Topic: Percentage Composition

Do Now:

Percent Composition

use chemical formula & assume 1 mole

Calculate formula mass then divide mass each element by formula mass (FM) of compound

Percent Composition of H2O2 H’s = 2 x 1.0 g = 2.0 g1 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 = 72g12 H’s = 12 x 1g = 12g6 O’s = 6 x 16g = 96gMolar 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 = contain water 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:

When heated, water is driven off = anhydrate

What can say about CuSO45H2O? 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

mole ratio in formula can be used to predict how much water found in any size sample

If had 2 moles of CuSO45H2O, how much water would you lose on heating? 10 moles

CuSO45H2O

REVIEW: Counting atoms!

1 Cu1 S4 O + 5x1 O = 9 O5x2 H =10 H

total = 21 atoms

5CuSO45H2O

Count up the atoms!

5 = Cu5 = S5x4 O + 5x5x1 O = 45 O5x5x2 H = 50 H

total = 105 atoms

FIND THE PERCENT WATER CuSO45H2O

1 Cu = 1 x 64 = 641 S = 1 x 32 = 329 O = 9 x 16 = 14410 H = 10 x 1 = 10

total = 240g

Just water5 O = 5 x 16 = 8010 H = 10 x 1 = 10

total = 90g

90/240 x 100 = 37.5% water