Chemistry Chapter 2 (foundation)

8/13/2019 Chemistry Chapter 2 (foundation)

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Chapter 2

Lecture Presentation

2012 Pearson Education, Inc.Chemistry, The Central Science, 12th Edition

Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

Atoms, Molecules,and Ions

John D. Bookstaver

St. Charles Community College

2012 Pearson Education, Inc.


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Atomic Theory of Matter

The theory that atoms arethe fundamental buildingblocks of matter


Theodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.

reemerge n t e ear ynineteenth century,championed by John

Dalton.


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Dalton's Postulates

Each element is composedof extremely smallparticles called atoms.




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Dalton's Postulates

All atoms of a givenelement are identical toone another in mass and



ot er propert es, ut t eatoms of one element aredifferent from the atomsof all other elements.


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Dalton's Postulates

Atoms of an element arenot changed into atoms ofa different element by



c em ca react ons; atomsare neither created nordestroyed in chemicalreactions.


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Dalton's Postulates

Compounds are formedwhen atoms of more thanone element combine; a



has the same relativenumber and kind ofatoms.


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Law of Conservation of Mass

The total mass of substances present at the end ofa chemical process is the same as the mass ofsubstances present before the process took place.




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The Electron



Streams of negatively charged particles were found to emanatefrom cathode tubes, causing fluorescence.

J. J. Thomson is credited with their discovery (1897).


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The Electron



Thomson measured the charge/mass ratio of theelectron to be 1.76 108 coulombs/gram (C/g).


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Millikan Oil-Drop Experiment

Once the charge/massratio of the electron wasknown, determination of



e t er t e c arge or t emass of an electronwould yield the other.


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Millikan Oil-Drop Experiment

Robert Millikan(University of Chicago)



the electron in 1909.


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Radioactivity

Radioactivity is the spontaneous emission ofradiation by an atom.

It was first observed by Henri Becquerel.

2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward 2012 Pearson Education, Inc.

Marie and Pierre Curie also studied it.


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Radioactivity Three types of radiation were discovered byErnest Rutherford: particles particles rays



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The Atom, circa 1900

The prevailing theory wasthat of the plum puddingmodel, put forward by


omson. It featured a positive sphereof matter with negative

electrons imbedded in it.


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Discovery of the Nucleus

Ernest Rutherfordshot particles ata thin sheet of


go o anobserved thepattern of scatterof the particles.


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Sample Exercise 2.1 Atomic Size

The unknown is the number of silver (Ag) atoms. Using the relationship 1 Ag atom = 2.88 as a conversionfactor relating number of atoms and distance, we start with the diameter of the dime, first converting this distance

into angstroms and then using the diameter of the Ag atom to convert distance to number of Ag atoms:

Solution

The diameter of a US dime is 17.9 mm, and the diameter of a silver atom is 2.88 . How many silver atoms couldbe arranged side by side across the diameter of a dime?

2012 Pearson Education, Inc.Chemistry, The Central Science, 12th EditionTheodore L. Brown; H. Eugene LeMay, Jr.; Bruce E. Bursten; Catherine J. Murphy; and Patrick Woodward

The diameter of a carbon atom is 1.54 . (a) Express this diameter in picometers. (b) How many carbon atomscould be aligned side by side across the width of a pencil line that is 0.20 mm wide?

Answer:

Practice Exercise


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Solution



That is, 62.2 million silver atoms could sit side by side across a dime!


Answer:

Practice Exercise


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Solution



That is, 62.2 million silver atoms could sit side by side across a dime!


Answer: (a) 154 pm, (b) 1.3 106 C atoms

Practice Exercise


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The Nuclear AtomSince some particleswere deflected at largeangles, Thomsons


correct.


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The Nuclear Atom

Rutherford postulateda very small, densenucleus with the


e ectrons aroun t eoutside of the atom.

Most of the volume of

the atom is emptyspace.


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Other Subatomic Particles

Protons were discovered by Rutherford in1919.

Neutrons were discovered b James Chadwick


in 1932.


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Subatomic Particles Protons and electrons are the only particles that have

a charge. Protons and neutrons have essentially the same mass.

The mass of an electron is so small we ignore it.



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Symbols of Elements

Elements are symbolized by one or two letters.



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Symbols of Elements


All atoms of the same element have the samenumber of protons, which is called the atomic

number, Z.


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Symbols of Elements


The mass of an atom in atomic mass units (amu) isthe total number of protons and neutrons in the

atom.


