2. Stoichiometry and the mole Chapter 8

3. What is stoichiometry?  Quantitative aspects of chemistry  Stoicheon Greek root (element)  Metron Greek root( to measure)  Calculate how much of a reactant needed to produce a product or how much product could be expected

4. Stoichiometry  How much do I want?  How much do I have?  How much will I get?

5. The Mole Defined as the number of carbon atoms in exactly 12 grams of carbon- 12. 1 mole is 6.02 x 10 23 particles. Treat it like a very large dozen 6.02 x 10 23 is called Avogadro's number.

6. The Mole  One mole = 6.022 x 10 23 (Avogadro’s number)

7. Avogadro’s hypothesis When all of the following are the same:  Volumes of containers  Temperature of the gases  The pressure exerted by and on each gas Then, the number of molecules in each container will be the same, too. However, the masses are not equal.

8. N 2(g) + 3H 2(g)  2NH 3(g) 132 1 dozen 1 dozen(12) 3 dozen (36) 2 dozen (24) 1 gross (144) 3 gross (432) 2 gross (288) 1mole (6.02x10 23 ) 3 moles (18.06x10 23 ) 2 moles (12.04 x10 23 )

9. Representative particles The smallest pieces of a substance. For an element it is an atom. –Unless it is diatomic For a molecular compound it is a molecule. For an ionic compound it is a formula unit.

10. Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO 2 = 44.01 grams per mole H 2 O = 18.02 grams per mole Ca(OH) 2 = 74.10 grams per mole

11. Chemical Equations Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances. C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O reactantsproducts 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water When the equation is balanced it has quantitative significance:

12. Mole Relations

13. Calculating Masses of Reactants and Products 1.Balance the equation. 2.Convert mass to moles. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired substance. 5.Convert moles to grams, if necessary.

14. Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed. 1. Identify reactants and products and write the balanced equation. Al+O2O2 Al 2 O 3 b. What are the reactants? a. Every reaction needs a yield sign! c. What are the products? d. What are the balanced coefficients? 43 2

15. Working a Stoichiometry Problem 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O 2  2Al 2 O 3 = 6.50 g Al ? g Al 2 O 3 1 mol Al 26.98 g Al 4 mol Al 2 mol Al 2 O 3 1 mol Al 2 O 3 101.96 g Al 2 O 3 6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 =12.3 g Al 2 O 3

17. Percent yield  The % yield is calculated from: (actual yield / theoretical yield) x 100 (actual yield / theoretical yield) x 100  The theoretical yield is how much product is predicted from balanced chemical equation.  The actual yield is how much is recovered when actual experiment is conducted.

18. Limiting Reactant The limiting reactant is the reactant The limiting reactant is the reactant that is consumed first, limiting the amounts of products formed. that is consumed first, limiting the amounts of products formed.

19. How to solve stoichiometry problems?  Write the balanced chemical equation  Convert to moles information on reactants and products  Use mole ratios from equation to determine number of moles of unknown  Convert moles to unit desired  g x  moles x  moles y  g y

20. Limiting reactant  Reactant present in short supply Excess reactant Reactant in excess relative to limiting reactant

21. Formulas  molecular formula = (empirical formula) n [n = integer]  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

22. Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

23. Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 O C 12 H 22 O 11

24. Empirical Formula Determination 1.Base calculation on 100 grams of compound. 2.Determine moles of each element in 100 grams of compound. 3.Divide each value of moles by the smallest of the values. 4.Multiply each number by an integer to obtain all whole numbers.

25. Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

26. Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

27. Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x 2 352 Empirical formula: C3H5O2C3H5O2

28. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

29. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 2. Divide the molecular mass by the mass given by the emipirical formula.

30. Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

31. Combustion Analysis Technique that requires burning of an unknown substance and trap the gases from burning.  Used to determine molecular formula of an unknown  Combustion= burning  Using oxygen

32. Points about combustion  Element that makes unknown almost always contain carbon and hydrogen. Oxygen is often involved and nitrogen is involved sometimes.  Must know mass of the unknown substance before burning it  Unknown will be burnt in pure oxygen, present in excess  Carbon dioxide and water are the products  All the carbon winds up as carbon dioxide and all the hydrogen winds up as water  The end result will be to determine the empirical formula of the substance