes. Because they are based on averages of atoms and of isotopes. can figure out the average atomic mass from the mass of the isotopes and their relative abundance. can figure out the average atomic mass from the mass of the isotopes and their relative abundance. add up the percent as decimals times the masses of the isotopes. add up the percent as decimals times the masses of the isotopes.

22. Examples There are two isotopes of carbon 12 C with a mass of 12.00000 amu(98.892%), and 13 C with a mass of 13.00335 amu (1.108%). There are two isotopes of carbon 12 C with a mass of 12.00000 amu(98.892%), and 13 C with a mass of 13.00335 amu (1.108%).

23. The Mole The mole is a number. The mole is a number. A very large number, but still, just a number. A very large number, but still, just a number. 6.022 x 10 23 of anything is a mole 6.022 x 10 23 of anything is a mole A large dozen. A large dozen. The number of atoms in exactly 12 grams of carbon-12. The number of atoms in exactly 12 grams of carbon-12.

24. The Mole Makes the numbers on the table the mass of the average atom. Makes the numbers on the table the mass of the average atom.

25. More Stoichiometry

26. Molar mass Mass of 1 mole of a substance. Mass of 1 mole of a substance. Often called molecular weight. Often called molecular weight. To determine the molar mass of an element, look on the table. To determine the molar mass of an element, look on the table. To determine the molar mass of a compound, add up the molar masses of the elements that make it up. To determine the molar mass of a compound, add up the molar masses of the elements that make it up.

27. Find the molar mass of CH 4 CH 4 Mg 3 P 2 Mg 3 P 2 Ca(NO 3 ) 2 Ca(NO 3 ) 2 Al 2 (Cr 2 O 7 ) 3 Al 2 (Cr 2 O 7 ) 3 CaSO 4 · 2H 2 O CaSO 4 · 2H 2 O

28. Percent Composition Percent of each element a compound is composed of. Percent of each element a compound is composed of. Find the mass of each element, divide by the total mass, multiply by a 100. Find the mass of each element, divide by the total mass, multiply by a 100. Easiest if you use a mole of the compound. Easiest if you use a mole of the compound. Find the percent composition of CH 4 Find the percent composition of CH 4 Al 2 (Cr 2 O 7 ) 3 Al 2 (Cr 2 O 7 ) 3 CaSO 4 · 2H 2 O CaSO 4 · 2H 2 O

29. Working backwards From percent composition, you can determine the empirical formula. From percent composition, you can determine the empirical formula. Empirical Formula the lowest ratio of atoms in a molecule. Empirical Formula the lowest ratio of atoms in a molecule. Based on mole ratios. Based on mole ratios. A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O what is its empirical formula. A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O what is its empirical formula.

30. Pure O 2 in CO 2 is absorbed H 2 O is absorbed Sample is burned completely to form CO 2 and H 2 O

31. More Stoichiometry

32. Empirical To Molecular Formulas Empirical is lowest ratio. Empirical is lowest ratio. Molecular is actual molecule. Molecular is actual molecule. Need Molar mass. Need Molar mass. Ratio of empirical to molar mass will tell you the molecular formula. Ratio of empirical to molar mass will tell you the molecular formula. Must be a whole number because... Must be a whole number because...

33. Example A compound is made of only sulfur and oxygen. It is 69.6% S by mass. Its molar mass is 192 g/mol. What is its formula? A compound is made of only sulfur and oxygen. It is 69.6% S by mass. Its molar mass is 192 g/mol. What is its formula?

34. Stoichiometry Given an amount of either starting material or product, determining the other quantities. Given an amount of either starting material or product, determining the other quantities. use conversion factors from use conversion factors from molar mass (g - mole) molar mass (g - mole) balanced equation (mole - mole) balanced equation (mole - mole) keep track. keep track.

35. Examples One way of producing O 2 (g) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. How many moles of O 2 (g) are produced? One way of producing O 2 (g) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. How many moles of O 2 (g) are produced? How many grams of potassium chloride? How many grams of potassium chloride? How many grams of oxygen? How many grams of oxygen?

36. Examples A piece of aluminum foil 5.11 in x 3.23 in x 0.0381 in is dissolved in excess HCl(aq). How many grams of H 2 (g) are produced? A piece of aluminum foil 5.11 in x 3.23 in x 0.0381 in is dissolved in excess HCl(aq). How many grams of H 2 (g) are produced? How many grams of each reactant are needed to produce 15 grams of iron form the following reaction? Fe 2 O 3 (s) + Al(s)  Fe(s) + Al 2 O 3 (s) How many grams of each reactant are needed to produce 15 grams of iron form the following reaction? Fe 2 O 3 (s) + Al(s)  Fe(s) + Al 2 O 3 (s)

37. Examples K 2 PtCl 4 (aq) + NH 3 (aq)  Pt(NH 3 ) 2 Cl 2 (s)+ KCl(aq) K 2 PtCl 4 (aq) + NH 3 (aq)  Pt(NH 3 ) 2 Cl 2 (s)+ KCl(aq) what mass of Pt(NH 3 ) 2 Cl 2 can be produced from 65 g of K 2 PtCl 4 ? what mass of Pt(NH 3 ) 2 Cl 2 can be produced from 65 g of K 2 PtCl 4 ? How much KCl will be produced? How much KCl will be produced? How much from 65 grams of NH 3 ? How much from 65 grams of NH 3 ?

38. Yield How much you get from an chemical reaction

39. Limiting Reagent Reactant that determines the amount of product formed. Reactant that determines the amount of product formed. The one you run out of first. The one you run out of first. Makes the least product. Makes the least product. Book shows you a ratio method. Book shows you a ratio method. It works. It works. So does mine So does mine

40. Limiting reagent To determine the limiting reagent requires that you do two stoichiometry problems. To determine the limiting reagent requires that you do two stoichiometry problems. Figure out how much product each reactant makes. Figure out how much product each reactant makes. The one that makes the least is the limiting reagent. The one that makes the least is the limiting reagent.

41. Example Ammonia is produced by the following reaction N 2 + H 2  NH 3 Ammonia is produced by the following reaction N 2 + H 2  NH 3 What mass of ammonia can be produced from a mixture of 100. g N 2 and 500. g H 2 ? What mass of ammonia can be produced from a mixture of 100. g N 2 and 500. g H 2 ? How much unreacted material remains? How much unreacted material remains? First thing you must do to solve the problem is balance the equation! First thing you must do to solve the problem is balance the equation!

42. Excess Reagent The reactant you don’t run out of. The reactant you don’t run out of. The amount of stuff you make is the yield. The amount of stuff you make is the yield. The theoretical yield is the amount you would make if everything went perfect. The theoretical yield is the amount you would make if everything went perfect. The actual yield is what you make in the lab. The actual yield is what you make in the lab.

43. Percent Yield % yield = Actual x 100% Theoretical % yield = Actual x 100% Theoretical % yield = what you got x 100% what you could have got % yield = what you got x 100% what you could have got

44. Examples Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the three types of yield. Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the three types of yield.

45. Examples Years of experience have proven that the percent yield for the following reaction is 74.3% Hg + Br 2  HgBr 2 If 10.0 g of Hg and 9.00 g of Br 2 are reacted, how much HgBr 2 will be produced? Years of experience have proven that the percent yield for the following reaction is 74.3% Hg + Br 2  HgBr 2 If 10.0 g of Hg and 9.00 g of Br 2 are reacted, how much HgBr 2 will be produced? If the reaction did go to completion, how much excess reagent would be left? If the reaction did go to completion, how much excess reagent would be left?