Empirical Formula

1 / 25

Empirical Formula - PowerPoint PPT Presentation

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 C 2 H 2 acetylene CH C 6 H 6 benzene CO 2 CO 2 carbon dioxide

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

PowerPoint Slideshow about ' Empirical Formula' - zoe-shaw

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

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

• In 1 mole of CO2there 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
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

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

Solution EF-1

A. What is the empirical formula for C4H8?

2) CH2

B. What is the empirical formula for C8H14?

1) C4H7

C. What is a molecular formula for CH2O?

1) CH2O 2) C2H4O2 3) C3H6O3

Solution 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

If the actual formula has 4 atoms of N, and S is related 1:1, then there must also be 4 atoms of S.

Empirical and Molecular Formulas

molar mass = a whole number = n

simplest 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

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) C9H12O9

Solution EF-3

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

2)C6H8O6

C3H4O3 = 88.0 g/EF

176.0 g = 2.00

88.0

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

Solution EF-4

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

3) C21H18O12

192 g O = 3 x O4 or 3 x C7H6O4

64.0 g O in EF

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.

Solution EF-5

60.0 g C x ___________= ______ mol C

4.5 g H x ___________ = _______mol H

35.5 g O x ___________ = _______mol O

Solution EF-5

60.0 g C x 1 mol C= 5.00 mol C

12.0 g C

4.5 g H x 1 mol H = 4.5 mol H

1.01 g H

35.5 g O x 1mol O = 2.22 mol O

16.0 g O

Divide by the smallest # of moles.

5.00 mol C = ________________

______ mol O

4.5 mol H = ________________

______ mol O

2.22 mol O = ________________

______ mol O

Are are the results whole numbers?_____

Divide by the smallest # of moles.

5.00 mol C = ___2.25_=2 ¼ = 9/4

2.22 mol O

4.5 mol H = ___2.00__

2.22 mol O

2.22 mol O = ___1.00__

2.22 mol O

Are the results whole numbers? no

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 x2 = 1

(1/3) 0.333 x 3 = 1

(1/4) 0.25 x4 = 1

(3/4) 0.75 x 4 = 3

(1/5) 0.20 x 5 = 5

Multiply everything x 4 to clear the denominator

C: 9/4 mol C x 4 = 9 mol C

H: 2.0 mol H x 4 = 8 mol H

O: 1.00 mol O x 4 = 4 mol O

Use the whole numbers of mols as the subscripts in the simplest formula

C9H8O4

Homework
• Worksheet C: #1-7