1. Chapter 11 : Matter Notes

2. Mole (mol) is equal to 6.02x10 23 The mole was named in honor of Amedeo Avogadro. He determined the volume of one mole of gas.

3. Molar Mass: mass in grams of 1 mole of any pure substance. Equal to the atomic mass for an element. Unit is g/mol. Converting moles to Mass: Number of moles number of grams = mass(g) 1 mole

4. Converting moles to particles. Number of moles_ 6.02x10 23 _ particles = particle 1 mole Example: How many molecules are in 3.50 moles of sucrose? 3.50 mol sucrose 6.20 x 10 23 molecules of sucrose 1 mole sucrose = 2.11 x10 23 molecules of sucrose

5. Converting Particles to Moles: Number of particles 1 mole = number of moles 6.02 x10 23 particles Example: How many moles are in 4.5 x 10 23 atoms of Zinc? 4.5 x 10 23 atoms Zn 1 mole Zn 6.02 x10 23 atoms Zn = 7.48 moles Zn

6. Converting Moles to Mass: Element: How many grams are in 3 moles of Manganese? 3.00 mol Mn54.9g Mn= 165g Mn 1 mol Mn

7. Converting Moles to Mass: Compound: How many grams are in 3 moles of K 2 CrO 4 ? 1. Find molar mass of compound. 2 mol K 39.10 g K = 78.20 g K 1 mol K 1mol Cr 52.00g Cr = 52.00 g Cr 1 mol Cr 4 mol O 16.00 g O = 64.00 g O 1mol O 194.20 g K 2 CrO 4

8. Converting Moles to Mass: Compound: 1. Multiply moles by molar mass: 3 moles K 2 CrO 4 194.20 g K 2 CrO 4 mol K 2 CrO 4 = 582.6 g K 2 CrO 4

9. Converting Mass to Moles: Number of grams1 mole = moles mass (g) Element: How many moles are in 525g of Ca? 525g Ca1 mol Ca = 13.1 mol Ca 40.06g Ca

10. Converting Mass to Moles: Compound: How many moles are in 325g of Calcium Hydroxide [Ca(OH) 2 ]? 1. Find the molar mass of compound: 1mol Ca 40.08g Ca = 40.08 g 1mol Ca 2mol O 16.00g O = 32.00 g 1mol O 2mol H 1.008g H = 2.026g 1mol H 74.096 g Ca(OH) 2

11. Converting Mass to Moles: Compound: 1. Convert grams to moles: 325g Ca(OH) 2 1mol Ca(OH) 2 74.096 g Ca(OH) 2 = 4.39 mol Ca(OH) 2

12. Converting Mass to Particles: Number of grams 1 mol 6.02 x 10 23 particles mass(g) 1 mol = particles Element: How many atoms are in 25 g of Gold? 25g Au 1 mol Au 6.02 x 10 23 atoms Au 196.97g Au 1 mol Au = 7.65 x 10 23 atoms Au

13. Converting Mass to Particles: Compound: How many formula units are in 35.6 g of aluminum chloride? [AlCl 3 ] 1. Find the molar mass of compound: 1 mol Al 26.98 g Al = 26.96 g 1mol Al 3 mol Cl 35.45g Cl= 106.35g 1 mol Cl 133.33g AlCl 3

14. Converting Mass to Particles: Compounds: 1. Convert grams to formula units: 35.6 g AlCl 3 1 mol AlCl 3 6.02 x 10 23 formula units 133.33g AlCl 3 1 mol AlCl 3 = 1.61 x10 23 formula units AlCl 3

15. Chapter 11 Matter Notes Section 11.4

16. Percent Composition: Percent composition is the percent by mass of any element in a compound. Mass of element X 100% = percent composition Mass of compound

17. Percent Composition: Example: Calculate the percent composition of water. H 2 O 2 mol H 1.01 g H =2.02 g H 1 mol H 18.02 g H 2 O 1 mol O 16.00 g O = 16.00 g O 1 mol O 18.02 g H 2 O

18. Percent Composition: con’t. 2.02 g H H X 100% = 11.2% H 18.02g H 2 O 16.00 g O 88.8% O O X 100% = 88.8% O 18.02 g H 2 O

19. Empirical Formula Empirical formula is a formula with the smallest whole-number ratio of the element. Example: Determine the empirical formula for methyl acetate which has 48.64% Carbon, 8.16% Hydrogen, 43.20 % Oxygen.

20. Empirical Formula: con’t. 1. Find the number of moles of each element 48.64g C 1 mol C = 4.05 mol C 12.01 g C 8.16 g H 1 mol H = 8.10 mol H 1.008 g H 43.20 g O 1 mol O = 2.7 mol O 16.00 g O

21. Empirical Formula: con’t. 1. Divide each mole by the smallest number: 4.05 mol C = 1.5 mol C 2.7 8.10 mol H = 3 mol H 2.7 2.7 mol O = 1 mol O 2.7

22. Empirical Formula: con’t. 1. Multiply each mole by the smallest number that will produce whole numbers: 1.5 mol C x 2 = 3 C 3 mol H x 2 = 6 H 1 mol O x 2 = 2 O 1. Use the new whole numbers as subscripts for each element in the compound. C 3 H 6 O 2

23. Molecular Formula: Experimental molar mass = n empirical formula mass Determine the molecular formula for a compound composed of 40.68% carbon, 5.08 % Hydrogen, and 54.24 % Oxygen which has a molar mass of 118.1 g/mol.

24. Molecular Formula: 1. Find the empirical formula: 40. 68 g C 1 mol C = 3.387 mol C 12.01 g C 5.08 g H 1 mol H = 5.04 mol H 1.008 g H 54.24 g O 1 mol O = 3.39 mol O 16.00 g O

25. Molecular Formula: ( finding the empirical formula con’t.) 3.387 mol C = 1 mol C x 2 = 2 3.387 5.04 mol H = 1.5 mol H x 2 = 3 3.387 3.39 mol O = 1 mol O x 2 = 2 3.387 empirical formula: C 2 H 3 O 2

26. Molecular Formula: 1. Find the molar mass of the empirical formula: 2 mol C 12.01 g C = 24.2 g 1 mol C 3 mol H 1.008 g H = 3.024 g 1 mol H 2 mol O 16.00 g O = 32.00g 1 mol O 59.04 g C 2 H 3 O 2

27. Molecular Formula: 1. Divide the molar mass given in the problem by the molar mass of the empirical formula to find n: n = 118.1 g 59.04g n = 2

28. Molecular Formula: 1. Multiply all subscripts of the empirical formula by n: (C 2 H 3 O 2 ) 2 molecular formula: C 4 H 6 O 4