2. The Mole 6.02 X 10 23

3. The Mole 1 pair = 2 representative particles1 pair = 2 representative particles 1 dozen = 12 representative particles1 dozen = 12 representative particles 1 mole = 6.02 x 10 23 representative particles1 mole = 6.02 x 10 23 representative particles  The Mole (mol): A unit to count numbers of particles –1 dozen cookies = 12 cookies –1 mole of cookies = 6.02 X 10 23 cookies –1 dozen Al atoms = 12 Al atoms –1 mole of Al atoms = 6.02 X 10 23 atoms Note that the NUMBER is always the same, but the MASS is very different!

4. A Mole of Particles A Mole of Particles Contains 6.02 x 10 23 particles Representative Particle:Representative Particle: –Refers to the species present. –Usually atoms, molecules, or formula units. –Example: Atomic Nitrogen = NAtomic Nitrogen = N Nitrogen gas = N 2Nitrogen gas = N 2 Water = H 2 OWater = H 2 O Calcium ion = Ca 2+Calcium ion = Ca 2+ Glucose = C 6 H 12 O 6Glucose = C 6 H 12 O 6 1 mole of all of these = 6.02 x 10 23 rep. particles

5. 6.02 x 10 23 particles 1 mole or 1 mole 6.02 x 10 23 particles Note that a particle could be an atom OR a molecule! Avogadro’s Number as Conversion Factor

6. 1. Number of atoms in 0.500 mole of Al a) 0.500 mole of Al b) 6.02 x 10 23 Al atoms in 1 mole c) So, in 0.5 mole, 3.01 x 10 23 Al atoms 2.Number of moles of S in 1.8 x 10 24 S atoms a) 6.02 x 10 23 S atoms in 1.0 mole of S atom b) So, 3 x 6.02 x 10 23 = 1.8 x 10 24 S atoms in 3 x 1 mole= 3.0 mole S atoms c) 3.0 mole S atoms Learning Check

7. Gram Atomic Mass Gram Atomic Mass –Gram Atomic Mass – –Atomic mass of an element expressed in grams. –The gram atomic mass is the mass of 1 mol of atoms of any element. 1 mole of C atoms= 12.0 g 1 mole of Mg atoms =24.3 g 1 mole of Cu atoms =63.5 g # of C atoms in12.0 g of C atoms = # of Mg atoms in24.3 g of Mg atoms = # of O atoms in 16.0 g How many is that?How many is that? 6.02 x 10 23 !!

8. Other Names Gram Molecular Mass/Molecular Weight Gram Formula Mass/Formula Weight What is the gram atomic mass (GAM) of 1 mol of the following compounds?What is the gram atomic mass (GAM) of 1 mol of the following compounds? –Gram Molecular Mass of SO 3 and H 2 O = ? SO 3 = 80.1gH 2 O = 18 gSO 3 = 80.1gH 2 O = 18 g –Gram Formula Mass of NaCl and CaI 2 = ? NaCl = 58.5 g CaI 2 = 293.9 gNaCl = 58.5 g CaI 2 = 293.9 g

9. Molar Mass( atoms, molecules and compounds) A.1 mole of Br atoms B.1 mole of Sn atoms =79.9 g/mole = 118.7 g/mole Molar Mass: mass of one mol of any substance( Molar Mass: mass of one mol of any substance( usually we round to the tenths place). Mass in grams of 1 mole equal numerically to the sum of the atomic masses 1 mole of CaCl 2 = 111.1 g/mol 1 mole Ca x 40.1 g/mol + 2 moles Cl x 35.5 g/mol = 111.1 g/mol CaCl 2 1 mole of N 2 O 4 = 92.0 g/mol

10. A.Molar Mass of K 2 O = ? Grams/mole B. Molar Mass of antacid Al(OH) 3 = ? Grams/moleExample: 1.How many grams are in 9.45 mol dinitrogen trioxide? 718.20 g N 2 O 3 2. How many moles are in 92.2 g iron (III) oxide? 0.58 mol Fe 2 O 3 0.58 mol Fe 2 O 3 Learning Check!

11. 10 What is the formula mass of Ca 3 (PO 4 ) 2 ? 1 formula unit of Ca 3 (PO 4 ) 2 3 Ca 3 x 40.08 2 P2 x 30.97 8 O + 8 x 16.00 310.18 amu

12. Volume of a Mol Standard temperature and pressure:Standard temperature and pressure: –STP –0°C and 101.3 kPa or 1 atm –At these conditions, 1 mol of any gas = 22.4 L 1. What is the volume, in L, at 0.60 mol SO 2 gas at STP?1. What is the volume, in L, at 0.60 mol SO 2 gas at STP? 13.44 L SO 2 (g) 2. The density of a gaseous compound containing carbon and oxygen is 1.964 g/L at STP. What is the molar mass?2. The density of a gaseous compound containing carbon and oxygen is 1.964 g/L at STP. What is the molar mass? 43.99 g/mol

13. Problems, cont. 3. A gaseous compound composed of sulfur and oxygen is linked in the formation of acid rain and has a density of 3.58 g/L at STP. What is the molar mass?3. A gaseous compound composed of sulfur and oxygen is linked in the formation of acid rain and has a density of 3.58 g/L at STP. What is the molar mass? 80.19 g/mol 80.19 g/mol 4. What is the density of Kr gas at STP?4. What is the density of Kr gas at STP? 3.74 g/L 3.74 g/L

