chapter 6 chemical composition
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Chapter 6 Chemical Composition. 2006, Prentice Hall. CHAPTER OUTLINE. Why is Knowledge of Composition Important?. everything in nature is either chemically or physically combined with other substances - PowerPoint PPT PresentationTRANSCRIPT
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