LECTURE 6 CHM 151 ©slg

TOPICS:

1. Ionic Nomenclature, Completion2. Naming Binary Molecular Compounds 3. “The Mole”: Introduction

•Cr3+ CO32-

•Ni2+ CN-

•Zn2+ NO2-

•Bi3+ H2PO4-

•Pb2+ N3-

Cr2(CO3)3 Chromium(III) carbonate

Ni(CN)2 Nickel(II) cyanide

Zn(NO2)2 Zinc Nitrite

Bi(H2PO4)3 Bismuth(III) dihydrogen phosphate

Pb3N2 Lead(II) Nitride

Tues. Group Work:

Polyatomic Anions of O, S (6A) Cr (6B)

Oxygen: OH- hydroxide

Remember also: O2- oxide

Sulfur: SO42- sulfate

SO32- sulfite

HSO4 - hydrogen sulfate

HSO3 - hydrogen sulfite

Remember also: S2-, sulfide Chromium: CrO4

2- chromate

The “hydroxides” and “oxides” of the metallic elements are referred to as “bases”; all other ionic combinations are referred to as “salts”

“BASES” and “SALTS”

Bases:

Mg(OH)2

NaOH

CaO

Fe(OH)3

Salts:

MgCl2 MgHSO4 MgCO3

Na3PO4 NaNO2 Na2SO3

Ca(NO3)2 Ca3N2 CaSO4

Fe(CN)2 Fe(CH3CO2)3 Fe(H2PO4)2

Polyatomic Anions of Cl, Br, I (7A) Mn (7B)

Fluorine, F forms only the monatomic anion F-; Bromine, Br and Iodine, I form the same ions as chlorine, Cl:

Chlorine: ClO- hypochlorite ClO2

- chlorite ClO3

- chlorate ClO4

- perchlorate

Remember also: Cl-, Chloride

Manganese: MnO4- permanganate

SUMMARY, NAMING IONIC SALTS AND BASES

State name of the cation, then name of the anion.

Cations with a variable charge are named by adding a Roman numeral

Monoatomic anions are named by changing

theirelemental name to end in “ide”

Polyatomic anions (memorized) end in “ite” or “ate”...

GROUP WORK

FORMULA

NH4ClO

Cd(BrO2)2

Co(IO3)3

Ba(ClO4)2

KMnO4

Ag2CrO4

NAME

FORMULA

NH4ClO

Cd(BrO2)2

Co(IO3)3

Ba(ClO4)2

KMnO4

Ag2CrO4

NAME

Ammonium hypochlorite

Cadmium bromite

Cobalt(III) Iodate

Barium perchlorate

Potassium permanganate

Silver chromate

Naming Binary Molecular Compounds

All compounds beginning with a metal or ammonium are named as “ ionic compounds.”

Compounds containing only two elements (“binary”) in which both elements in the formula are a non-metal or metalloid are named in a different manner...

The change in nomenclature reflects the fact that these compounds are “molecular” and not “ionic” in nature!

• Name the first element in the formula

• Name the second element in the formula to end in “ide”:

carbide, nitride, phosphide, oxide, sulfide, fluoride, bromide, chloride, iodide •Add numerical prefixes to indicate more than one atom of the element in the formula:

di (2), tri (3), tetra (4), penta (5), hexa (6), hepta (7), octa(8)

Binary Molecular Nomenclature Method:

Typical Nomenclature

• NO2

• SF6

• ICl5

• N2O5

• CBr4

• SO3

• P2O3

• nitrogen dioxide• sulfur hexafluoride• iodine pentachloride• dinitrogen pentoxide• carbon tetrabromide• sulfur trioxide• diphosphorus trioxide

• BH3

• CH4

• SiH4

• NH3

• PH3

• borane

• methane*

• silane

• ammonia*

• phosphine

COMMON NAMES, BINARY MOLECULES ENDING IN H

From PARTICULATE (“too small to touch”)

to MACROSCOPIC (amounts we can handle):

THE MOLETHE MOLE

Kotz & Treichel, Chapter 3, 3.6-3.8

Many different items we encounter in our daily lives come packaged in set amounts described by various“counting terms”.

Let’s consider a few of them:

Many different items we encounter in our daily lives come packaged in set amounts described by various“counting terms”.

Let’s consider a few of them:

•shoes and socks and earrings come in pairs (2),

• eggs come in dozens (12),

•pencils are wholesaled by the gross (144), •donuts and sweet corn are often sold as “the baker’s dozen” or “the farmer’s dozen” (13),

•and diet pop and beer by the 6- pack or case (24).....

Chemists deal in atoms, molecules and ions, which need to be counted and measured as well.

BUT:

The mass of one atom of the 19F isotope is 3.156X10-23 g.

The radius of a nucleus is about .001 pm and the radius of an atom about 100 pm. (1012 picometers or 1,000,000,000,000 pm = 1 m)

THEREFORE......

Chemists need their own unit for counting and weighing amounts of substances which come in particle size too tiny to be seen or weighed on any balance.

