2
Chemical Formulas and
Composition Stoichiometry
化學計量法
1
Chapter Goals
 Chemical Formulas 化學式
 Ions and Ionic Compounds 離子及離子化合物
 Names and Formulas of Some Ionic Compounds
 Atomic Weights 原子重
 The Mole 莫耳數
 Formula Weights, Molecular Weights, and Moles
 Percent Composition and Formulas of Compounds
 Derivation of Formulas from Elemental Composition
 Determination of Molecular Formulas 分子式
 Some Other Interpretations of Chemical Formulas
 Purity of Samples
2
Chemical Formulas 化學式
• Chemical formula shows the chemical
composition of the substance.
– ratio of the elements present in the molecule
or compound
• He, Ne, Na – monatomic elements
• O2, H2, Cl2 – diatomic elements
• O3, S4, P8 - more complex elements
• H2O, C12H22O11 – compounds
• Substance consists of two or more elements
3
Chemical Formulas
Allotropic modifications (allotropes) 同素異形物:
different forms of the same element in the
same physical state (chapter 13)
4
Chemical Formulas
• Compound contain two or more elements in
chemical combination in fixed proportions
-Law of Definite Proportions 定組成定律
2 H atoms
1 O atom
1 H atom
1 Cl atom
1 N atom
3 H atoms
丙烷
3 C atoms
8 H atoms
4 C atoms
10 H atoms
1 O atom
乙醚
Organic compounds: contain C―C or C―H bonds or both, often
in combination with nitrogen, oxygen, sulfur and
other elements
Inorganic compounds: do not contain C―H bonds
5
Chemical Formulas
Chemical formula
(球棍模型)
(比例模型)
the number of atoms of each
type in the molecule
Structural Formula
the order in which atoms are
connected
(chemical bonds between atoms)
 Ball-and-Stick
Model
three-dimensional shape of
molecules
四氯化碳
 Space-Filling Model
Relative size of atoms and the
shapes of molecules
Fig 2-1 Formula and models for some molecules
6
Example 2-1 Chemical Formula
Look at each of following molecular models. For each one, write
the structural formula and the chemical formula.
(color code: black=carbon; white=hydrogen; red=oxygen;
blue=nitrogen; light green=fluorine; dark green=chloride.)
a)1-butanol丁醇
Occurs in some fruits, dried
beans, cheese, and nuts; used
as an additive in certain
plastics, detergents, and some
medicinal formulations
b)Freon-12 二氯二氟代甲烷
A highly
toxic substance,
used
as
Formerly
used as a(氟氯烷)
refrigerant;
a chemotherapy
drug
in the
implicated in
atmospheric
treatment ofozone
Hodgkin’s
disease
depletion
and of some forms of chronic
leukemia
c)Nitrogen mustard HN1
C H O N F Cl
H H H H
H C C C C O H C4H10O
H H H H
Cl
F C Cl
F
CF2Cl2
H H H H H
Cl C C N C O Cl
H H
H H
C4H9NCl2
Ions and Ionic Compounds
•Ions are atoms or groups of atoms
that possess an electric charge
•Two basic types of ions:
Positive ions or cations 正離子
one or more electrons less than neutral
Na+, Ca2+, Al3+
NH4+ - polyatomic cation
Negative ions or anions 負離子
one or more electrons more than neutral
F-, O2-, N3SO42-, PO43- - polyatomic anions
Formula Unit: the small repeat of a substancefor non-ionic substances, the
molecule NaCl, CaCl2
Fig 2-2 The arrangement of ions in NaCl
8
Ions and Ionic Compounds
Polyatomic ion
(NH4)2SO4
Polyatomic compound
Metals form
more than one
kind of ion
9
Names and Formulas of
Some Ionic Compounds
• Formulas of ionic compounds are determined by
the charges of the ions
Charge on the cations = the charge on the anions
(add to zero)
– The compound must be neutral
NaCl
sodium chloride
(Na1+ & Cl1-)
KOH
potassium hydroxide (K1+ & OH1-)
CaSO4
calcium sulfate
(Ca2+ & SO42-)
Al(OH)3 aluminum hydroxide (Al3+ & 3 OH1-)
10
Names and Formulas of
Some Ionic Compounds
You must know all of
the molecular
compounds from
Table 2-2.
