2-1

물질의 구성

Chapter 2

Unobtainium - $20 million per kg

2-2

1. 원소 , 화합물 , 혼합물 : 원자 관점에서의 개요2. 원자설의 배경3. 돌턴의 원자설4. 원자핵 , 전자5. 현대의 원자론6. 원소 : 주기율표7. 화합물과 결합8. 화합물 : 화학식 , 명칭 , 질량9. 혼합물의 분류 

제 2 장 물질의 구성

2-3

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

2-4

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

2-5

Compound - a substance

composed of two or more elements

which are chemically combined.

Mixture - a group of two or more elements and/or compounds that are physically intermingled.

Definitions for Components of Matter

Figure 2.1

2-6

2-7

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)

2-8

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 Mass Conservation:

2-9

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)

Mass Fraction(parts/1.00 part)

Percent by Mass(parts/100 parts)

8.0 g calcium2.4 g carbon9.6 g oxygen

20.0 g

40% calcium12% carbon48% oxygen

100% by mass

0.40 calcium0.12 carbon0.48 oxygen

1.00 part by mass

Law of Definite (or Constant) Composition:

2-10

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:

= 86.5 kg uranium

= 102 kg pitchblende x

mass(kg) pitchblende xmass(kg) uranium in pitchblende

mass(kg) pitchblende

71.4kg uranium

84.2kg pitchblende

mass (kg) of uranium =

86.5 kg uranium x1000g

kg= 8.65 x 104g uranium

2-11

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 & BCarbon Oxide I : 57.1% oxygen and 42.9% carbonCarbon 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

g C=

57.1

42.9= 1.33

= g O

g C

72.7

27.3= 2.66

2.66 g O/g C in II

1.33 g O/g C in I

2

1=

Law of Multiple Proportions:

2-12

Dalton’s Atomic Theory

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.

The Postulates

2-13

Dalton’s Atomic Theory

Mass conservation

Atoms cannot be created or destroyed

or converted into other types of atoms.

postulate 1

postulate 2

Since every atom has a fixed mass,

during a chemical reaction atoms are combined differently and therefore there is no mass change overall.

postulate 3

2-14

Dalton’s Atomic Theory

Definite composition

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.

postulate 3

postulate 4

2-15

Dalton’s Atomic Theory

Multiple proportions

Atoms of an element have the same mass

and atoms are indivisible.

So when different numbers of atoms of elements combine, they must do so in ratios of small, whole numbers.

postulate 3postulate 1

Figure 2.3

2-16

J.J. Thomson, "Cathode Rays," The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, Fifth Series, October 1897. p. 295

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.

Thomson's first experiment

2-17

J.J. Thomson, "Cathode Rays," The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, Fifth Series, October 1897. p. 296

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.

Thomson's second experiment

2-18

J.J. Thomson, "Cathode Rays," The London, Edinburgh, and Dublin Philosophical Magazine and Journal of Science, Fifth Series, October 1897. p. 301

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.

Thomson's third experiment

2-19

J.J. Thomson, measured mass/charge of e-

(1897)

(1906 Nobel prize in physics)

with external magnetic field

with external electric field

with external magnetic and electric field

< 플레밍의 왼손법칙 >

Bi

F

S

2e

e SE

m lLB

l

L

2-20

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

2-21

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

Measured of e- charge

(1923 Nobel Prize in physics)

(1909)

2-22

Millikan used his findings to also calculate the mass of an electron.

mass of electron =mass

charge× charge

= (-5.686×10-12 kg/C) ×(-1.602×10-19C)

= 9.109×10-31kg = 9.109×10-28g

determined by J.J. Thomson and others

2-23

1. atoms positive charge is concentrated in the nucleus2. proton (p) has opposite (+) charge of electron (-)3. mass of p is 1840 x mass of e- (1.67 x 10-24 g)

a particle velocity ~ 1.4 x 107 m/s(~5% speed of light)

Geiger–Marsden experiment(1909)and Rutherford’s interpretation(1911)

2-24

(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

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"

Geiger–Marsden experiment(1909)and Rutherford’s interpretation(1911)

2-25

Chadwick’s Experiment (1932)(1935 Noble Prize in physics)

H atoms - 1 p; He atoms - 2 p

mass He/mass H should = 2

measu4 mass He/mass H = 4

a + 9Be 1n + 12C + energy

neutron (n) is neutral (charge = 0)

n mass ~ p mass = 1.67 x 10-24 g

2.2

2-26

Helium Atom

“ 원자가 반지름이 100m 인 운동장 만하다면 , 핵은 운동장 중심에 있는 반지름이 수 mm 인

작은구슬 정도의 크기이다 .”