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Isotopes Isotopes are atoms of the same element with

different masses. Isotopes have different numbers of neutrons.



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Sample Exercise 2.2 Determining the Number of SubatomicParticles in Atoms

Solution

How many protons, neutrons, and electrons are in (a) an atom of 197Au, (b) an atom of strontium-90?

How many protons, neutrons, and electrons are in (a) a 138Ba atom, (b) an atom of phosphorus-31?Practice Exercise


Answer:


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(a) The superscript 197 is the mass number (protons + neutrons). According to the list of elements given on theinside front cover, gold has atomic number 79. Consequently, an atom of 197Au has 79 protons, 79 electrons, and197 79 = 118 neutrons. (b) The atomic number of strontium (listed on inside front cover) is 38. Thus, all atoms

of this element have 38 protons and 38 electrons. The strontium-90 isotope has 90 38 = 52 neutrons.

Solution




Answer:


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(a) The superscript 197 is the mass number (protons + neutrons). According to the list of elements given on theinside front cover, gold has atomic number 79. Consequently, an atom of 197Au has 79 protons, 79 electrons, and197 79 = 118 neutrons. (b) The atomic number of strontium (listed on inside front cover) is 38. Thus, all atoms

of this element have 38 protons and 38 electrons. The strontium-90 isotope has 90 38 = 52 neutrons.

Solution




Answer:(a) 56 protons, 56 electrons, and 82 neutrons, (b) 15 protons, 15 electrons, and 16 neutrons


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Sample Exercise 2.3 Writing Symbols for Atoms

Solution

Magnesium has three isotopes with mass numbers 24, 25, and 26. (a) Write the complete chemical symbol(superscript and subscript) for each. (b) How many neutrons are in an atom of each isotope?

Give the complete chemical symbol for the atom that contains 82 protons, 82 electrons, and 126 neutrons.

Practice Exercise


Answer:


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(a) Magnesium has atomic number 12, so all atoms of magnesium contain 12 protons and 12 electrons. Thethree isotopes are therefore represented by 24/12Mg, 25/12Mg, and 26/12Mg. (b) The number of neutrons

in each isotope is themass number minus the number of protons. The numbers of neutrons in an atom of each isotope aretherefore 12, 13, and 14, respectively.

Solution



Practice Exercise


Answer:


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(a) Magnesium has atomic number 12, so all atoms of magnesium contain 12 protons and 12 electrons. Thethree isotopes are therefore represented by 24/12Mg, 25/12Mg, and 26/12Mg. (b) The number of neutrons

in each isotope is themass number minus the number of protons. The numbers of neutrons in an atom of each isotope aretherefore 12, 13, and 14, respectively.

Solution



Practice Exercise


Answer:


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Atomic Mass

Atomic and molecularmasses can bemeasured with great


accuracy us ng a massspectrometer.


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Average Mass

Because in the real world we use large amounts ofatoms and molecules, we use average masses incalculations.


Average mass is calculated from the isotopes ofan element weighted by their relative abundances.


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Sample Exercise 2.4 Calculating the Atomic Weight of an Elementfrom Isotopic Abundances

Solution

Naturally occurring chlorine is 75.78% 35Cl (atomic mass 34.969 amu) and 24.22% 37Cl (atomic mass 36.966 amu).Calculate the atomic weight of chlorine.


Three isotopes of silicon occur in nature: 28Si (92.23%), atomic mass 27.97693 amu; 29Si (4.68% ), atomicmass 28.97649 amu; and 30Si (3.09%), atomic mass 29.97377 amu. Calculate the atomic weight of silicon.

Answer:

Practice Exercise


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We can calculate the atomic weight by multiplying the abundance of each isotope by its atomic mass and summingthese products. Because 75.78% = 0.7578 and 24.22% = 0.2422, we have

Atomic weight = (0.7578)(34.969 amu) + (0.2422)(36.966 amu)= 26.50 amu + 8.953 amu= 35.45 amu

Solution



, ,two isotopes and is closer to the value of 35Cl, the more abundant isotope.


Answer:

Practice Exercise


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We can calculate the atomic weight by multiplying the abundance of each isotope by its atomic mass and summingthese products. Because 75.78% = 0.7578 and 24.22% = 0.2422, we have

Atomic weight = (0.7578)(34.969 amu) + (0.2422)(36.966 amu)= 26.50 amu + 8.953 amu= 35.45 amu

Solution



, ,two isotopes and is closer to the value of 35Cl, the more abundant isotope.


