2. SOL R EVIEW 5 The Mole

3. A M OLE OF P ARTICLES A M OLE OF P ARTICLES C ONTAINS 6.02 X 10 23 PARTICLES = 6.02 x 10 23 C atoms = 6.02 x 10 23 H 2 O molecules = 6.02 x 10 23 NaCl formula units So one mole of NaCl contains: 6.02 x 10 23 Na + ions and 6.02 x 10 23 Cl – ions (2 moles of ions altogether!) 1 mole C 1 mole H 2 O 1 mole NaCl

4. 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! A VOGADRO ’ S N UMBER AS C ONVERSION F ACTOR

5. 1. Find the number of atoms in 0.500 mole of Al. 0.500 mol x 6.02 x10 23 atoms = 3.01 x 10 23 atoms 1 mol 2.Find the number of moles in 1.204 x 10 24 S atoms 1.204 x 10 24 atoms x ______1 mol____ = 2 mol 6.02 x 10 23 atoms U SING A VOGADRO ’ S NUMBER AS A CONVERSION FACTOR

6. The mass of 1 mole of a substance (in grams) Molar mass of an element is equal to the numerical value of the atomic mass (get from periodic table) 1 atom of Mg= 24.3 amu 1 mole of Mg atoms =24.3 g (the molar mass of magnesium is 24.3 g) M OLAR M ASS

7. To find the molar mass of a compound, add up the number of grams of each element in one mole of the compound. Molar mass of Ca Cl 2 : 1 mole Ca x 40.1 g/mol = 40.1 g + 2 moles Cl x 35.5 g/mol= 71.0 g 1 mole of CaCl 2 = 111.1 g The molar mass of CaCl 2 is 111.1 g/mol M OLAR M ASS OF M OLECULES AND C OMPOUNDS

8. U SING M OLAR M ASS AS A C ONVERSION F ACTOR How many moles are in 25.6 g of NaCl? 25.6 g x 1mol NaCl =.438 mol NaCl 58.5 g

9. T HE M OLE -V OLUME R ELATIONSHIP According to Avogadro’s hypothesis: At standard temperature and pressure (STP), 1 mol of any gas occupies a volume of 22.4 L STP is defined as 0 o C and 1 atm pressure (101.3 kPa)

10. U SING M OLAR V OLUME AS A CONVERSION FACTOR What is the volume, in liters, of 1.5 mol Cl 2 at STP? 1.5 mol Cl 2 x 22.4 L = 33.6 L Cl 2 1 mol

11. 22.4 L 1 mol 22.4 L 1 mol 6.02 x 10 23 particles 1 mol Molar mass 1 mol MASS

12. A TOMS /M OLECULES AND G RAMS 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

13. P ERCENT C OMPOSITION What is the percent composition of propane (C 3 H 8 )? Mass of C in 1 mol propane = 3 mol C x 12.0 g/mol = 36.0 g Mass of H in 1 mol propane = 8 mol H x 1.0 g/mol = 8.0 g Molar mass of propane = 44.0 g % C = __mass of C __ x 100% = 36 g = 81.8% C mass of propane 44.0 g % H = ___mass of H__ x 100% = 8.0 g = 18.2 %H mass of propane 44.0 g

14. T YPES OF F ORMULAS Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Note: Ionic formulas are always empirical formulas Note: Ionic formulas are always empirical formulas Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

15. D ETERMINING E MPIRICAL F ORMULA What is the empirical formula of a compound that is 48.38% C, 8.12% H, and 43.5 % O? 48.38g C x 1 mol = 4.03 mol C ÷ 2.72 = 1.5 mol C 12 g 8.12g H x 1 mol = 4.06 mol H ÷ 2.72 = 1.5 mol H 2 g 43.5g O x 1 mol = 2.72 mol O ÷ 2.72 = 1 mol O 16 g The ratio is C 1.5 H 1.5 O 1 but we must multiply this by 2 to get whole numbers in our empirical formula: C 3 H 3 O 2

16. H OW CAN WE DETERMINE MOLECULAR FORMULA FROM EMPIRICAL FORMULA AND MOLAR MASS ? What is the molecular formula of a compound that has an empirical formula of CH 2 O and a molecular mass of 180.0 g/mole? 1. The mass of this empirical formula is: 1C x 12.0 g = 12.0 g 2H x 1.0 g = 2.0 g 1O x 16.0 g = 16.0g 30.0 g 2. How many times does the empirical mass go into the molecular mass? 180.0 ÷ 30.0 = 6 times 3. The molecular formula is 6 time the empirical formula: 6(CH 2 O) = C 6 H 12 O 6