물질의
구성
제
2장
물질의
구성
. 원소, 화합물, 혼합물: 원자
관점에서의
개요
. 원자설의
배경
. 돌턴의
원자설
. 원자핵, 전자
. 현대의
원자론
. 원소: 주기율표
. 화합물과
결합
. 화합물: 화학식, 명칭, 질량
. 혼합물의
분류
Chapter 2: The Components of Matter
2.1 Elements, Compounds, and Mixtures: An Atomic Overview
2.2 The Observations That Led to an Atomic View of Matter
2.3 Dalton’s Atomic Theory
2.4 The Observations That Led to the Nuclear Atom Model
2.5 The Atomic Theory Today
2.6 Elements: A First Look at the Periodic Table
2.7 Compounds: Introduction to Bonding
2.8 Compounds: Formulas, Names, and Masses
2.9 Classification of Mixtures
Definitions for Components of Matter
Element -the simplest type of substance with unique physical and
chemical properties. An element consists of only one type of atom. It
cannot be broken down into any simpler substances by physical or
chemical means.
Molecule -a structure that consists of two or
more atoms that are chemically bound together
and thus behaves as an independent unit.
Figure 2.1
Definitions for Components of Matter
Compound -a substance
composed of two or more elements
which are chemically combined.
Figure 2.1
Mixture -a group of two or more
elements and/or compounds that
are physically intermingled.
원자설의
배경
.
고전적
견해원자설
이전
.
Law of Conservation of Mass
In Reflexions sur le Phlogistique (1783),
Antoine Lavoisier
.
Law of the Definite Proportions
Joseph-Louise Proust
Proust's Law(1799) <=> Pierre Berthollet
.
Law of Multiple Proportions
Dalton(1803)
Law of Mass Conservation:
The total mass of substances does not change during a chemical
reaction.
reactant 1 + reactant 2 product
total mass
=
total mass
calcium oxide + carbon dioxide
calcium carbonate
CaO + CO2
CaCO3
56.08g + 44.00g
100.08g
Law of Definite (or Constant) Composition:
No matter the source, a particular compound is
composed of the same elements in the same parts
(fractions) by mass.
Calcium carbonate
Analysis by Mass
(grams/20.0g)
8.0 g calcium
2.4 g carbon
9.6 g oxygen
20.0 g
Mass Fraction
(parts/1.00 part)
0.40 calcium
0.12 carbon
0.48 oxygen
1.00 part by mass
Percent by Mass
(parts/100 parts)
40% calcium
12% carbon
48% oxygen
100% by mass
Calculating the Mass of an Element in a Compound
Pitchblende is the most commercially important compound of uranium.
Analysis shows that 84.2 g of pitchblende contains 71.4 g of uranium, with
oxygen as the only other element. How many grams of uranium can be
obtained from 102 kg of pitchblende?
SOLUTION:
mass (kg) of uranium =
mass(kg) uranium in pitchblende
mass(kg) pitchblende x
mass(kg) pitchblende
= 86.5 kg
84.2kg pitchblende
71.4kg uranium
= 102 kg pitchblende x
uranium
86.5 kg uranium x
1000gkg
= 8.65 x 104g uranium
Law of Multiple Proportions:
If elements A and B react to form two compounds, the different
masses of B that combine with a fixed mass of A can be expressed
as a ratio of small whole numbers.
Example: Carbon Oxides A & B
Carbon Oxide I : 57.1% oxygen and 42.9% carbon
Carbon Oxide II : 72.7% oxygen and 27.3% carbon
Assume that you have 100g of each compound.
In 100 g of each compound: g O = 57.1 g for oxide I & 72.7 g for oxide II
g C = 42.9 g for oxide I & 27.3 g for oxide II
g O 57.1
= = 1.33
g C 42.9
g O 72.7
= = 2.66
g C 27.3
2.66 g O/g C in II 2
=
1.33 g O/g C in I 1
Dalton’s Atomic Theory Dalton’s Atomic Theory
The Postulates
1. All matter consists of atoms.
2. Atoms of one element cannot be converted into
atoms of another element.
