1. 1 Unit 9

2. 2 How do we measure how much of something we have? How do we measure how much of something we have? n Mass (g) - how much stuff n Volume (L) – space the stuff occupies n Count pieces or particles

3. 3 Representative particles n The smallest pieces of a substance n For an element it is an atom (Fe) –Unless it is diatomic n For a molecular compound it is a molecule (O 2 or CO 2 ). n For an ionic compound it is a formula unit (NaCl).

4. 4 But we have a problem counting those pieces! n Most samples have LOTS of atoms, molecules, or formula units n We want things to be simple! n Scientists created a new unit

5. 5 The MOLE n Defined as the number of carbon atoms in exactly 12 g of carbon-12. n Avogadro’s number = the number of particles in one mole of a substance n Avogadro’s number = 6.022 x 10 23 particles

6. 6 The Mole n 1 mole Fe = 6.022 x 10 23 atoms n 1 mole CO 2 = 6.022 x 10 23 molecules n 1 mole NaCl = 6.022 x 10 23 formula units

7. 7 Avogadro as a conversion factor n 0.30 mol F = _________ atoms F 0.30 mol F mol F atoms F 6.022 x 10 23 1

8. 8 Calculation question n How many molecules of CO 2 are there in 4.56 moles of CO 2 ?

9. 9 Calculation question n How many moles of water is 5.87 x 10 22 molecules?

10. 10 Calculation question n How many atoms of carbon are there in 1.23 moles of C 6 H 12 O 6 ? n Umm…this one isn’t as easy… n How many atoms of C are in one molecule of C 6 H 12 O 6 ? n 1 molecule = 6 atoms 1.23 moles mole molecules 6.022 x 10 23 1 molecule atoms C 6161

11. 11 Molar Mass n n The mass of one mole (g/mol) n n One mole is always 6.022 x 10 23 pieces n n But those pieces can have different masses – –1 dozen eggs vs. 1 dozen elephants n n AKA – –gram atomic mass (atoms) – –gram molecular mass (molecules) – –gram formula mass (compounds)

12. 12 Molar Mass/ Gram Atomic Mass n The mass of 1 mole of an element in grams n Equal to the average atomic mass found on the periodic table n We can write this as 12.01 g C = 1 mole C n And since 1 mole = 6.022 x 10 23 atoms, we can also say that 6.022 x 10 23 atoms C = 12.01 g C 6.022 x 10 23 atoms C = 12.01 g C

13. 13 Molar mass/gram molecular mass n The mass of one mole of a molecule (formed by covalent bonds) n In 1 mole of H 2 O molecules there are 2 moles of H atoms and 1 mole of O atoms n To find the mass of one mole of a molecule, add the masses of the parts of that molecule

14. 14 Example n What is the mass of 1 mole of CH 4 ? n 1 mole of C = 12.01 g n 4 mole of H x 1.01 g = 4.04g n 1 mole CH 4 = 12.01 + 4.04 = 16.05g n The Gram Molecular mass or Molar Mass of CH 4 is 16.05g

15. 15 Molar mass/ gram formula mass n The mass of one mole of an ionic compound (formed by ionic bonds) n Calculated the same way n What is the molar mass of Fe 2 O 3 ? n 2 moles of Fe x 55.85 g = 111.70 g n 3 moles of O x 16.00 g = 48.00 g n Molar mass = 111.70 g + 48.00 g = 159.70g

16. 16 Molar mass as a conversion factor n 3.50 mol Cu = _________ g Cu 3.50 mol Cu mol Cu g Cu63.55 1

17. 17 Molar mass as a conversion factor n How many moles is 5.69 g of NaOH?

18. 18 Gases and the Mole

19. 19 Gases n It’s difficult to find the mass of a sample of gas, so we usually measure volume. n Two things affect the volume of a gas –Temperature and pressure n Must compare samples of gases at the same temperature and pressure

20. 20 Standard Temperature and Pressure n Avogadro's Hypothesis - at the same temperature and pressure, equal volumes of gas have the same number of particles n 0ºC and 1 atm pressure n abbreviated STP n At STP, 1 mole of gas occupies 22.4 L n Called the molar volume n Same volume but can still have different masses! (remember the density blocks…)

21. 21 Examples n Remember - 1 mole = 22.4 L n What is the volume of 4.59 moles of CO 2 gas at STP? n How many moles is 5.67 L of O 2 at STP? n What is the volume of 8.8g of CH 4 gas at STP?

