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Percentage Composition. Percent. Part. X 100%. Whole. Percent by mass =. Mass of element. X 100%. Mass of compound. Percent Composition. use chemical formula & assume 1 mole divide mass each element by formula mass (FM) of compound. Percent Composition of H 2 O.

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## Percentage Composition

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**Percent**Part X 100% Whole Percent by mass = Mass of element X 100% Mass of compound**Percent Composition**use chemical formula & assume 1 mole divide mass each element by formula mass (FM) of compound**Percent Composition of H2O**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 = 18.0g/mol % H = 2.0g x 100% = 11.11% 18.0g % O = 16.0g x 100% = 88.89% 18.0g**Percent Composition of C6H12O6**6 C’s = 6 x 12g = 72g 12 H’s = 12 x 1g = 12g 6 O’s = 6 x 16g = 96g Molar mass = 180.0 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! • 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: CuSO45H2O means “is associated with” or “included” Does NOT refer to multiplication Not true chemical bond:**What can say about CuSO45H2O?**It’s hydrated salt – hydrate = water 1 mole CuSO4 contains 5 moles water 1 molecule of CuSO4 contains 5 molecules of water When heated, water is driven off & anhydrous salt is left: CuSO4 – anhydrous = without water If had 2 moles of CuSO45H2O, how much water would you lose on heating?**CuSO45H2O**Count up the atoms! 1 Cu 1 S 4 O + 5x1 O = 9 O 5x2 H =10 H total = 21 atoms**5CuSO45H2O**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 CuSO45H2O**Just water 5 O = 5 x 16 = 80 10 H = 10 x 1 = 10 total = 90 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 90/240 x 100 = 37.5% water

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