Percent Composition

It’s just like finding out your test score!!! 97 correct out of 100 questions = 97 x 100 = 97% 100 Total of “something” x 100 = % TOTAL

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 C3H8. Assume you have one mole of the substance. 3 C = 3(12.01) = 36.03 g 8 H = 8(1.01) = 8.08 g Total MW = 44.11 g Carry out % mass to the tenths place.

Example: Determine the percent composition of iron(III) sulfate. iron(III) sulfate = Fe2(SO4)3 2 Fe = 2(55.85) = 111.70 g 3 S = 3(32.07) = 96.21 g 12 O = 12(16.00) = 192.00 g Total MW = 399.91 g 111.70 g Fe 399.91 g Fe2(SO4)3 ×100=27.9 % Fe 96.21 g S 399.91 g Fe2(SO4)3 ×100=24.1 % S 192.00 g O 399.91 g Fe2(SO4)3 ×100=48.0 % O Carry out % mass to the tenths place.

Some compounds exist in a “hydrated” state. Hydrated Compounds Some compounds exist in a “hydrated” state. Some specific number of water molecules are present for each molecule of the compound.

the dot does not mean to multiply! Example: oxalic acid (COOH)2 can be obtained in the laboratory as (COOH)2•2H2O. note: the dot in (COOH)2•2H2O shows that the crystals of oxalic acid contain 2 water molecules per (COOH)2 molecule. the dot does not mean to multiply!

Example: oxalic acid (COOH)2 can be obtained in the laboratory as (COOH)2•2H2O. The molar mass of (COOH)2 = 90.04 g/mol The molar mass of (COOH)2•2H2O = 126.08 g/mol Water can be driven out of a hydrated compound by heating it to leave an “anhydrous” (without water) compound.

Example: A 7.0 g sample of calcium nitrate, Ca(NO3) 2•4H2O, is heated to constant mass. How much anhydrous salt remains? % mass anhydrous salt = Mass of anhydrous salt remaining =

(magnesium sulfate heptahydrate) Hydration Number 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 MgSO47H2O (magnesium sulfate heptahydrate)

Anhydrides BaCl22H2O barium chloride dihydrate 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 a hydrate 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 Moles Water Molar ratio… Hydration Number is 4, Sr(NO3)24H2O strontium nitrate tetrahydrate