Section 9.3—Analysis of a Chemical Formula

How can we determine a chemical formula?

Percent Composition

What is Percent Composition?

100Whole

PartPercent

100samplewholeofmass

elementofmassncompositioPercent

Example #1

Example:A 8.5 g sample is composed of carbon and

hydrogen. If 5.5 g of the sample is found to be carbon, what is the percent composition by mass of

the sample?

Example #1

If total = 8.5 g

and C = 5.5 g

then H = 3.0 g

-

1005.8

5.5%

g

gC 100

5.8

0.3%

g

gH

%7.64% C %3.35% H

Example:A 8.5 g sample is composed of carbon and

hydrogen. If 5.5 g of the sample is found to be carbon, what is the percent composition by mass of

the sample?

Let’s Practice #1

A sample is 57.0% by mass chlorine, how many grams of chlorine are present in a 27.5 g sample?

Let’s Practice #1

1005.27

?%0.57

g

gCl

gClg

?%0.57100

5.27

27.5 g 27.5 g

100 100

gClg ?7.15

A sample is 57.0% by mass chlorine, how many grams of chlorine are present in a 27.5 g sample?

Example #2

Percent composition can also be determined from a chemical formula

Example:Find the percent composition, by mass, of CaCl2

Example #2

Percent composition can also be determined from a chemical formula

total = 110.98 g

Ca = 40.08 g

Cl = 70.90 g

10098.110

08.40%

g

gCa 100

98.110

90.70%

g

gCl

%1.36% Ca %9.63% Cl

CaCl

12

40.08 g/mole35.45 g/mole

= 40.08 g/mole= 70.90 g/mole +

110.98 g/mole

So for 1 mole:

Example:Find the percent composition, by mass, of CaCl2

Let’s Practice #2

Find the percent composition, by mass, of NaNO3

Let’s Practice #2

10000.85

99.22%

g

gNa 100

00.85

01.14%

g

gN

%0.27% Na %5.16% N

Find the percent composition, by mass, of NaNO3

NaN

11

22.99 g/mole14.01 g/mole

= 22.99 g/mole= 14.01 g/mole

+

85.00 g/mole

O 3 16.00 g/mole = 48.00 g/mole

10000.85

00.48%

g

gO

%5.56% O

Empirical Formulas

What’s an Empirical Formula?

Empirical – from data

Empirical Formula – Chemical formula determined from lab data. Lowest possible ratio of atoms

CH2 is the lowest ratio (and empirical formula) of the molecule C3H6

Ratio of Atoms in a Molecule

Subscripts in a chemical formula show the ratio of atoms (or ions) in a molecule

a sample of CaCl2 has 1 calcium ion : 2 chlorine ions

We can use the unit “mole” to count things

Atoms and ions can be counted by “moles”

If the subscripts give the ratio of atoms, then they also give the ratio of moles of atoms

a sample of CaCl2 has 1 mole of calcium ions : 2 moles of chlorine ions

Bmoles

Amoles

Batoms

Aatoms

#

#

Using Mole Ratio of Atoms in a Molecule

Therefore, if the ratio of moles of each atom is found…

then the subscripts of the chemical formula are known

1 mole C2 mole H

CH2

Example #3If given percents, use those percents as grams (as if you assume you have a 100 g sample)—Remember %’s add up to 100!

Change grams to moles for each atom

Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles)

Use the ratio as subscripts for writing the chemical formula

1

2

3

4

Example:Find the empirical formula if a sample contains Ca

and Cl and is 36.1% Ca

Example #3If given percents, use those percents as grams (as if you assume you have a 100 g sample)—Remember %’s add up to 100!