Answer: 28.09 amu

Practice Exercise


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Periodic Table

The periodic tableis a systematiccatalog of the


e ements.

Elements arearranged in order ofatomic number.


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Periodic Table The rows on the

periodic chart areperiods.

Columns are groups.


Elements in thesame group havesimilar chemical

properties.


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Periodicity


When one looks at the chemical properties ofelements, one notices a repeating pattern of

reactivities.


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Groups


These five groups are known by their names.


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Periodic Table

Nonmetals are onthe right side ofthe periodic table


(with the exceptionof H).


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Periodic TableMetalloidsborder the stair-step line (with


Al, Po, and At).


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Periodic TableMetals are onthe left side ofthe chart.



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Sample Exercise 2.5 Using the Periodic Table

Which two of these elements would you expect to show the greatest similarity in chemical and physical properties:B, Ca, F, He, Mg, P?

Locate Na (sodium) and Br (bromine) in the periodic table. Give the atomic number of each and classify each asmetal, metalloid, or nonmetal.

Practice Exercise

Solution


Answer:


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Practice Exercise

Elements in the same group of the periodic table are most likely to exhibit similar properties. We therefore expect Caand Mg to be most alike because they are in the same group (2A, the alkaline earth metals).

Solution


Answer:


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Practice Exercise

Elements in the same group of the periodic table are most likely to exhibit similar properties. We therefore expect Caand Mg to be most alike because they are in the same group (2A, the alkaline earth metals).

Solution



Answer: Na, atomic number 11, is a metal; Br, atomic number 35, is a nonmetal.


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Chemical FormulasThe subscript to the right ofthe symbol of an element

tells the number of atoms ofthat element in one




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Chemical FormulasMolecular compounds arecomposed of molecules and

almost always contain onlynonmetals.




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Diatomic Molecules These seven elements occur naturally asmolecules containing two atoms:

N2o O2ne Cl2eans F2loors I2n Br2ians H2ouse Hydrogen



Oxygen Fluorine Chlorine

Bromine Iodine


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Types of Formulas

Empirical formulas give the lowest whole-number ratio of atoms of each element in acompound.




Molecular formulas give the exact number ofatoms of each element in a compound.


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Sample Exercise 2.6 Relating Empirical and Molecular Formulas

Solution

Write the empirical formulas for (a) glucose, a substance also known as either blood sugar or dextrose, molecular

formula C6H12O6; (b) nitrous oxide, a substance used as an anesthetic and commonly called laughing gas, molecularformula N2O.

Practice Exercise



Give the empirical formula for diborane, whose molecular formula is B2H6.

Answer:

S l E i


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(a) The subscripts of an empirical formula are the smallest whole-number ratios. The smallest ratios areobtained by dividing each subscript by the largest common factor, in this case 6. The resultant empirical

formula for glucose is CH2O.(b) Because the subscripts in N2O are already the lowest integral numbers, the empirical formula for nitrousoxide is the same as its molecular formula, N2O.

Solution



Practice Exercise




Answer:

S l E i 2 6 R l ti E i i l d M l l F l


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(a) The subscripts of an empirical formula are the smallest whole-number ratios. The smallest ratios areobtained by dividing each subscript by the largest common factor, in this case 6. The resultant empirical

formula for glucose is CH2O.(b) Because the subscripts in N2O are already the lowest integral numbers, the empirical formula for nitrousoxide is the same as its molecular formula, N2O.

Solution



Practice Exercise




Answer: BH3


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Types of Formulas Structural formulas show the

order in which atoms arebonded.




Perspective drawings alsoshow the three-dimensionalarray of atoms in a compound.

Ions


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Ions




When atoms lose or gain electrons, they becomeions.

Cations are positive and are formed by elements on theleft side of the periodic chart.

Anions are negative and are formed by elements on theright side of the periodic chart.

Sample Exercise 2 7 Writing Chemical Symbols for Ions


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Sample Exercise 2.7 Writing Chemical Symbols for Ions

Solution

Give the chemical symbol, including superscript indicating mass number, for (a) the ion with 22 protons, 26neutrons, and 19 electrons; (b) the ion of sulfur that has 16 neutrons and 18 electrons.



How many protons, neutrons, and electrons does the 79Se2 ion possess?

