Transcript percent comp and hydrates

It’s just like finding out your test
score!!!
97 correct out of 100 questions = 97 x 100 = 97%
100
Total of “something”
TOTAL
x 100 = %
The chemical composition can be
expressed as the mass percent of
each element in the compound.
Example: Determine the percent
composition of C3H8.

Assume you have one mole of the substance.
3mol C = 3(12.01g/mol) = 36.03 g
%C 
36.03g
 100  81 . 7 %
44.11g
8mol H = 8(1.01 g/mol) = 8.08 g
%H 
8.08g
44.11g
Total MW = 44.11 g
 100  18 . 3 %
Example: Determine the percent
composition of iron(III) sulfate.

Iron(III) sulfate = Fe2(SO4)3
2mol Fe = 2(55.85g/mol) = 111.70 g
% Fe 
 100  27.9%
399.91g
3mol S = 3(32.07g/mol) = 96.21 g
%S 
12mol O = 12(16.00g/mol) = 192.00 g
%O 
Total MW = 399.91 g
111.70g
96.21g
 100  24.1%
399.91g
192.00g
399.91g
 100  48.0%
Hydrated Compounds
Some compounds
exist in a “hydrated”
state.
Some specific number of water molecules are
present for each molecule of the compound.
Dessicants

You’ve probably noticed that some consumer
goods contain a small packet labeled “Silica
gel: Do not eat”. What’s that packet for,
anyway?

As you know, many fragile goods can be easily
damaged by moisture. The silica gel in each
packet is used to soak up water from the
atmosphere. This minimizes moisture that
causes damage during shipping.

Many ionic compounds can be used to soak up
water. Before they absorb water, they’re
referred to as “anhydrous”, which means
“without water”.

After they’ve soaked up the maximum amount
of water, they’re called “hydrates”, making that
water molecules are stuck to them


If you heat hydrates to very high
temperatures, they “dehydrate”, meaning that
the water is lost.
Once all of the water is lost, these compounds
are again referred to as “anhydrous”.


In this lab, we will be dehydrating the hydrate
of magnesium sulfate.
Using the data from this lab, you will determine
the percent composition
– and the empirical formula of the
hydrate.
–
EXAMPLES
Sodium carbonate
decahydrate
Calcium chloride dihydrate
Magnesium sulfate
heptahydrate
Iron (III)phosphate
tetrahydrate
FORMULAS





A hydrate can be
analyzed by driving off
the water with heat.
The remaining substance
is called the Anhydrous
salt… meaning “without
water”
Some hydrates are a
color different than their
anhydrous salt…
Cobalt (II) chloride
hexahydrate is pink
Without water it is blue.
Example:

A 7.0 g sample of hydrated calcium nitrate is
heated to constant mass. Find the percent
water in the hydrate and the formula for it.




4.9 grams remains after heating
This is the anhydrous salt.
How much water was driven off from the
hydrate?
7.0 g sample of hydrated calcium nitrate - 4.9 g
anhydrous salt = 2.1 grams of water
To write the formula



Covert the mass of anhydrous salt to moles.
Convert the mass of water to moles.
Write the formula for the hydrate.
MgSO4* ?H2O
1.
2.
3.
4.
5.
Heat the sample to drive off all water.
Mass the anhydrous compound.
mass of water = Mass of hydrate – mass of
anhydrous salt
Convert these masses to moles.
Calculate the mole ratio between the compound
and the water molecules.
Hydration Number
Some molecules attach themselves to water
molecules. This is done in set numbers, depending on
the molecule. For example, copper sulfate attaches to
5 water molecules. We say its hydration number is 5
CuSO45H2O
(copper (II) Sulfate Pentahydrate)
Anhydrides
A compound that is normally a hydrate and
has lost its hydration water is said to be
anhydrous and is called an anhydride.
BaCl22H2O
Barium Chloride Dihydrate
BaCl2
Barium Chloride Anhydride
-or-
Anhydrous Barium Chloride
Finding the Hydration Number
The hydration number can be conveniently
found by heating the compound and measuring its
mass loss. This mass loss is usually due to the
hydration water molecules being driven off.
For example…
A 15.35 g sample of Strontium nitrate,
Sr(NO3)2nH2O, is heated to a constant mass of
11.45 g. Calculate the hydration number.
Sample Data:
Mass Hydrate
15.35g
Mass Anhydride
11.45g
Mass of Water (mass loss)
3.90g
Calculations
Moles Anhydride
11 . 45 g Sr(NO
3
)2
1 mol Sr(NO
3
)2
211.64 g Sr(NO
3

 0 . 05410 mol Sr(NO
)2
3
)2
Moles Water
3 . 90 g H 2 O

1 mol H 2 O
18.02 g H 2 O
 0 . 216 mol H 2 O
Molar ratio…
mol H 2 O
mol Sr(NO
3

)2
0 . 216 H 2 O
0.05410 mol Sr(NO
 4 . 01  4
3
)2
Hydration Number is 4, Sr(NO3)24H2O
(Strontium Nitrate Tetrahydrate)