Some examples are:
H2SO4 - sulfuric acid
FeBr2 - iron(II) bromide
C2H5OH - ethanol
H+
1+
hydrogen
11
Example 2-2 Formulas for Ionic Compounds
Write the formula of the following ionic compounds: (a) sodium
fluoride, (b) calcium fluoride, (c) iron(II) sulfate, (d) zinc
phosphate
(a) Na+ F(b) Ca2+ F- (c) Fe2+ SO42(d) Zn2+ PO43CaF2
NaF
FeSO4
Zn3(PO4)2
Example 2-3 Name for Ionic Compounds
Name the following ionic compounds: (a) (NH4)2S, (b)
Cu(NO3)2, (c) ZnCl2, (d) Fe2(CO3)3
(a) NH4+ S2ammonium sulfide
(c) Zn2+ ClZnic Chloride
(b) Cu2+ NO3copper(II) nitrate
(d) Fe3+ CO32iron(III) carbonate
Names and Formulas of
Some Ionic Compounds
You do it!
• What is the formula of nitric acid?
HNO3
• What is the name of FeBr3?
iron(III) bromide
• What is the name of K2SO3?
potassium sulfite
• What is charge on sulfite ion?
SO32- is sulfite ion
13
Names and Formulas of
Some Ionic Compounds
You do it!
• What is the formula of ammonium sulfide?
(NH4)2S
• What is charge on ammonium ion?
NH41+
• What is the formula of aluminum sulfate?
Al2(SO4)3
• What is charge on both ions?
Al3+ and SO42-
14
Atomic Weights 原子量
Atomic weights (AW)
– an early observation was that
carbon and hydrogen have relative
atomic masses, of approximately 12
and 1
Atomic mass unit (amu)
– exactly 1/12 of the mass of a
particular kind of carbon atom,
called carbon-12
H Hydrogen
1.00794 amu
Na Sodium
22.989768 amu
Mg Magnesium 24.305 amu
15
The Mole 莫耳數
1 mole = 6.022 x 1023 particles
Avogadro’s number (NA) = 6.022 x 1023
亞佛加厥常數
Helium exists as discrete He atom
 1 mole of He consist of 6.022x1023 He atoms
Hydrogen commonly exists as diatomic (two-atom) H2
 1 mole of Hydrogen is 6.022x1023 H2 molecule
(2x(6.022x1023) H atoms)
Molar mass (g/mol) 莫耳質量
the mass of 1 mole of atoms of a pure element in grams =
the atomic weight of the element in atomic mass unit
H has an atomic weight of 1.00794 g
1.00794 g of H atoms = 6.022 x 1023 H atoms
Mg has an atomic weight of 24.3050 g
24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms
16
The Mole
C 12
Ti 47.9
Au 197
H1
S 32
17
Table 2-3, p. 57
The Mole
One mole of atoms
Mercury (200.6g)
Bromine (79.9g)
Aluminum (27.0g)水銀
溴
鋁
Copper (63.5g)
銅
Sulfur (32.1g)
硫
Znic (65.4g)
鋅
iron (55.8g)
鐵
18
Fig. 2-4, p. 57
Example 2-4; 2-5 Moles of atoms; Numbers of atoms
How many moles of atoms does 136.9g of iron metal
contain? And How many atoms?
1 mole atoms
55.85g Fe
55.85g Fe
1 mole atoms
1mole Fe atoms
? Mole Fe atoms= 136.9g Fe x 55.85g Fe
=2.451 mol Fe atoms
6.022x1023 atoms and 1 mole atoms
1 mole atoms
6.022x1023 atoms
23 atoms
6.022x10
? Fe atoms= 2.451 mole atoms x
1 mole atoms
=1.476x1024 Fe atoms
Exercise 32, 40 and 42
Example 2-6 Masses of Atoms
Calculate the average mass of one iron atom in
grams.
? g Fe
55.85g Fe
1 mole Fe atoms
Fe atom = 1 mole atomsx 6.022x1023 Fe atoms
=9.274x10-23 g (the average mass of 1 Fe atom)
Example 2-7 Avogadro’s Number
A stack of 500 sheets of typing paper is 1.9 inches thick.
Calculate the thickness, in inches and in miles, of a stack
of typing paper that contains one mole of sheets.
23
? In.= 1 mole atoms sheets x 6.022x10 atoms x 1.9 in.
500 sheets
1 mole sheets
21
=2.3x10 in.
1ft x 1 mi.
? mi.= 2.3x1021 in. x 12
in. 5,280ft
=3.6x1016 mi.
