Chapter #6 Chemical Composition. Chapter Contents 6-1 Chemical Composition 6-2 Counting Units 6-3 Counting Atoms By the Gram 6-4 Counting Molecules by

Chapter #6Chemical Composition

Chapter Contents

• 6-1 Chemical Composition• 6-2 Counting Units• 6-3 Counting Atoms By the Gram• 6-4 Counting Molecules by the Gram• 6-5 Mole Conversions• 6-6 Mass Percent Composition• 6-8 Empirical Formula Calculation• 6-9 Molecular Formula Calculation

6-1 The Chemical Package

• The baker uses a package called the dozen. All dozen packages contain 12 objects.

• The stationary store uses a package called a ream, which contains 500 sheets of paper.

• So what is the chemistry package?

About Packages

6-2 The Chemical Package

• The baker uses a package called the dozen. All dozen packages contain 12 objects.

• The stationary store uses a package called a ream, which contains 500 sheets of paper.

• So what is the chemistry package? Well, it is called the mole (Latin for heap).

About Packages

Each of the above packages contain a number of objects that are convenient to work with, for that particular discipline.

The atomic weights listed on the periodic chart are the weights of a mole of atoms. For example a mole of hydrogen atoms weighs 1.00797 g and a mole of carbon atoms weighs 12.01 g.

6-3 The MoleA mole contains 6.022X1023 particles, which is the number of carbon-12 atoms that will give a mass of 12.00 grams, which is a convenient number of atoms to work with in the chemistry laboratory.

6-4 Moles of ObjectsSuppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows?

6-4 Moles of Objects

Suppose we order a mole of marshmallows for a chemistry party. How much space here at Central would be required to store the marshmallows?Would cover the entire 50 states 60 miles deep



How about a mole of computer paper instead of a ream of computer paper, how far would that stretch?



How about a mole of computer paper instead of a ream of computer paper, how far would that stretch? Way past the planet Pluto!

To calculate the molar mass of a compound we sum together the atomic weights of the atoms that make up the formula of the compound. This is called the formula weight (MW, M).

Formula weights are the sum of atomic weights of atoms making up the formula.

The following outlines how to find the formula weight of water

symbol weight numberHO

1.0116.0

21X

X== 2.02

16.018.0 g/mole

6-5 Formula Weight Calculation

6-7 Percent Composition

Find the formula weight and the percent composition of

glucose (C6H12O6)

symbol weight number

HO

C

16.01.01

12.0

6

12

6

x

x

x

=

=

=

72.0

12.1296.0

180.1 g/mole

%C =

%H =

%O =

72.0

12.12

96.0

180.1

180.1

180.1

X 100 =

X 100 =

X 100 =

40.0 %C

6.73 %H

53.3 %O

6-7 Percent ConversionsSeawater contains approximately 3.5% NaCl by mass and has a density of 1.02 g/mL. What volume of seawater would be required to produce 1.0 g of NaCl when evaporated?


1.02 g seawater

mL seawater


1.02 g seawater

mL seawater

3.5 g NaCl100 g seawater


1.02 g seawater

mL seawater

3.5 g NaCl100 g seawater 1.00 g NaCl


1.02 g seawater

mL seawater

3.5 g NaCl100 g seawater 1.00 g NaCl = 28 mL

A mole of glucose (C6H12O6) contains 6.022 X 1023 molecules of glucose. And 6 X 6.022 X 1023 atoms of C. Since a mole is 6.022 X 1023 particles then a mole of glucose must contain 6 moles of C atoms.

How many moles of hydrogen atoms are contained in a mole of glucose?

In 5 moles of H2SO4 how many moles of oxygen atoms is there?

6-8 Mole Concepts


How many moles of hydrogen atoms are contained in a mole of glucose? 12 Moles of hydrogen.

How many moles of oxygen and hydrogen are in one mole of H2O contains:

6-8 Mole Concepts



How many moles of oxygen and hydrogen are in one mole of H2O contains:One mole of oxygen atoms Two moles of hydrogen atoms

6-8 Mole Concepts




In 5 moles of H2SO4 how many moles of oxygen atoms are there?

6-8 Mole Concepts




In 5 moles of H2SO4 how many moles of oxygen atoms is there?

20 moles of O atoms.

6-8 Mole Concepts

In 50.0g of H2SO4 how many moles of sulfuric acid are there?

6-8 Mole Conversions

50.0g of H2SO4



50.0g of H2SO4 =98.0g of H2SO4

mole H2SO4



50.0g of H2SO4 = 0.510 mole H2SO498.0g of H2SO4

mole H2SO4

In 50.0g of H2SO4 how many moles of oxygen atoms are there?




50.0g of H2SO4 =



50.0g of H2SO4 =98.0g of H2SO4

mole H2SO4



50.0g of H2SO4 =mole H2SO4

4mole O98.0g of H2SO4

mole H2SO4



50.0g of H2SO4 =mole H2SO4

4mole O 2.04 mole O98.0g of H2SO4

mole H2SO4

In 5 moles of H2SO4 how many atoms of oxygen are present?