3. Atoms of an element are identical in mass and other
properties and are different from atoms of any other
element.
4. Compounds result from the chemical combination of
a specific ratio of atoms of different elements.
Dalton’s Atomic Theory Dalton’s Atomic Theory
Mass conservation
postulate 1
postulate 2
postulate 3
Atoms cannot be created or destroyed
or converted into other types of atoms.
Since every atom has a fixed mass,
during a chemical reaction atoms are combined
differently and therefore there is no mass change
overall.
Dalton’s Atomic Theory Dalton’s Atomic Theory
Definite composition
postulate 3
postulate 4
Atoms are combined in compounds in
specific ratios
and each atom has a specific mass.
So each element has a fixed fraction of the total mass
in a compound.
Dalton’s Atomic Theory Dalton’s Atomic Theory
Multiple proportions
Atoms of an element have the same mass
and atoms are indivisible.
postulate 3
postulate 1
So when different numbers of atoms of elements
combine, they must do so in ratios of small, whole
numbers.
Figure 2.3
Thomson's first experiment
Cathode rays pass from
the tube in the upper left
into the larger bulb, where
they are deflected with a
magnetic field. When they
are bent so as to enter the
slits in the cylinders, the
electrometer measures
the charge transferred to
the cylinder.
J.J. Thomson, "Cathode Rays," The London,
Edinburgh, and Dublin Philosophical Magazine
and Journal of Science, Fifth Series, October
1897. p. 295
Thomson's second experiment
Rays from the cathode (C)
pass through a slit in the
anode (A) and through a slit
in a grounded metal plug
(B). An electrical voltage is
established between
aluminum plates (D and E),
and a scale pasted on the
outside of the end of the
tube measures the
deflection of the rays.
J.J. Thomson, "Cathode Rays," The London,
Edinburgh, and Dublin Philosophical Magazine
and Journal of Science, Fifth Series, October
1897. p. 296
Thomson's third experiment
Rays originate at the cathode
on the left and pass through a
slit in the anode into a bell jar
containing gas at low
pressure. The deflected paths
of the rays are photographed
against a ruled glass plate.
J.J. Thomson, "Cathode Rays," The London,
Edinburgh, and Dublin Philosophical Magazine
and Journal of Science, Fifth Series, October
1897. p. 301
with external magnetic and
electric field
<플레밍의
왼손법칙
>
with external electric field
with external magnetic field
B
i
F
S
l
L
2eeSEmlLB
.
J.J. Thomson, measured mass/charge of e-(1897)
(1906 Nobel prize in physics)
Experiments to Determine the Properties of Cathode Rays
OBSERVATION CONCLUSION
1. Ray bends in magnetic field. consists of charged particles
2. Ray bends towards positive
plate in electric field.
consists of negative particles
3. Ray is identical for any
cathode.
particles found in all matter
Thomson’s plum
pudding model
of the atom
(1909) (1909)
Measured of e-charge
(1923 Nobel Prize in physics)
e-charge = -1.60 x 10-19 C
Thomson’s charge/mass of e-= -1.76 x 108 C/g
e-mass = 9.10 x 10-28 g
Millikan used his findings to also calculate the mass of an
electron.
determined by J.J. Thomson
and others
mass
mass of electron = ×
charge
charge
= (-5.686×10-12 kg/C) ×(-1.602×10-19C)
= 9.109×10-31kg = 9.109×10-28g
Geiger.Marsden experiment(1909)
and Rutherford’s interpretation(1911)
.
particle velocity ~ 1.4 x 107 m/s
(~5% speed of light)
1. atoms positive charge is concentrated in the nucleus
2. proton (p) has opposite (+) charge of electron (-)
3. mass of p is 1840 x mass of e-(1.67 x 10-24 g)
Geiger.Marsden experiment(1909)
(a)
The expected result of the
Rutherford's α particle
scattering experiment,
assuming the Thomson’s
model.
(b) The result of the Rutherford's
α particle scattering
experiment
and Rutherford’s interpretation(1911)
It was quite the most incredible event that has ever happened to me in my life. It was almost
as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit
you. ... It was then that I had the idea of an atom with a minute massive centre, carrying a
charge. .Ernest Rutherford, Nobel prize in chemistry, 1908 for "for his investigations
into the disintegration of the elements, and the chemistry of radioactive substances"
Chadwick’s Experiment (1932)
(1935 Noble Prize in physics)
H atoms -1 p; Heatoms -2 p
mass He/mass H should = 2
measu4 mass He/mass H = 4
.
+ 9Be
1n + 12C + energy
neutron (n) is neutral (charge = 0)
n mass ~ p mass = 1.67 x 10-24 g
Helium Atom
“원자가반지름이100m인운동장만하다면, 핵은운동장중심에있는반지름이수mm인작은구슬정도의크기이다.”
Properties of the Three Key Subatomic Particles
Name
(Symbol)
Proton (p+)
Neutron (n0)
Electron (e -)
Charge
Mass
(unit of e)
(C)
(amu) †
(kg)
1+
+1.60218x10-19
1.00727
1.67262×10-27
0
0
1.00866
1.67493×10-27
1
-1.60218x10-19
5.4858×10-4
9.10939×10-31
Location
in the Atom
Nucleus
Nucleus
Outside
Nucleus
†
The atomic mass unit (amu) equals 1.66054x10-27 kg.
Atomic number, Mass number and Isotopes
Atomic number (Z) = number of protons in nucleus
Mass number (A) = number of protons + number of neutrons
= atomic number (Z) + number of neutrons
Isotopes are atoms of the same element (X) with different
numbers of neutrons in their nuclei
Mass Number
A
Element Symbol
ZX
Atomic Number
12 3
1H 1H (D) 1H (T)
235U 238U
92 92
The Mass Spectrometer and Its Data
20 Ne
21 Ne
22 Ne
222222EHFeEmaeElvmeVvmvFevBmRmvmeVVmReBeBmBe
..
.
.
..
...
0 -V
acceleration
R
Mass Selection
M+
Ion Generation
The Mass Spectrometer and Its Data
20 Ne
21 Ne
22 Ne
222222EHFeEmaeElvmeVvmvFevBmRmvmeVVmReBeBmBe
..
.
.
..
...
0 -V
acceleration
R
Mass Selection
M+
Ion Generation
Determining the Number of Subatomic Particles in the Isotopes of an
Element
Silicon(Si) is essential to the computer industry as a major component
of semiconductor chips. It has three naturally occurring isotopes:
28Si, 29Si, and 30Si.
Determine the number of protons, neutrons, and electrons in each
silicon isotope.
SOLUTION:
The atomic number of silicon is 14. Therefore
28Si has 14p+, 14e -and 14n0 (28-14)
29Si has 14p+, 14e -and 15n0 (29-14)
30Si has 14p+, 14e -and 16n0 (30-14)
Calculating the Atomic Mass of an Element
Silver(Ag: Z = 47) has 46 known isotopes, but only two occur naturally,
107Ag and 109Ag. Given the following mass spectrometric data,
calculate the atomic mass of Ag:
Isotope Mass(amu) Abundance(%)
107Ag 106.90509 51.84
109Ag 108.90476 48.16
SOLUTION:
weighted average of the isotopic masses
106.90509amu ×0.5184 + 108.90476amu ×
0.4816 =107.87amu
The Modern Reassessment of the Atomic Theory
1. All matter is composed of atoms. The atom is the smallest body that
retains the unique identity of the element.
2. Atoms of one element cannot be converted into atoms of another
element in a chemical reaction. Elements can only be converted
into other elements in nuclear reactions.
3. All atoms of an element have the same number of protons and
electrons, which determines the chemical behavior of the element.
Isotopes of an element differ in the number of neutrons, and thus
in mass number. A sample of the element is treated as though its
atoms have an average mass.
4. Compounds are formed by the chemical combination of two or more
elements in specific ratios.
Figure 2.10
The modern periodic table.
The formation of an ionic compound The formation of an ionic compound
An ion is an atom, or group of atoms, that has a net
positive or negative charge.
cation .
ion with a positive charge
If a neutral atom loses one or more electrons
it becomes a cation.
Na
11 p+
11 p+
11 e .→
Na+ + 1e.
10 e.
sodium ion
sodium
anion .
ion with a negative charge
If a neutral atom gains one or more electrons
it becomes an anion.
17 p+
17 p+
Cl + 1e.→
17 e.
Cl.
18 e.
chlorine
chloride ion
Factors that influence the strength of ionic bonding.
coulomb attraction
12014qqEr..
.
Pridicting the Ion and Element Forms
What monatomic ions do the following elements form?
(a) Iodine (Z = 53) (b) Calcium (Z = 20) (c) Aluminum (Z = 13)
SOLUTION:
I
Iodine is a nonmetal in Group 7A(17). It gains one electron to
have the same number of electrons as 54Xe.
Ca2+ Calcium is a metal in Group 2A(2). It loses two electrons to
have the same number of electrons as 18Ar.
Al3+ Aluminum is a metal in Group 3A(13). It loses three electrons
to have the same number of electrons as 10Ne.
Figure 2.13 Formation of a covalent bond between two H atoms.
Covalent bonds form when elements share electrons, which usually
occurs between nonmetals.
Elements that occur as molecules.
H2
N2 O2 F2
P4 S8 Cl2
Se8 Br2
I2
diatomic molecules
tetratomic molecules
octatomic molecules
1A 2A 3A 4A 5A 6A 7A 8A
(1) (2) (13) (14) (15) (16) (17) (18)
Figure 2.15
Elements that are polyatomic.
A polyatomic ion
Types of Chemical Formulas
A chemical formula is comprised of element symbols and numerical
subscripts that show the type and number of each atom present in the
smallest unit of the substance.
An empirical formula indicates the relative number of atoms of
each element in the compound. It is the simplest type of formula.
The empirical formula for hydrogen peroxide is HO.
A molecular formula shows the actual number of atoms of
each element in a molecule of the compound.
The molecular formula for hydrogen peroxide is H2O2.
A structural formula shows the number of atoms and the
bonds between them, that is, the relative placement and
connections of atoms in the molecule.
The structural formula for hydrogen peroxide is H-O-O-H.
Figure 2.16 Some common monatomic ions of the elements. Figure 2.16 Some common monatomic ions of the elements.
Can you see any patterns?
Alkali MetalAlkali Earth MetalNoble Gas
Halogen
Table 2.3 Common Monoatomic Ions
Cations
Common ions are in 2.
Anions
Charge Formula Name
+1
H+
Li+
Na+
K+
Cs+
Ag+
hydrogen
lithium
sodium
potassium
cesium
silver
+2
Mg2+
Ca2+
Sr2+
Ba2+
Zn2+
Cd2+
magnesium
calcium
strontium
barium
zinc
cadmium
+3 Al3+ aluminum
Char
ge
For
mula
Name
-1
H.
F.
Cl.
Br.
I.
hydride
fluoride
chloride
bromide
iodide
수소화
플루오르화
염화
브롬화
아이오딘화
-2
O2.
S2.
oxide
sulfide
산화
황화
-3 N3.
nitride 질화
Chemical Nomenclature
Ionic Compounds
. cation + anion
. anion (nonmetal), add suffix “-ide” to element root
BaCl2 barium chloride 염화
바륨
K2O potassium oxide 산화
포타슘
Mg(OH)2 magnesium hydroxide 수산화
마그네슘
KNO3 potassium nitrate 질산
포타슘
. 음이온 (비금속), 접미사 “ -화”
. 음이온 + 양이온
Binary Ionic Compounds
Write empirical formulas for the compounds formed from the following pairs
of elements, and name them:
(a) magnesium and nitrogen (b) iodine and cadmium
(c) strontium and fluorine (d) sulfur and cesium
SOLUTION:
(a) Mg2+ and N3.; three Mg2+(6+) and two N3.
(6-); Mg3N2(a) magnesium nitride
(b) Cd2+ and I .
; one Cd2+(2+) and two I .
(2-); CdI2 (b) cadmium iodide
(c) Sr2+ and F .
; one Sr2+(2+) and two F .
(2-); SrF2
(c) strontium fluoride
(d) Cs+ and S2.
; two Cs+(2+) and one S2.
(2-); Cs2S
(d) cesium sulfide
Metals With Several Oxidation States
Element Ion Formula Systematic Name Common Name
Copper
Cobalt
Iron
Manganese
Tin
Cu+ copper(I) cuprous
Cu2+ copper(II) cupric
Co2+ cobalt(II)
Co3+ cobalt (III)
Fe2+ iron(II) ferrous
Fe3+ iron(III) ferric
Mn2+ manganese(II)
Mn3+ manganese(III)
Sn2+ tin(II) stannous
Sn4+ tin(IV) stannic
Determining Names and Formulas of Ionic Compounds of Elements
That Form More Than One Ion
Give the systematic names for the formulas or the formulas for the names of
the following compounds:
(a) tin(II) fluoride (b) CrI3
(d) CoS
(c) ferric oxide
SOLUTION: (a) Tin (II) is Sn2+; fluoride is F .; so the formula is SnF2.
(b) The anion I is iodide(I.); 3I-means that Cr(chromium) is +3.
CrI3 is chromium(III) iodide
(c) Ferric is a common name for Fe3+; oxide is O2., therefore the
formula is Fe2O3.
(d) Co is cobalt; the anion S is sulfide(S2.); the compound is
cobalt (II) sulfide.
Some Common Polyatomic Ions
Cations
Formula Name
NH4
+ ammonium 암모늄
H3O+ hydronium 하이드로늄
Common Anions
Formula Name
CH3COO.acetate 아세트산Formula Name
CO3
2.carbonate 탄산
OH.hydroxide 수산화CN.
cyanide 시안화
CrO42.
chromate 크로뮴산
Cr2O7
2.dichromate 다이크로뮴산
.
NO2 nitrite 아질산.
ClO3 chlorate 염소산
O22.
peroxide 과산화
SO4
2.sulfate 황산
3.
NO3
.
nitrate 질산
PO4 phosphate 인산
MnO4
.permanganate 과망가니즈산
Don’t confuse with
ion
Bond
length
Ox
no
example
O2 1.21 A 0 oxygen
O2
-super-
oxide
1.33 A -1/2 HO2
hydrogen superoxide
O2
2peroxide
1.49 A -1 H2O2
hydrogen peroxide
O2oxide
-2 H2O water(dihydrogen oxide)
Molecular compounds
.nonmetals or nonmetals + metalloids
.common names
.
H2O, NH3, CH4, C60
.element further left in periodic table is 1st
.element closest to bottom of group is 1st
.if more than one compound can be formed from
the same elements, use prefixes to indicate
number of each kind of atom
.last element ends in -ide
HI hydrogen iodide
NF3 nitrogen trifluoride
SO2 sulfur dioxide
N2Cl4 dinitrogen tetrachloride
NO2 nitrogen dioxide
N2O dinitrogen monoxide
Determining Names and Formulas of Ionic Compounds Containing
Polyatomic Ions
Give the systematic names or the formula or the formulas for the names of
the following compounds:
(a) Fe(ClO4)2 (b) sodium sulfite (c) Ba(OH)2 8H2O
.
SOLUTION:
(a) ClO4 is perchlorate; iron must have a 2+ charge. This is
iron(II) perchlorate.
2.
(b) The anion sulfite is SO3 therefore you need 2 sodiums per
sulfite. The formula is Na2SO3.
(c) Hydroxide is OH.
and barium is a 2+ ion. When water is
included in the formula, we use the term “hydrate” and a prefix
which indicates the number of waters. So it is barium hydroxide
octahydrate.
Recognizing Incorrect Names and Formulas of Ionic Compounds
Something is wrong with the second part of each statement. Provide
the correct name or formula.
(a) Ba(C2H3O2)2 is called barium diacetate.
