# Percentage Composition Percent Part Whole X 100 Percent

- Slides: 11

Percentage Composition

Percent Part Whole X 100% Percent by mass = Mass of element Mass of compound X 100%

Percent Composition use chemical formula & assume 1 mole divide mass each element by formula mass (FM) of compound

Percent Composition of H 2 O 2 H’s = 2 x 1. 0 g = 2. 0 g 1 O = 1 x 16. 0 g = 16. 0 g Molar mass = 2. 0 g + 16. 0 g = 18. 0 g/mol % H = 2. 0 g x 100% = 11. 11% 18. 0 g % O = 16. 0 g x 100% = 88. 89% 18. 0 g

Percent Composition of C 6 H 12 O 6 6 C’s = 6 x 12 g = 72 g 12 H’s = 12 x 1 g = 12 g 6 O’s = 6 x 16 g = 96 g Molar mass = 180. 0 grams/mole % % % C = (72 g/180 g) X 100% = 40. 00% H = (12 g/180 g) X 100% = 6. 67% O = (96 g/180 g) X 100% = 53. 33% 100%

Percentage Composition Remember to calculate FM! • Nowhere in word problem will it tell you that! • Sum of individual element %’s must add up to exactly 100%

Hydrates group salts that have water molecules stuffed in their empty spaces Formulas are distinctive Ex: Cu. SO 4 5 H 2 O means “is associated with” or “included” Does NOT refer to multiplication Not true chemical bond:

What can say about Cu. SO 4 5 H 2 O? It’s hydrated salt – hydrate = water • 1 mole Cu. SO 4 contains 5 moles water • 1 molecule of Cu. SO 4 contains 5 molecules of water When heated, water is driven off & anhydrous salt is left: Cu. SO 4 – anhydrous = without water If had 2 moles of Cu. SO 4 5 H 2 O, how much water would you lose on heating?

Cu. SO 4 5 H 2 O Count up the atoms! 1 Cu 1 S 4 O + 5 x 1 O = 9 O 5 x 2 H =10 H total = 21 atoms

5 Cu. SO 4 5 H 2 O Count up the atoms! 5 = Cu 5=S 5 x 4 O + 5 x 5 x 1 O = 45 O 5 x 5 x 2 H = 50 H total = 105 atoms

FIND THE PERCENT WATER Cu. SO 4 5 H 2 O 1 Cu = 1 x 64 = 64 1 S = 1 x 32 = 32 9 O = 9 x 16 = 144 10 H = 10 x 1 = 10 total = 240 Just water 5 O = 5 x 16 = 80 10 H = 10 x 1 = 10 total = 90 90/240 x 100 = 37. 5% water

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