Percent Composition 10.3. Percent Composition I think you already know how to do this? What did Bob score on his test? How do you figure it out? 87

Percent Composition

10.3

Percent Composition

I think you already know how to do this?

What did Bob score on his test?

How do you figure it out? 87 / 100

Bob

What did Bob score on his test?

How do you figure it out?

Part X 100 = 87 Whole

87 / 100 Bob

What did Sue score on her paper?Sue

174 / 205

What did Sue score on her paper?

Part x 100 = 83 Whole

.8341 x 100 = 83

Sue

171 / 205

Percent Composition

= the relative amount of elements in a compound or substance.

The percent by mass of an element in a compound is the number of grams of the element divided by the mass in grams of the compound, multiplied by 100%.

% mass of element = mass of element x 100% mass of compound

Sample Problem 10.9 on pg. 306

When 13.60 gram sample of a compound containing only oxygen and magnesium is decomposed, 5.40 g of oxygen is obtained. What is the percent composition of this compound? Knowns Unknowns Mass of compound= % Mg = Mass of oxygen= % O =Mass of magnezium=

Analyze Calculate Evaluate

% Composition from Chemical Formula

What is the percent composition of propane?C3H8

Knowns Unknowns

Calculate the percent composition of H2S?

Calculate the percent composition of Mg(OH)2?

10.2 HW

25. 22.4 L26. 567 g CaCO3

27. 11.0 mol C2H6O

28. 33.6 L Cl2

29. 39.9 g/mol30. Gas A: 28.0 g, nitrogen 31. The balloons have the same number of molecules. Each has 1 mole of gas or Avogadro’s number of particles. They will have different masses.

Percent Composition as a Conversion Factor

If you know the % composition you can use it to determine how much of a specific element you have.

Percent Composition as a Conversion Factor

If you know the % composition you can use it to determine how much of a specific element you have.

Propane is 82% C and 18% H. If you have 300 g of C3H8, how much C and H do you have?

Percent Composition as a Conversion Factor

Propane is 82% C and 18% H. If you have 300 g of C3H8, how much C and H do you have?

300 g C3H8 x 82 g C =

100 g C3H8

300 g C3H8 x 18 g H =

100 g C3H8

Analyze

Percent Composition as a Conversion Factor

Propane is 82% C and 18% H. If you have 300 g of C3H8, how much C and H do you have?

300 g C3H8 x 82 g C = 246 g C

100 g C3H8

300 g C3H8 x 18 g H = 54 g H

100 g C3H8

Calculate

Percent Composition as a Conversion Factor

Propane is 82% C and 18% H. If you have 300 g of C3H8, how much C and H do you have?

300 g C3H8 x 82 g C = 246 g C

100 g C3H8

300 g C3H8 x 18 g H = 54 g H

100 g C3H8

Evaluate

246

+ 54 _______ 300

Empirical Formula

= the empirical formula of a compound shows the smallest whole number ratio of the elements in the compound.

C2H2 Acetylene Empirical Formula

C8H8 Styrene

Empirical Formula and % Composition

A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?

Empirical Formula and % Composition

A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula?

Empirical Formula and % Composition

A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown% N = 25.9 % N N?O?

% O = 74.1 % O

Because percent means parts per 100, you can assume the 100.0g of the compound contains 25.9g of N and 74.1g of O. Use these values and convert to moles.

Empirical Formula and % Composition

A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown% N = 25.9 % N N?O?

% O = 74.1 % O

25.9 g N X 1 mol N = 1.85 mol N 14.0 g N

74.1 g O X 1mol O = 4.63 mol O 16.0 g O

Empirical Formula and % Composition

A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown% N = 25.9 % N N?O?

% O = 74.1 % O

25.9 g N X 1 mol N = 1.85 mol N 14.0 g N

74.1 g O X 1mol O = 4.63 mol O 16.0 g O

Remember …. The

amount of N to O is a

ratio between the two

Empirical Formula and % Composition

A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown% N = 25.9 % N N?O?

% O = 74.1 % O

25.9 g N X 1 mol N = 1.85 mol N 14.0 g N

74.1 g O X 1mol O = 4.63 mol O 16.0 g O

1.85 mol N = 1 mol N : 4.63 mol O = 2.5 mol O 1.85 1.85

Empirical Formula and % Composition

A compound is analyzed and found to contain 25.9% nitrogen and 74.1% oxygen. What is the empirical formula? Know Unknown% N = 25.9 % N N?O?

% O = 74.1 % O

25.9 g N X 1 mol N = 1.85 mol N 14.0 g N74.1 g O X 1mol O = 4.63 mol O 16.0 g O

1.85 mol N = 1 mol N ; 4.63 mol O = 2.5 mol O 1.85 1.85

1 mol N X 2 = 2 mol N 2.5 mol O X 2 = 5 mol O

N2O5 is the empirical formula