1 Warm-up Determine the molar mass of the following compounds: Magnesium sulfide Potassium fluoride Carbon tetrabromide Dinitrogen hexachloride

2 Warm-up How many moles are in… 15 grams of water? 500 molecules of sodium chloride?

3 Unit Five, Day Four Kimrey 15 October 2012

4 The “Mole Hill”

5 Mole Conversions Remember the conversions we’ve done!!!! Start with the given Write what you want over what you have. Solve Ex. How many moles of water are there in 25.0 grams? 14.2 moles of H 2 O = ? Molecules

6 2 Step Conversions. Notice on the mole hill there is no way to get to grams from particles directly. You must go through moles! Step 1 convert what you have to moles Step 2 convert moles to what you want.

7 Example 5o grams of H 2 O = ? Molecules 1.6 x 10 40 molecules of MgCl 2 = ? Grams

8 Practice How many grams of NaCl are in 45.3 moles? How many atoms of Mg in 35.2 grams? How many moles are in 50.0 grams of oxygen?

9 Percent Composition The percent by mass of each element in a compound is called the percent composition. Formula: (Mass of one element/Molar mass of compound) x 100

10 Steps 1. Find the formula and formula mass for the compound. 2. Determine the masses of each element in the compound. 3. Plug into the formula and solve. 4. Check yourself!!! The Percentages should add to 100%

11 Example Bicarbonate of soda (sodium hydrogen carbonate) is used in many commercial preparations. Its formula is NaHCO 3. Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate. First, look up the atomic masses for the elements. The atomic masses are found to be: Na 22.99 H 1.01 C 12.01 O 16.00

12 Next, determine how many grams of each element are present in one mole of NaHCO 3 : 22.99 g (1 mol) of Na 1.01 g (1 mol) of H 12.01 g (1 mol) of C 48.00 g (3 mole x 16.00 gram per mole) of O The mass of one mole of NaHCO 3 is: 22.99 g + 1.01 g + 12.01 g + 48.00 g = 84.01 g

13 So, the mass percentages of the elements are mass % Na = 22.99 g / 84.01 g x 100 = 27.36 % mass % H = 1.01 g / 84.01 g x 100 = 1.20 % mass % C = 12.01 g / 84.01 g x 100 = 14.30 % mass % O = 48.00 g / 84.01 g x 100 = 57.14 %