1. The Mole Introduction to Chemistry

2. The Mole Atoms and molecules are too small to count out individually Avogadro’s Number = 6.02 x 10 23 particles / mole One mole is equal to exactly 6.02 x 10 23 particles of any given compound or element 6.02 x 10 23 particles of “X”= 1 mole of “X”

3. The Mole 6.02 x 10 23 particles / mole means that any of the following are true: 6.02 x 10 23 atoms / mole 6.02 x 10 23 molecules / mole 6.02 x 10 23 ions / mole 6.02 x 10 23 formula units / mole

4. The Mole Example: How many molecules of water are there in 3.0 moles of water? 1.8 x 10 24 molecules of water

5. The Mole Example: How many atoms are there in 5.21 moles of carbon? 3.14 x 10 24 atoms of C

6. The Mole Example: If we have 9.28 x 10 33 formula units of BaCl 2, how many moles of barium chloride do we have? 1.54 x 10 10 moles of BaCl 2

7. The Mole Example: How many moles of magnesium are in 4.21 x 10 23 atoms of magnesium? 0.699 moles of magnesium

8. The Mole Two-step problems: Example: How many atoms of oxygen are in 3.9 moles of carbon dioxide? 4.7 x 10 24 atoms of O

9. The Mole Example How many hydroxide ions are there in 13.2 moles of Mn(OH) 3 ? 2.38 x 10 25 hydroxide ions

10. The Mole Example How many sulfate ions in 7.25 moles of Al 2 (SO 4 ) 3 ? 1.31 x 10 25 sulfate ions

11. Converting from Grams to Moles Using Molar Mass Molar Mass the mass of one mole of a given compound or element Elements the molar mass of an element is simply it’s atomic mass with grams/mole as units Example: Find the molar mass of Nitrogen 14.0 g/mol

12. Molar Mass Converting from Grams to Moles Example: How many moles of sodium are in 56.32 grams of sodium? 2.45 mol Na

13. Molar Mass Compounds use the “SSMT” method to find the molar mass Examples: H 2 O Na 2 CO 3 Pb(NO 3 ) 2 331.2 g/mol 106.0 g/mol 18.0 g/mol

14. Using Molar Mass We can use molar mass to find the mass of an amount of substance. Example: What is the mass (in grams) of 1.50 mole of potassium sulfate? 261 grams of K 2 SO 4

15. Using Molar Mass Example: What is the mass of 2.90 moles of Cu(NO 3 ) 2 ? 544 grams of copper nitrate

16. Using Molar Mass We can use molar mass to do the following conversions: Particles → moles → grams Grams → moles → particles Note: you can never go straight from particles to grams! You must always convert to moles first (using molar mass)

17. Using Molar Mass Example: If you have 39 grams of water in a beaker, how many molecules of water are present? 1.3 x 10 24 molecules of water

18. Using Molar Mass Example: If you have 4.21 x 10 23 atoms of gold, how many grams of gold do you have? 138 grams of gold

19. Molar Volume Molar volume – the volume of one mole of a gas at standard temperature and pressure (22.4 L/mol) Standard temperature and pressure (STP) temp. = 0ºC, pressure = 1 atmosphere

20. Molar Volume Example How many moles of gas are present in 12.5 liters of gas at STP? 0.558 moles of gas

21. Molar Volume Example How many moles of gas are present in 82 liters of gas at STP? 3.7 moles of gas

22. Molar Volume Example How many liters of gas are present in 2.1 moles of gas at STP? 47 liters of gas

23. Percent Composition the percentage of mass that each element contributes to the total mass of the compound % Comp. = Molar mass of element x 100 Molar mass of entire compound

24. Percent Composition Example: Find the percent composition for each element in Al 2 (SO 4 ) 3 % Al = 15.8 % % S = 28.1 % % O = 56.1 %

25. Percent Composition Example: Find the percent composition for each element in Mg(OH) 2 % Mg = 41.7 % % O = 54.9 % % H = 3.4 %

26. Empirical Formula Molar mass can be used to determine a compound’s empirical formula Steps: 1. Convert each element’s mass to moles 2. Divide by the smallest number of moles 3. Round to the nearest whole number and use as subscripts

27. Empirical Formula Example: What is the empirical formula for a compound that has 1.80 g of carbon and 4.98 g of oxygen? CO 2

28. Empirical Formula Example: What is the empirical formula for a compound that has 38.67 g of carbon, 16.22 g hydrogen, and 45.11 g of nitrogen? CH 5 N

29. Empirical Formula Using Percent Composition Steps 1. Assume a 100 g sample 2. Convert from grams to moles 3. Divide by the smallest number of moles 4. Round to the nearest whole number and use as subscripts

30. Empirical Formula Example: What is the empirical formula for a compound that has a percent composition of 40.0% carbon, 6.71% hydrogen, and 53.3% oxygen? CH 2 O

31. Empirical Formula Example: If the empirical formula for a compound is CH 2 Cl and the molar mass for the molecular formula is 98.96 g/mol, what is the molecular formula? molecular = C 2 H 4 Cl 2

32. Empirical/Molecular Formula Example: If empirical formula for a compound is CH 2 O and the molar mass of the molecular formula is 60.06 g/mol, what is the molecular formula? molecular = C 2 H 4 O 2

33. SOL covered during lesson CH 4 a, b CH 3 c

34. Extra Slides

35. Empirical/Molecular Formula Example: Caffeine has a percent composition of 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen. If the molar mass of the compound is 194.2 g/mol, what is the molecular formula of caffeine? C 8 H 10 N 4 O 2

36. Empirical Formula Example: What is the empirical formula for a compound that has a percent composition of 71.65% chlorine, 24.27% carbon, and 4.07% hydrogen? If the molar mass for the molecule is 98.96 g/mol, what is the molecular formula? empirical = CH 2 Cl molecular = C 2 H 4 Cl 2

37. Empirical/Molecular Formula Example: What is the empirical formula for a compound that has a percent composition of 40.0% carbon, 6.71% hydrogen, and 53.3% oxygen? If the molar mass of the compound is 60.06 g/mol, what is the molecular formula? empirical = CH 2 O molecular = C 2 H 4 O 2