Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
Introductory Chemistry, 3rd Edition
Nivaldo Tro
Chapter 6
Chemical
Composition
2009, Prentice Hall
Chapter opening figure
Outline• 6.1 - How much Sodium?
• 6.2 - Counting Nails by the Pound.
• 6.3 – Counting Atoms by the Gram.
• 6.4 – Counting Molecules by the Gram.
• 6.5 – Chemical Formulas as Conversion Factors
• 6.6 – Mass Percent Composition of Compounds
• 6.7 – Mass Percent Composition and Chemical Formula
• 6.8 – Calculating Empirical Formulas for Compounds
• 6.9 – Calculating Molecular Formulas for Compounds
Tro's "Introductory Chemistry",
Chapter 6
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6.1 - How much Sodium?
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Chapter 6
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Tro's "Introductory Chemistry",
Chapter 6
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Why Is Knowledge of
Composition Important?
• It is important to know the fraction (or %) of
material in a substance.
• Some Applications:
sodium in foods for diet.
iron in iron ore for steel production.
hydrogen in water for hydrogen fuel.
chlorine in freon to estimate ozone depletion.
Figures of
Salt
Water
And
Freon
6.2 - Counting Nails by the
Pound.
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Chapter 6
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Tro's "Introductory Chemistry",
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Counting Nails by the Pound
• Nails are not counted individually.
• Consider nails 500 nails per lb.
• Need 1500 nails need 3.000 lb
• What if you but 0.750 lb. How
many nails?
• 500 nails/1 lb or 1 lb/500 nails can be
used as a conversion factor.
• 0.75 lb nails x 500 nails = 375 nails
1lb nails
nails
Tro's "Introductory Chemistry",
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Counting Nails by the Pound,
Continued• What if we bought a different size nail?
A smaller nail would have a different number of
nails per pound, for example 1500 nails per lb
0.750 lb of these smaller nails.
• 0.75 lb nails x 1500 nails = 1125 nails
1lb nails
• Figure 6.1 big nails and small nails
6.3 – Counting Atoms by the
Gram.
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Chapter 6
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Tro's "Introductory Chemistry",
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Counting Atoms by Moles
• In the same way that we count nails by the
pounds we count atoms by the mole
• 6.022 x 1023 atoms is equal to a mole.1 mole = 6.022 x 1023 things.Like 1 dozen = 12 things.
Avogadro’s number.
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Chemical Packages—Moles
• Mole = equal to the number of atoms in 12 g of C-12.
1 atom of C-12 weighs exactly 12 amu.
1 mole of C-12 weighs exactly 12 g.
• In 12 g of C-12 there are 6.022 x1023 C-12 atoms.
• Atomic mass in amu is numerically equal to molar mass in
grams
mole 1
atoms 10022.6 23atoms 10022.6
mole 123
Molar Mass and Avogadro’s
Number as Conversion Factors• Molar mass has units of g
mole
conversions grams moles
• Avagadro’s number units 6.022 x 1023 particles
1 mole
conversions particles moles
Particles may be atoms, molecules, ions, protons, etc.
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How Many mol are There in 10.0 g
of Al?
• g Al mol Al
• Atomic mass is the conversion factor
• 10.0 g Al x 1 mol Al = 0.371 mol Al
26.983 g Al
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How Many Al atoms are There in
10.0 g of Al?
Tro's "Introductory Chemistry",
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• g Al mol Al number Al atoms
•The moles was determine on the previous slide.
•0.371 mol Al x 6.022 x 1023 Al atoms = 2.23 x 1023 Al atoms
1 mol Al
Summary of conversions
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g of A mol of A # particles of A
÷ molar mass
x molar mass
x 6.022 x 1023
÷ 6.022 x 1023
Need 3.75 mol of S for a reaction.
How many g of S is this?
How many S atoms is this
• mol S g S and mol S number S atoms
• Atomic mass is the conversion factor
• 3.75 mol S x 32.07 g S = 120 g S
1 mol S3.75 mol S x 6.022 x 1023 S atoms = 2.26 x 1024 S atoms
1 mol S
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6.4 – Counting Molecules by the
Gram.
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Counting molecules by the gram
• We can count molecules by the gram
• The molar mass is the conversion factor
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Example counting molecules by
the gram
• Dihydrogren sulfide, H2S, is the smell of
rotten eggs
• 12.0 g of H2S is how many g and now many
molecules.