Answer:

Practice Exercise

Sample Exercise 2 7 Writing Chemical Symbols for Ions


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(a) The number of protons is the atomic number of the element. A periodic table or list of elements tells usthat the element with atomic number 22 is titanium (Ti). The mass number (protons plus neutrons) of thisisotope of titanium is 22 + 26 = 48. Because the ion has three more protons than electrons, it has a net

charge of 3+: 48Ti3+.(b) The periodic table tells us that sulfur (S) has an atomic number of 16. Thus, each atom or ion of sulfurcontains 16 protons. We are told that the ion also has 16 neutrons, meaning the mass number is 16 + 16 =32. Because the ion has 16 protons and 18 electrons, its net charge is 2 and the ion symbol is 32S2.

In eneral we will focus on the net char es of ions and i nore their mass numbers unless

Solution




the circumstances dictate that we specify a certain isotope.


Answer:

Practice Exercise



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(a) The number of protons is the atomic number of the element. A periodic table or list of elements tells usthat the element with atomic number 22 is titanium (Ti). The mass number (protons plus neutrons) of thisisotope of titanium is 22 + 26 = 48. Because the ion has three more protons than electrons, it has a net

charge of 3+: 48Ti3+.(b) The periodic table tells us that sulfur (S) has an atomic number of 16. Thus, each atom or ion of sulfurcontains 16 protons. We are told that the ion also has 16 neutrons, meaning the mass number is 16 + 16 =32. Because the ion has 16 protons and 18 electrons, its net charge is 2 and the ion symbol is 32S2.

In eneral we will focus on the net char es of ions and i nore their mass numbers unless

Solution




the circumstances dictate that we specify a certain isotope.


Answer: 34 protons, 45 neutrons, and 36 electrons

Practice Exercise

Sample Exercise 2.8 Predicting Ionic Charge


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Solution

Predict the charge expected for the most stable ion of barium and the most stable ion of oxygen.

Practice Exercise



Predict the charge expected for the most stable ion of (a) aluminum and (b) fluorine.

Answer:



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p g g

We will assume that these elements form ions that have the same number of electrons as the nearest noble-gas atom. From the periodic table, we see that barium has atomic number 56. The nearest noble gas isxenon, atomic number 54. Barium can attain a stable arrangement of 54 electrons by losing two electrons,

forming the Ba2+ cation.Oxygen has atomic number 8. The nearest noble gas is neon, atomic number 10. Oxygen can attain this

stable electron arrangement by gaining two electrons, forming the O2 anion.

Solution


Practice Exercise




Answer:



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p g g

We will assume that these elements form ions that have the same number of electrons as the nearest noble-gas atom. From the periodic table, we see that barium has atomic number 56. The nearest noble gas isxenon, atomic number 54. Barium can attain a stable arrangement of 54 electrons by losing two electrons,

forming the Ba2+ cation.Oxygen has atomic number 8. The nearest noble gas is neon, atomic number 10. Oxygen can attain this

stable electron arrangement by gaining two electrons, forming the O2 anion.

Solution


Practice Exercise




Answer: (a) 3+, (b) 1


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Ionic BondsIonic compounds (such as NaCl) are generallyformed between metals and nonmetals.




Sample Exercise 2.9 Identifying Ionic and Molecular Compounds


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Solution

Which of these compounds would you expect to be ionic: N2

O, Na2

O, CaCl2

, SF4

?

Which of these compounds are molecular: CBr4, FeS, P4O6, PbF2?

Answer:

Practice Exercise





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We predict that Na2O and CaCl2 are ionic compounds because they are composed of a metal combined witha nonmetal. We predict (correctly) that N2O and SF4 are molecular compounds because they are composedentirely of nonmetals.

Solution

Which of these compounds would you expect to be ionic: N2O, Na2O, CaCl2, SF4?


Answer:

Practice Exercise





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We predict that Na2O and CaCl2 are ionic compounds because they are composed of a metal combined witha nonmetal. We predict (correctly) that N2O and SF4 are molecular compounds because they are composedentirely of nonmetals.

Solution

Which of these compounds would you expect to be ionic: N2O, Na2O, CaCl2, SF4?


Answer: CBr4 and P4O6

Practice Exercise



Writing Formulas


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Writing Formulas

Because compounds are electrically neutral, onecan determine the formula of a compound this




The charge on the cation becomes the subscript on theanion.

The charge on the anion becomes the subscript on thecation.

