1. AP Chemistry- Review 2

2.  Many times in chemistry is more useful to represent a compound by its percent make- up.  Example: Fe 2 O 3 also known as Iron III Oxide Molar mass is: 2 (Fe) + 3 (O)  2 (56) + 3 (16) = 160g % of Iron would be (112/160) x 100 = 70.00% % of oxygen would be (48/160) x 100 = 30.00%

3. Another example in which % formula is useful is found when we deal with hydrates. Hydrates are ionic compounds that are found as crystals and contain water. Example: Copper II Nitrate Pentahydrate Cu +2 NO 3 -1 So formula becomes; Cu(NO 3 ) 2 5H 2 O So what is the percent of water in this hydrate? Molar Mass 1 (Cu) + 2 (N) + 6 (O) + 5 (H 2 O) 1 (64) + 2 (14) + 6 (16) + 5 (18) = 278g % of water= (90/278) x100 = 32.37%

4.  An anhydrate would be the ionic compound that is left after you have evaporated the water from a hydrate. Obviously this only works if the boiling point of the ionic compound is HIGHER than the boiling point of water? Can you explain why?  One very common lab is to identify the moles of water in a hydrate.  We want to find the moles of water in a hydrate. MgCO 3 X H 2 O  Mass of hydrate + beaker10.80 g +/- 0.01g  Mass of beaker5.00 g +/- 0.01g  Mass after first heating9.55g +/- 0.01g  Mass after second heating9.21 g +/- 0.01g  Mass after third heating9.20g +/- 0.01g

5.  Mass of the hydrate = 10.80 – 5.00 = 5.80 g +/- 0.02 g  Mass of anhydrate = 9.20- 5.00 = 4.20g +/- 0.02 g  Mass of water = 5.80- 4.20 = 1.60 g +/- 0.04 g anhydrate: MgCO 3 Molar mass of anhydrate: 1( Mg) + 1(C ) + 3 (O)= 1(24) + 1(12) + 3 (16)= 84g Moles of anhydrate: mass of anhydrate/molar mass of anhydrate 4.20/ 84 = 0.0500 moles +/-.48% Water: H 2 O Molar mass of water = 2(H) + 1(O) = 2 (1) + 1(16) = 18g Moles of water= mass of water/molar mass of water= 1.60/18=.0889 moles +/- 3%

6.  Finally we divide the moles of water by the moles of anhydrate:  0.0889/0.0500 = 1.78 moles +/3.48% or 1.78 moles +/- 3%  So our hydrate was MgCO 3 1.78 H 2 O +/- 3%  Now in reality the number in front of the water should be a whole number, so most likely our hydrate was MgCO 3 2H 2 O  So we can do a % error [( 2-1.78)/ 2] x100= 11% error  Once again we can compare our percent error 11% and our uncertainty 3% and discuss what type of error we had and where did it most likely come from.

7.  Empirical formula has the lowest possible ratio between the atoms in a compound.  Example: Mg 2 O 2 is NOT an empirical formula because we can reduce it to:  MgO  ALL ionic compounds are written as empirical formulas. Lowest ratio.  Covalent formulas because they are not reduced are NOT written as empirical formulas:  Example:  Benzene: C 6 H 6 and Ethyne C 2 H 2 both have the same empirical formula;  CH

8.  We start by being given either grams or % of each element in the formula  Example  C= 27.27% O= 72.73%  First we get rid of %, think of them as grams and divide by mass of each element to fond moles of each:  C = 27.27/12 = 2.27molesO= 72.73/16 = 4.55 moles  Secondly divide all the answers in first step by lowest number of moles;  C = 2.27/2.27 = 1 O = 4.55/2.27 = 2  So Empirical Formula is CO 2

9.  Sometimes the problems are a bit more difficult, and after dividing by the lowest number of moles we don’t get whole numbers.  Example:  C = 1.80gH= 0.25gO= 1.60g  C = 1.80/ 12 = 0.15molesH= 0.25/1= 0.25molesO=1.60/16= 0.10  C= 0.15/0.10= 1.5H= 0.25/0.10= 2.5O=.10/.10=1  Since the C is not a whole # answer we must multiply by 2 to make it a whole #. What we do to one we do to all of them, so;  C=3H=5O=2 so Empirical Formula is: C 3 H 5 O 2

10.  When you are thinking what to multiply by, remember this hint  Ending of.5 multiply by 2  Ending of.33 multiply by 3  Ending of.25 multiply by 4  Ending of.20 multiply by 5  Those above are the most common multiplication factors:  Remember the rules  #1 Divide by atomic masses  #2 Divide by smallest mole answer from step 1  #3 If all are not close enough to whole numbers, multiply by factors shown above.

11.  Remember that MOLECULAR formulas are only used for covalent compounds. They exist as molecules and we do not reduce them to empirical formulas  Example  C=9.23gH= 0.77g Molecular mass is 78g  Step #1  C= 9.23/12= 0.77 molesH= 0.77/1= 0.77  Step #2  C= 0.77/0.77= 1H= 0.77/0.77= 1 Empirical Formula: CH  Empirical Formula Mass: 1(C ) + 1 (H) = 1(12) + 1(1)= 13g  Molecular mass/Empirical Mass= 78/13= 6  So Molecular Mass is C 6 H 6