1. Chp 6 The Mole

2. Counting by Weighing  Every item has a certain average mass.  We can use that mass to calculate how much we would expect 50, 1000 or 1000000 of that item to weigh.  Then we could weigh out that amount rather than count out 1000000 of that item. –This is how we can count atoms. –The atomic mass is the amount that 1 mole of that element weighs.

3. The Mole  Just like the term “a dozen” represents the number 12, a mole represents the number 6.023 x 10 23. –That’s 602,300,000,000,000,000,000,000!!! –A number that big is hard to comprehend so a few facts:  If you distributed a mole of pennies between everyone alive on Earth, we’d each get a trillion $.  If you stacked a mole of paper, it would reach the moon and back 80 billion times.  A mole of marshmallows would completely cover the Earth in a pile 12 miles deep.

4. The Practical Application  So 12.01 g of carbon contains 6.023 x 10 23 atoms of carbon.  Therefore 24.02 g of carbon contains 1.2046 x 10 24 atoms (2 times 6.023 x 10 23 ).  We can use this in dimensional analysis to figure out fractions of this: –Atomic mass = 1 mole –1 mole = 6.023 x 10 23 atoms

5. An Example  How many moles are in 0.5 g of hydrogen? –1 mole H = 1.007 g  How many atoms is this? –1 mole = 6.023 x 10 23 atoms 0.5 g H 1.007 g H 1 mole H = 0.496 mol H 0.496 mol H = 2.99 x 10 23 atoms H 6.023 x 10 23 atoms H 1 mol H

6. What About Compounds?  We can do the same thing for a compound by first adding the individual masses to get the molar mass of the compound. –Ex. CH 4 C = 12.01 g/mol H = 1.007 g/mol x 4 So CH 4 = 16.04 g/mol –How many moles are in 0.5 g of CH 4 ? 0.5 g CH 4 16.04 g CH 4 1 mol CH 4 = 0.03 mol CH 4

7. Mass Percent  How much of a particular element makes up a compound.  Mass % = mass element x 100 molar mass of compound molar mass of compound  Ex. What is the mass % of C in CH 4 ? 12.01 x 100 = 74.9% 16.04 What about H? What about H? 4.028 x 100 = 25.1% 16.04 ** Should add to 100%

8. Empirical Formulas  The simplest, whole number ratio of atoms in a compound.  Rules: 1.Get the mass of each element present. 2.Convert grams to moles. 3.Divide all answers by the smallest number from step 2. 4.Multiply by an integer, if necessary, to get whole numbers. 5.Those numbers are the subscripts of the empirical formula.

9. An Example What is the empirical formula of a compound containing 4.151 g Al and 3.692 g O? 1.Masses are listed above for us. 2.4.151 g Al1 mol Al 26.98 g Al 3.692 g O1 mol O 15.99 g O = 0.1539 mol Al = 0.2308 mol O

10. An Example 3.0.1539 is smallest so: 0.1539 = 1 mol Al 0.1539 0.2308 = 1.5 mol O 0.1539 4.Since we don’t have whole #s, we need to multiply by an integer. Let’s try 2. 1 mol Al x 2 = 2 mol Al 1.5 mol O x 2 = 3 mol O 5.Now we have whole numbers, so the empirical formula is Al 2 O 3.

11. Molecular Formulas  The actual ratio of atoms in a compound.  May be the same as the empirical or a multiple of it. –Ex. CH 4 is an empirical formula CH 4, C 2 H 8, C 3 H 12 could all be molecular formulas based on that E.F.  Found by dividing the actual molar mass by the molar mass of the empirical compound. If the answer isn’t 1, you multiply the E.F. subscripts by that number.

12. An Example  What is the molecular formula of a compound with a molar mass of 283.33 g/mol if the empirical formula is P 2 O 5 ? The mass of P 2 O 5 is: 2 x 30.97 = 61.94 g 5 x 15.99 = 79.95 g 141.89 g/ mol 141.89 g/ mol

13. An Example So, now we divide the real M.M. by the M.M. of the empirical formula. 283.33 g = 2 141.89 g Now we multiple the subscripts of the empirical formula by that 2. 2 (P 2 O 5 ) = P 4 O 10 is our answer!