If these subscripts are not in the lowest whole-numberratio, divide them by the greatest common factor.

Common Cations


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Common Anions


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Common Anions




Sample Exercise 2.10 Using Ionic Charge to Write EmpiricalFormulas for Ionic Compounds


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Formulas for Ionic Compounds

Solution

Write the empirical formula of the compound formed by (a) Al3+ and Cl ions, (b) Al3+ and O2 ions, (c) Mg2+ andNO3 ions.



Write the empirical formula for the compound formed by (a) Na+ and PO43, (b) Zn2+ and SO42, (c) Fe3+and CO32.

Answer:

Practice Exercise



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(a) Three Cl ions are required to balance the charge of one Al3+ ion, making the formula AlCl3.(b) Two Al3+ ions are required to balance the charge of three O2 ions. That is, the total positive charge is 6+,

and the total negative charge is 6. The formula is Al2O3.(c) Two NO3 ions are needed to balance the charge of one Mg2+, yielding Mg(NO3)2. Note that the formula

for the polyatomic ion, NO3, must be enclosed in parentheses so that it is clear that the subscript 2applies to all the atoms of that ion.

Solution





Answer:

Practice Exercise



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(a) Three Cl ions are required to balance the charge of one Al3+ ion, making the formula AlCl3.(b) Two Al3+ ions are required to balance the charge of three O2 ions. That is, the total positive charge is 6+,

and the total negative charge is 6. The formula is Al2O3.(c) Two NO3 ions are needed to balance the charge of one Mg2+, yielding Mg(NO3)2. Note that the formula

for the polyatomic ion, NO3, must be enclosed in parentheses so that it is clear that the subscript 2applies to all the atoms of that ion.

Solution





Answer: (a) Na3PO4, (b) ZnSO4, (c) Fe2(CO3)3

Practice Exercise


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Inorganic Nomenclature Write the name of the cation. If the anion is an element, change its ending to

-ide; if the anion is a polyatomic ion, simplywrite the name of the polyatomic ion.




t e cat on can ave more t an one poss e

charge, write the charge as a Roman numeralin parentheses.


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Patterns in OxyanionNomenclature When there are two oxyanions involving the same

element: The one with fewer oxygens ends in -ite.




The one with more oxygens ends in -ate. NO2 : nitrite; SO32 : sulfite NO3 : nitrate; SO42 : sulfate

Patterns in Oxyanion Nomenclature


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Central atoms on the second row have bond to atmost three oxygens; those on the third row take up

to four. Charges increase as you go from rightto left.



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The one with the second fewest oxygens ends in -ite.

ClO2

: chlorite

The one with the second most oxygens ends in -ate.

ClO3 : chlorate






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y The one with the fewest oxygens has the prefix

hypo- and ends in -ite.

ClO

: hypochlorite The one with the most oxygens has the prefixper-

-




.

ClO4

: perchlorate

Sample Exercise 2.11 Determining the Formula of an Oxyanionfrom Its Name


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Based on the formula for the sulfate ion, predict the formula for (a) the selenate ion and (b) the selenite ion. (Sulfurand selenium are both in group 6A and form analogous oxyanions.)

Solution

The formula for the bromate ion is analogous to that for the chlorate ion. Write the formula for the hypobromite

Practice Exercise



.

Answer:



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(a) The sulfate ion is SO42. The analogous selenate ion is therefore SeO42.(b) The ending -ite indicates an oxyanion with the same charge but one O atom fewer than the corresponding

oxyanion that ends in -ate. Thus, the formula for the selenite ion is SeO32.

Solution


Practice Exercise



.

Answer:



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(a) The sulfate ion is SO42. The analogous selenate ion is therefore SeO42.(b) The ending -ite indicates an oxyanion with the same charge but one O atom fewer than the corresponding

oxyanion that ends in -ate. Thus, the formula for the selenite ion is SeO32.

Solution


Practice Exercise



.

Answer: BrO

and BrO2

Sample Exercise 2.12 Determining the Names of Ionic Compoundsfrom Their Formulas


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In naming ionic compounds, it is important to recognize polyatomic ions and to determine the charge of cationswith variable charge.

Solution

Name the ionic compounds (a) K2SO4,(b) Ba(OH)2,(c) FeCl3.



Name the ionic compounds (a) NH4Br,(b) Cr2O3, (c) Co(NO3)2.

