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Chemistry 111

Chemistry 111. Chapter 7. 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. Formulas 1. C 2 H 6 O

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Chemistry 111

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  1. Chemistry 111 Chapter 7

  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. Formulas 1 • C2H6O • 2 carbon atoms • 6 hydrogen atoms • 1 oxygen atom • Sometimes written as CH3CH2OH

  4. Mass of a Formula • C2H6O • 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. Moles • 1 Dozen = 12 items • useful for eggs, cookies, etc. • 1 Mole = 6.02 × 1023 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 × 1023) was chosen so we don’t have to do math to convert.

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

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

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

  10. Formula of a Compound • Ratio (moles/moles) of Cu:S gave a formula: • Cu1.75S1.00 = “Experimental Formula” • Round it off • Cu2S1 or Cu2S = “Empirical Formula” • Is Copper +1 or +2? • Matches Figure 6.3, pg 141.

  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. Solution: (use tabular method) • Write out Element Names & Percent • Divide by Atomic Weight to get Moles • Choose Smallest # of moles Element%÷ AW= Moles Na 32.4 ÷22.99 1.409 S 22.6 ÷32.07 0.7047 O 45.0 ÷16.00 2.813

  13. Solution: (use tabular method) • Divide all moles by that # • Round off. Element= Moles÷ Smallest Na 1.409 ÷ 0.7047 = 1.999  2 S 0.7047 ÷ 0.7047 = 1.000  1 O 2.813 ÷ 0.7047 = 3.992  4 • Write Formula: Na2SO4

  14. You Try One: • 25.8 % P, 26.7 % S, 47.5 % F • F3PS Element%÷ AW= Moles÷ SmallestRound P 25.8 ÷30.97 0.8331÷0.8326 =1.001  1 S 26.7 ÷32.07 0.8326 ÷0.8326 =1.000  1 F 47.5 ÷19.00 2.500÷0.8326 =3.003  3

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