Section 7.3 - Percent Composition and

Chemical Formulas

Calculating the Percent Composition of a

compound

The relative amounts of each element (Part) in a compound (Whole) are

expressed as the Percent Composition or the Percent by Mass of each element in

the compound.

100xWholePart

%

The percent composition of a

compound has as many percent values as there are different elements

in the compound.

The percent composition of CaCO3 is:

Ca = 40.1 %

C = 12.0 %

O = 47.9 %

To calculate the percent composition of a

compound, use the chemical formula to

determine which elements and how many moles of

each are in the compound.

Divide the total mass of one of the elements in the compound by the molar mass of the compound then multiply by 100 to

convert to a percent.

Remember that the total mass of the element is the molar mass of the

element multiplied by its subscript in the formula.

Example: What is the percent composition of

FeSO4?

Fe = 100xg151.9

g55.8

S = 100xg151.9

g32.1

O = 100xg151.9

g64.0

= 36.7 %

= 21.1 %

= 42.2 %

The Percent By Mass in a compound is the

number of grams of the element divided by the number of grams of the compound multiplied by

100.

Sample Problem 7 - 10 p 189

An 8.20 g piece of magnesium combines with 5.40 g of

oxygen to form a compound. What is the Percent

Composition (Percent By Mass) of this compound?

% Mg = 100xcompoundofmass

Mgofmass

% Mg = %60.3100xg13.60g8.20

% O = 100xcompoundofmass

Oofmass

% O = %39.7100xg13.6g5.40

Using Percent Composition as a

Conversion Factor

You can use the percent composition to determine the mass of

each element in a sample of a compound.

Example:Aluminum makes up

52.9 % of Al2O3. Calculate the mass of Aluminum in a 50.0 g

sample of Al2O3.

We first assume that we have a 100.0g sample of Al2O3. 52.9 % of 100.0 g

is 52.9 g. We can use 52.9 g divided by 100.0 g sample as a conversion

factor.

Alg26.5g100.0

g52.9x

1

g50.0

Calculating Empirical Formula

Empirical Formula

EMPIRICAL FORMULAAn empirical formula is the smallest whole

number ratio of atoms in a formula.

EMPIRICAL FORMULAIf the atoms are in a small whole number ratio, then the moles of atoms of each element in the substance will also be in a small whole number ratio.

Two Sources of Data From Which to Calculate the

Empirical FormulaExperimental Analysis - What

is the empirical formula of a compound if a 2.50 gram sample contains 0.900 g of calcium and 1.60 g of chlorine?

Two Sources of Data From Which to Calculate the

Empirical Formula

Percentage Composition - A compound has a percentage composition of 40.0% C, 6.71% H, and 53.3% O. What is the empirical Formula?

The following is an Example of the Empirical Formula Calculation from Experimental Analysis.

Determine the number of moles of each element in the compound.

What is the empirical formula for a compound if a 2.50 g sample contains 0.900 g of calcium and 1.60 g of chlorine?

Cag40.1Camole1

x1

Cag0.900

0. 0224 moles Ca

ClgClmole

xClg

5.351

160.1

0.0450 moles Cl

To obtain the smallest whole number ratio, divide both numbers of moles by the smaller number of moles.

The next step is to determine the smallest whole number ratio.

20224.0

0450.0

molesClmoles

10224.0

0224.0

molesCamoles

The mole ratio is 1 mole of Calcium to 2 moles of Chlorine.

The Empirical formula is therefore CaCl2.

Example of Empirical Formula Calculation From

Percentage Composition:A compound has a percentage composition of 40.0% C, 6.71% H, and 53.3% O. What is the empirical formula?

Example of Empirical Formula Calculation From

Percentage Composition To calculate the ratio of

moles of these elements, we assume a 100 g sample . 40.0% of 100 g is 40.0 g.

Example of Empirical Formula Calculation From

Percentage Composition

Next we calculate the number of moles of each element present.

Example of Empirical Formula Calculation From

Percentage Composition

Cg12.0Cmol1

x1

Cg40.0

1.01gH1molH

x1

6.71gH

16.0gO1molO

x1

53.3gO

Mol C 3.33

Mol H 6.67

Mol O 3.33

The number of moles of each element calculated in step 2 were: 3.33 mol C, 6.64 mol H, and 3.33 mol O. Dividing by the smallest value we obtain the smallest whole number ratio of moles.

4 Determine the smallest Whole Number Ratio.

For C 1mol3.33mol3.33

For H 2mol3.33mol6.64

For O 1mol3.33mol3.33

The empirical formula is therefore --- CH2O.

With a ratio of 1-C to 2-H to 1-O

What is the empirical formula for a compound

if a 40.0 g sample contains 29.7 g of sodium

and 10.3 g of oxygen?

Solution:

Step 1 Determine the number of moles of each element.

Namoles1.29Nag23.0Namole1

x1

Nag29.7

Omoles.644Og16.0Omole1

x1

Og10.3

Step 2 Divide each number of moles by the smallest.

This will give the smallest whole number

ratio.

20.6441.29

10.6440.644

This ratio shows that there are 2 moles of

sodium for one mole of oxygen.

The empirical formula is therefore Na2O.

Calculating Molecular Formula

The Empirical Formula tells us the ratio of each

kind of atom present in a compound.

The Molecular Formula tells us the actual number of each kind of atom in a molecule of a compound.

If the empirical formula and the molar mass are known, the molecular

formula can be calculated.

Example:

A certain gas, having a molar mass of 78 g, has an empirical formula of CH. What is the molecular formula?

If we multiply the mass of the empirical formula

by an integer, the product must equal the

molar mass.

(CH)x = 78 g

Where x represents an integer.

(13)x = 78 g

X = 6

(CH)6 gives C6H6

By multiplying each subscript in the

empirical formula by the integer (6), we obtain the

molecular formula.

Problem:

The empirical formula of a compound is CHOCl and the molar mass is 129 g. What is the molecular formula?