1. Avogadro’s number is written as conversion factors. 6.02 x 10 23 particles and 1 mole 1 mole 6.02 x 10 23 particles The number of molecules in 0.50 mole of CO 2 molecules is calculated as 0.50 mole CO 2 molecules x 6.02 x 10 23 CO 2 molecules 1 mole CO 2 molecules = 3.0 x 10 23 CO 2 molecules Avogadro’s Number

2. A. Calculate the number of atoms in 2.0 moles of Al. 1) 2.0 Al atoms 2) 3.0 x 10 23 Al atoms 3) 1.2 x 10 24 Al atoms B. Calculate the number of moles of S in 1.8 x 10 24 S. 1) 1.0 mole S atoms 2) 3.0 mole S atoms 3) 1.1 x 10 48 mole S atoms Learning Check

3. A. Calculate the number of atoms in 2.0 moles of Al. 3) 1.2 x 10 24 Al atoms 2.0 moles Al x 6.02 x 10 23 Al atoms 1 mole Al B. Calculate the number of moles of S in 1.8 x 10 24 S. 2) 3.0 mole S atoms 1.8 x 10 24 S atoms x 1 mole S 6.02 x 10 23 S atoms Solution

4. Molar Mass of CaCl 2 For a compound, the molar mass is the sum of the molar masses of the elements in the formula. We calculate the molar mass of CaCl 2 to the nearest 0.1 g as follows. ElementNumber of Moles Atomic MassTotal Mass Ca140.1 g/mole40.1 g Cl 2 235.5 g/mole71.0 g CaCl 2 111.1 g

5. Molar Mass of K 3 PO 4 Determine the molar mass of K 3 PO 4 to 0.1 g. ElementNumber of Moles Atomic MassTotal Mass in K 3 PO 4 K339.1 g/mole 117.3 g P131.0 g/mole 31.0 g O416.0 g/mole 64.0 g K 3 PO 4 212.3 g

6. Methane CH 4 known as natural gas is used in gas cook tops and gas heaters. 1 mole CH 4 = 16.0 g The molar mass of methane can be written as conversion factors. 16.0 g CH 4 and 1 mole CH 4 1 mole CH 4 16.0 g CH 4 Molar Mass Factors

7. A glucose solution with a volume of 2.0 L contains 72 g glucose (C 6 H 12 O 6 ). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1)0.20 M 2)5.0 M 3)36 M Learning Check

8. 1) 0.20 M 72 g x 1 mole x 1 = 0.20 moles 180. g 2.0 L 1 L = 0.20 M Solution

9. The units in molarity can be used to write conversion factors. Molarity Conversion Factors

10. Stomach acid is 0.10 M HCl solution. How many moles of HCl are present in 1500 mL of stomach acid? 1) 15 moles HCl 2) 1.5 moles HCl 3) 0.15 mole HCl Learning Check

11. The mass percent (%m/m) Is the g of solute in 100 g of solution. mass percent = g of solute 100 g of solution mass percent = g of solute x 100% g of solution Mass Percent

12. Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution). g of KCl = 8.0 g g of solvent (water) = 42.0 g g of KCl solution = 50.0 g 8.0 g KCl (solute) x 100 = 16% (m/m) KCl 50.0 g KCl solution Make 50.0 g 16% (m/m) KCl Solution

13. A solution is prepared by mixing 15 g Na 2 CO 3 and 235 g of H 2 O. Calculate the mass percent (%m/m) of the solution. 1) 15% (m/m) Na 2 CO 3 2) 6.4% (m/m) Na 2 CO 3 3) 6.0% (m/m) Na 2 CO 3 Learning Check

14. 3) 6.0% (m/m) Na 2 CO 3 mass solute = 15 g Na 2 CO 3 mass solution= 15 g + 235 g = 250 g mass %(m/m) = 15 g Na 2 CO 3 x 100 250 g solution = 6.0% Na 2 CO 3 solution Solution