Percent Composition
Empirical Formula
Naming
Chapter 8
1

• An ionic compound that contains water inside its crystal structure.
✓ These compounds do not look or feel wet.
✓ The water molecules are loosely attached and can be removed by heating.
• Generic Formula
✓ M x
Nm y
• n H
2
O
✓ M x
Nm y
• n H
2
O hydrate
✓ M x
Nm y
• n H
2
O anhydrate
✓ M x
Nm y
• n H
2
O water
2

• The water molecules are loosely attached and can be removed by heating.
• M x
Nm y
• n H
2
O → M x
Nm y
+ n H
2
O
✓ The remaining anhydrate (anhydrous crystals) may look the same, may have a different texture or may even be a different color
✓ M x
– Metal Ion; Nm y
– Nonmetal Ion; n = prefix #
3

Law of Constant Composition
•
•
•
• Determining the mass ratio of unknown “SG”
SG was a hydrate, G was the water S was the anhydrate Per 5
SG 14.505
Look at this Lab Data
S 7.773
G (water) 6.732
SG was actually
% water 46.4
✓ Potassium aluminum sulfate dodecahydrate
✓ KAl(SO
4
)
2
• 12H
2
O
✓
‣
‣
Calculate the theoretical % of water in this compound water
12(18.02) [216.2] x100 =
39.1+27+2(32.07)+8(16) + 12(18) [474.4] hydrate
‣ 45.6% water 4

•
•
•
•
•
•
•
Measure the mass of the hydrate
✓ 8.986 g hydrate to start with
Heat the compound to drive off the water
Measure the mass of the remaining anhydrate
✓ 5.746 g anhydrate
Calculate the mass of water removed, H
2
O
✓ 8.986 - 5.746 = 3.243 g water removed
Calculate moles of water
✓ 3.243 * 1mol/18g = 0.180 mol H
2
O
Calculate moles of anhydrate, formula
✓ 5.764 * 1mol/159.6 = 0.036 mol CuSO
4
Determine the mole ratio water / anhydrate
✓
✓
0.180 / 0.036 = 4.999/1
Thus CuSO
4
• 5 H
2
O
5

•
•
•
•
•
•
•
✓
Lets test barium chloride hydrate
Measure the mass of the hydrate
2.11 g hydrate to start with
Heat the compound to drive off the water
Measure the mass of the remaining anhydrate
✓ 1.78 g anhydrate
Calculate the mass of water removed, H
2
O
✓ 2.11 – 1.78 = 0.33 g water removed
Calculate moles of water
✓ 0.33 * 1mol/18g = 0.0183 mol H
2
O
Calculate moles of anhydrate, formula
✓ 1.78 * 1mol/208.2 = 0.00855 mol BaCl
2
Determine the mole ratio water / anhydrate
✓
✓
0.0183mol H
2
O/ 0.00855 = 2.1
Thus BaCl
2
• 2 H
2
O
6

A 344 gram sample of hydrated calcium sulfate and the sample is heated to evaporate the water. The dry sample of calcium sulfate has a mass of 272 grams. What is the mole ratio between the calcium sulfate, CaSO the hydrate?
4 and water, H
2
O? What is the formula of
Step 1: Calculate the difference between the hydrated sample and the dry sample.
 344 grams (hydrate) - 272 grams (dry sample) = 72.0 grams of water
Step 2: Convert mass to moles for each sample.
 272g CaSO
4
(1 mol/136.14g) = 1.9979 mol CaSO
4
 72.0g H
2
O(1mol/18.02g) = 3.996 mol H
2
O
Step 3: Divide by lowest molar value
 1.99/1.99 = 1 mol CaSO
4
 3.99/1.99 = 2 mol H
2
O
CaSO
4
• 2 H
2
O
7

A 87.2 gram sample of a hydrate of MgI
2 was heated thoroughly in a porcelain crucible, until its weight remained constant. After heating, 57.4 grams of the anhydrous compound remained. What is the formula of the hydrate?
Step 1: Calculate the difference between the hydrated sample and the dry sample.
 87.2 grams (hydrate) – 57.4 grams (dry sample) = 29.8 grams of water
Step 2: Convert mass to moles for each sample.
 57.4g MgI
2
(1 mol/278.11g) = 0.107 mol MgI
2
 29.8g H
2
O(1mol/18.02g) = 3.996 mol H
2
O
Step 3: Divide by lowest molar value
 1.99/1.99 = 1 mol MgI
2
 3.99/1.99 = 2 mol H
2
O
MgI2 • 2 H
2
O
8

A hydrate was analyzed and determined to be 34% water.
Further the anhydrate was analyzed to be 23.96% nickel, 17.16% nitrogen, and 58.88% oxygen. Determine the formula for this hydrate
9

A hydrate was analyzed and determined to be 34% water.
Further the anhydrate was analyzed to be 23.96% nickel,
17.16% nitrogen, and 58.88% oxygen. Determine the formula for this hydrate
10