14. Big Helper… moles Representative particles Liters Grams Molar Mass (Molecules, Atoms, Ions, Formula units) N A =6.02 x 10 23 22.4 liters (STP) Molar volume Gram atomic mass/ Gram molecular mass/ formula mass

15. Aluminum is often used for the structure of light-weight bicycle frames. How many grams of Al are in 3.00 moles of Al? 3.00 moles Al ? g Al Converting Moles and Grams

16. 1. Molar mass of Al1 mole Al = 27.0 g Al 2. Conversion factors for Al 27.0g Al or 1 mol Al 1 mol Al 27.0 g Al 3. Setup3.00 moles Al x 27.0 g Al 1 mole Al Answer = 81.0 g Al

17. Atoms/Molecules and Grams Since 6.02 X 10 23 particles = 1 mole AND 1 mole = molar mass (grams) You can convert atoms/molecules to moles and then moles to grams! (Two step process) You can’t go directly from atoms to grams!!!! You MUST go thru MOLES.

18. molar mass Avogadro’s number Grams Moles particles Everything must go through Moles!!! Calculations

19. Atoms/Molecules and Grams How many atoms of Cu are present in 35.4 g of Cu? 35.4 g Cu 1 mol Cu 6.02 X 10 23 atoms Cu 63.5 g Cu 1 mol Cu = 3.4 X 10 23 atoms Cu

20. Learning Check! How many atoms of K are present in 78.4 g of K?

21. Learning Check! How many atoms of O are present in 78.1 g of oxygen? 78.1 g O 2 1 mol O 2 6.02 X 10 23 molecules O 2 2 atoms O 32.0 g O 2 1 mol O 2 1 molecule O 2

22. Percent Composition Percent composition:Percent composition: –The percent by mass of each element in a compound. % mass of element = grams of element grams of compound 1. Calculate the percent composition of each element in propane, C 3 H 8.1. Calculate the percent composition of each element in propane, C 3 H 8. C = 81.82%C = 81.82% H = 18.18%H = 18.18% X 100%

23. Types of Formulas Empirical FormulaEmpirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Ionic formula are always empirical formula Molecular FormulaMolecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

24. Empirical Formula Empirical Formulas:Empirical Formulas: The lowest whole number ratio of atoms of the elements in a compound.The lowest whole number ratio of atoms of the elements in a compound. –Example: CO (could be C 2 O 2, C 4 O 4, but not CO 2 ) *Empirical Formula may or may not be the molecular formula.*Empirical Formula may or may not be the molecular formula. Examples:Examples: –Hydrogen Peroxide = H 2 O 2 but the empirical formula is HO –Carbon dioxide = CO 2 and it is the empirical formula

25. To obtain an Empirical Formula 1.Determine the mass in grams of each element present, if necessary. 2.Calculate the number of moles of each element. 3.Divide each by the smallest number of moles to obtain the simplest whole number ratio. 4.If whole numbers are not obtained * in step 3), multiply through by the smallest number that will give all whole numbers * Be careful! Do not round off numbers prematurely

26. Empirical Problems 1. What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?1. What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen? Looks hard, but follow the three steps…Looks hard, but follow the three steps…

27. Problem, cont. Step 1:Step 1: –Convert % to moles. 1.85 mol N1.85 mol N 4.63 mol O4.63 mol O Step 2:Step 2: –Divide both answers by the lowest one. 1.85 mol N is lowest.1.85 mol N is lowest. 1 mol N –2.5 mol O

28. Problem, cont. Step 3:Step 3: –If one answer is a decimal, multiply both by a number so it is converted to a whole number. 2.5 mol O becomes 5 mol2.5 mol O becomes 5 mol 1 mol N becomes 2 mol1 mol N becomes 2 mol Answer: N 2 O 5Answer: N 2 O 5

29. Chemical Formulas of Compounds Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions).Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions). NO 2 2 atoms of O for every 1 atom of N NO 2 2 atoms of O for every 1 atom of N 1 mole of NO 2 : 2 moles of O atoms to every 1 mole of N atoms If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

30. A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine a formula for this substance. require mole ratios so convert grams to moles moles of N = 2.34g of N = 0.167 moles of N 14.01 g/mole 14.01 g/mole moles of O = 5.34 g = 0.334 moles of O 16.00 g/mole 16.00 g/mole Formula: Formula:

31. Determine Empirical Formula 1.75% carbon, 25% hydrogen 2.52.7% potassium, 47.3% chlorine 3.22.1% aluminium, 25.4% phosphorous, 52.5% oxygen 4.13% magnesium, 87% bromine 5.32.4% sodium, 22.5% sulfur, 45.1% oxygen

32. Molecular Formulas Molecular Formulas:Molecular Formulas: –Either the same as the empirical formula or a simple whole number multiple of it. Known molar mass of molecule = Molar mass of empirical form. the number of empirical formula units needed (EFU). Multiply empirical formula with EFU to get the molecular formula

33. Calculation of the Molecular Formula A compound has an empirical formula of NO 2. The colourless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance?