1. The Mole & Chemical Quantities

2. The Mole Mole-the number of particles equal to the number of atoms in exactly 12.0 grams of carbon-12. 1 mol = 6.02 x 10 23 particles The mole is also called Avogadro’s number.

3. Particles can be: Atoms = single elements Formula units = ionically bonded compounds Molecules = covalently bonded compounds

4. Particle-Mole Problems Don’t Forget!!! The number of particles in 1 mole of any substance is always the same. 1 mol = 6.02 x 10 23 particles How many molecules are in 2.2 moles of water? 1 mol H 2 O = 6.02 x 10 23 molecules H 2 O 1 mol H 2 O 6.02 x 10 23 molecules 6.02 x 10 23 molecules or 1 mol H 2 O

5. 2.2 mol H 2 0 6.02 x 10 23 molecules 1 X 1 mol H 2 O = 1.3244 x 10 24 1.3 x 10 24 molecules H 2 O How many moles of sodium carbonate contain 7.9 x 10 24 formula units? 7.9 x 10 24 f.u. Na 2 CO 3 1 mol Na 2 CO 3 1 X 6.02 x 10 23 f.u. Na 2 CO 3 = 13.1229235 mol 13 mol Na 2 CO 3

6. Mole-Mass and Mole-Volume Relationships Mass-Mole Problems Change mass to moles or visa versa using the molar mass. You need to know the mass of 1 mole of the substance, or the molar mass.

7. How many moles are in 11.2 g of NaCl? 1. Determine the molar mass. Na: 1 x 22.99 = 22.99g Cl: 1 x 35.45 = 35.45g = 58.44 g/mol 1 mol of NaCl = 58.44 g it can be written as: 1 mol NaCl58.44g NaCl 58.44 gNaCl or 1 mol NaCl

8. 2. Write what you have been given over 1 and multiply by the equality that will cancel the starting unit (on bottom) and keep the desired unit (on top). 11.2 g NaCl 1 mol NaCl 1 X 58.44 g NaCl = 0.1916495551 mol NaCl 0.192 mol NaCl

9. What is the mass of 2.50 mol of NaCl? Find the molar mass. 1 mol of NaCl = 58.44 g 1 mol NaCl58.44g NaCl 58.44 g NaCl or 1 mol NaCl 2.50 mol NaCl58.44 g NaCl 1 X 1 mol NaCl = 146.1 g NaCl 146 g NaCl

10. Mole-Volume Problems Equal volumes of gases at the same temperature and pressure contain the same number of particles. Molar volume-the volume of 1 mol of a gas at standard condition (STP). STP-standard temperature ( 0 ºC) and pressure ( 1 atm).

11. What is the volume of 0.35 moles of helium gas at STP? 1 mol = 22.4 liters 0.35 mol He22.4 L He 1 X 1 mol He = 7.84 L He 7.8 L He

12. Multistep Conversions Mass-Particle g  mol  particles Particle-Mass particles  mol  g What is the mass of 8.2 x 10 22 atoms of calcium? 1 mol Ca = 6.02 x 10 23 atoms Ca 1 mol Ca = 40.08 g Ca

13. What is the mass of 8.2 x 10 22 atoms of calcium? 8.2 x 10 22 atoms Ca x 1 mol Ca x 40.08 g Ca 1 6.02 x 10 23 atoms 1 mol Ca = 5.459401993 g Ca 5.5 g Ca

14. Empirical and Molecular Formulas Remember: Percent composition -the mass of each element in a compound compared to the entire mass of the compound multiplied by 100. What is the percent of hydrogen in water? Molar mass of water is 18.02 g. Mass of hydrogen in water is 2.02 g. 2.02 g 18.02 g X 100 = 11.20976693% 11.2 %

15. Empirical Formula- formula that gives the simplest whole-number ratio of atoms. To solve for the empirical formula from percentages: 1. Assume there is a 100 g sample. 2. Convert masses to moles. 3. Find the smallest whole-number ratio between moles. This means divide each number by the smallest moles.

16. Determine the empirical formula for a compound made of 80 % carbon and 20 % hydrogen. 80. g of carbon 20. g of hydrogen 80. g C 1 mol C 1 x 12.01 g C = 6.661115737 mol C 6.7 mol C 20. g H1 mol H 1 x 1.01 g H = 19.8019802 mol H 20. mol H

17. C: 6.7  6.7 = 1 H: 20.  6.7 = 2.98  3 1:3 Ratio Empirical formula = CH 3

18. Molecular Formula -formula that gives the actual number of atoms of each element in a molecular compound. -whole number multiple of the empirical formula C 6 H 12 O 6 (molecular formula for glucose) CH 2 0 (empirical formula)

19. The molecular formula can be determined by comparing the empirical formula mass to the molecular formula mass. To solve for molecular formula from masses: 1. Solve for the empirical formula. 2. Solve for empirical formula mass. 3. Compare the empirical formula mass to the molecular formula mass.

20. A compound has the following composition: 76.54 % carbon, 12.13 % hydrogen, and 11.33 % oxygen. Its molar mass is 282.45 g/mol, what is its molecular formula. Solve for empirical formula. 76.54 g carbon 12.13 g hydrogen 11.33 g oxygen

21. 76.54 g C 1 mol C 1 x 12.01 g C = 6.373022481 mol C = 6.373 mol C 12.13 g H 1 mol H 1 x 1.01 g H = 12.00990099 mol H = 12.0 mol H 11.33 g O 1 mol O 1 x 16.00 g O = 0.708125 mol O = 0.7081 mol O

22. 6.373 mol C  0.7081 = 8.9  9 12.0 mol H  0.7081 = 16.9  17 0.7082 mol O  0.7081 = 1 9:17:1 Ratio Empirical Formula = C 9 H 17 O C: 9 x 12.01 =108.09g H: 17 x 1.01 = 17.17 g O: 1 x 16.00 = 16.00 g Empirical formula mass = 141.26 g

23. 282.45 141.26 = 1.99  2 2(C 9 H 17 O) = C 18 H 34 O 2