chapter 6 chemical composition 2006, prentice hall

Chapter 6 Chemical Composition

2006, Prentice Hall

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CHAPTER OUTLINE

The Mole Concept Molecular & Molar Mass Calculations Using the Mole Percent Composition Chemical Formulas Calculating Empirical Formulas Molecular Formulas

Tro's Introductory Chemistry, Chapter 6

Why is Knowledge of Composition Important?

• everything in nature is either chemically or physically combined with other substances

• to know the amount of a material in a sample, you need to know what fraction of the sample it is

• Some Applications:– the amount of sodium in sodium chloride for diet– the amount of iron in iron ore for steel production– the amount of hydrogen in water for hydrogen fuel– the amount of chlorine in freon to estimate ozone

depletion

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THE MOLECONCEPT

Chemists find it more convenient to use mass relationships in the laboratory, while chemical reactions depend on the number of atoms present.

In order to relate the mass and number of atoms, chemists use the SI unit mole (abbreviated mol).

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THE MOLECONCEPT

The number of particles in a mole is called Avogadro’s number and is 6.02x1023.

1 mole equals to 6.02 x 1023

Avogadro’s number

(NA)

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THE MOLECONCEPT

A mole is a very large quantity

6.02x1023

If 10,000 people started to count Avogadro’s number and counted at the rate of 100 numbers per minute

each minute of the day, it would take over 1 trillion years to count

the total number.

MOLE

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THE MOLECONCEPT

1 mole H2 molecules = 6.02x1023 H2 molecules

= 2 x (6.02x1023) H atoms

1 mole Na+ ions

1 mole H2O molecules = 6.02x1023 H2O molecules

= 2 x (6.02x1023) H atoms

= 6.02x1023 O atoms

= 6.02x1023 Na+ ions

1 mole H atoms = 6.02x1023 H atoms

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THE MOLECONCEPT

The atomic mass of one atom expressed in amu is numerically the same as the mass of 1 mole of atoms of the element expressed in grams.

Mass of 1 H atom = 1.008 amu

Mass of 1 mol H atoms = 1.008 grams

Mass of 1 Mg atom = 24.31 amu

Mass of 1 mol Mg atoms = 24.31 g

Mass of 1 Cl atom = 35.45 amu

Mass of 1 mol Cl atoms = 35.45 g

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MOLEDAY

Chemists and chemistry students celebrate two days in the year in honor of the Mole and call them Mole Days.

October 23rd

Jun 2nd

6:02 a.m.

10:23 a.m.

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MOLECULAR MASS

The sum of atomic masses of all the atoms in one molecule of a substance is called molecular mass, and is measured in amu.

1 O atom =

2 H atoms =

Mass of one molecule of H2O

2 (1.008 amu) = 2.016 amu1 (16.00 amu) = 16.00 amu

18.02 amuMolecular mass

Relationship Between Moles and Mass

• The mass of one mole of atoms is called the molar mass

• The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu

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MOLAR MASS

1 mol O atom =

2 mol H atoms =

Mass of one mole of H2O

2 (1.008 g) = 2.016 g

1 (16.00 g) = 16.00 g

18.02 gMolar mass

The mass of one mole of a substance is called molar mass, and is measured in grams.

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CALCULATIONSUSING THE MOLE

Conversions between mass, mole and particles can be done using molar mass and Avogadro’s number.

Mass of a substance

Moles of a substanceMM

Particles of a

substance

NA

Molar mass

Avogadro’s number

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What is the mass of 5.00 mol of water?

Example 1:

3 significant figures

2 2

g5.00 mol H O x g H O

mol=

118.02

90.1

Molar mass

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How many Mg atoms are present in 5.00 g of Mg?

Example 2:

mol5.00 g Mg x

g1

24.3

mass mol atoms

atomsx =

mol

Avogadro’s number

1

6.02 x 1023

1.24x1023 atoms MgMolar mass3 significant figures

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How many molecules of HCl are present in 25.0 g of HCl?

Example 3:

mol25.0 g HCl x

g

1

36.45

mass mol molecules

moleculesx =

mol1

6.02 x 1023

4.13 x 1023 molecules HCl

3 significant figures

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PERCENTCOMPOSITION

The percent composition of a compound is the mass percent of each element in the compound.

(no. of X in formula) x (molar mass of X)Mass % X = x 100

molar mass of compound

Total mass of element

Total mass of compound

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Calculate the percent composition of sodium chloride (NaCl).

Example 1:

Step 1: determine molar mass of NaCl

1 mol Na atoms =

1 (22.99 g) = 22.99 g1 mol Cl atom

=

1 (35.45 g) = 35.45 g

58.44 g/mol

Molar mass

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Example 1:

22.99 g x 100 = 39.34%

58.44 g

35.45 g x 100 = 60.66%

58.44 g

Step 2: calculate the mass % of each element

% Na =

% Cl =

Sum = 100%

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1.63 g of zinc combines with 0.40 g of oxygen to form zinc oxide. Determine the % composition of the compound formed.

