1. Chemistry 103 Lecture 14

2. Outline I. Empirical/Molecular Formulas II. Chemical Reactions - basic symbols - balancing - classification III. Stoichiometry

3. The molecular formula Is the true or actual number of the atoms in a molecule The empirical formula Is the simplest whole number ratio of the atoms (this is the formula for ionic compounds) H 2 O 2 HO molecular formula empirical formula Types of Formulas

4. Empirical & Molecular Formulas Ionic Compounds - Only need Empirical formula. Molecules - Need information on both to determine exact make-up of your system Molecular Formula Empirical Formula C 6 H 6 CH

5. A molecular formula Is a multiple (or equal) of its empirical formula Has a molar mass that is the empirical mass multiplied by a whole number molar mass = a whole number empirical mass Is obtained by multiplying the empirical formula by a whole number Relating Molecular and Empirical Formulas

6. Determine the molecular formula of compound that has a molar mass of 78.11 g/mole and an empirical formula of CH. Finding the Molecular Formula

7. A compound is 24.27% C, 4.07% H, and 71.65% Cl. The molar mass is known to be 99.0 g. What are the empirical and molecular formulas? Molecular Formula

8. Chemical Reactions

9. Physical Change In a physical change, The identity and composition of the substance do not change The state can change or the material can be torn into smaller pieces Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

10. Chemical Change In a chemical change, Reacting substances form new substances with different compositions and properties A chemical reaction takes place Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

11. Chemical Reaction In a chemical reaction, Old bonds are broken and new bonds are formed Atoms in the reactants are rearranged to form one or more different substances Fe and O 2 form rust (Fe 2 O 3 )

12. Chemical Reaction In a chemical reaction, A chemical change produces one or more new substances There is a change in the composition of one or more substances Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

13. Evidence of a Chemical Reaction Changes that can be seen are evidence of a chemical reaction.

14. Writing a Chemical Reaction Chemists use a shorthand approach when writing the specifics of a chemical reaction. This approach is called the chemical equation. Reactants -----> Products

15. Chemical Equations A chemical equation, Gives the chemical formulas of the reactants on the left of an arrow and the products on the right Reactants Product C(s) O 2 (g) CO 2 (g)

16. Symbols Used in Equations Symbols used in chemical equations show: The states of the reactants The states of the products The reaction conditions

17. Chemical Equations Are Balanced In a balanced chemical reaction, Atoms are not gained or lost

18. Chemical Equations Are Balanced In a balanced chemical reaction, The number of reactant atoms are equal to the number of product atoms

19. Chemical Equations Chemical equations: symbolic descriptions of chemical reactions. Two parts to an equation: reactants and products H 2 + O 2  H 2 O A Chemical Equation must also be “balanced”. 2H 2 + O 2 --> 2H 2 O

20. Balanced Chemical Equations Chemical Equations must be balanced  There must be equal numbers of atoms of each element on both sides of the equation (both sides of the arrow) 1. Write the correct symbols and formulas for all of the reactants and products. 2. Count the number of each type of atom on BOTH sides of the equation. 3. Insert coefficients until there are the equal numbers of each kind of atom on both sides of the equation.

21. Example - making ammonia Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state.

22. Example - making ammonia Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state. H 2 (g) + N 2 (g) NH 3 (g)

23. Example - making ammonia Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state. H 2 (g) + N 2 (g) NH 3 (g)

24. Example - making ammonia Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state. H 2 (g) + N 2 (g) NH 3 (g)

25. Example - making ammonia Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state. H 2 (g) + N 2 (g) 2 NH 3 (g)

26. Example - making ammonia Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state. 3 H 2 (g) + N 2 (g) 2 NH 3 (g)

27. Example - making ammonia Hydrogen gas and Nitrogen gas combine to make ammonia in the gaseous state. BEFORE AFTER 3 H 2 (g) + N 2 (g) 2 NH 3 (g)

28. Balancing Equations Methane reacts with oxygen (combustion reaction) to form carbon dioxide and water. Write a properly balanced chemical equation

29. The Numbers in Chemical Equations

30. More Practice: Balancing Reactions C 2 H 6 + O 2  CO 2 + H 2 O C 3 H 6 + O 2  CO 2 + H 2 O NH 3 + O 2  NO + H 2 O

31. Stoichiometry Chemical Stoichiometry: using mass and quantity relationships among reactants and products in a chemical reaction to make predictions about how much product will be made.

32. 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Four molecules NH 3 react with five molecules O 2 to produce four molecules NO and six molecules H 2 O Quantities in a Chemical Reaction

33. 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Four molecules NH 3 react with five molecules O 2 to produce four molecules NO and six molecules H 2 O and Four mol NH 3 react with five mol O 2 to produce four mol NO and six mol H 2 O Quantities in a Chemical Reaction

34.  We can read the equation in “moles” by placing the word “mole” or “mol” between each coefficient and formula. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) 4 mol Fe + 3 mol O 2 2 mol Fe 2 O 3 Moles in Equations

35. Conservation of Mass The Law of Conservation of Mass indicates: No change in total mass occurs in a reaction Mass of products is equal to mass of reactants

36. In an ordinary chemical reaction, Matter cannot be created or destroyed The number of atoms of each element are equal The mass of reactants equals the mass of products H 2 (g) + Cl 2 (g) 2HCl(g) 2 mol H, 2 mol Cl = 2 mol H, 2 mol Cl 2(1.008) + 2(35.45) = 2(36.46) 72.92 g = 72.92 g Law of Conservation of Mass

37. Conservation of Mass 2 mol Ag + 1 mol S = 1 mol Ag 2 S 2 (107.9 g) + 1(32.07 g) = 1 (247.9 g) 247.9 g reactants = 247.9 g product

38. A mole-mole factor is a ratio of the moles for two substances in an equation. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Fe and O 2 4 mol Fe and 3 mol O 2 3 mol O 2 4 mol Fe Fe and Fe 2 O 3 4 mol Fe and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 4 mol Fe O 2 and Fe 2 O 3 3 mol O 2 and 2 mol Fe 2 O 3 2 mol Fe 2 O 3 3 mol O 2 Writing Mole-Mole Factors

39. How many moles of Fe 2 O 3 can form from 6.0 mol O 2 ? 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) Relationship:3 mol O 2 = 2 mol Fe 2 O 3 Write a mole-mole factor to determine the moles of Fe 2 O 3. 6.0 mol O 2 x 2 mol Fe 2 O 3 = 4.0 mol Fe 2 O 3 3 mol O 2 Calculations with Mole Factors

40. How many moles of Fe are needed to react with 12.0 mol O 2 ? 4Fe(s) + 3O 2 (g) 2 Fe 2 O 3 (s) A) 3.00 mol Fe B) 9.00 mol Fe C) 16.0 mol Fe Mole relations