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.

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Presentation transcript:

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

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

5 Moles 1 Dozen = 12 items –useful for eggs, cookies, etc. 1 Mole = 6.02 × items –useful for atoms, molecules, etc. Mass of Nitrogen = (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 × ) 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: –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 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) 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 ÷ S22.6 ÷ O45.0 ÷

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

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