Change grams to moles for each atom

Find the lowest possible whole number ratio of the atom (divide all moles by the smallest # of moles)

Use the ratio as subscripts for writing the chemical formula

1

2

3

4

36.1 g Ca= _____ mol Cag Ca

mol Ca 140.08

0.901

63.9 g Cl= _____ mol Clg Cl

mol Cl 135.45

1.80

0.901 mol Ca = 1 mol Ca 0.901

1.80 mol Cl = 2 mol Cl 0.901

CaCl2

Example:Find the empirical formula if a sample contains Ca

and Cl and is 36.1% Ca

Let’s Practice #3

40.92 g C= _____ mol Cg C

mol C 112.01

3.41

4.58 g H= _____ mol Hg H

mol H 11.01

4.53

3.41 mol C = 1 mol C 3.41

4.53 mol H = 1.33 mol H 3.41

C3H4O3

54.5 g O= _____ mol Og O

mol O 116.00

3.41

3.41 mol O = 1 mol O 3.41

Multiply the ratio (1 : 1.33 : 1) by 3to make a whole number ratio (3 : 4: 3)

Find the empirical formula if a sample contains

40.92 g C, 4.58 g H and 54.5 g O

Molecular Formulas

What’s a Molecular Formula?

Empirical Formula – Chemical formula determined from lab data. Lowest possible ratio of atoms

Molecular Formula – Actual ratio of atoms in a molecule

Empirical versus Molecular Formula

The empirical formula is the lowest possible ratio.

A molecule with the empirical formula:

Could have one of the following molecular

formulas:

NO2 NO2, N2O4, N4O8…

CH2 CH2, C2H4, C4H8…

The molecular formula is the actual ratio

Example #4

Find the empirical formula, if not given

Find the molar mass of the empirical formula

Find the ratio of the molecular formula’s molar mass (must be given to you) to the empirical formula’s molar mass

Multiply the empirical formula’s subscripts by the ratio found in step 3.

1

2

3

4

Example:Empirical formula = C3H4O3. The molecular

formula’s molar mass = 176.14 g/mole.Find the molecular formula.

Example #4

Find the empirical formula, if not given

Find the molar mass of the empirical formula

Find the ratio of the molecular formula’s molar mass (must be given to you) to the empirical formula’s molar mass

Multiply the empirical formula’s subscripts by the ratio found in step 3.

1

2

3

4

C3H4O3

176.14 g/mole = 2 88.07 g/mole

3

3

12.01 g/mole

16.00 g/mole

= 36.03 g/mole

= 48.00 g/mole+

88.07 g/mole

4 1.01 g/mole = 4.04 g/moleC

OH 2

C6H8O6

Example:Empirical formula = C3H4O3. The molecular

formula’s molar mass = 176.14 g/mole.Find the molecular formula.

Hydrate Formulas

What’s a Hydrate?

Hydrate – Molecule that has water physically attached to it

It’s not dissolved in water…hydrates can be solid, liquid or gas!

Hydrates are inorganic salts "containing water molecules combined in a definite ratio as an integral part of the crystal“

either bound to a metal center or have crystallized with the metal

When writing a hydrate, a dot is used to separate the formula compound from the number of water moleculesExamples)

magnesium sulfate heptahydrate

cobalt (II) chloride hexahydrate

tin (II) chloride dihydrate

Hydrate & anhydride

Hydrate = molecule with water molecules physically attached

anhydride = molecule with water removedThe water can be removed by heating the

hydrate

Hydrate anhydride + waterheat

Finding the ratio of anhydride molecules to water molecules gives you the hydrate formula

Example #5

Find the mass of anhydride & water if not given

Change mass of anhydride & water to moles

Find the ratio of the moles water to moles anhydride

Write the hydrate formula

1

2

3

4

Example:2.46 g MgSO4 hydrate is heated and 1.20 g MgSO4 anhydride is left. Find the hydrate formula.

Example #5

Find the mass of anhydride & water if not given

Change mass of anhydride & water to moles

Find the ratio of the moles water to moles anhydride

Write the hydrate formula

1

2

3

4

Example:2.46 g MgSO4 hydrate is heated and 1.20 g MgSO4 anhydride is left. Find the hydrate formula.

1.20 g MgSO4 = _______ mol MgSO4g MgSO4

mol MgSO4 1120.38

0.00997

1.26 g H2O = _______ mol H2Og H2Omol H2O 1

18.020.0699

0.0699 mol H2O = 7.01 0.00997 mole MgSO4

MgSO4 7 H2O

Hydrate = anhydride + water2.46 g = 1.20 g + water

Water = 1.26 g

Calculating % of Water

Determine the mass of the hydrated compound

Determine the mass of JUST the waterThen divide

% H2O = Mass of water x 100

Mass of hydrate

Example)

CuSO4 5H⋅ 2O CuSO4 + 5H2O

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