1. Stoichiometry: Chapter 9 Lesson 1

2. Law of Conservation of Mass “We may lay it down as an incontestable axiom that, in all the operations of art and nature, nothing is created; an equal amount of matter exists both before and after the experiment. Upon this principle, the whole art of performing chemical experiments depends.” --Antoine Lavoisier, 1789

3. What is Stoichiometry? Stoichiometry is the quantitative study of reactants and products in a chemical reaction.

4. Chemical Equations Concise representations of chemical reactions

5. Anatomy of a Chemical Equation CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

6. Anatomy of a Chemical Equation Reactants appear on the left side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

7. Anatomy of a Chemical Equation Products appear on the right side of the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

8. Anatomy of a Chemical Equation The states of the reactants and products are written in parentheses to the right of each compound. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

9. Anatomy of a Chemical Equation Coefficients are inserted to balance the equation. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (g)

10. Moles

11. Avogadro’s Number 6.02 x 10 23 1 mole of 12 C has a mass of 12 g

12. Molar Mass By definition, these are the mass of 1 mol of a substance (i.e., g/mol) –The molar mass of an element is the mass number for the element that we find on the periodic table –The formula weight (in amu’s) will be the same number as the molar mass (in g/mol)

13. Using Moles Moles provide a bridge from the molecular scale to the real-world scale

14. Mole Relationships One mole of atoms, ions, or molecules contains Avogadro’s number of those particles One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound

15. Finding Empirical Formulas

16. Calculating Empirical Formulas One can calculate the empirical formula from the percent composition

17. Calculating Empirical Formulas The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of carbon (61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen (23.33%). Find the empirical formula of PABA.

18. Calculating Empirical Formulas Assuming 100.00 g of para-aminobenzoic acid, C:61.31 g x = 5.105 mol C H: 5.14 g x= 5.09 mol H N:10.21 g x= 0.7288 mol N O:23.33 g x = 1.456 mol O 1 mol 12.01 g 1 mol 14.01 g 1 mol 1.01 g 1 mol 16.00 g

19. Calculating Empirical Formulas Calculate the mole ratio by dividing by the smallest number of moles: C:= 7.005  7 H:= 6.984  7 N:= 1.000 O:= 2.001  2 5.105 mol 0.7288 mol 5.09 mol 0.7288 mol 1.458 mol 0.7288 mol

20. Calculating Empirical Formulas These are the subscripts for the empirical formula: C 7 H 7 NO 2

21. Combustion Analysis Compounds containing C, H and O are routinely analyzed through combustion in a chamber like this –C is determined from the mass of CO 2 produced –H is determined from the mass of H 2 O produced –O is determined by difference after the C and H have been determined

22. Stoichiometric Calculations From the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant)

23. Stoichiometric Calculations Starting with 1.00 g of C 6 H 12 O 6 … we calculate the moles of C 6 H 12 O 6 … use the coefficients to find the moles of H 2 O… and then turn the moles of water to grams C 6 H 12 O 6 + 6 O 2  6 CO 2 + 6 H 2 O

24. 1 mol of a gas=22.4 L at STP C. Molar Volume at STP S tandard T emperature & P ressure 0°C and 1 atm

25. C. Molar Volume at STP Molar Mass (g/mol) 6.02  10 23 particles/mol MASS IN GRAMS MOLES NUMBER OF PARTICLES LITERS OF SOLUTION Molar Volume (22.4 L/mol) LITERS OF GAS AT STP Molarity (mol/L)

26. What You Should Expect Given : Amount of reactants Question: how much of products can be formed. Example 2 A + 2B 3C Given 20.0 grams of A and sufficient B, how many grams of C can be produced?

27. What do you need? You will need to use i.molar ratios, ii.molar masses, iii.balancing and interpreting equations, and iv. conversions between grams and moles. Note: This type of problem is often called "mass-mass."

28. Steps Involved in Solving Mass-Mass Stoichiometry Problems Balance the chemical equation correctly Using the molar mass of the given substance, convert the mass given to moles. Construct a molar proportion (two molar ratios set equal to each other) Using the molar mass of the unknown substance, convert the moles just calculated to mass.

29. Mole Ratios A mole ratio converts moles of one compound in a balanced chemical equation into moles of another compound.

30. Example Reaction between magnesium and oxygen to form magnesium oxide. ( fireworks) 2 Mg(s) + O2(g) 2 MgO(s) Mole Ratios: 2 : 1: 2

31. D. Stoichiometry Problems How many moles of KClO 3 must decompose in order to produce 9 moles of oxygen gas? 9 mol O 2 2 mol KClO 3 3 mol O 2 = 6 mol KClO 3 2KClO 3  2KCl + 3O 2 ? mol9 mol

32. How many grams of KClO 3 are req’d to produce 9.00 L of O 2 at STP? 9.00 L O 2 1 mol O 2 22.4 L O 2 = 32.8 g KClO 3 2 mol KClO 3 3 mol O 2 122.55 g KClO 3 1 mol KClO 3 ? g9.00 L D. Stoichiometry Problems 2KClO 3  2KCl + 3O 2

33. D. Stoichiometry Problems How many grams of silver will be formed from 12.0 g copper? 12.0 g Cu 1 mol Cu 63.55 g Cu = 40.7 g Ag Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 2 mol Ag 1 mol Cu 107.87 g Ag 1 mol Ag 12.0 g? g

34. 63.55 g Cu 1 mol Cu D. Stoichiometry Problems How many grams of Cu are required to react with 1.5 L of 0.10M AgNO 3 ? 1.5 L.10 mol AgNO 3 1 L = 4.8 g Cu Cu + 2AgNO 3  2Ag + Cu(NO 3 ) 2 1 mol Cu 2 mol AgNO 3 ? g 1.5L 0.10M