2-27

Properties of the Three Key Subatomic Particles

† The atomic mass unit (amu) equals 1.66054x10-27 kg.

Name(Symbol)

Charge MassLocation

in the Atom(unit of e) (C) (amu) † (kg)

Proton (p+) 1+ +1.60218x10-19 1.00727 1.67262×10-27 Nucleus

Neutron (n0) 0 0 1.00866 1.67493×10-27 Nucleus

Electron (e-) 1- -1.60218x10-19 5.4858×10-4 9.10939×10-31 Outside Nucleus

2-28

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

H11 H (D)2

1 H (T)31

U23592 U238

92

XAZ

Mass Number

Atomic NumberElement Symbol

2.3

Atomic number, Mass number and Isotopes

2-29

The Mass Spectrometer and Its Data

20 N

e2

1N

e2

2N

e

2

2

2

2

2 2

E

H

F eE ma

eEl v

m

eVv

m

vF evB m

R

mv m eV V mR

eB eB m B e

0 -Vacceleration

RMass Selection

M+

Ion Generation

2-30

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)

2-31

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(%)

107Ag109Ag

106.90509

108.90476

51.84

48.16

weighted average of the isotopic masses

SOLUTION:

106.90509amu ×0.5184 + 108.90476amu × 0.4816 =107.87amu

2-32

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.

2-33

Figure 2.10 The modern periodic table.

2-34

An ion is an atom, or group of atoms, that has a net positive or negative charge.

cation – ion with a positive chargeIf a neutral atom loses one or more electronsit becomes a cation.

anion – ion with a negative chargeIf a neutral atom gains one or more electronsit becomes an anion.

2.5

Na 11 p+

11 e – Na+11 p+

10 e– + 1e– →

Cl 17 p+

17 e– Cl–17 p+

18 e–+ 1e– →

chloride ion

sodium ion

chlorine

sodium

The formation of an ionic compound

2-35

1 2

0

1

4

q qE

r

coulomb attraction

Factors that influence the strength of ionic bonding.

2-36

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)

Ca2+ Calcium is a metal in Group 2A(2). It loses two electrons to have the same number of electrons as 18Ar.

SOLUTION:

I- Iodine is a nonmetal in Group 7A(17). It gains one electron to have the same number of electrons as 54Xe.

Al3+ Aluminum is a metal in Group 3A(13). It loses three electrons to have the same number of electrons as 10Ne.

2-37

Formation of a covalent bond between two H atoms.Figure 2.13

Covalent bonds form when elements share electrons, which usually occurs between nonmetals.

2-38

Elements that occur as molecules.

1A 2A 3A 4A 5A 6A 7A 8A

(1) (2) (13) (14) (15) (16) (17) (18)

H2

N2 O2 F2

P4 S8 Cl2

Se8 Br2

I2

diatomic molecules tetratomic molecules octatomic molecules

2-39

A polyatomic ion

Figure 2.15

Elements that are polyatomic.

2-40

Types of Chemical Formulas

An empirical formula indicates the relative number of atoms of each element in the compound. It is the simplest type of formula.

A molecular formula shows the actual number of atoms of each element in a molecule of the compound.

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.

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.

The empirical formula for hydrogen peroxide is HO.

The molecular formula for hydrogen peroxide is H2O2.

The structural formula for hydrogen peroxide is H-O-O-H.

2-41

Figure 2.16 Some common monatomic ions of the elements.

Can you see any patterns?

Alkali M

etalA

lkali Earth

Metal

No

ble G

asH

alog

en

2-42

Charge Formula Name

+1

H+ hydrogen

Li+ lithium

Na+ sodium

K+ potassium

Cs+ cesium

Ag+ silver

+2

Mg2+ magnesium

Ca2+ calcium

Sr2+ strontium

Ba2+ barium

Zn2+ zinc

Cd2+ cadmium

+3 Al3+ aluminum

Common Monoatomic IonsTable 2.3 Common ions are in 2.

