Percent Composition, Empirical and Molecular Formulas

Percent Composition, Empirical and Molecular Formulas

What Is Percent Composition?A relative measure of the mass

of each different element present in a compound.Obviously, it is expressed as a percent

(%).It can be determined regardless of the #

of elements in the compound.The sum of all the % values MUST

equal 100.

Why Is % Comp Important?

Sometimes it is important to know what percent of a compound’s mass is due to a certain element.A mining company may wish to know

what % of 400 tons of Fe2O3 is due to iron. They can then work out if it is worth trying to recover the iron from the sample of iron oxide.

STEPS TO Calculate % Comp

1. Calculate the mass of an element

2. Calculate the molar mass of the compound.

3. Divide the mass of the element by the molar mass of the substance.

4. Multiply by 100 to make it a percentage.

Calculating % CompositionCalculate the % composition of magnesium carbonate, MgCO3

Molar mass of magnesium carbonate:24.31 g + 12.01 g + 3(16.00 g) =

84.32 g

100.00

%83.2810032.84

31.24

Mg

%24.1410032.84

01.12

C

%93.5610032.84

00.48

O

Percent Composition

Percent = mass of element in 1 mole x 100

by mass molar mass of compound

Let’s try some

1. What is the percent composition by mass of carbon in carbon dioxide (CO2)?

C = 12.011 x 1 atom = 12.011 gO = 15.999 x 2 atoms = 31.998 g

Total Mass = 44.009 g

Percent comp of carbon = 12.011 g x 100% 44.009 g

= 27.292 % of carbon

2. What is the percent composition of both hydrogen and oxygen in water?

H = 1.008 x 2 atoms = 2.016 g

O = 15.999 x 1 atoms = 15.999 g Total Mass = 18.015 g

Percent Comp H: 2.016/18.015 = 11.19 %

Percent Comp O: 15.999/18.015 = 88.81 %

Your percentage for the entire compound should always add to 100 percent!

Let’s practice

Complete Percent Comp Worksheet

Complete Percent composition POGIL

Determining Formulas from Percent Composition

Definitions of Formulas

molecular formula = C6H6 = (CH)6

empirical formula = CH

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.

Formulas (continued)

Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio).

Examples:NaCl

MgCl2

Al2(SO4)3

K2CO3

Formulas (continued)

Formulas for covalent (molecular) compounds MIGHT be empirical (lowest whole number ratio).

Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

Let’s start with Calculating Empirical Formulas from Percent Composition!

Empirical Formula DeterminationIn the problem, you will be given the % comp of the compound.

1.Base your calculation on 100 grams of compound.

2.Determine the # of moles of each element in 100 grams of the compound. (use molar mass)

3.Divide each value of moles by the smallest of the mole values.

4.Multiply each number by an integer to obtain all whole numbers.

Rhyme to help you remember

1.Percent to mass

2.Mass to mole

3.Divide by small

4.Multiple ‘til whole

Problem 1: Empirical Formula Calculation

1.Cinnamon contains cinnamaldehyde. A molecule of cinnamaldehyde contains 81.79% carbon, 6.10% hydrogen, and 12.11% oxygen. Determine the molecules empirical formula.

Empirical Formula Calculation

Step 1: Percent to mass

C: 81.79% 81.79 g

H: 6.10% 6.10 g

O: 12.11% 12.11 g


Step 2: Mass to Mole

C 81.79g x 1 mole/12g = 6.61 mol

H 6.10g x 1 mole/1g = 6.05 mol

O 12.11g x 1 mole/16g =0.757 mol


Step 3: Divide by small

C 6.61 mol/ 0.757 mol = 9

H 6.05 mol/ 0.757 mol = 8

O 0.757 mol/ 0.757 mol = 1


Step 4: Multiple ‘til whole

C9H8O

(in this case, we didn’t have to)

Problem 2: Empirical Formula Calculation

2. Methyl acetate is a solvent commonly used in some paints, inks, and adhesives. Determine the empirical formula for methyl acetate, which has the following chemical analysis: 48.64% carbon, 8.16 hydrogen, and 43.20% oxygen.


Step 1: Percent to mass

C: 48.64%

H: 8.16 O: 43.20%


Step 2: Mass to Mole

C 48.64g x 1 mole/12g =

H 8.16g x 1 mole/1g =

O 43.20g x 1 mole/16g =


Step 3: Divide by small

C 4.050 mol / 2.70 mol = 1.5

H 8.10 mol / 2.70 mol = 3

O 2.70 mol / 2.70 mol = 1


Step 4: Multiple ‘til whole

C 1.5 x 2 = 3

H 3 x 2 = 6

O 1 x 2 = 2

C3H6O2

Problem 3:Empirical Formula Determination3. Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

Problem 3:Empirical Formula Determination

carbonmolcarbong

carbonmolcarbong107.4

01.12

132.49

hydrogenmolhydrogeng

hydrogenmolhydrogeng78.6

01.1

185.6

oxygenmoloxygeng

oxygenmoloxygeng74.2

00.16

184.43

1. Percent to mass and 2. mass to mole

Empirical Formula Determination3. Divide by small

Carbon:

Hydrogen:

Oxygen:

50.174.2

107.4

mol

carbonmol

47.274.2

78.6

mol

hydrogenmol

00.174.2

74.2

mol

oxygenmol

Empirical Formula Determination4. Multiple ‘til whole

Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00x 2 x 2 x 2

3 5 2

Empirical formula:C3H5O2

STOP

Work with a partner to complete worksheet:

Determining Empirical Formulas

Now let’s calculate Molecular Formulas from Percent Composition!

Molecular Formula Determination

In these problems, either you will be given the empirical formula or you will be asked to calculate the empirical formula. You will also be given the molar mass of the compound.

1. Calculate the molar mass of the empirical formula.

2. Divide the given molar mass by the molar mass of the empirical formula.

3. Multiply the empirical formula by this number to get the molecular formula.

Given molar mass = number of times

Empirical molar mass

Molecular Formula Determination

Finding the Molecular Formula

1. The empirical formula for adipic acid is C3H5O2. The molar mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

1. Find the molar mass of C3H5O2

3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

Finding the Molecular FormulaThe empirical formula for adipic acid is C3H5O2. The molar mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g

2. Divide the molar mass by the mass given by the empirical formula

273

146

Finding the Molecular Formula

The empirical formula for adipic acid is C3H5O2. The molar mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid?

3. Multiply the empirical formula by this number to get the molecular formula

(C3H5O2) x 2 = C6H10O4273

146

Pulling it all together

2. Naphthalene, commonly found in mothballs, is composed of 93.7% carbon and 6.3% hydrogen. The molar mass of naphthalene is 128 g/mole. Determine the empirical and molecular formulas for naphthalene.

Pulling it all together

3. Succinic acid is a substance produced by lichens. Chemical analysis indicates it is composed of 40.68% carbon, 5.08% hydrogen, and 54.24% oxygen and has a molar mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid.

STOP

Work on completing worksheet:

Determining Molecular Formulas (True)

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Percent Composition, Empirical and Molecular Formulas