1. 1 Chemistry 111 Chapter 7

2. 2 Chemical Formulas Atoms in Formulas Adding up weights (Formula Weights) Mole Concept –amu  g/mol –Chemist’s Dozen Moles  Molecules Percent Composition –Learn my table technique (for empirical formulas!) Empirical Formulas

3. 3 Formulas 1 C 2 H 6 O –2 carbon atoms –6 hydrogen atoms –1 oxygen atom –Sometimes written as CH 3 CH 2 OH

4. 4 Mass of a Formula C 2 H 6 O Carbon atoms are 12.01 amu Hydrogen atoms are 1.008 amu Oxygen atoms are 16.00 amu (12.01 × 2) + (1.008 × 6) + (16.00 × 1) = 46.068  46.07 amu

5. 5 Moles 1 Dozen = 12 items –useful for eggs, cookies, etc. 1 Mole = 6.02 × 10 23 items –useful for atoms, molecules, etc. Mass of Nitrogen = 14.01 (peridic table) –14.01 amu = mass of 1 Nitrogen atom –14.01 g/mol = mass of 1 mole of Nitrogen atoms Avogadro’s Number (6.02 × 10 23 ) was chosen so we don’t have to do math to convert.

6. 6 Converting Between Moles & Molecules Avogadro’s number is a conversion factor: 2.1 moles of Carbon is: atoms of carbon. A BIG number!

7. 7 Percent Composition I will show you a tabular way to get % composition: –USE IT TODAY (practice) Percent composition is a ratio of:

8. 8 Empirical & Molecular Formulas “Formula of a Compound” Tabular Method (Homework Problems) Molecular Formulas

9. 9 Formula of a Compound Knew Mass of Copper  Moles of Cu Knew added mass  had to be Sulfur  Moles of S

10. 10 Formula of a Compound Ratio (moles/moles) of Cu:S gave a formula: –Cu 1.75 S 1.00 = “Experimental Formula” Round it off –Cu 2 S 1 or Cu 2 S= “Empirical Formula” Is Copper +1 or +2? –Matches Figure 6.3, pg 141.

11. 11 Tougher Problems (More Elements) This is done all the time: –Make a new chemical –Send a pure sample off for “elemental analsis” –Get 6 sig. fig. answer with percents to verify exact formula. Sample: An unknown chemical was analyzed and found to contain: –32.4 % Sodium –22.6 % Sulfur –45.0 % Oxygen

12. 12 Solution: (use tabular method) 1.Write out Element Names & Percent 2.Divide by Atomic Weight to get Moles 3.Choose Smallest # of moles Element % ÷ AW= Moles Na 32.4 ÷22.991.409 S22.6 ÷32.070.7047 O45.0 ÷16.002.813

13. 13 Solution: (use tabular method) 4.Divide all moles by that # 5.Round off. Element= Moles ÷ Smallest Na 1.409÷ 0.7047 = 1.999  2 S0.7047÷ 0.7047 = 1.000  1 O2.813÷ 0.7047 = 3.992  4 6.Write Formula: Na 2 SO 4

14. 14 You Try One: 25.8 % P, 26.7 % S, 47.5 % F F 3 PS Element % ÷ AW= Moles÷ SmallestRound P 25.8 ÷30.970.8331÷ 0.8326=1.001  1 S26.7 ÷32.070.8326 ÷ 0.8326 =1.000  1 F47.5 ÷19.002.500÷ 0.8326 =3.003  3