Chapter 8Percentage Composition

Water is made of hydrogen and oxygen. How many grams of each are in the sample of water below?

Percent Composition

Mass of elementMass of elementFormula Mass of compoundFormula Mass of compound X 100X 100% =% =

Percent composition is the percent by mass of an element within a compound.

• Divide the mass of each element by the formula mass of the compound

Percent Composition of H2O2 X H = 2 X 1.0 g = 2.0 g1 X O = 1 X 16.0 g = 16.0 g

gram formula mass = 2.0 g + 16.0 g = 18.0 g

% H = 2.0g X 100 = 11.2% 18.0g

% O = 16.0g X 100 = 88.8% 18.0g

Percent Composition of H2O2 X H = 2 X 1.0 g = 2.0 g1 X O = 1 X 16.0 g = 16.0 g

gram formula mass = 2.0 g + 16.0 g = 18.0 g

% H = 2.0g X 100 = 11.2% 18.0g

% O = 16.0g X 100 = 88.8% 18.0g

What does this mean?

Pure water’s mass is 11.2% hydrogen and 88.8% oxygen.

(0.112)(128.98g) = 14.45g H

(.888)(128.98g) = 114.53g O

or

128.98g – 14.45g = 114.53g O

11.2% H

88.8% O

Percentage Composition

• Remember that you have to calculate the formula mass!

Determine the Percent Composition of C6H12O6

Determine the Percent Composition of C6H12O6

6 X C = 6 X 12.0 g = 72.0g12 X H = 12 X 1.0g = 12.0g6 X O = 6 X 16.0g = 96.0ggfm = 180.0 grams/mole

Determine the Percent Composition of C6H12O6

6 X C = 6 X 12.0 g = 72.0g12 X H = 12 X 1.0g = 12.0g6 X O = 6 X 16.0g = 96.0ggfm = 180.0 grams/mole

% C =% H = % O =

(72.0g/180.0g) X 100 =(72.0g/180.0g) X 100 = 40.0%40.0%(12.0g/180.0g) X 100 =(12.0g/180.0g) X 100 = 6.67%6.67%(96.0g/180.0g) X 100 =(96.0g/180.0g) X 100 = 53.3%53.3%

How many grams of carbon are in 156g C6H12O6

6 (156g)(0.400) = 62.4g carbon126 Molar

% C =% H = % O =

(72.0g/180.0g) X 100 =(72.0g/180.0g) X 100 = 40.0%40.0%(12.0g/180.0g) X 100 =(12.0g/180.0g) X 100 = 6.67%6.67%(96.0g/180.0g) X 100 =(96.0g/180.0g) X 100 = 53.3%53.3%

What is this?

a

Iron Pyrite “Fool’s Gold” (FeS2)

a

How many grams of sulfur are in this 28.8g sample of iron pyrite FeS2?

a

How many grams of sulfur are in this 28.8g sample of iron pyrite FeS2?

a

15.4g S

How many grams of sulfur are in this 28.8g sample of iron pyrite FeS2?

a

15.4g S

Fe + 2(S)

55.8 + 2( 32.1) = 120.0 g/mol

%S = 64.2/120.0 x 100= 53.5%S

28.8g x 0.535 = 15.4g S

Homework

Molarity Worksheet (Due Tomorrow). Percentage Composition Worksheet (Due

in 2 Days).