Answer:

Practice Exercise



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Solution


(a) The cation is K+, the potassium ion, and the anion is SO42, the sulfate ion, making the name potassiumsulfate. (If you thought the compound contained S2 and O2 ions, you failed to recognize the polyatomicsulfate ion.)

(b) The cation is Ba2+, the barium ion, and the anion is OH, the hydroxide ion: barium hydroxide.




Answer:

Practice Exercise

.Because the compound contains three chloride ions, Cl, the cation must be Fe3+, the iron(III), or ferric, ion.

Thus, the compound is iron(III) chloride or ferric chloride.



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Solution


(a) The cation is K+, the potassium ion, and the anion is SO42, the sulfate ion, making the name potassiumsulfate. (If you thought the compound contained S2 and O2 ions, you failed to recognize the polyatomicsulfate ion.)

(b) The cation is Ba2+, the barium ion, and the anion is OH, the hydroxide ion: barium hydroxide.




Answer: (a) ammonium bromide,(b) chromium(III) oxide, (c) cobalt(II) nitrate

Practice Exercise

.Because the compound contains three chloride ions, Cl, the cation must be Fe3+, the iron(III), or ferric, ion.

Thus, the compound is iron(III) chloride or ferric chloride.

Sample Exercise 2.13 Determining the Formulas of IonicCompounds from Their Names


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Solution

Write the chemical formulas for (a) potassium sulfide, (b) calcium hydrogen carbonate, (c) nickel(II) perchlorate.

In going from the name of an ionic compound to its chemical formula, you must know thecharges of the ions to determine the subscripts.



Give the chemical formulas for (a) magnesium sulfate, (b) silver sulfide, (c) lead(II) nitrate.

Answer:

Practice Exercise



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(a) The potassium ion is K+, and the sulfide ion is S2. Because ionic compounds are electrically neutral, two K+ions are required to balance the charge of one S2 ion, giving K2S for the empirical formula.

(b) The calcium ion is Ca2+. The carbonate ion is CO32, so the hydrogen carbonate ion is HCO3. Two HCO3

ions are needed to balance the positive charge of Ca2+, giving Ca(HCO3)2.

Solution






Answer:

. 4 . 4 one Ni2+ ion, giving Ni(ClO4)2.

Practice Exercise



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(a) The potassium ion is K+, and the sulfide ion is S2. Because ionic compounds are electrically neutral, two K+ions are required to balance the charge of one S2 ion, giving K2S for the empirical formula.

(b) The calcium ion is Ca2+. The carbonate ion is CO32, so the hydrogen carbonate ion is HCO3. Two HCO3

ions are needed to balance the positive charge of Ca2+, giving Ca(HCO3)2.

Solution






Answer: (a) MgSO4, (b) Ag2S, (c) Pb(NO3)2

. 4 . 4 one Ni2+ ion, giving Ni(ClO4)2.

Practice Exercise

Acid Nomenclature


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If the anion in the acid

ends in -ide, change theending to -ic acidandadd the refixh dro- .




HCl: hydrochloric acid HBr: hydrobromic acid

HI: hydroiodic acid

Acid Nomenclature


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ends in -ite, change theending to -ous acid. HClO: h ochlorous acid




HClO2: chlorous acid

Acid Nomenclature


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ends in -ate, change theending to -ic acid. HClO : chloric acid




HClO4: perchloric acid

Sample Exercise 2.14 Relating the Names and Formulas of Acids

Name the acids (a) HCN, (b) HNO3, (c) H2SO4, (d) H2SO3.


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Solution



Give the chemical formulas for (a) hydrobromic acid, (b) carbonic acid.

Answer:

Practice Exercise




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(a) The anion from which this acid is derived is CN, the cyanide ion. Because this ion has an -ide ending, theacid is given a hydro- prefix and an -ic ending: hydrocyanic acid. Only water solutions of HCN are referredto as hydrocyanic acid. The pure compound, which is a gas under normal conditions, is called hydrogencyanide. Both hydrocyanic acid and hydrogen cyanide are extremely toxic.

(b) Because NO3 is the nitrate ion, HNO3 is called nitric acid (the -ate ending of the anion is replaced with an-ic ending in naming the acid).

(c) Because SO42 is the sulfate ion, H2SO4 is called sulfuric acid.(d) Because SO32 is the sulfite ion, H2SO3 is sulfurous acid (the -ite ending of the anion is replaced with an

Solution



-ous en ng .


