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## PowerPoint Slideshow about ' Formula Calculations' - sequoia-lopez

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A. Percentage Composition

EMPIRICAL and MOLECULAR FORMULAS

- the percentage by mass of each element in a compound

159.17 g Cu2S

32.07 g S

159.17 g Cu2S

A. Percentage Composition- Find the % composition of Cu2S.

100 =

%Cu =

79.852% Cu

%S =

100 =

20.15% S

36 g

8.0 g

36 g

A. Percentage Composition- Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.

100 =

78% Fe

%Fe =

100 =

22% O

%O =

A. Percentage Composition

- How many grams of copper are in a 38.0-gram sample of Cu2S?

Cu2S is 79.852% Cu

(38.0 g Cu2S)(0.79852) = 30.3 g Cu

147.02 g H2O

A. Percentage Composition- Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O?

24.51%

H2O

%H2O =

100 =

B. Empirical Formula

1. Find mass (or %) of each element.

2. Find moles of each element.

3. Divide moles by the smallest # to find subscripts.

4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.

What ??????????How can I remember?

- There is a poem to help
- Percent to Mass
- Mass to Mole
- Divide by Small
- Multiply ‘til Whole
- Sometimes the mole values are still not whole numbers so multiply by the smallest factor that will make them whole numbers

1.85 mol

B. Empirical Formula- Find the empirical formula for a sample of 25.9% N and 74.1% O.

25.9 g N

1 mol N

14.01 g N

= 1.85 mol N

= 1 N

74.1 g

1 mol

16.00 g

= 4.63 mol O

= 2.5 O

Example of Determining Empirical Formula

- Ex. What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% Oxygen?
- 25.9g N 1 mol N = 1.85 mol N
- 14.0g N
- 74.1g O 1 mol O = 4.63 mol O
- 16.0g O
- Divide each element by the lowest number of
- moles in the compound:
- 1.85 mol N = 1 mol N
- 1.85
- 4.63 mol O = 2.50 mol O
- 1.85
- Now find the lowest common denominator.
- 1 mol N x 2= N2 2.5 mol O x 2 = O5
- Therefore, N2O5 is the empirical formula

Practice Problems

Calculate the empirical formula for a compound that is 94.1% O and 5.90%H

Calculate the empirical formula for a compound that is 79.8% C and 20.2% H

C. Molecular Formula

- “True Formula” - the actual number of atoms in a compound

CH3

empirical

formula

?

molecular

formula

C. Molecular Formula

1. Find the empirical formula.

2. Find the empirical formula mass.

3. Divide the molecular mass by the empirical mass.

4. Multiply each subscript by the answer from step 3.

14.03 g/mol

C. Molecular Formula- The empirical formula for ethylene is CH2. Find the molecular formula if the molecular mass is 28.1 g/mol?

empirical mass = 14.03 g/mol

= 2.00

2(CH2) C2H4

Example of Molecular Formula

Calculate the molecular formula of the compound whose Molar mass is 60.0g and empirical formula is CH4N.

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