AP Chemistry- Review 2
PERCENT FORMULAS, EMPIRICAL AND
MOLECULAR FORMULAS
PERCENT FORMULAS
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Many times in chemistry is more useful to represent a compound by its
percent make- up.
Example: Fe2O3 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%
HYDRATES
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 NO3-1 So formula becomes; Cu(NO3)2 5H2O
So what is the percent of water in this hydrate?
Molar Mass
1 (Cu) + 2 (N) + 6 (O) + 5 (H2O)
1 (64) + 2 (14) + 6 (16) + 5 (18) = 278g
% of water= (90/278) x100 = 32.37%
ANHYDRATE
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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. MgCO3 X H2O
Mass of hydrate + beaker
Mass of beaker
Mass after first heating
Mass after second heating
Mass after third heating
10.80 g +/- 0.01g
5.00 g +/- 0.01g
9.55g +/- 0.01g
9.21 g +/- 0.01g
9.20g +/- 0.01g
LAB CALCULATIONS
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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: MgCO3
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: H2O
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%
RESULTS OF LAB
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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%
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So our hydrate was MgCO3 1.78 H2O +/- 3%
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Now in reality the number in front of the water should be a whole number, so
most likely our hydrate was MgCO3 2H2O
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So we can do a % error [( 2-1.78)/ 2] x100= 11% error
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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.
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EMPIRICAL FORMULA
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Empirical formula has the lowest possible ratio between the atoms in a
compound.
Example: Mg2O2 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: C6H6 and Ethyne C2H2 both have the same empirical formula;
CH
FINDING AN EMPIRICAL FORMULA
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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 = 2.27 moles O= 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
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So Empirical Formula is CO2
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EMPIRICAL FORMULA
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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 = 1.80g
H= 0.25g
O= 1.60g
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C = 1.80/ 12 = 0.15moles
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C= 0.15/0.10= 1.5 H= 0.25/0.10= 2.5
O= .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;
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C=3 H=5
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H= 0.25/1= 0.25moles O=1.60/16= 0.10
O=2 so Empirical Formula is: C3H5O2
EMPIRICAL FORMULA
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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
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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.
MOLECULAR FORMULAS
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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=9.23g
H= 0.77g
Molecular mass is 78g
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Step #1
C= 9.23/12= 0.77 moles
H= 0.77/1= 0.77
Step #2
C= 0.77/0.77= 1
H= 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
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So Molecular Mass is C6H6
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