The Mole
Example 2-1 Calculate the mass of a single Mg
atom in grams to 3 significant figures.
24.3g Mg
1
mole
Mg
atoms
? g Mg = 1 Mg atom x
x 1mole Mg atoms
23
6.022x10 Mg atoms
= 4.04 x10-23 g Mg
21
The Mole
Example 2-2 Calculate the number of atoms in
one-millionth of a gram of Mg to 3 significant
figures.
? Mg atoms =
1.00x10-6g
1mole Mg x 6.022x1023 Mg
Mg x
1 mole Mg
24.3g Mg
= 2.48 x10-16 Mg atoms
22
The Mole
Example 2-3 How many atoms are contained in 1.67
moles of Mg?
6.022x1023 Mg atoms
? Mg atoms = 1.67 mol Mg x
1 mol Mg
= 1.0 x1024 Mg atoms
23
The Mole
Example 2-4 How many moles of Mg atoms are
present in 73.4 g of Mg?
? mol Mg = 73.4g Mg x
1mole Mg
24.3g Mg
= 3.02 mol Mg
IT IS IMPERATIVE THAT YOU KNOW
HOW TO DO THESE PROBLEMS
24
Formula Weights, Molecular
Weights, and Moles 式量,分子量
及莫耳數
• How do we calculate the molar mass of a
compound?
– add atomic weights of each atom
Formula Weight (FW)
• The molar mass of propane (丙烷), C3H8, is:
3 x C = 3 x 12.01 amu = 36.03 amu
8 x H = 8 x 1.01 amu = 8.08 amu
Molar Mass
= 44.11 amu
25
Formula Weights, Molecular
Weights, and Moles
• The molar mass of calcium nitrate, Ca(NO3)2 , is:
You do it!
1 x Ca = 1 x 40.08 amu = 40.08 amu
2 x N = 2 x 14.01 amu = 28.02 amu
6 x O = 6 x 16.00 amu = 96.00 amu
Molar Mass
= 164.10 amu
26
Formula Weights, Molecular
Weights, and Moles
Molar mass of the substance
= the formula weight of the substance
27
Formula Weights, Molecular
Weights, and Moles
28
29
Formula Weights, Molecular
Weights, and Moles
• One Mole of
Contains
– Cl2 or 70.90g
– C3H8
6.022 x 1023 Cl2 molecules
2(6.022 x 1023 ) Cl atoms
You do it!
44.11 g
6.022 x 1023 C3H8 molecules
3 (6.022 x 1023 ) C atoms
8 (6.022 x 1023 ) H atoms
30
Formula Weights, Molecular
Weights, and Moles
Example 2-9 Masses of Molecules
What is the mass in grams of 10.0 million SO2 molecules?
? g SO2 = 1.0x106 SO2 molecules x
=1.06x10-15 g SO2
64.1g SO2
6.022x1023 SO2 molecule
Exercise 44
31
Formula Weights, Molecular
Weights,
and
Moles
Example 2-10 Moles
How many (a) moles of O2 (b) O2 molecules, and (c) O atoms are contained
in 40g of oxygen gas (dioxygen) at 25oC?
One mole of O2 contains 6x1023 O2 molecules, and
its mass is 32.0g
(a) ? Mol O2 = 40.0g O2 x 1 mol O2 = 1.25 mol O2
32.0g O2
23 O molecules
6.02x10
2
(b) ? O2 molecules =1.25 mol O2 x
1mol O2
23
= 7.52x10 O2 molecules
23 O molecules
atoms
6.02x10
2
x 1 O2 O
(c) ? O atoms = 40.0g O2 x
2 molecule
32.0g O2
=1.5x1024 O atoms
Exercise 36
32
Formula Weights, Molecular
Weights, and Moles
Example 2-11 Numbers of atoms
Calculate the number of hydrogen atoms in 39.6g of
ammonium sulfate, (NH4)2SO4
One mole of (NH4)2SO4 is 6x1023 formula units
and has a mass of 132.1g
g of
mol of
Formula units of
H atoms
(NH4)2SO4
(NH4)2SO4
(NH4)2SO4
? H atoms = 39.6g (NH4)2SO4 x 1 mol (NH4)2SO4
132.1g (NH4)2SO4
23 formula units (NH ) SO
8 H atoms
6.02x10
4
2
4
x
x
1 formula units (NH4)2SO4
1mol (NH4)2SO4
=1.44x1024 H atoms
Exercise 34
33
Formula Weights, Molecular
Weights, and Moles
Example 2-5: Calculate the number of C3H8
molecules in 74.6 g of propane
? C3H8 molecules = 74.6g C3H8 x
1mole C3H8
44.11g C3H8
23 C H molecules
6.022x10
3 8
x
1 mole C3H8
= 1.02 x1024 molecules
34
Formula Weights, Molecular
Weights, and Moles
Example 2-6. What is the mass of 10.0 billion
propane molecules?