22. 22 Density of a gas n D = m /V n For a solid, the units were g/cm 3 n For a gas, the units will be g / L n To find the density, we need to know the mass and the volume. n Assume you have 1 mole –At STP the volume is 22.4 L –The mass is the molar mass

23. 23 Examples n Find the density of CO 2 at STP. n Find the density of CH 4 at STP.

24. 24 Going the other way n Given the density, we can find the molar mass of the gas n Again, assume you have 1 mole at STP, so V = 22.4 L. n What is the molar mass of a gas (at STP) with a density of 1.964 g/L? n D = m/V n What is the molar mass of a gas with a density of 2.86 g/L?

25. 25 Percent Composition n Describes the composition by mass of a substance n Like all percents n Part x 100 whole n To calculate - –Find the mass of each component –Divide by the total mass

26. 26 Example n Calculate the percent composition of silver in a compound that is made of 29.0 g of Ag and 4.30 g of S.

27. 27 % Composition from the Formula n If given the formula, assume you have 1 mole. 1. Calculate the total molar mass 2. Calculate the total mass of each element present 3. Calculate the percent composition: = (total mass of element present ÷ total molar mass) x 100 = (total mass of element present ÷ total molar mass) x 100

28. 28 Examples n Calculate the percent composition of each element in C 2 H 4 n Calculate the percent composition of aluminum in aluminum carbonate.

29. 29 Percent to Mass n Multiply % composition by the total mass to find the mass of that component n How much aluminum is in 450 g of aluminum carbonate?

30. 30 Empirical Formula n The lowest whole number ratio of elements in a compound n Molecular formula - the actual ratio of elements in a compound n C 2 H 4 molecular formula n CH 2 empirical formula n C 3 H 6 molecular formula n H 2 O molecular formula n H 2 O empirical formula

31. 31 Finding Empirical Formulas n Just find the lowest whole number ratio if given the molecular formula n C 6 H 12 O 6 = n CH 4 N 2 n CH 4 N 2 = n It is not just the ratio of atoms, it is also the ratio of moles

32. 32 Calculating Empirical Formulas n We can find empirical formula from the percent composition n Assume you have a 100 g sample n The percentages become grams. n Convert grams to moles. n Find lowest whole number ratio by dividing everything by the smallest number of moles. n Use the whole number ratio to write the formula

33. 33 Example n Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. n Assume 100 g so… n 38.67 g C x 1mol C = 3.220 mole C 12.01 g C n 16.22 g H x 1mol H = 16.1 mole H 1.01 g H n 45.11 g N x 1mol N = 3.220 mole N 14.01 g N

34. 34 Example n The ratio is 3.220 mol C = 1 mol C 3.220 mol N 1 mol N n The ratio is 16.1 mol H = 5 mol H 3.220 mol N 1 mol N n 1C: 5H: 1N n C 1 H 5 N 1 or CH 5 N

35. 35 Empirical to Molecular Formula n Since the empirical formula is the lowest ratio, the actual molecule could have the same mass or a higher mass n If given the empirical formula & the molar mass, calculate the molecular formula by… –Dividing the given molar mass by the mass of one mole of the empirical formula. –You will get a whole number. –Multiply the empirical formula by this whole number

36. 36 Example n n A compound has an empirical formula of ClCH 2 and a molar mass of 98.96 g/mol. What is its molecular formula? n Find the mass of one mole of the empirical formula or “1 empirical mole” n Divide the molar mass given by the mass of one empirical mole. n Multiply the empirical formula by that whole number

37. 37 Percent comp  molecular formula n Multiply the percent by the molar mass –This gives you the mass of that element in 1 mole of the compound n Change this to moles –You will get whole numbers –These are the subscripts

38. 38 Percent comp  Molecular Formula n Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. It has a molar mass of 194 g. What is its molecular formula? n 49.48% C x 194 g = n Change that to moles n 5.15% H x 194 g = n 28.87% N x 194 g = n 16.49% O x 194 g =