Percent Composition Empirical Formulas Molecular Formulas

Courtesy: www.lab-initio.com

•There are two common ways to describe the composition of a compound: the numbers of its constituents atoms (chemical formula) or by percentages (by mass) of its elements (percent composition).•The mass percents of the elements in a compound can be determined by comparing the mass of each element present in 1 mole of the compound to the total mass of 1 mole of the compound.

%xcompoundofmolofmass

compoundofmolinelementofmass

elementofpercentMass

1001

1=

Example: Find the percent composition of the compound N2O5.

Solution: First determine the mass of each element in the compound.

We know that 1 mol of N2O5 must contain 2 mol N and 5 mol O. The number of grams of N and O are found as follows:

Total mass of compound = 28.02 g N + 80.00 g O = 108.02 g N2O5

Og.mol.

xmolOofMass

Ng.mol

g.xmolNofMass

008010016

5

022810114

2

=g

=

==

The mass percent of N and O in N2O5 can be computed by comparing the mass of N and O in 1 mole of N2O5 to the total mass of 1 mole of N2O5.

Notice the percents add up to 100.00%; this provides a check that the calculations are correct.

O%.%xg.g.

OofpercentMass

N%.%xg.g.

NofpercentMass

0674100021080080

9425100021080228

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Example:A sample of a liquid with a mass of 8.657 g was decomposed into its elements and gave 5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the percent composition of this compound?

Solution:We can apply an equation for the percent by mass of an element to each element. The “mass of whole sample” here is 8.657 g so we can take each element in turn and perform the calculations.

%xsamplewholeofmass

elementofmasselementofmassby% 100=

For C:

For H:

For O:

The percentages must add up to 100%, allowing for small differences caused by rounding.Sum of percentages: 99.99%

C%.%xg.g.

266010065782175

=

H%.%xg.g.

1111100657896200

=

O%.%xg.g.

622810065784782

=

•When a new compound is prepared, one of the first items of interest is the formula of the compound. •For example, the compound that forms when phosphorus burns in oxygen consists of molecules with the formula P4O10.•When a formula gives the composition of one molecule, it is called a molecular formula.•Molecular formula: the exact formula of a molecule, giving the types of atoms and the number of each type.

•Notice that in the formula P4O10 that both the subscripts 4 and 10 are divisible by 2, so the smallest numbers that tell us the ratio of P to O are 2 and 5.•In a simpler kind of formula, the empirical formula, the subscripts are the smallest whole numbers that describe the ratios of the atoms in the substance.•Empirical formula: the simplest whole number ratio of atoms in a compound.•It is quite common for the empirical and molecular formulas to be different; some examples are shown below.

•Example: Calculating an Empirical Formula from Percent CompositionA white powder used in paints, enamels, and ceramics has the following percent composition: Ba, 69.58%, C, 6.090%, and O, 24.32%. What is its empirical formula?

Solution:Since mass percentage gives the number of grams of a particular element per 100 grams of compound, base the calculation on 100 grams of compound. Each percent will then represent the mass in grams of that element.

If we have 100 g of the above compound, based on the percentages we would have 69.58 g of Ba, 6.090 g of C, and 24.32 g of O.

Determine the number of moles of each element present in 100 grams of compound using the atomic masses of the elements present.

Ba:

C:

O: Omol.Og.Omol

xOg.

Cmol.Cg.Cmol

xCg.

Bamol.Bag.Bamol

xBag.

52010016

13224

507100112

10906

5067033137

15869

=

=

=

Divide each value of the number of moles by the smallest of the values. If each resulting number is a whole number (after appropriate rounding), these numbers represent the subscripts of the elements in the empirical formula.

If the numbers obtained in the previous process are not whole numbers, multiply each number by an integer so that the results are all whole numbers.

30003506705201

100115067050700

15067050670

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==

=

...

:O

...

:C

.

.:Ba The coefficients are acceptably

close to whole numbers, so the empirical formula is

BaCO3.

Example: One compound of mercury with a molar mass of 519 contains 77.26% Hg, 9.25% C, and 1.17% H (with the balance being O). Calculate the empirical and molecular formulas, arranging the symbols in the order HgCHO.

Solution:First obtain the empirical formula.Assume a 100 g sample therefore, we have 77.26 g Hg, 9.25 g C, 1.17 g H, and 12.32 g O.

The amount of oxygen was determined by subtracting the total amounts of the other three elements from the total assumed mass of 100 g.

Convert each of these masses into moles.

Omol.Og.Omol

xOg.:O

Hmol.Hg.Hmol

xHg.:H

Cmol.Cg.Cmol

xCg.:C

Hgmol.Hgg.Hgmol

xHgg.:Hg

770000016

13212

1610111

171

77000112

1259

3852059200

12677

=

=

=

=

Divide each number of moles by the smallest number of moles.

O...

:O

H...

:H

C...

:C

Hg...

:Hg

299913852077000

301338520161

2002385207700

100013852038520

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==Empirical Formula:HgC2H3O2

Compute the mass corresponding to the empirical formula.

(1 Hg x 200.59 amu) + (2 C x 12.01 amu) + (3 H x 1.01 amu) + (2 O x 16.00 amu) = 259.64 g/mol

Calculate the ratio:

The integer from the previous step represents the number of empirical formula units in one molecule. When the empirical formula subscripts are multiplied by this integer, the molecular formula results.Molecular formula = 2(HgC2H3O2) = Hg2C4H6O4

00264259

519.

mol/g.mol/g

massformulaEmpiricalmassMolar

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