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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

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empirical formula

Empirical Formula

From percentage to formula

types of formulas
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

slide4

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

  • 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
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
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
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
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
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
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
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
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
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
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
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
slide17

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
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
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 51
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

slide21
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?_____

slide22
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
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

slide24
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
Homework
  • Worksheet C: #1-7
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