1. C H E M I S T R Y Chapter 3 Mass Relationships in Chemical Reactions

2. Chemical Symbols on Different Levels 2H 2 O(l)2H 2 (g) + O 2 (g) 0.56 kg of hydrogen gas react with 4.44 kg of oxygen gas to yield 5.00 kg of liquid water. macroscopic: 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to yield 2 molecules of liquid water. microscopic: How can we relate the two to each other?

3. Avogadro’s Number and the Mole 2(12.0 amu) + 4(1.0 amu) = 28.0 amuC2H4:C2H4: HCl: 1.0 amu + 35.5 amu = 36.5 amu Molecular Mass: Sum of atomic masses of all atoms in a molecule. Formula Mass: Sum of atomic masses of all atoms in a formula unit of any compound, molecular or ionic.

4. Molar Mass of K 3 PO 4 Calculate the molar mass of K 3 PO 4. 4 ElementNumber of Moles Atomic MassTotal Mass in K 3 PO 4 K P O

5. Stoichiometry: Chemical Arithmetic Stoichiometry: The relative proportions in which elements form compounds or in which substances react. aA + bB cC + dD Moles of A Grams of A Moles of B Grams of B Mole Ratio Between A and B (Coefficients) Molar Mass of BMolar Mass of A

6. Stoichiometry: Chemical Arithmetic Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/6 How many grams of NaOH are needed to react with 25.0 g Cl 2 ? 2NaOH(aq) + Cl 2 (g) NaOCl(aq) + NaCl(aq) + H 2 O(l) Aqueous solutions of sodium hypochlorite (NaOCl), best known as household bleach, are prepared by reaction of sodium hydroxide with chlorine gas: Moles of Cl 2 Grams of Cl 2 Moles of NaOH Grams of NaOH Mole Ratio Molar Mass

7. Stoichiometry: Chemical Arithmetic The comercial production of iron from iron ore involves the reaction of Fe 2 O 3 with CO to yield iron metal plus carbon dioxide: Fe 2 O 3 (s) + 3CO(g)  2Fe(s) + 3CO 2 (g) Predict how many grams of CO will react with 0.500 moles Fe 2 O 3 ? Copyright © 2008 Pearson Prentice Hall, Inc.Chapter 3/7

8. Yields of Chemical Reactions The amount actually formed in a reaction. The amount predicted by calculations. Actual Yield: Theoretical Yield: actual yield theoretical yield x 100% Percent Yield =

9. Reactions with Limiting Amounts of Reactants Limiting Reactant: The reactant that is present in limiting amount. The extent to which a chemical reaction takes place depends on the limiting reactant. Excess Reactant: Any of the other reactants still present after determination of the limiting reactant.

10. Reactions with Limiting Amounts of Reactants C 2 H 4 O(aq) + H 2 O(l) C2H6O2(l)C2H6O2(l) If 3 moles of ethylene oxide react with 5 moles of water, which reactant is limiting and which reactant is present in excess? At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers:

11. Reactions with Limiting Amounts of Reactants At a high temperature, ethylene oxide reacts with water to form ethylene glycol which is an automobile antifreeze and a starting material in the preparation of polyester polymers: C 2 H 4 O(aq) + H 2 O(l) C2H6O2(l)C2H6O2(l)

12. Example The reaction between aluminum and iron (III) oxide can generate temperatures approaching 3000oC and is used in welding metals: 2 Al(s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(s) In one process, 124 g Al are reacted with 601 g of Fe 2 O 3. a.Which reactant is limited? b.What is the theoretical yield of Al 2 O 3 (s)?

13. Reactions with Limiting Amounts of Reactants Li 2 O(s) + H 2 O(g) 2LiOH(s) Lithium oxide is used aboard the space shuttle to remove water from the air supply according to the equation: If 80.0 g of water are to be removed and 65.0 g of Li 2 O are available, which reactant is limiting? How many grams of excess reactant remain? How many grams of LiOH are produced?

14. Percent Composition and Empirical Formulas Percent Composition: Expressed by identifying the elements present and giving the mass percent of each. Empirical Formula: It tells only the ratios of the atoms in a compound. Molecular Formula: It tells the actual numbers of atoms in a compound. It can be either the empirical formula or a multiple of it. multiple = molecular mass empirical formula mass

15. Steps in determine the Empirical formula Step 1: Obtain the mass of each element (in grams) Step 2: Determine the numbers of atom present Step 3: Divide the smallest moles by numbers of each atoms to obtain the closet integer as possible. Step 4: If the result ended with 0.5, 0.33, 1.125, 1.50 etc… then multiply with a factor to get the nearest integer as possible.

16. Example An unknown sample gives the following mass percent: 17.5% Na, 39.7% Cr and 42.8% O. What is the empirical formula? Moles Mass percents Mole ratios Subscripts Relative mole ratios Molar masses

17. Example A compound of nitrogen and oxygen is analyzed, and a sample weighing 1.587 g is found 0.483 g N and 1.104 g O. What is the empirical formula of the compound?

18. Examples Determine the empirical formula of the compound made when 8.65 g of iron combines with 3.72 g of oxygen..

19. Percent Composition and Empirical Formulas A colorless liquid has a composition of 84.1 % carbon and 15.9 % hydrogen by mass. Determine the empirical formula. Also, assuming the molar mass of this compound is 114.2 g/mol, determine the molecular formula of this compound.

20. Determining Empirical Formulas: Elemental Analysis Combustion Analysis: A compound of unknown composition (containing a combination of carbon, hydrogen, and possibly oxygen) is burned with oxygen to produce the volatile combustion products CO 2 and H 2 O, which are separated and weighed by an automated instrument called a gas chromatograph. hydrocarbon + O 2 (g) xCO 2 (g) + yH 2 O(g) carbon hydrogen

21. Example The molecular formula of trioxane, which contains carbon, hydrogen, and oxygen, can be determined using the data from two different experiments. In the first experiment, 17.471 g of trioxane is burned in the apparatus shown above, and 10.477 g H 2 O and 25.612 g CO 2 are formed. In the second experiment, the molecular mass of trioxane is found to be 90.079amu.

22. Example A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of compound yields 16.01 mg CO 2 and 4.37 mg H 2 O. The molar mass of the compound is 176.1 g/mol.