?g C3H8 molecules =
1.00x1010
1 mole C3H8
molecules x
6.022x1023 C3H8 molecules
44.11g C3H8
x
1mole C3H8
= 7.32 x10-13 g of C3H8
35
Formula Weights, Molecular
Weights, and Moles
Example 2-7. How many (a) moles, (b) molecules, and
(c) oxygen atoms are contained in 60.0 g of ozone,
O3? The layer of ozone in the stratosphere is very
beneficial to life on earth.
(a) ? moles O3 = 60.0g O3 x
1mole = 1.25 moles
48.0 gO3
23 molecules
6.022x10
(b) ? molecules O3= 1.25 moles x
1 mole
= 7.53 x1023 molecules O3
(c) ? O atoms= 7.53x1023 molecules O3 x 3O atoms
1 O3 molecule
= 2.26 x1024 atoms O
36
Formula Weights, Molecular
Weights, and Moles
Example 2-8. Calculate the number of O atoms in
26.5 g of Li2CO3.
? O atoms = 26.5g Li2CO3 x
1 mol Li2CO3
73.8g Li2CO3
6.022x1023 form. units Li2CO3 x
3 O atoms
x
1 mole Li2CO3
1 form. unit Li2CO3
= 6.49 x1023 O atoms
37
Formula Weights, Molecular
Weights, and Moles
• Occasionally, we will use millimoles
– Symbol - mmol
milli10-3
– 1000 mmol = 1 mol 1 mmol = 10-3 mole
• For example: oxalic acid 草酸 (COOH)2
– 1 mol = 90.04 g
– 1 mmol = 0.09004 g or 90.04 mg
38
Formula Weights, Molecular
Weights, and Moles
Example 2-9 Calculate the number of mmol in
0.234 g of oxalic acid, (COOH)2.
? Mmol (COOH)2 = 0.234g (COOH)2 x
1mmole (COOH)2
0.09004g (COOH)2
= 2.6mmol (COOH)2
39
Percent Composition and
Formulas of Compounds
• % composition = mass of an individual
element in a compound divided by the total
mass of the compound x 100%
• Determine the percent composition of C in
C3H8.
Mass C
x 100%
%C=
Mass C3H8
3x12.01g
x 100% = 81.68%
=
3x12.01+8x1.01g
40
Percent Composition and
Formulas of Compounds
• What is the percent composition of H in C3H8?
Mass H x 100%
% H = Mass
C3H8
8x1.01g
x 100% = 18.32%
=
3x12.01+8x1.01g
41
Percent Composition and
Formulas of Compounds
Example 2-12 Percent Composition
Calculate the percent composition by mass of HNO3
1xH=1x1.0g=1.0g 1xN=1x14.0g=14.0g 3xO=3x16g=48g
Mass of 1 mol of HNO3= 1+14.0+48.0 = 63.0g
Mass C
%H=
x 100% =
Mass HNO3
Mass N
%N=
x 100% =
Mass HNO3
Mass O
%O=
x 100% =
Mass HNO3
1.0g x 100%
= 1.6% H
63.0g
14.0g x 100%
= 22.2% N
63.0g
48.0g x 100% = 76.2% O
63.0g
Total =100.0%
Exercise 62
42
Percent Composition and
Formulas of Compounds
Example 2-10 Calculate the percent composition of
Fe2(SO4)3 to 3 significant figures.
2x55.8g
2Fe
x 100% = 2x55.8+3x32.1+12x16.0 g x 100%
Fe2(SO4)3
= 27.9%
3S
x 100% = 2x32.1g x 100% = 24.1%
%S =
Fe2(SO4)3
399.9g
12O x 100% = 12x16g x 100% = 48.0%
%O=
Fe2(SO4)3
399.9g
% Fe =
43
Derivation of Formulas from
Elemental Composition
• Empirical Formula (實驗式) - smallest wholenumber ratio of atoms present in a compound
– CH2 is the empirical formula for alkenes
– No alkene exists that has 1 C and 2 H’s
• Molecular Formula - actual numbers of atoms of
each element present in a molecule of the
compound
– Ethene – C2H4
– Pentene – C5H10
(CH2)2
(CH2)5
• We determine the empirical and molecular
formulas of a compound from the percent
composition of the compound.