Example 2:

Step 1: determine total mass of sample

1.63 g + 0.40 g = 2.03 g mass of sample =

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Example 2:

1.63 g x 100 = 80.3%

2.03 g

Step 2: calculate the mass % of each element

% Zn =

% O = 0.40 g

x 100 = 19.7% 2.03 g

Sum = 100%

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Calculate the percent composition of sodium hydroxide (NaOH).

Example 3:

Step 1: determine molar mass of NaOH

1 mol Na atoms = 1 (23.0 g) = 23.0 g

1 mol O atom = 1 (16.0 g) = 16.0 g

1 mol H atoms = 1 (1.01 g) = 1.01 g

40.0 g/molMolar mass

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Calculate the percent composition of sodium hydroxide (NaOH).

Example 3:

Step 1: determine molar mass of NaOH

1 mol Na atoms = 1 (23.0 g) = 23.0 g

1 mol O atom = 1 (16.0 g) = 16.0 g

1 mol H atoms = 1 (1.01 g) = 1.01 g

40.0 g/molMolar mass

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CHEMICALFORMULAS

Molecular formula

Empirical formula

Shows the actual number of atoms in a compound

Shows the simplest ratio of atoms in a

compound

Can be written for molecular

compounds only

Can be written for molecular and

ionic compounds

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CHEMICALFORMULAS

Molecular Empirical Multiplier

H2O2

H2O

C6H12O6

C6H6

C2H2

C6H12

HO 2

H2O 1

CH2O 6

CH 6

CH 2

CH2 6

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CHEMICALFORMULAS

Several compounds may possess the same percent composition and empirical formula, but different molecular formulas.

FormulaComposition

% C % H

Molar mass

(g/mol)

CH 92.3 7.7

C2H2 92.3 7.7

C6H6 92.3 7.7

13.02

26.04 (2x13.02)

78.12 (6x13.02)

Finding an Empirical Formula1) convert the percentages to grams

a) skip if already grams

2) convert grams to molesa) use molar mass of each element

3) write a pseudoformula using moles as subscripts

4) divide all by smallest number of moles5) multiply all mole ratios by number to make all

whole numbersa) if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67,

multiply all by 3, etc. b) skip if already whole numbers

If, after dividing by the smallest number of moles, the subscripts are not whole numbers, multiply all the

subscripts by a small whole number to arrive at whole-number subscripts.

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CALCULATINGEMPIRICAL FORMULAS

Arsenic (As) reacts with oxygen (O) to form a compound that is 75.7% As and 24.3% oxygen by mass. What is the empirical formula for this compound?

Step 1: Percent to mass

75.7% As 75.7 g As

24.3% O 24.3 g O

Assume 100 g

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CALCULATINGEMPIRICAL FORMULAS

1 molx = 1.01 mol As

74.9 g

Step 2: Mass to mole

75.7 g As

24.3 g O1 mol

x = 1.52 mol O16.0 g

Use atomic mass of oxygen

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CALCULATINGEMPIRICAL FORMULAS

1.01 mol= 1.00

1.01 mol

Step 3: Divide by small

1.52 mol= 1.50

1.01 mol

As =

O =

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CALCULATINGEMPIRICAL FORMULAS

Step 4: Multiply till Whole

As1.00O1.50

x 2 = 2 x 2 = 3

As2O3

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Determine the empirical formula for a compound containing 11.2% H and 88.8% O.

Example 1:

11.2 g1 mol

x 1.01 g

= 11.1 mol H

88.8 g1 mol

x 16.0 g

= 5.55 mol O

11.1= 2.0

5.55mol H =

mol O =5.55

= 1.05.55

H2O

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Determine the empirical formula for a compound with the following percent composition: 52.14% C, 13.12% H, 34.73% O.

Example 2:

52.14 g1 mol

x 12.0 g

= 4.345 mol C

34.73 g1 mol

x 16.0 g

= 2.17 mol O

4.345= 2.0

2.17mol C =

mol H =

2.17= 1.0

2.17

C2H6O

mol O =

13.12 g1 mol

x 1.01 g

= 13.0 mol H13.0

= 6.02.17

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MOLECULARFORMULAS

Molecular formula can be calculated from empirical formula if molar mass is known.

Molecular formula = (empirical formula) n

Molar massn (multiplier) =

Mass of empirical formula

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A compound of N and O with a molar mass of 92.0 g, has the empirical formula of NO2. What is its molecular formula?

Example 1:

Mass of empirical formula = 14.0 + 2(16.0) = 46.0

92.0 gn = = 2

46.0 g

Molecular formula = 2 x (NO2) = N2O4

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Calculate the empirical and molecular formulas of a compound that contains 80.0% C and 20.0% H, and has a molar mass of 30.00 g.

Example 2:

80.0 g1 mol

x 12.0 g

= 6.667 mol C6.667

= 1.06.667

mol C =

mol H =

CH3

20.0 g1 mol

x 1.01 g

= 19.8 mol H19.8

= 3.06.667

Empirical formula

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Example 2:

Mass of empirical formula = 12.0 + 3(1.01) = 15.0

30.0 gn = = 2

15.0 g

Molecular formula = 2 x (CH3) = C2H6

Empirical formula = CH3

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THE END