Answer:

Practice Exercise




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(a) The anion from which this acid is derived is CN, the cyanide ion. Because this ion has an -ide ending, theacid is given a hydro- prefix and an -ic ending: hydrocyanic acid. Only water solutions of HCN are referredto as hydrocyanic acid. The pure compound, which is a gas under normal conditions, is called hydrogencyanide. Both hydrocyanic acid and hydrogen cyanide are extremely toxic.

(b) Because NO3 is the nitrate ion, HNO3 is called nitric acid (the -ate ending of the anion is replaced with an-ic ending in naming the acid).

(c) Because SO42 is the sulfate ion, H2SO4 is called sulfuric acid.(d) Because SO32 is the sulfite ion, H2SO3 is sulfurous acid (the -ite ending of the anion is replaced with an

Solution



-ous en ng .


Answer: (a) HBr, (b) H2CO3

Practice Exercise

Nomenclature of Binary Compounds


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The less electronegative

atom is usually listed first. A prefix is used to denote




element in the compound(mono- is not used on thefirst element listed,however) .



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The ending on the more

electronegative element ischanged to -ide.




CO2: carbon dioxide CCl4: carbon tetrachloride



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If the prefix ends with a or

o and the name of theelement begins with a




,vowels are often elided intoone.

N2

O5

: dinitrogen pentoxide

Sample Exercise 2.15 Relating the Names and Formulas of BinaryMolecular Compounds

Name the compounds (a) SO2, (b) PCl5, (c) Cl2O3.


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Solution

Name the compounds SO , PCl , Cl O .

Practice Exercise



ve e c em ca ormu as or a s con e ra rom e, su ur c or e.

Answer:




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The compounds consist entirely of nonmetals, so they are molecular rather thanionic. Using the prefixes in Table 2.6, we have (a) sulfur dioxide, (b) phosphorus

pentachloride, and (c) dichlorine trioxide.

Solution

p , ,

Practice Exercise




Answer:




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The compounds consist entirely of nonmetals, so they are molecular rather thanionic. Using the prefixes in Table 2.6, we have (a) sulfur dioxide, (b) phosphorus

pentachloride, and (c) dichlorine trioxide.

Solution

p

Practice Exercise




Answer: (a) SiBr4, (b) S2Cl2

Nomenclature of Organic Compounds


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Organic chemistry is the study of carbon.

Organic chemistry has its own system of

nomenclature.



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The simplest hydrocarbons (compoundscontaining only carbon and hydrogen) arealkanes.



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The first part of the names just listed correspondto the number of carbons (meth- = 1, eth- = 2,

prop- = 3, etc.).



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When a hydrogen in an alkane is replaced withsomething else (a functional group, like -OH in thecompounds above), the name is derived from the nameof the alkane.

The ending denotes the type of compound. Analcohol ends in -ol.

Assuming the carbon atoms in pentane are in a linear chain, write (a) the structural formula and (b) the molecularformula for this alkane.

Sample Exercise 2.16 Writing Structural and Molecular Formulasfor Hydrocarbons


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Solution



Assuming the carbon atoms in pentane are in a linear chain, write (a) the structural formula and (b) the molecularformula for this alkane.



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(a) Alkanes contain only carbon and hydrogen, and each carbon is attached to four other atoms. The name

pentane contains the prefixpenta-

for five (Table 2.6), and we are told that the carbons are in a linear chain. Ifwe then add enough hydrogen atoms to make four bonds to each carbon, we obtain the structural formula

Solution



Continued



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This form of pentane is often called n-pentane, where then- stands for normal because all five carbon atomsare in one line in the structural formula.(b) Once the structural formula is written, we determine the molecular formula by counting the atoms present.Thus, n-pentane has the molecular formula C5H12.

(a) What is the molecular formula of butane, the alkane with four carbons? (b) What are the name and molecular

Practice Exercise



Answer:

Continued



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This form of pentane is often called n-pentane, where then- stands for normal because all five carbon atomsare in one line in the structural formula.(b) Once the structural formula is written, we determine the molecular formula by counting the atoms present.Thus, n-pentane has the molecular formula C5H12.

(a) What is the molecular formula of butane, the alkane with four carbons? (b) What are the name and molecular

Practice Exercise



Answer: (a) C4

H10

, (b) butanol, C4

H10

O or C4

H9

OH

Chemistry Chapter 2 (foundation)

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Transcript of Chemistry Chapter 2 (foundation)