Charge

Formula

Name

-1

H– hydride 수소화F– fluoride 플루오르화Cl– chloride 염화Br– bromide 브롬화I– iodide 아이오딘화

-2O2– oxide 산화S2– sulfide 황화

-3 N3– nitride 질화

Cations Anions

2-43

Chemical Nomenclature

– 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

2.7

Ionic Compounds

– 음이온 ( 비금속 ), 접미사 “ - 화”– 음이온 + 양이온

2-44

Binary Ionic Compounds

Write empirical formulas for the compounds formed from the following pairs of elements, and name them:

SOLUTION:

(a) Mg2+ and N3–; three Mg2+(6+) and two N3– (6-); Mg3N2

(b) Cd2+ and I–; one Cd2+(2+) and two I– (2-); CdI2

(c) Sr2+ and F–; one Sr2+(2+) and two F– (2-); SrF2

(d) Cs+ and S2–; two Cs+(2+) and one S2– (2-); Cs2S

(a) magnesium and nitrogen (b) iodine and cadmium

(c) strontium and fluorine (d) sulfur and cesium

(a) magnesium nitride

(b) cadmium iodide

(c) strontium fluoride

(d) cesium sulfide

2-45

Metals With Several Oxidation States

Element Ion Formula Systematic Name Common Name

Copper Cu+

Cu2+

copper(I)

copper(II)

cuprous

cupric

CobaltCo2+

Co3+

cobalt(II)

cobalt (III)

ferrousIron

Fe2+ iron(II)

Fe3+ iron(III) ferric

ManganeseMn2+ manganese(II)

Mn3+ manganese(III)

TinSn2+ tin(II)

Sn4+ tin(IV)

stannous

stannic

2-46

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

(c) ferric oxide (d) CoS

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.

2-47

Some Common Polyatomic Ions

Cations

Common Anions

Formula Name

CH3COO– acetate 아세트산

CN– cyanide 시안화OH– hydroxide 수산화

ClO3– chlorate 염소산

NO2– nitrite 아질산

NO3– nitrate 질산

MnO4– permanganate 과망가니즈산

Formula Name

CO32– carbonate 탄산

CrO42– chromate 크로뮴산

Cr2O72– dichromate 다이크로뮴산

O22– peroxide 과산화

SO42– sulfate 황산

PO43– phosphate 인산

Formula Name

NH4+ ammonium 암모늄

H3O+ hydronium 하이드로늄

2-48 2.7

2-49

Don’t confuse with

ionBondlength

Ox no

example

O2 1.21 Å 0 oxygen

O2- super-ox-

ide1.33 Å -1/2 HO2

hydrogen superoxide

O22- peroxide 1.49 Å -1 H2O2

hydrogen peroxide

O2- oxide -2 H2O water(dihydrogen oxide)

2-50

nonmetals or nonmetals + metalloidscommon 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 atomlast element ends in -ide

HI hydrogen iodide

NF3 nitrogen trifluoride

SO2 sulfur dioxide

N2Cl4 dinitrogen tetrachloride

NO2 nitrogen dioxide

N2O dinitrogen monoxide

Molecular compounds

2-51

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

SOLUTION: (a) ClO4– is perchlorate; iron must have a 2+ charge. This is

iron(II) perchlorate.

(b) The anion sulfite is SO32– 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.

(c) Ba(OH)2 8H2O

2-52

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.

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. Sulfide is S2–, not SO3

2–. 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.

(c) Iron(II) sulfate has the formula Fe2(SO4)3.

(d) Cesium carbonate has the formula Cs2(CO3).

2-53

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.

2-54

Determining Names and Formulas of Anions and Acids

Name the following anions and give the names and formulas of the acids derived from them:

(a) Br – (b) IO3 – (c) CN– (d) SO4

2– (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.

2-55

Determining Names and Formulas of Binary Covalent Compounds

(a) What is the formula of carbon disulfide?

(c) Give the name and formula of the compound whose molecules each consist of two N atoms and four O atoms.

(b) What is the name of PCl5?