(b) Sodium sulfide has the formula (Na)2SO3.
(c) Iron(II) sulfate has the formula Fe2(SO4)3.
(d) Cesium carbonate has the formula Cs2(CO3).
SOLUTION:
(a) Barium is always a +2 ion and acetate is -1. The “di-” is
unnecessary.
(b) An ion of a single element does not need parentheses.
2.
Sulfide is S2., not SO3 . The correct formula is Na2S.
(c) Since sulfate has a 2-charge, only 1 Fe2+ is needed. The
formula should be FeSO4.
(d) The parentheses are unnecessary. The correct formula is
Cs2CO3.
Naming Acids
1) Binary acids solutions form when certain gaseous compounds
dissolve in water.
For example, when gaseous hydrogen chloride(HCl) dissolves in
water, it forms a solution called hydrochloric acid. Prefix hydro-+
anion nonmetal root + suffix -ic + the word acid -hydrochloric acid
2) Oxoacid names are similar to those of the oxoanions, except for
two suffix changes:
Anion “-ate” suffix becomes an “-ic” suffix in the acid. Anion “-ite”
suffix becomes an “-ous” suffix in the acid.
The oxoanion prefixes “hypo-” and “per-” are retained. Thus, BrO4
.
is perbromate, and HBrO4 is perbromic acid; IO2
. is iodite, and
HIO2 is iodous acid.
Determining Names and Formulas of Anions and Acids
Name the following anions and give the names and formulas of the
acids derived from them:
..
2..
(a) Br (b) IO3 (c) CN.
(d) SO4 (e) NO2
SOLUTION:
(a) The anion is bromide; the acid is hydrobromic acid, HBr.
(b) The anion is iodate; the acid is iodic acid, HIO3.
(c) The anion is cyanide; the acid is hydrocyanic acid, HCN.
(d) The anion is sulfate; the acid is sulfuric acid, H2SO4.
(e) The anion is nitrite; the acid is nitrous acid, HNO2.
Determining Names and Formulas of Binary Covalent Compounds
(a) What is the formula of carbon disulfide?
(b) What is the name of PCl5?
(c) Give the name and formula of the compound whose molecules
each consist of two N atoms and four O atoms.
SOLUTION:
(a) Carbon is C, sulfide is sulfur S and di-means 2 -CS2.
(b) P is phosphorous, Cl is chloride, the prefix for 5 is penta-.
Phosphorous pentachloride.
(c) N is nitrogen and is in a lower group number than O (oxygen).
Therefore the formula is N2O4 -dinitrogen tetraoxide.
Recognizing Incorrect Names and Formulas of Binary Covalent
Compounds
Explain what is wrong with the name of formula in the second part
of each statement and correct it:
(a) SF4 is monosulfur pentafluoride.
(b) Dichlorine heptaoxide is Cl2O6.
(c) N2O3 is dinitrotrioxide.
SOLUTION:
(a) The prefix mono-is not needed for one atom; the prefix for
four is tetra-. So the name is sulfur tetrafluoride.
(b) Hepta-means 7; the formula should be Cl2O7.
(c) The first element is given its elemental name so this is
dinitrogen trioxide.
Calculating the Molecular Mass of a Compound
Using the data in the periodic table, calculate the molecular (or formula)
mass of the following compounds:
(a) tetraphosphorous trisulfide
(b) ammonium nitrate
SOLUTION:
(a) P4S3 (b) NH4NO3
molecular
= (4xatomic mass of P)
molecular = (2xatomic mass of N)
mass
mass
+ (3xatomic mass of S)
+ (4xatomic mass of H)
=
(4x30.97amu) + (3x32.07amu) + (3xatomic mass of O)
= (2x14.01amu)+ (4x1.008amu) +
= 220.09amu
(3x16.00amu)
= 80.05amu
Figure 2.19 The distinction between mixtures and compounds.
S
Fe
Physically mixed therefore can be Allowed to react chemically
separated by physical means; in therefore cannot be separated by
this case by a magnet. physical means.
Determining Formulas and Names from Molecular Depictions
Each box contains a representation of a binary compound. Determine
its formula, name, and molecular (formula) mass.
