Empirical formula
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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 CHC 2 H 2 acetylene CHC 6 H 6 benzene CO 2 CO 2 carbon dioxide

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Empirical Formula

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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

CHC2H2 acetylene

CHC6H6benzene

CO2CO2carbon dioxide

CH2OC5H10O5ribose


Empirical formula

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


Empirical formula

  • 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


Empirical formula

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


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) C4H72) C6H12 3) C8H14

C. What is a molecular formula for CH2O?

1) CH2O2) 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) CH2O2) 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


Empirical formula

  • 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


Empirical formula

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?_____


Empirical formula

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


Empirical formula

Multiply everything x 4 to clear the denominator

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

H: 2.0 mol Hx 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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