2. Chapter 7 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

3. Introduction In the margin of your notes record two things a chemical formula can tell us A chemical formula indicates: 1. elements present in a compound 2. relative number of atoms/ions of each element present in a compound

4. Introduction Chemical formulas also allow chemists to calculate a number of other characteristic values for a compound:  formula mass  molar mass  percentage composition

5. SO 2 1S32.07 amu 2O+ 2 x 16.00 amu SO 2 64.07 amu 1 molecule SO 2 = 64.07 amu Formula mass is the sum of the atomic masses of all the atoms in a compound, in atomic mass units (amu) Formula mass for SO 2 How do we find the atomic mass of an element? recall masses on P.T. are based on 12 C

6. NO!! Need to be able to relate amu to a macroscale (grams)

8. Discovery of the MOLE Amedeo Avogadro  Italian Scientist The mole: the amount of a substance that contains as many particles as there are atoms in exactly 12.00 grams of 12 C 1 mol = 6.02 x 10 23 atoms, formula units, molecules

9. How do we measure an amount? 1 mol = 6.02 x 10 23 particles 1 mole C atoms = 12.00 g C 1 mole C atoms = 6.022 x 10 23 atoms 1 C atom = 12.00 amu

10. Molar Mass the mass in grams of one mole (or approx. 6.02 × 10 23 particles) of a substance Relating Mass to Numbers of Particles calculated by adding the masses of the elements present in a mole (same as recorded values on P.T. compound’s molar mass is numerically equal to its formula mass

11. SO 2 1S32.07 grams/mol 2O+ 2 x 16.00 grams/mol SO 2 64.07 grams/mol 1 molecule SO 2 = 64.07 amu 1 mole SO 2 = 64.07 g SO 2 For any element atomic mass (amu) = molar mass (grams/mol)

12. It is often useful to know the percentage by mass of a particular element in a chemical compound. Mining for Bauxite: Al(OH) 3 Mining for Diamonds! Chemical Formulas Express Composition

13. Percentage Composition of Iron Oxides The mass percentage of an element in a compound is the same regardless of the sample’s size.

14. There are 2 ways to determine percent composition of a compound: From a chemical formula From experimental data

15. Percent Composition: The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100 percent.

16. Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n = the number of moles of the element in 1 mole of the compound

17. H: 2 x 1.01 = 2.02 O: 1 x 16.00 = 16.00 18.02 g/mol 2 x1.01 g/mol H 18.02 g/mol H 2 O X 100 = 11.21 % H 16.00 g/mol O 18.02 g/mol H 2 O X 100 = 88.79 % O Find the percent composition of the elements in water (H 2 O). H : O :

18. Calculate Total Mass: Al: 2 x 26.98 = 53.96 S: 3 x 32.07 = 96.21 O: 12 x 16.00 = 192.00 342.17 g/mol Find the percent composition of the elements in Al 2 (SO 4 ) 3.

19. Find the percent composition of the elements in Al 2 (SO 4 ) 3. - continued. 53.96 g/mol Al 342.17 g/mol Al 2 (SO 4 ) 3 X 100 = 15.77 % Al 192.00 g/mol O 342.17 g/mol Al 2 (SO 4 ) 3 X 100 = 56.11 % O 96.21 g/mol S 342.17 g/mol Al 2 (SO 4 ) 3 X 100 = 28.12 % S Al : S : O :

20. Find the percent composition of a compound that contains 0.9480 g of C, 0.1264 g of O, and 0.0158 g H. C = 0.9480 g O = 0.1264 g H = 0.0158 g 1.0902 g

21. 0.9480 g 1.0902 g C : X 100 = 86.96 % 0.1264 g 1.0902 g O : X 100 = 11.59 % 0.0158 g 1.0902 g H : X 100 = 1.45 %

22. Mole Conversions

23. 11.2 g NaCl x Mass to Mole Conversions 1 mol NaCl 58.44 g NaCl 0.192 mol NaCl Na: 1 X 22.99 = 22.99 Cl: 1 X 35.45 = 34.45 58.44 g/mol MOLAR MASS

24. 3.2 mol Zn(NO 3 ) 2 x Mole to Mass Conversions 189.53 g Zn(NO 3 ) 2 1 mol Zn(NO 3 ) 2 606 g Zn(NO 3 ) 2 Zn: 1 X 65.39 = 65.39 N: 2 X 14.07 = 28.14 O: 6 X 16.00 = 96.00 189.53g/mol MOLAR MASS

25. 8.74 x 10 23 atoms CaCO 3 x Particle to Mole Conversions 1 mol CaCO 3 6.02 x 10 23 atoms CaCO 3 1.45 mol CaCO 3 AVOGADRO’S NUMBER

26. 0.36 mol Al x Mole to Particle Conversions 6.02 x 10 23 atoms Al 1 mol Al 2.2 x 10 23 atoms Al AVOGADRO’S NUMBER

27. Mole Map PARTICLES MASS MOLE MOLAR MASS 6.02 x 10 23 What conversion factor do I use???

28. 250 g C 12 H 22 O 11 X Multistep Conversions 1 mol C 12 H 22 O 11 342.34 g C 12 H 22 O 11 C: 12 X 12.01 = 144.12 H: 22 X 1.01 = 22.22 O: 11 X 16.00 = 176.00 342.34 g/mol

29. 4.4 x 10 23 molecules C 12 H 22 O 11 6.02 x 10 23 molec. C 12 H 22 O 11 1 mol C 12 H 22 O 11 1 mol C 12 H 22 O 11 X 342.34 g C 12 H 22 O 11

30. Remember Empirical Formula: A chemical formula that gives the simplest whole-number ratio of the elements in the formula. - Subscripts are used for these ratios.

31. Determine the empirical formula of a compound found to have 13.5 g of Ca, 10.8 g of O, and 0.675 g of H. Example Problem 1.Assume 100 g 2.Convert to mole 3.Divide by the smallest 4.Multiply ‘til whole

32. 1 mol 40.08 g Ca : 1 mol 16.00 g O : 1 mol 1.01 g H : 13.5 g Ca x 0.675 g H x 10.8 g O x = 0.337 mol Ca = 0.668 mol H = 0.675 mol O

33. Writing the complete formula: a)Round to whole numbers b)Put parentheses around polyatomic ions c)Re-write the final formula. Ca(OH) 2

34. Remember Molecular Formula: A chemical formula that gives the actual number of the elements in the molecular compound. Example:C 2 H 4 NOT CH 2 C 6 H 12 O 6 NOT CH 2 O

35. Comparing Empirical and Molecular Formulas

36. There is a direct relationship between empirical and molecular formulas. There is a direct relationship between the empirical formula mass and the molecular formula mass. FIND THE COMMON MULTIPLE! The correct ratio can be found by: dividing the experimental formula mass by the empirical formula mass

37. The empirical formula of a compound of phosphorus and oxygen was found to be P 2 O 5. Experimentation shows that the molar mass of this actual compound is 283.89 g/mol. What is the compound’s molecular formula? Example Problem

38. P: 2 x 30.97 = 61.94 O: 5 x 16.00 = 80.00 141.94 g/mol 1.Find Molar Mass of empirical formula. 2. Divide the experimental formula mass by the empirical formula mass. 283.89 g/mol 141.94 g/mol = 2.00

39. 3. Multiply the subscripts by the common multiple. 2 × (P 2 O 5 ) = P 4 O 10