(a) (b)
SOLUTION:
(a) There is 1 sodium (1) for every fluorine (5), so the formula is NaF.
formula mass = (1x atomic mass of Na) + (1x atomic mass of F)
= 22.99 amu + 19.00 amu = 41.99 amu
(b) There are 3 fluorines (5) for every nitrogen (2), so the formula is
NF3.
molecular mass =
(3x atomic mass of F) + (1x atomic mass of N)
= (3x 19.00 amu) + 14.01 amu = 71.01 amu
Classifications of Matter
Matter
Mixtures
Heterogeneous
mixtures
Homogeneous
mixtures
Pure Substances
Compounds
Separation by
physical methods
Separation by
chemical methods
Elements
Solution
(용액)
Mixtures
Heterogeneous mixtures : has one or more visible boundaries
between the components.
Homogeneous mixtures : has no visible boundaries because the
components are mixed as individual atoms, ions, and molecules.
Solutions : A homogeneous mixture is also called a solution.
Solutions in water are called aqueous solutions, and are very
important in chemistry. Although we normally think of solutions as
liquids, they can exist in all three physical states.
extra data
section
CODATA Recommended Values of the Fundamental Physical
Constants:2006.
extra data
section
basic mass
mass(amu)
1H 1.007 825 032 07(10) neutral atom
n 1.008 664 915 74(56)
p 1.007 276 466 77(10)
e 0.000 548 579 9111(12)
p 1.007 276 466 77(10)
e 0.000 548 579 9111(12)
= 1.007 825 046 68 ≠ 1H why?
Law of Multiple Proportion
문제: 다음의 주어진 자료에서 배수비례의 법칙을 설명하라 . 단 원소의
원자량을 이용하지 않고 문제를 해결하시오 .
화합물
1: 질량으로
47.5%의
황과
52.5%의
염소
.
화합물
2: 질량으로
31.1%의
황과
68.9%의
염소
.
해답: 같은 질량의 황에 대하여 염소의 질량비
화합물
1: 52.5/47.5 =1.1053 A
화합물
2: 68.9/31.1 =2.2154 B
A : B = 1 : 2.004 = 1 : 2 → 간단한 정수비 성립
Naming, Molecular formula, Empirical Formula
문제: 다음
화합물의
이륽
, 분자식, 실험식읁
쓰시오
:
해답: disulfur dichloride( 이염화이황), S2Cl2, SCl.
tetraphosphorus hexaoxide( 육산화사인), P4O6, P2O3.
Chemical reaction and the atomic hypothesis
문제:
다음
그림은
용기
안에서의
어떤
반응읁
나타낸
것이다
. 질량보존,
일정성분비, 배수
비례의
법칙
중
어떤
법칙읁
가장
잘
나타내고
있는가? 이유를
설명하시오
.
해답: 질량보존과
일정성분비의
법칙
Chemical properties of elements
문제:
설명과
일치하는
원소가
옆의
주기율표에서
색이
있는
칸
중에서
어디에
있을까
? 번호를
쓰고, 이유를
설명하시오
.
(a) 비금속인
원소
4개
(b) 금속인
원소
2개
(c) 상온에서
기체인
원소
3개
(d) 상온에서
고체인
원소
3개
(e) 공유결합
화합물읁
형성함직한
원소의
쌍
(3 쌍)
(f) 실험식이
MX인
이온성
화합물읁
형성함직한
원소의
쌍
(두쌍
)
(g) 실험식이
M2X인
이온성
화합물읁
형성함직한
원소의
쌍
(두쌍
)
(h) 실험식이
MX2인
이온성
화합물읁
형성함직한
원소의
쌍
(i) 화합물읁
형성하지
않음직한
원소
a) 3, 4, 5, and 6 b)1 and 2
c) 4, 5, and 6
해답:
d)1, 2, and 3 e)3 and 4 or3 and 5 or4 and 5 f) 1 and 5 or 2 and 4
g)1 and 4 or 2 and 3 h)2 and 5 i)6