For convenience in describing amounts of atoms, molecules, and ions , chemists have a unique unit of measure,

THE MOLE

The MOLE

•The chemist’s counting number

•Comes from the Latin meaning “whole heap or pile of”

•SI base unit for measuring amount of substance

•Defined as the number of atoms in exactly 12 grams of 12C, 6.022 X 1023

1 Mole always contains the same number of particles of whatever is being described, that is, Avogadro’s number of particles:

1 Mole = Avogadro’s number of particles

= 6.022136736 X 1023 particles

= 602,213,673,600,000,000,000,000 particles

If one used A’s number of particles to describe macroscopic objects, one would be overwhelmed:

I mole of green peas would cover the entire United States to a depth of 3 miles!

How do we get “a mole” of a substance ???

One mole is defined as the number of particlesin exactly 12 g of the 12C isotope of carbon.

Carbon was used as the standard for the amu scale, where the mass of one atom of 12C was defined as 12 amu.

The mole answers the question: “How many atomswould you have if you took the amu scale (which describesmass of one atom) and use it as grams instead?

Mass, amu, = 1 atom

Mass, g = ? atoms = 6.02 X 10 23 atoms (1 mole)

So, How to Get a Mole:

•We consult the periodic table, obtain the atomic mass of an element in amu’s, the relative mass of one atom.....

•We weigh this amount out in grams.....

•We now have one mole of atoms, A’s number, 6.02 X 10 23 atoms, a convenient “package of atoms”.....

•We have gone into the chemist’s counting system and can deliver not a dozen eggs but a mole of atoms....

This system works because of the relative nature of the atomic mass unit scale, in which all atoms were assigned a mass relative to 12C, the mole standard:

“One mole is the number of atoms in exactly 12 g of 12C”

The mole “pile or heap” of atoms for each element will weigh more or less than the mole “pile” for carbon, depending on whether the individual atoms weigh more or less than carbon.

If one mole of carbon atoms weighs 12.0 g, then one mole of oxygen atoms, which weighs 1.33 times more than carbon, would be: 1.33 X 12.0 g =16.0 g = 1 mol O

Since H atom is 1/12 the mass of a carbon atom, the matching pile of hydrogen atoms would be

1/12 X 12.0 g = 1.00 g = 1 mol H

•For any element, the molar mass, M, is the mass in grams of a mole of atoms, “#g/mole”

•M , molar mass, is NUMERICALLY equal to the mass of one atom in amu’s as given on your PT.

•If, however, one weighs out the molar mass, one has 6.022 x 10 23 atoms every time

The “Molar Mass, M”

MOLES of ATOMS: the MOLAR MASS

PT: amu’s / 1 atom M, g /mol, A’s # atoms

Li, 6.941 amu/atom Li, 6.941 g/ mol

Pb, 207.2 amu/atom Pb, 207.2 g/ mol

Zn, 65.39 amu/atom Zn, 65.39 g/ mol

Cr

51.9961

24

atomic mass, one atom,atomic mass units,relative to C

molar mass, 6.022 x 1023 atoms, in grams

ONE MOLE

A's NUMBER6.02 x 1023 UNITS

M,molar massin grams

Using this knowledge, the chemist can interconvertgrams, moles, and atoms of any element.

The molar mass, “g/mol”, like density, “g/cm3”, is a convenient conversion factor: For any element:

1mole = atomic weight, grams = 6.022 X 101mole = atomic weight, grams = 6.022 X 102323 atoms atoms

Suppose you weighed out 35.89 g of aluminum metal.How many moles and how many atoms of aluminum would be contained in this sample?

35.89 g Al=_ ?____mol Al

#1

Question: 35.89 g Al = ? mol Al = ? atoms AlRelationships: 1 mol Al = 26.98 g Al = 6.022 X 1023 atoms AlSetup and Solve: g ---> mol

35.89 g Al=_ ?____atoms Al#2

g mol atoms

35.89 g Al

26.98 g Al

1 mol Al=_ ?____mol Al = 1.330 mol Al#1

g mol

ans.

35.89 g Al=_ ?____atoms Al#2

26.98 g Al

1 mol Al

1 mol Al

6.022 x 1023 atoms Al

=atoms Al8.008 x 1023

ans

g --------> mol ---------> atoms

What would 9.00 x 1024 atoms of mercury weigh in grams?

Question: 9.00 x 1024 atoms Hg = ? g HgRelationships: 1 mol Hg = 200.59 g Hg = 6.022 X 1023 atoms HgSetup and Solve: atoms -----> mol -----> g

= 9.00 x 1024

atoms Hg g Hg

6.022 x 1023 atoms Hg

=

9.00 x 1024 atoms Hg 1 mol Hg

1 mol Hg

200.59 g Hg

g Hg2998

= 2.998 X 103 = 3.00 x 103 g Hg

atoms mol g

Mercury is a liquid metal with a density of 13.534 g/cm3. If you measured out 75.0 mL of Hg into a graduated cylinder, how many atoms of Hg would be in the sample?