– percent composition is determined experimentally
44
Determination of Molecular
Formulas
• Butane C4H10  Empirical formula C2H5
2x (C2H5) = C4H10 Molecular Formula
• Benzene C6H6  Empirical formula CH
6x (CH) = C6H6
Molecular formula = n x simplest formula
Molecular weight = n x simplest formula weight
Molecular weight
n=
Simplest formula weight
45
Derivation of Formulas from
Elemental Composition
Example 2-13 Simplest Formulas
Compounds of containing sulfur and oxygen are serious air
pollutants; they represent the major cause of acid rain.
Analysis of a simple of a pure compound reveals that it
contains 50.1% sulfur and 49.9% oxygen by mass. What
is the simplest formula of the compound?
? mol S atoms = 50.1g S x 1 mol S atoms = 1.56 mol S atoms
32.1g S
? mol O atoms = 49.9g O x 1 mol O atoms = 3.12 mol O atoms
16.0g O
S mol : O mol = 1.56 mol S atoms : 3.12 mol O atoms
=1:2
SO2
Exercise 54
Derivation of Formulas from
Elemental Composition
Example 2-14 Simplest Formulas
A 20.882g sample of an ionic compound is found to
contain 6.072g of Na, 8.474g of S, and 6.336g of O.
What is its simplest formula?
? mol Na = 6.072g Na x 1 mol Na = 0.264 mol Na
23.0g Na
? mol S = 8.474g S x 1 mol S = 0.264 mol S
32.1g S
? mol O = 6.336g O x 1 mol O = 0.396 mol O
16.0g O
Na mol: S mol : O mol = 0.264 : 0.264 : 0.396
=2 : 2 : 3
Na2S2O3
Exercise 56
Derivation of Formulas from
Elemental Composition
Example 2-11 A compound contains 24.74% K, 34.76% Mn,
and 40.50% O by mass. What is its empirical formula?
Let the compound weight 100g
1 mol K
? mol K = 24.74g K x 39.10g K = 0.6327 mol K
1 mol Mn
? mol Mn = 34.76g Mn x 32.1g Mn = 0.6327 mol Mn
1 mol O
? mol O = 40.50g O x 16.0g O = 2.531 mol O
K mol: Mn mol : O mol = 0.6327 : 0.6327 : 2.531
=1 : 1 : 4
KMnO4
48
Derivation of Formulas from
Elemental Composition
Example 2-12: A sample of a compound contains 6.541g of
Co and 2.368g of O. What is the empirical formula for
this compound?
1 mol Co
? mol Co = 6.541g Co x 58.93g Co = 0.111 mol Co
? mol O = 2.638g O x 1 mol O = 0.148 mol O
16.0g O
Co mol: O mol = 0.111 : 0.148
=3:4
Co3O4
49
Derivation of Formulas from
Elemental Composition
Magnesium perchlorate 過氯酸鎂
Mg(ClO4)2 H
Sodium hydroxid  C
O??
A combustion train used for carbon-hydrogen analysis
燃燒反應器
50
Example 2-15 Percent Composition
Hydrocarbons are organic compounds composed
entirely of hydrogen and carbon. A 0.1647g sample
of pure hydrocarbon was burned in a C-H
combustion train to produce 0.4931g of CO2 and
0.2691g of H2O. Determine the masses of C and H in
the sample and the percentage of these elements in
this hydrocarbon. 12.01g C
= 0.1346 g C
? g C = 0.4931g C x
44.01g CO2
? g H = 0.2691g H x 1.01g H = 0.03010g H
18.1g H2O
? % C = 0.1346g C x 100% = 81.72% C
0.1647g sample
? % H = 0.03010g H x 100% = 18.28% H
0.1647g sample
Exercise 66
51
Example 2-16 Percent Composition
A 0.1014g sample of purified glucose was burned in a CH combustion train to produce 0.1486g of CO2 and
0.0609g of H2O. An elemental analysis showed that
glucose contains only carbon, hydrogen and oxygen.
Determine the masses of C,H, and O in the sample
and the percentage of these elements in glucose.
