1. Molar Mass & Molarity

2. Molar Mass Mass in grams of one mole of an element or compound. Numerically equal to the atomic weight of the element or the sum of all the atomic weights in the formula. Round all atomic masses to keep 2 decimals.

3. Molar Mass Examples NaCl = 22.99 + 35.45 = 58.44g/mole CuSO 4. 5H 2 O (this is a hydrate; it means there are 5 water molecules added for each copper sulfate molecule) = 63.55 + 32.07 + 4(16.00) + 5(18.02) = 159.62 + 90.10 = 249.72 g/mole

4. Practice 1) KCl 2) Li 2 SO 4 3) (NH 4 ) 2 C 2 O 4. H 2 O

5. Answers 1) KCl= 39.10 + 35.45 = 74.55g/mole 2) Li 2 SO 4 = 2(6.94) + 32.07 + 4(16.00) = 109.95g/mole 3) (NH 4 ) 2 C 2 O 4. H 2 O = 2(14.01)+8(1.01)+2(12.01)+4(16.00)+18.02 =142.14g/mole

6. Practice Again 4)potassium hydroxide 5) copper (II) bromide 6) magnesium phosphate

7. Answers 4) KOH = 39.10 + 16.00 + 1.01 = 56.11g/mole 5) CuBr 2 = 63.55 + 2(79.90) = 383.15g/mole 6) Mg 3 (PO 4 ) 2 = 3(24.31)+ 2(30.97) + 8(16.00) = 262.87g/mole

8. Molarity, M Moles of solute (compound) per liter of solution. Moles divided by volume (in liters). –You will have to convert mL to L.

9. 1.5M HCl This means there are 1.5 moles of HCl in each liter of HCl How many grams of HCl are in 0.45 liters of 1.5M HCl? –You must first calculate the number of moles in that volume. –Then multiply the moles by the molar mass of HCl

10. 0.45L of 1.5 M HCl 1.5moles/L x 0.45L =0.675moles; round for sig. figs.=0.68moles (notice the L cancels out) Next, multiply the moles by the molar mass of HCl 0.68moles x 36.46g/mole= 24.7928g =24g (notice the moles cancel out)

11. 0.500M NaOH How many grams of NaOH are in 25.0mL of this solution? First, convert the volume to L 25.0mL /1000= 0.0250mL (keep all sig. figs. from original number) Next multiply the molarity by the volume 0.500mol/L x 0.0250L= 0.0125moles Finally, multiply the moles by the molar mass 0.0125moles x 40.00g/mol= 0.500g