75.0 mL Hg= ? atoms Hg

Question: 75.0 mL Hg = ? Atoms HgRelationships: 13.534 g Hg = 1cm3 or mL Hg 200.59 g Hg = 1 mol Hg 1 mol Hg = 6.022 x 1023 atoms HgSetup and solve: mL---> g ---> mol --->atoms

75.0 mL Hg

1 mL Hg

13.534 g Hg

200.59 g Hg

1mol Hg

1 mol Hg

6.02 x1023 atoms Hg

= 30.46 x 1023 atoms = 3.05 X 1024 atoms Hg

mL g mol atoms

Molecules, Compounds, and the Mole

Let us now extend the use of molar mass, M, to include all particles chemists need to measure: not just atoms but also especially ions and molecules....

The basic principle is this: whenever you weigh out the“formula weight” of any substance or species in grams, you have A’s number of particles of that species, and the molar mass of that species...

Molar Mass of Molecules

The formula of any molecule describes the number of atoms making up one unit of that molecule:

Br2 The diatomic bromine molecule, as bromine is found in nature: the formula tells us that 2 atoms of bromine are contained in every molecule.

By extension, 2 moles of bromine atoms are contained in every 1 mole of bromine molecules. The calculation of the molar mass of molecular bromine then looks like this:

The atomic weight of Br, from the PT, is 79.904 amu’s.

Therefore:

2 moles of Br= 2 X 79.904 g = 159.808 g

And the molar mass, M, of Br2 is 159.808 g/mol

Now let’s try the molar mass of CH3CH2OH, ethylalcohol:

Molar Mass, M, CH3CH2OH

Element # of atoms M, g/mol total

C 2 12.01 24.02

H 6 1.008 6.048

O 1 16.00 16.00

Total 46.068

MM, CH, CH33CHCH22OH, 46.07 g/molOH, 46.07 g/mol

Molar Mass of Ionic Compounds

The formula of an ionic compound indicates the simplest ratio of ions present in any sample of the compound. It is this “formula unit” that we use for calculating the molar mass.

Actually, we needn’t ask what kind of compound we are getting the M for; we simply calculate for all atoms found in the formula of any species!

Molar Mass, M, NaCl

Element # of atoms M, g/mol Total

Na(as Na+)

1 22.99 22.99

Cl(as Cl-)

1 35.45 35.45

Total 58.44

M, NaCl = 58.44 g/mol

MM, CH, CH33CHCH22OH, 46.07 g/mol, use in OH, 46.07 g/mol, use in

problems:problems:

Given a mass, or volume and density, Given a mass, or volume and density, solve for:solve for:

a) moles of compound or individual a) moles of compound or individual atomsatoms

b) grams of individual atomsb) grams of individual atoms

c) number of molecules or atomsc) number of molecules or atoms

MM, CH, CH33CHCH22OH, 46.07 g/mol, use in OH, 46.07 g/mol, use in

problems:problems:

Given a mass, or volume and density, Given a mass, or volume and density, solve for:solve for:

a) moles of compound or individual a) moles of compound or individual atomsatoms

b) grams of individual atomsb) grams of individual atoms

c) number of molecules or atomsc) number of molecules or atoms

How many moles of ethyl alcohol are contained in a sample that weighs 33.95 g? (CH3CH2OH, 46.07 g/mol).

Question: 33.95 g CH3CH2OH = ? mol CH3CH2OHRelationship: 46.07 g CH3CH2OH = 1 molSetup and Solve: ( g ---> mol)

33.95 g CH3CH2OH= ? mol CH3CH2OH

33.95 g CH3CH2OH = ? mol CH3CH2OH

46.07 g

1 mol

= .7369 mol CH3CH2OH

g ----------> mol

How many moles of hydrogen atoms are contained in33.95 g CH3CH2OH?

Question: 33.95 g CH3CH2OH = ? mol HRelationship: 46.07 g CH3CH2OH = 1 mol 1 mol CH3CH2OH = 6 mol HSetup and Solve: ( g ---> mol CH3CH2OH ---> mol H)

33.95 g CH3CH2OH= ? mol H

33.95 g CH3CH2OH = ? mol H

46.07 g CH 3CH 2OH

CH 3CH 2OH

CH 3CH 2OH

1 mol 6 mol

1 mol

H

= 4.422 mol H

g mol alcohol mol H

How many grams of hydrogen are contained in33.95 g CH3CH2OH?

Question: 33.95 g CH3CH2OH = ? g H

Relationship: 46.07 g CH3CH2OH = 1 mol 1 mol CH3CH2OH = 6 mol H 1 mol H = 1.008 g

Setup and Solve: ( g CH3CH2OH ---> mol CH3CH2OH ---> mol H -----> g H)

33.95 g CH3CH2OH = ? g H

g alcohol mol alcohol mol H

33.95 g CH3CH2OH

= ? g H

46.07 g CH 3CH 2OH

CH 3CH 2OH

CH 3CH 2OH

1 mol 6 mol

1 mol

H

= 4.457 g H

g H

1 mol H

1.008 g H

END, Lecture 6