Percent Composition. The chemical composition can be expressed as the mass percent of each element in the compound. Carry out % mass to the tenths place. Example: Determine the percent composition of C 3 H 8. Assume you have one mole of the substance. 3 C = 3(12.0) = 36.0 g
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Carry out % mass to the tenths place.
3 C = 3(12.0) = 36.0 g
8 H = 8(1.0) = 8.0 g
Total MW = 44.0 g
2 Fe = 2(55.8) = 111.6 g
3 S = 3(32.1) = 96.3 g
12 O = 12(16.0) = 192.0 g
Total MW = 399.9 g
Some compounds exist in a “hydrated” state.
Some specific number of water molecules are present for each molecule of the compound.
Note: the dot in (COOH)2•2H2O shows that the crystals of oxalic acid contain 2 water molecules per (COOH)2 molecule.
Some molecules attach themselves to water molecules. This is done in set numbers, depending on the molecule. For example, Magnesium sulfate attaches to 7 water molecules. We say it’s hydration number is 7
(Magnesium Sulfate Heptahydrate)
A compound that is normally a hydrate and has lost its hydration water is said to be anhydrous and is called an anhydride.
BaCl22H2O Barium Chloride Dihydrate
BaCl2 Barium Chloride Anhydride
Anhydrous Barium Chloride
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.
A 15.35 g sample of Strontium nitrate, Sr(NO3)2nH2O, is heated to a constant mass of 11.45 g. Calculate the hydration number.
Sample Data: sulfate.
Mass Hydrate 15.35g
Mass Anhydride 11.45g
Mass of Water (mass loss) 3.90g
Hydration Number is 4, Sr(NO3)24H2O
(Strontium Nitrate Tetrahydrate)