I II III

Formula Calculations

Ch. 7 – The Mole

A. Percentage Composition the percentage by mass of each

element in a compound

100mass total

element of massncompositio %

EMPIRICAL and EMPIRICAL and MOLECULAR MOLECULAR FORMULASFORMULAS

100 =

A. Percentage Composition

%Cu =127.10 g Cu

159.17 g Cu2S 100 =

%S =32.07 g S

159.17 g Cu2S

79.852% Cu

20.15% S

Find the % composition of Cu2S.

%Fe =28 g

36 g 100 =78% Fe

%O =8.0 g

36 g 100 =22% O

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

A. Percentage Composition

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

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

Cu2S is 79.852% Cu

A. Percentage Composition

100 =%H2O = 36.04 g H2O

147.02 g H2O

24.51%H2O

Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O?

A. Percentage Composition

B. Empirical FormulaB. Empirical Formula

C2H6

CH3

reduce subscripts

Smallest whole number ratio of atoms in a compound

B. Empirical Formula1. 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 help1. Percent to Mass2. Mass to Mole 3. Divide by Small 4. Multiply ‘til Whole

Sometimes the mole values are still not whole numbers so multiply by the smallest factor that will make them whole numbers

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

74.1 g 1 mol

16.00 g = 4.63 mol O

1.85 mol N

1.85 mol

= 1 N

= 2.5 O

B. Empirical Formula

N1O2.5Need to make the subscripts whole

numbers multiply by 2

N2O5

Example of Determining Empirical Example of Determining Empirical FormulaFormulaEx. 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 N14.0g N

74.1g O 1 mol O = 4.63 mol O16.0g O

Divide each element by the lowest number of moles in the compound:

1.85 mol N = 1 mol N1.85

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 ProblemsPractice Problems

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

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

C. Molecular Formula “True Formula” - the actual number

of atoms in a compound

CH3

C2H6

empiricalformula

molecularformula

?

C. Molecular Formula1. 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.

nmass EF

mass MF nEF

C. Molecular Formula The empirical formula for ethylene is CH2. Find the molecular

formula if the molecular mass is 28.1 g/mol?

28.1 g/mol

14.03 g/mol = 2.00

empirical mass = 14.03 g/mol

2(CH2) C2H4

Example of Molecular Example of Molecular FormulaFormula

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

Molar Mass of

Compound

Empirical formula

Empirical formula

mass(efm)

Molar Massefm

Molecular Formula

60.0g CH4N 12+4+14 = 30.0g CH4N

60.0/30.0= 2

CH4N x 2

= C2H8N2