• Need the molar mass
• (2 x H) + (1 x S)
• (2 x 1.008 g) + (1 x 32.07 g) = 34.09 g/mol
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Example counting molecules by
the gram (cont.)• 12.0 g H2S x 1 mol H2S = 0.352 mol H2S
34.09 g H2S0.352 mol H2S x 6.022 x 1023 H2S molecules = 2.12 x 1023 H2S
1 mol H2S molecules
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6.5 – Chemical Formulas as
Conversion Factors
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Chemical Formulas can be Used
to Convert to Amount of an
Element
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•1 molecule H2O
•2 atoms of H
•1 atom of O
•1 mole H2O
•2 mole H
•1 mole O
•1 mole C3H6O3
•3 mole C
•6 mole H
•3 mole O
In 25.0 g of glucose, C6H12O6,
How Many g of Oxygen is there?
• g glucose mol glucose mol O g O
• Molar mass glucose = 180.16 g/mol• 25.0 g glucose x 1 mol glucose = 0.139 mol glucose
180.16 g glucose
• 0.139 mol glucose x 6 mol O = 0.834 mol O
1 mol glucose
• 3.75 mol O x 16.00 g O = 13.3 g O
1 mol O
•
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6.6 – Mass Percent Composition
of Compounds
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Definition of Mass Percent
• Mass % mass of element in compound
of = X 100
element total mass of the compound
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6.7 – Mass Percent Composition
and Chemical Formula
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Mass % from formula
• Mass % may be calculated conveniently
from the formula of a compound.
• Mass % mass of element in one mole
of = X 100
element molar mass of compound
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What is the mass % of each
element in Na2CrO4
• Molar mass = (2 x Na) + (1 x Cr) + (4 x O)• (2 x 22.99) + (52.00) + (4 x 16.00) g = 161.98 g
(45.98 g Na) (64.00 g O) mol Na2CrO4
• % Na = 45.98 g Na x 100 = 28.39 % Na
161.98 g total
• % Cr = 52.00 g Cr x 100 = 32.10 % Cr
161.98 g total
• % Na = 64.00 g O x 100 = 39.51 % O
161.98 g total
Note the sum of the % is equal to 100.00 %27
6.8 – Calculating Empirical
Formulas for Compounds
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Tro's "Introductory Chemistry",
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Empirical Formulas
• The simplest, whole-number ratio of atoms in a molecule is called the empirical formula.
Can be determined from percent composition or combining masses.
• The molecular formula is a whole number multiple of the empirical formula. (usually called n)
% A mass A (g) moles A100g MMA
% B mass B (g) moles B100g MMB
moles A
moles B
30
Empirical Formulas, Continued
Benzene
Molecular formula = C6H6
Empirical formula = CH
Glucose
Molecular formula = C6H12O6
Empirical formula = CH2O
Hydrogen Peroxide
Molecular formula = H2O2
Empirical formula = HOMolecules shown
On page 179
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Practice—Determine the Empirical
Formula of Benzopyrene, C20H12.
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Practice—Determine the Empirical Formula of
Benzopyrene, C20H12, Continued
• Find the greatest common factor (GCF) of the
subscripts:
20 and 12 are both divisible by 4.
• Divide each subscript by 4 to get the empirical
formula.
C20H12 = (C5H3) x4
Empirical formula = C5H3
33
Finding an Empirical Formula
from Experimental Data1. Convert the percentages to grams. (assume 100 g to
make the math easy.
a. Skip if already grams.
2. Convert grams to moles.
a. Use molar mass of each element.
3. Write a pseudoformula using moles as subscripts.
4. Divide all by smallest number of moles At least the smallest one will be a whole number.
5. Multiply all mole ratios by number to make all whole numbers, if necessary.
a. If in the ratio fractional part is ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc.
b. Skip if already whole numbers after Step 4.
Examples from Experimental
Data•An oxide of Nickel, Ni, is analyzed and found to
contain 0.2636 g of Ni and 0.0718 g of O.
