Percentage Composition. Percent PartWhole X 100% Mass of element Mass of compound X 100% X 100% Percent by mass =

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Presentation transcript:

Percentage Composition

Percent PartWhole X 100% Mass of element Mass of compound X 100% X 100% Percent by mass =

Percent Composition use chemical formula & assume 1 mole use chemical formula & assume 1 mole divide mass each element by formula mass (FM) of compound 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 g = 18.0g/mol % H = 2.0g x 100% = 11.11% 18.0g 18.0g % O = 16.0g x 100% = 88.89% 18.0g 18.0g

Percent Composition of C 6 H 12 O 6 6 C’s = 6 x 12g = 72g 12 H’s = 12 x 1g = 12g 6 O’s = 6 x 16g = 96g Molar mass = grams/mole % C = % H = % O = (72g/180g) X 100% = 40.00% (12g/180g) X 100% = 6.67% (96g/180g) X 100% = 53.33% 100%

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

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

What can say about CuSO 4  5H 2 O? It’s hydrated salt – hydrate = water It’s hydrated salt – hydrate = water 1 mole CuSO 4 contains 5 moles water1 mole CuSO 4 contains 5 moles water 1 molecule of CuSO 4 contains 5 molecules of water1 molecule of CuSO 4 contains 5 molecules of water When heated, water is driven off & anhydrous salt is left: CuSO 4 – anhydrous = without water When heated, water is driven off & anhydrous salt is left: CuSO 4 – anhydrous = without water If had 2 moles of CuSO 4 5H 2 O, how much water would you lose on heating? If had 2 moles of CuSO 4 5H 2 O, how much water would you lose on heating?

CuSO 4  5H 2 O Count up the atoms! Count up the atoms! 1 Cu 1 S 4 O + 5x1 O = 9 O 5x2 H =10 H total = 21 atoms

5CuSO 4  5H 2 O Count up the atoms! Count up the atoms! 5 = Cu 5 = S 5x4 O + 5x5x1 O = 45 O 5x5x2 H = 50 H total = 105 atoms

FIND THE PERCENT WATER CuSO 4  5H 2 O FIND THE PERCENT WATER CuSO 4  5H 2 O 1 Cu = 1 x 64 = 64 1 S = 1 x 32 = 32 9 O = 9 x 16 = H = 10 x 1 = 10 total = 240 Just water 5 O = 5 x 16 = H = 10 x 1 = 10 total = 90 90/240 x 100 = 37.5% water