? g C = 0.1486g C x 12.01g C = 0.04055g C
44.01g CO2
? g H = 0.0609g H x 1.01g H = 0.00681g H
18.1g H2O
? g O = 0.1014 – [0.04055g C + 0.00681g H] = 0.0540g O
0.04055g C
? % C = 0.1014g sample x 100% = 39.99% C
0.00681g H
? % H = 0.1014g sample x 100% = 6.72% H
0.0540 g O
? % O = 0.1014g sample x 100% = 53.2% O Exercise 68
52
Determination of Molecular
Formulas
Example 2-17 Molecular Formula
In example 2-16, we found the elemental composition of
glucose. Other experiments show that its molecular
weight is approximately 180 amu. Determine the
simplest formula and the molecule formula.
Relative Mass Relative Number of Atoms
Element Of element
(divide mass by AW) Divide by Smallest
C
H
O
0.0405 = 0.003376 0.003376 = 1.00 C
12.01
0.003376
0.0068 = 0.00676 0.00676
0.00681
1.008
0.003376 = 2.00 H
0.0540 = 0.00338 0.00338
0.0540
= 1.00 O
16.00
0.003376
CH2O formula weight = 30.03 amu
The molecular weight of glucose weight 180 amu
n=180/30.03=6
Molecular weight = (CH2O)x6
0.04055
C6H12O6
CH2O
Exercise 49 and 50
53
Determination of Molecular
Formulas
Example 2-13: A compound is found to contain 85.63%
C and 14.37% H by mass. In another experiment its
molar mass is found to be 56.1 g/mol. What is its
molecular formula?
? mol C = 85.63g C x 1 mol C = 7.1 mol C
12.0g C
? mol H = 14.37g H x 1 mol H = 14.2 mol H
1.01g H
C mol: H mol = 7.1 : 14.2 = 1:2
Simplest formula CH2
1 mol = 56.1g/mol
Molecular formula = (Simplest formula) x
56.1 = (C+2xH)xX =(12+2x1.01)xX
X = 4  (CH2)4  C4H8
54
Determination of Molecular
Formulas
Law of Multiple Proportions 倍比定律
• When two elements, A and B, form more than one compound,
the ratio of the masses of element B that combine with a given
mass of element A in each of the compound can be expressed
by small whole numbers.
H2O H2O2 (1:2 oxygen ratio)
SO2 SO3 (2:3 oxygen ratio)
Example 2-18 Law of Multiple Proportions
What is the ratio of the numbers of oxygen atoms
that are combined with a given number of
nitrogen atoms in the compound N2O3 and NO?
Oxygen ratio =
N 2O 3
3O/2N 3O 3
= 2O/2N = 2O =
2
2x(NO)
Exercise 71 and 72
55
Some Other Interpretations of
Chemical Formulas
Example 2-19 Composition of Compounds
What mass of Chromium is contained in 35.8g of (NH4)2Cr2O7?
Mass (NH4)2Cr2O7
Mol (NH4)2Cr2O7
Mol Cr
Mass Cr
? mol (NH4)2Cr2O7 = 35.8g (NH4)2Cr2O7 x 1 mol (NH4)2Cr2O7
252.0g (NH4)2Cr2O7
= 0.142 mol (NH4)2Cr2O7
2 mol Cr atoms
? mol Cr atoms = 0.142mol (NH4)2Cr2O7 x
1 mol (NH4)2Cr2O7
= 0.284 mol Cr atoms
52.0g Cr
? g Cr =0.284 mol Cr atoms x
1 mol Cr atoms
= 14.8g Cr
Exercise 76
56
Some Other Interpretations of
Chemical Formulas
Example 2-20 Composition of Compounds
What mass of potassium chlorate, KClO3, would contain 40.0g
of oxygen?
Mass O
Mol O
Mol KClO3
Mass KClO3
1 mol KClO3 122.6g KClO3
x 1 mol KClO
? g KClO3 = 40g O x 1 mol O atoms x
16.0g O atoms 3 mol O atoms
3
= 102g KClO3
Exercise 78
57
Example 2-21 Composition of Compounds
(a) What mass of sulfur dioxide, SO2, would contain the same
mass of oxygen as is contained in 33.7g of arsenic pentoxide,
As2O5?
(b) What mass of calcium chloride, CaCl2, would contain the same
number of chloride ions as are contained in 48.6g of sodium
chloride, NaCl?
