% composition and hydrates
What percent of the boxes are orange? Blue? Purple? Yellow?
Percentages
• Percentage = part / whole
• %=p
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1st Qtr
2nd Qtr
3rd Qtr
4th Qtr
w
Think of this like an algebraic equation and you can solve
for %, parts, and wholes!
KMnO4
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K= 39.10g
Mn=54.94g
O= 16.00 (4)g
Total Mass= 158.04 g
% K= 39.10g/158.04g = 24.74 %
%Mn=54.94/158.04 =34.76%
%O=4(16.00)/158.04 = 40.50%
HCl
• H= 1.01 g
%H= 2.77 %
• Cl= 35.45 g
%H= 97.23 %
• Total Mass= 36.46 g
Mg(NO3)2
• Mg= 24.31 g x 1
• N= 14.01 g x 2
• O= 16.00 g x 6
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Total Mass=148.33 g
Mg= 24.31 / 148.33 g = 16.39%
N= 14.01(2) / 148.33 g =18.89 %
O = 16.00(6) / 148.33 = 64.72 %
(NH4)3PO4
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N= 14.01 g x 3
H= 1.01 g x 12
P= 30.97 g x 1
O= 16.00 g x 4
Total Mass= 149.12 g
%N = 14.01(3) / 149.12 g = 28.19%
%H= 1.01(12) / 149.12 g =8.13%
%P=30.97(1) / 149.12 g =20.77 %
%O=16.00 (4) / 149.12 g = 42.92 %
Al2(SO4)3
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Al= 26.98g x 2 =
S= 32.07g x 3 =
O= 16.00g x 12=
Total Mass= 342.17 g
%Al=
26.98g x 2 /342.17 g =15.77 %
%S=
32.07g x 3 / 342.17 g = 28.12 %
%O=
16.00g x 12 / 342.17 g = 56.11 %
% Water in a hydrate
• How is a hydrate different from other
compounds?
Hydrate  anhydrate + H2O
• An anhydrate is the chemical component of a
hydrate that does not have the water.
• A hydrate is a chemical compound that contains
water molecules bound to its structure.
• Dehydration- a term used to describe a reaction
that loses water.
Naming
• FeCl3-6H2O
– Iron (III) Chloride Hexahydrate
• CuSO4-5H2O
– Copper II Pentahydrate
• Barium Chloride Dihydrate
– BaCl2-2H2O
• Manesium Sulfate Heptahydrate
– MgSO4-7H2O
Copper II Sulfate Pentahydrate
Hydrate Lab Demonstration
• http://www.youtube.com/watch?v=5fZlLEBAf
EU
Naming Practice
• FeCl3 6H2O
• CuSO4 5H2O
Answer: iron (III) chloride hexahydrate
Answer: copper II sulfate
pentahydrate
• Barium Chloride Dihydrate
Answer: BaCl2 2H2O
• Magnesium Sulfate Heptahydrate
Answer: MgSO4 7H2O
Percent water in a hydrate
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MgSO4 7H2O
Mg= 24.31
S= 32.07
O=4 (16.00)
H2O = 7(18.02) % = 7(18.02)g/246.52g
• 51.17%
Practice
• If 125 g of MgSO4 7H2O is completely
dehydrated, how many grams of anhydrous
magnesium sulfate will remain?
– Hint: how many grams of water would there have
been driven off?
– Hint: what percent of the whole compound is
water?
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63.97 g of water, so 61.04 g of anhydrous salt will
remain.