Determine the empirical fomula.g of mol of pseudoformuladivide by clear
elements elements smallest fractions
0.2636 g Ni x 1 mol Ni = 0.004491 mol Ni
58.69 Ni
0.0718 g O x 1 mol O = 0.00449 mol O
16.00 OTro's "Introductory Chemistry",
Chapter 6
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Examples from Experimental
Data (cont)• A pseduoformula has the correct ratio of moles of
compound, but may not ne whole numbers. It is
not correct because a formula can not contain a
fraction of an atom
• Ni0.04491O0.0449
• Divide by smallest to at least convert one element
to a whole number 1
• Ni0.04491O0.0449 Ni1O1 NiO empirical
• formula
• 35
0.04490.0449
More Examples
• Cisplatin
65.02 % Pt, 9.34 % N, 2.02 % H, 23.63 % Cl
• Naproxen (active ingredient in Aleve)
73.03 % C, 6.13 % H, 20.85 % O,
Tro's "Introductory Chemistry",
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Cisplatin
• Cisplatin
65.02 % Pt, 9.34 % N, 2.02 % H, 23.63 % Cl
• We assume 100 g of compound to make the
conversion to g easy. In 100 g of Cisplatin there
are
• 65.02 % Pt
• 9.34 % N
• 2.02 % H
• 23.63 % Cl
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Cisplain (cont)
• Convert each of these mass to moles
• 65.02 g Pt x 1 mol Pt = 0.3333 mol Pt
195.1 g Pt
• 9.34 g N x 1 mol N = 0.667 mol Pt
14.01 g Pt
• 2.02 g H x 1 mol H = 2.00 mol Pt
1.008 g H
• 23.63 g Cl x 1 mol Cl = 0.6667 mol Pt
35.45 g ClTro's "Introductory Chemistry",
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Cisplain (cont)
• Pseudoformula
• Pt0.333N0.667H2.00Cl0.6667
• Divide by smallest 0.3333
• Pt0.3333N0.667 H2.00 Cl0.6667
Tro's "Introductory Chemistry",
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0.3333 0.3333 0.3333 0.3333
PtN2H6Cl2 empirical formula
Naproxen
• Naproxen (active ingredient in Aleve)
73.03 % C, 6.13 % H, 20.85 % O,
• Assume 100 g convert to moles
73.03 g C 6.08 mol C
6.13 g H 6.08 mol H
20.85 g O 1.302 mol O
• Pseudoformula C6.08H6.08O1.303
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Naproxen (cont)
• Dividc by the smallest
• C6.08H6.08O1.303 C4.67H4.67O1
• Not all are whole numbers but 4.67 = 4 and 2/3
• We can remove fractions by multiplying all
subscript by 3. --? C14H14O3 empirical formula
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1.303 1.3031.303
Fractions in empirical formula
calculations• When dividing by the smallest number of moles fractions often
occur
• Learn to recognize common fractions
• 0.5 = ½
• 0.33 = 1/3 0.67 = 2/3
• 0.25 = 1/4 0.75 = ¾
• 0.20 = 1/5 0.40 = 2/5 0.60 = 3/5 0.80 = 4/5
• In each case multiply by the denominator to
clear the fraction 42
6.9 – Calculating Molecular
Formulas for Compounds
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All These Molecules Have the Same
Empirical Formula. How Are the
Molecules Different?
Name Molecular
Formula
Empirical
Formula
Glyceraldehyde C3H6O3 CH2O
Erythrose C4H8O4 CH2O
Arabinose C5H10O5 CH2O
Glucose C6H12O6 CH2O
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All These Molecules Have the Same Empirical
Formula. How Are the Molecules Different?,
ContinuedName Molecular
Formula
Empirical
Formula
Molar
Mass, g
Glyceraldehyde C3H6O3 CH2O
Empirical
Mass = 30
90 =30 x 3
Erythrose C4H8O4 CH2O 120 = 30 x 4
Arabinose C5H10O5 CH2O 150 =30 x 5
Glucose C6H12O6 CH2O 180 = 30 x 6
Notice the molar mass is a multiple of the empirical mass.
Molecular Formula
• Compound has empirical formula C3H2N
and molar mass 312.39. What is the
molecular formula?
• Empirical mass (3 x C) + (2 x H) + (1 x N)
• (3 x 12.01) + (2 x 1.008) + (14.01) = 52.06 g
• Molar mass = n 312.39 g = 6.00
Empirical mass 52.06 g
Molecular formula = C3H2N x 6 = C18H12N6
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