5 moles H2SO4 =



5 moles H2SO4

mole H2SO4

4 mole O



5 moles H2SO4

mole H2SO4

4 mole O 6.02 x 1023 atoms O mole O

=

1.20 x 1025 atoms

6-8 Empirical FormulasEmpirical formula is the smallest whole number ratio between atoms and can be calculated from the percent composition.

Molecular formulas happen to be the exact number of atoms making up a molecule, and may or may no be the simplest whole number ratio. Molecular formulas are whole number multiples of the empirical formula.

6-8 Empirical Formula Steps1. Assume 100 g of compound.2. Convert percent to a mass number.3. Convert the mass to moles.4. Divide each mole number by the smallest mole

number.5. Rounding:

a. If the decimal is ≤ 0.1, then drop the decimalsb. If the decimal is ≥0.9, then round up.c. All other decimal need to be multiplied by a whole

number until roundable.

6-8 Empirical Formula ExampleA compound is composed of 75.0% C and 25.0% H. Find its empirical formula.

Step #1 Assume 100 g of compound

75.0 g C

25.0 g H


Step #2 Convert grams to moles.

75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole H


Step #3 Divide each mole number by the smallest.

75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole H

6.225 6.2256.22524.802

= 1.00 = 3.98


Step #4 Rounding; Decimal ≤ 0.1, drop decimals

75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole H

6.225 6.2256.22524.802

= 1.00 = 3.98


Step #4 Rounding; Decimal ≤ 0.1, drop decimals

75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole H

6.225 6.2256.22524.802

= 1 C = 3.98


Step #4 Rounding; Decimal ≥ 0.9, round up

75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole H

6.225 6.2256.22524.802

= 1 C = 3.98



75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole H

6.225 6.2256.22524.802

= 1 C = 3.98



75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole H

6.225 6.2256.22524.802

= 1 C = 4 H



75.0 g C

25.0 g H12.01 g

Mole C

1.008 g H

Mole H

= 6.225 mole C

= 24.802 mole H

6.225 6.2256.22524.802

= 1 C = 4 H

Empirical Formula = CH4

6-9 Molecular Formulas

Empirical formula, is the smallest ratio between atoms in a molecular or formula unit.

Molecular formula, is the exact number of atoms in a molecule; a whole number multiple of an empirical formula

6-9 Possible Molecular Formulas

Assume an empirical formula of C3H5O

Empirical formula Integer Molecular Formula

C3H5O 1

2

3

4

5

C3H5O

C3H5O

C3H5O

C3H5O

C3H5O




C3H5O 1

2

3

4

5

C3H5O

C3H5O

C3H5O

C3H5O

C3H5O

C6H10O2




C3H5O 1

2

3

4

5

C3H5O

C3H5O

C3H5O

C3H5O

C3H5O

C6H10O2

C9H15O3




C3H5O 1

2

3

4

5

C3H5O

C3H5O

C3H5O

C3H5O

C3H5O

C6H10O2

C9H15O3

C12H20O4

C15H25O5

6-9 Sample ProblemCalculate the molecular formula of a molecule composed of 83.7%C and 16.3% H, with a molar mass of 86.0 g/mole


Step #1 Assume 100g of compound

83.6 g C

16.3 g H


Step #2 Convert grams to moles

83.6 g C

16.3 g H12.01 g Cmole

mole1.008 g H


Step #2 Convert grams to moles

83.6 g C

16.3 g H12.01 g Cmole

mole1.008 g H

= 6.961 mole

= 16.17 mole



83.6 g C

16.3 g H12.01 g Cmole

mole1.008 g H

= 6.961 mole

= 16.17 mole



83.6 g C

16.3 g H12.01 g Cmole

mole1.008 g H

= 6.961 mole

= 16.17 mole

6.961 6.961

= 1.00

= 2.32


Step #4 Round if---Not Roundable

83.6 g C

16.3 g H12.01 g Cmole

mole1.008 g H

= 6.961 mole

= 16.17 mole

6.961 6.961

= 1.00

= 2.32

Step #4, Multiply by an integer until roundable

1.00 X 3 = 3

2.32 X 3 = 7Empirical formula C3H7

6-9 Molecular Formula IntegerDivide empirical weight into molecular weight

3x12 + 7x1 =43

43 862

Now multiply the empirical formula by 2

6-9 Molecular Formula IntegerDivide empirical weight into molecular weight

3x12 + 7x1 =43

43 862

Now multiply the empirical formula by 2

Molecular Formula is C6 H14

The End

Documents

Chapter #6 Chemical Composition. Chapter Contents 6-1 Chemical Composition 6-2 Counting Units 6-3 Counting Atoms By the Gram 6-4 Counting Molecules by