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Chapter 3

Percent Compositionsand

Empirical Formulas

AP Chemistry

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Calculating Percent Composition of a Compound

The following general formula is used to The following general formula is used to determine the percent composition of determine the percent composition of each element in a compound:each element in a compound:

Mass of element (in 1 mole of compound) X 100 Molar Mass of Compound

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What is the percent composition What is the percent composition of each element in Ca(NOof each element in Ca(NO33))22 ? ?

Step 1: Find the molar mass of Ca(NOStep 1: Find the molar mass of Ca(NO33). ). Ca 40.1 grams/moleCa 40.1 grams/mole = 40.1 g/mol = 40.1 g/mol N 2 x (14.0) grams/mole = 28.0g/molN 2 x (14.0) grams/mole = 28.0g/mol O 6 x (16.0) grams/mole = O 6 x (16.0) grams/mole = 96.0 g/mol96.0 g/mol 164.1 g/mole164.1 g/mole

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Step 2: Divide the mass of each element by the Step 2: Divide the mass of each element by the molar mass and multiply by 100 to get the percent.molar mass and multiply by 100 to get the percent.

% Ca = % Ca = 40.140.1g x 100% = 24.4 % Calciumg x 100% = 24.4 % Calcium 164.1g164.1g    % N = % N = 28.028.0g x 100% = 17.1% Nitrogeng x 100% = 17.1% Nitrogen 164.1 g164.1 g    % O = % O = 96.0g96.0g x 100% = 58.5 % Oxygen x 100% = 58.5 % Oxygen 164.1g164.1g

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*Note percent composition can also be done given masses *Note percent composition can also be done given masses

of a compound and each element in the compoundof a compound and each element in the compound..

Calculate the percent composition of a compound that is made of 29.0 grams of Ag with 4.30 grams of S.

29.0 g Ag

33.3 g totalX 100 = 87.1 % Ag

4.30 g S

33.3 g totalX 100 = 12.9 % S

Total = 100 %

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Formulas

• The empirical formula for CThe empirical formula for C33HH1515NN33 is is

CHCH55N. N.

• you are basically dividing by the greatest you are basically dividing by the greatest common factor , which in this case is 3, common factor , which in this case is 3, for each element subscript.for each element subscript.

Empirical formula: the lowest whole number ratio of atoms in a compound.

Molecular formula: the true number of atoms of each element in the formula of a compound.

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Formulas (continued)

Formulas for ionic compounds are ALWAYS empirical (the lowest whole number ratio = cannot be reduced).

Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

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Formulas (continued)

Formulas for molecular compounds

MIGHT be empirical (lowest whole number ratio).

Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

(Correct formula)

(Lowest whole number ratio)

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Calculating Empirical We can get a ratio from the percent

composition.

1) Assume you have a 100 g sample- the percentage become grams (75.1% = 75.1 grams)

2) Convert grams to moles.

3) Find lowest whole number ratio by dividing each number of moles by the smallest value.

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Example Calculate the empirical formula of a

compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.

Step 1 : Assume 100gram sample Step 2 : Convert grams to moles for each element

38.67 g C x 1mol C = 3.22 mole C 12.0 g C

16.22 g H x 1mol H = 16.22 mole H 1.0 g H

45.11 g N x 1mol N = 3.22 mole N 14.0 g N

Now divide each value by the smallest value

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Example

The ratio is 3.22 mol C = 1 mol C 3.22 mol N 1 mol N

The ratio is 16.22 mol H = 5 mol H 3.22 mol N 1 mol N

= C1H5N1 which is = CH5N

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Empirical to molecular Since the empirical formula is the

lowest ratio, the actual molecule would weigh more.By a whole number multiple.

Divide the actual molar mass by the empirical formula mass – you get a whole number to increase each coefficient in the empirical formula

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Caffeine has a molar mass of 194 g. and its and its empirical formula is Cempirical formula is C44HH55NN22O, O, what is its

molecular formula? Find the empirical formula molar mass.Find the empirical formula molar mass. 97 g/mole97 g/mole Set up ratio 194g = 2Set up ratio 194g = 2 97g 97g Multiply each subscript in the empirical Multiply each subscript in the empirical

formula by 2.formula by 2. CC88HH1010NN44OO2 2 is the molecular formula. is the molecular formula.

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Empirical formula from combustion reactions

Example: 7.321 grams of an organic compound containing carbon, hydrogen, and oxygen was analyzed by combustion. The amount of carbon dioxide produced was 17.873 grams and the amount of water produced was 7.316 grams. Determine the empirical formula of the compound.

Combustion reaction looks like the following:

CHO + O2 CO2 + H2O

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Assume all of the Carbon in the compound Assume all of the Carbon in the compound turned into carbon dioxide, and determine the turned into carbon dioxide, and determine the mass percent of carbon in the compoundmass percent of carbon in the compound

17.873 g CO17.873 g CO22 1mole 1 mol C 12g C 1mole 1 mol C 12g C

44g CO44g CO22 1 mol CO 1 mol CO2 2 1 mol1 mol

= 4.8744 grams Carbon= 4.8744 grams Carbon

4.8744 g x 100 = 66.58% Carbon4.8744 g x 100 = 66.58% Carbon

7.321 g 7.321 g

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Assume all of the hydrogen in the compound went to Assume all of the hydrogen in the compound went to water, and determine the mass percent of hydrogen water, and determine the mass percent of hydrogen in the compoundin the compound

7.316 g H7.316 g H22O 1mol HO 1mol H22O 2 mols H 1.0 g HO 2 mols H 1.0 g H

18g H18g H22O 1mol HO 1mol H22O 1 mol HO 1 mol H

= 0.8129 grams Hydrogen= 0.8129 grams Hydrogen

0.8129 x 100 = 11.10% Hydrogen0.8129 x 100 = 11.10% Hydrogen

7.3217.321

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With the mass percent of Carbon and With the mass percent of Carbon and Hydrogen, you can figure out the mass Hydrogen, you can figure out the mass percent of oxygen in the compound. percent of oxygen in the compound.

100 - (66.58% C + 11.10% H) =100 - (66.58% C + 11.10% H) =

22.32 % Oxygen22.32 % Oxygen

Now use these percents to find the Now use these percents to find the empirical formula!empirical formula!

CC44HH88OO