Some Other Interpretations of
Chemical Formulas
Mass As2O5
Mol As2O5
Mol O atoms
Mol SO2
Mass NaCl
Mol NaCl
Mol Cl- ions
Mol CaCl2
Mass SO2
Mass CaCl2
? g SO2 = 33.7g As2O5 x 1 mol As2O5 x5 mol O atoms
229.8g As2O5 1 Mol As2O5
64.1g SO2
1mol SO2
x
x
2 mol O atoms
1mol SO2 = 23.5g SO2
1
mol
Cl
1
mol
NaCl
? g CaCl2 = 48.6g NaCl x
x
58.4g NaCl
1 Mol NaCl
111.0g CaCl
1mol CaCl
x 2 mol Cl- 2 x 1mol CaCl 2 = 46.2g CaCl2
2
Exercise 78
58
Some Other Interpretations of
Chemical
Formulas
Hydrated copper (II)
Sulfate,
(CuSO4•5H2O)
1mol=249.7g
Anhydrous oxalic
acid, (COOH)2
1mol=90g
Mercury (II) oxide
1mol=216.6g
Water
H2O 18
Hydrated oxalic acid,
(COOH)2•2H2O)
1mol=126.1g
Hydrate : with water
Anhydrous: without water
59
Some Other Interpretations of
Chemical
Formulas
Hydrated copper (II) Sulfate,
copper (II) Sulfate,
(CuSO4•5H2O)
1mol=249.7g
(CuSO4)
1mol=159.6g
Blue
Gray
60
Example 2-22 Composition of Compounds
A reaction requires pure anhydrous calcium sulfate, CaSO4,
Only an unidentified hydrate of calcium sulfate,CaSO4•x
H2O, is available.
(a) We heat 67.5g of unknown hydrate until all the water has
been driven off. The resulting mass of pure CaSO4 is 53.4g.
What is the formula of the hydrate, and what is the
formula weight?
(b) Suppose we wish to obtain enough of this hydrate to
supply 95.5g of CaSO4 after heating. How many grams
should we weight out?
(a) ? g water driven off =67.5g CaSO4•x H2O - 53.4g CaSO4=14.1g H2O
Some Other
Interpretations of
Chemical Formulas
? mol H2O = 14.1g H2O x 1mol H2O x 136.2 g CaSO4 = 2
x = mol CaSO
18.0g H2O 1mol CaSO4
53.4g CaSO4
4
CaSO4 •2H2O
FW=136.2g/mol+2x(18.0g/mol)=172.2g/mol
172.2 g CaSO4•2H2 O
? g CaSO4 x 2H2O = 95.5g CaSO4 desired x
136.2g CaSO4
=121g CaSO4•2H2 O
61
Some Other Interpretations of
Chemical Formulas
Example 2-16 What mass of ammonium phosphate,
(NH4)3PO4, would contain 15.0 g of N?
Molar mass of (NH4)3PO4 = 149.0 g/mol
mol H = 1.07 mol N
? mol N = 15.0g N x 114.0g
N
1 mol (NH4)3PO4
1.07 mol N x
= 0.357 mole (NH4)3PO4
3mol N
149.0g (NH4)3PO4
0.357mol (NH4)3PO4 x
1 mol (NH4)3PO4 = 53.2g (NH4)3PO4
149.0g (NH4)3PO4
X g (NH4)3PO4
=
3x14 N
15g N
X =53.2g
62
Purity of Samples
• The percent purity of a sample of a substance is
always represented as
% purity =
Mass of pure substance
x100%
Mass of sample
Mass of sample includes impurities
Impurity: 1.8%
63
Purity of Samples
Example 2-23 Percent Purity
Calculate the masses of NaOH and impurities in 45.2g of
98.2% pure NaOH.
98.2 g NaOH
=44.4g NaOH
? g NaOH = 45.2g sample x
100g sample
? g impurities = 45.2g sample x 1.8 g impurities =0.81g impurities
100g sample
64
Purity of Samples
Example 2-18 A bottle of sodium phosphate,
Na3PO4, is 98.3% pure Na3PO4. What are the
masses of Na3PO4 and impurities in 250.0 g of
this sample of Na3PO4?
98.3 g Na3PO4
? g Na3PO4 = 250.0g sample x
=246g Na3PO4
100g sample
? g impurities = 250.0g sample x
1.8 g impurities
= 4g impurities
100g sample
or
? g impurities = 250.0g sample - 246g Na3PO4 = 4g impurities
65