Empirical Formula

From percentage to formula

Types of Formulas

The formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula.

Empirical Molecular (true) Name

CH C2H2 acetylene

CH C6H6 benzene

CO2 CO2 carbon dioxide

CH2O C5H10O5 ribose

• An empirical formula represents the simplest whole number ratio of the atoms in a compound.

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• It is not just the ratio of atoms, it is also the ratio of moles of atoms

• In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen

• In one molecule of CO2 there is 1 atom of C and 2 atoms of O

• The molecular formula is the true or actual ratio of the atoms in a compound.

Learning Check EF-1

A. What is the empirical formula for C4H8?

1) C2H4 2) CH2 3) CH

B. What is the empirical formula for C8H14?

1) C4H7 2) C6H12 3) C8H14

C. What is a molecular formula for CH2O?

1) CH2O 2) C2H4O2 3) C3H6O3

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Learning Check EF-1

A. What is the empirical formula for C4H8?

1) C2H4 2) CH2 3) CH

B. What is the empirical formula for C8H14?

1) C4H7 2) C6H12 3) C8H14

C. What is a molecular formula for CH2O?

1) CH2O 2) C2H4O2 3) C3H6O3

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Learning Check EF-2

If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.1) SN

2) SN4

3) S4N4

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Learning Check EF-2

If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.

3) S4N4

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Empirical and Molecular Formulas

molar mass = a whole number = nsimplest mass

n = 1 molar mass = empirical mass molecular formula = empirical formula

n = 2 molar mass = 2 x empirical mass molecular formula =

2 x empirical formula molecular formula = or > empirical formula

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Learning Check EF-3

A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?

1) C3H4O3

2) C6H8O6

3) C9H12O911

Learning Check EF-3

A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?

1) C3H4O3

2) C6H8O6

3) C9H12O912

Learning Check EF-4

If there are 192.0 g of O in the molecular formula, what is the true formula if the EF is C7H6O4?

1) C7H6O4

2) C14H12O8

3) C21H18O12

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Learning Check EF-4

If there are 192.0 g of O in the molecular formula, what is the true formula if the EF is C7H6O4?

1) C7H6O4

2) C14H12O8

3) C21H18O12

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Calculating Empirical Formula• Pretend that you have a 100 gram sample of the

compound.• That is, change the % to grams.• Convert the grams to mols for each element.• Write the number of mols as a subscript in a

chemical formula.• Divide each number by the least number.• Multiply the result to get rid of any fractions.

Example• Calculate the empirical formula of a compound

composed of 38.67 % C, 16.22 % H, and 45.11 %N.• Assume 100 g so• 38.67 g C x 1mol C = 3.220 mole C

12.01 gC • 16.22 g H x 1mol H = 16.09 mole H

1.01 gH• 45.11 g N x 1mol N = 3.219 mole N

14.01 gN

• If we divide all of these by the smallest• one it will give us the subscripts for the

empirical formula• 3.220 mol C = 1 3.219 mol N • 16.09 mol H = 5 3.219 mol N • 3.219 mole N = 1 3.219 mol N

• Empirical formula: CH5N

Learning Check EF-5

Aspirin is 60.0% C, 4.5 % H and 35.5 O. Calculate its simplest formula. In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O.

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Solution EF-5

60.0 g C x ___________= ______ mol C

4.5 g H x ___________ = _______mol H

35.5 g O x ___________ = _______mol O

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Divide by the smallest # of moles.5.00 mol C = ________________

______ mol O

4.5 mol H = ______________________ mol O

2.22 mol O = ______________________ mol OAre are the results whole numbers?_____

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Finding Subscripts

A fraction between 0.1 and 0.9 must not be rounded. Multiply all results by an integer to give whole numbers for subscripts.(1/2) 0.5 x 2 = 1(1/3) 0.333 x 3 = 1 (1/4) 0.25 x 4 = 1(3/4) 0.75 x 4 = 3

(1/5) 0.20 x 5 = 5

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Homework

• Worksheet C: #1-7