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.

2-56

Recognizing Incorrect Names and Formulas of Binary Covalent Compounds

(a) SF4 is monosulfur pentafluoride.

(c) N2O3 is dinitrotrioxide.

(b) Dichlorine heptaoxide is Cl2O6.

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.

Explain what is wrong with the name of formula in the second part of each statement and correct it:

2-57

Calculating the Molecular Mass of a Compound

SOLUTION:

(a) tetraphosphorous trisulfide

(b) ammonium nitrate

Using the data in the periodic table, calculate the molecular (or formula) mass of the following compounds:

(a) P4S3

molecular mass

= (4xatomic mass of P)

+ (3xatomic mass of S)

= (4x30.97amu) + (3x32.07amu)

= 220.09amu

(b) NH4NO3

molecular mass

= (2xatomic mass of N)

+ (4xatomic mass of H)

+ (3xatomic mass of O)

= (2x14.01amu)+ (4x1.008amu) + (3x16.00amu)

= 80.05amu

2-58

Allowed to react chemically therefore cannot be separated by physical means.

Figure 2.19 The distinction between mixtures and compounds.

S

Fe

Physically mixed therefore can be separated by physical means; in this case by a magnet.

2-59

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)

(1x atomic mass of Na) (1x atomic mass of F)+

SOLUTION:

(a) There is 1 sodium (1) for every fluorine (5), so the formula is NaF.

formula mass =

(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)+

= 22.99 amu + 19.00 amu = 41.99 amu

= (3x 19.00 amu) + 14.01 amu = 71.01 amu

2-60

Homogeneous mixtures

Classifications of Matter

Matter

Mixtures Pure Substances

Heterogeneous mixtures

Compounds Elements

Separation by physical methods

Separation by chemical methods

Solution(용액 )

2-61

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.

2-62

extra data section

CODATA Recommended Values of the Fundamental Physical Constants:2006∗

2-63

basic massextra data section

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?

2-64

Law of Multiple Proportion

화합물 1: 질량으로 47.5% 의 황과 52.5% 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

문제 : 다음의 주어진 자료에서 배수비례의 법칙을 설명하라 . 단 원소의 원자량을 이용하지 않고 문제를 해결하시오 .

화합물 2: 질량으로 31.1% 의 황과 68.9% 의 염소 .

2-65

Naming, Molecular formula, Empirical Formula

해답 : disulfur dichloride( 이염화이황 ), S2Cl2, SCl.

tetraphosphorus hexaoxide( 육산화사인 ), P4O6, P2O3.

문제 : 다음 화합물의 이름 , 분자식 , 실험식을 쓰시오 :

2-66

Chemical reaction and the atomic hypothesis

해답 : 질량보존과 일정성분비의 법칙

문제 : 다음 그림은 용기 안에서의 어떤 반응을 나타낸 것이다 . 질량보존 , 일정성분비 , 배수 비례의 법칙 중 어떤 법칙을 가장 잘 나타내고 있는가 ? 이유를 설명하시오 .

2-67

Chemical properties of elements

해답 : a) 3, 4, 5, and 6 b) 1 and 2 c) 4, 5, and 6d) 1, 2, and 3 e) 3 and 4 or 3 and 5 or 4 and 5 f) 1 and 5 or 2

and 4 g) 1 and 4 or 2 and 3 h) 2 and 5 i) 6

문제 : 설명과 일치하는 원소가 옆의 주기율표에서 색이 있는 칸 중에서 어디에 있을까 ? 번호를 쓰고 , 이유를 설명하시오 .

(a) 비금속인 원소 4 개

(b) 금속인 원소 2 개

(c) 상온에서 기체인 원소 3 개

(d) 상온에서 고체인 원소 3 개

(e) 공유결합 화합물을 형성함직한 원소의 쌍 (3 쌍 )

(f) 실험식이 MX 인 이온성 화합물을 형성함직한 원소의 쌍 ( 두 쌍 )

(g) 실험식이 M2X 인 이온성 화합물을 형성함직한 원소의 쌍 ( 두 쌍 )

(h) 실험식이 MX2 인 이온성 화합물을 형성함직한 원소의 쌍

(i) 화합물을 형성하지 않음직한 원소