Compounds & Molecules

Molecule: The smallest identifiable unit that retains the physical and chemical properties of the pure substances.

NaCl, salt Ethanol, C 2 H 6 O Buckyball, C 60

Compounds & Molecules

• COMPOUND

is a combination of 2 or more elements in definite ratios by mass.

•

The character of each element is lost when forming a compound (e.g., think of NaCl).

• MOLECULES

are the smallest units of a compound that retains the characteristics of the compound.

MOLECULAR FORMULAS

• •

Formula for glycine is C 2 H 5 NO 2 (description of the composition) In one molecule there are

–

2 C atoms

–

5 H atoms

–

1 N atom

–

2 O atoms

CONDENSED FORMULAS

• •

Formula for glycine is NH 2 CH 2 CO 2 H (composition and functional groups) In one molecule there are

–

1 NH 2

–

1 CH 2

–

1 CO 2 (amine group) group H group

STRUCTURAL FORMULAS

• •

Show how the atoms are attached within a molecule The lines between atoms represent chemical bonds that hold the atoms together.

WRITING FORMULAS

• •

Can also write glycine as the condensed formula H 2 NCH 2 COOH showing functional groups (atom ordering and connectivity) or in the form of a structural formula H H O H N C H C O H showing how atoms are attached to each other (bond orders)

MOLECULAR MODELING

An even higher level of structural detail H H N H O C H C O H Drawing of glycine Ball & stick Space-filling

Resources for Molecular Modeling

•

Modeling software

–

CAChe (General Chemistry Interactive

CD-ROM)

–

Rasmol

–

Molden

–

Gaussview

–

Maestro

MOLECULAR WEIGHT AND MOLAR MASS

Molecular weight = sum of the atomic weights of all atoms in the molecule.

Molar mass = molecular weight in grams per mol.

What is the molar mass of ethanol, C 2 H 6 O?

1 mol contains 2 moles of C (12.01 g C/1 mol) = 24.02 g C 6 moles of H (1.01 g H/1 mol) = 6.06 g H 1 mol of O (16.00 g O/1 mol) = 16.00 g O TOTAL = molar mass = 46.08 g/mol

Tylenol

• •

Formula = C 8 H 9 NO 2 Molar mass = 151.2 g/mol

Molar Mass

How many

moles

of alcohol (C 2 H 6 O) are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O?

(a) Molar mass of C 2 H 6 O = 46.08 g/mol (b) Calc. moles of alcohol 21.3 g • 1 mol 46.08 g = 0.462 mol

How many molecules of alcohol are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O?

We know there are 0.462 mol of C 2 H 6 O.

0.462 mol • 6.022 x 10 23 molecules 1 mol = 2.78 x 10 23 molecules

How many atoms of C are there in a “standard” can of beer if there are 21.3 g of C 2 H 6 O?

There are 2.78 x 10 23 molecules.

Each molecule contains 2 C atoms.

Therefore, the number of C atoms is 2.78 x 10 23 molecules • 2 C atoms 1 molecule = 5.57 x 10 23 C atoms

Molecular & Ionic Compounds Heme NaCl Molecular compounds consist of discrete molecules Ionic compounds consist of discrete ions

IONS AND IONIC COMPOUNDS

• • IONS

are atoms or groups of atoms with a positive or negative charge. Taking away an electron from an atom gives a CATION with a positive charge

•

Adding an electron to an atom gives an ANION with a negative charge .

Forming Cations & Anions

A CATION forms when an atom loses one or more electrons.

An ANION forms when an atom gains one or more electrons Mg --> Mg 2+ + 2 e oxidation F + e- --> F reduction

oxidation reduction

PREDICTING ION CHARGES

In general

•

metals (Mg) lose electrons ---> cations

•

nonmetals (F) gain electrons ---> anions

Charges on Common Ions

Cation charge=group # +1 Anion charge=group #-8 -4 -3 -2 -1 +2 +3 By losing or gaining e-, atom has same number of electrons as nearest Group 8A atom .

Predicting Charges on Monatomic Ions

METALS

M ---> n e- + M n+ where n = periodic group Na + Mg 2+ Al 3+ sodium ion magnesium ion aluminum ion Transition metals --> M 2+ or M 3+ are common Fe 2+ iron(II) ion Fe 3+ iron(III) ion

NONMETALS

NONMETAL + n e- ------> X n where n = 8 - Group no.

Group 4A Group 5A Group 6A

C 4 ,carbide N 3 , nitride O 2 , oxide

Group 7A

F , fluoride S 2 , sulfide Cl , chloride Br , bromide I , iodide

Ion Formation

Reaction of aluminum and bromine

POLYATOMIC IONS

CD Screen 3.6

Groups of atoms with a charge.

MEMORIZE the names and formulas of common polyatomic ions listed in Table 3.1, page 107 (next slide)

Polyatomic Ions

NH 4 + ammonium ion One of the few common polyatomic cations

Polyatomic Ions (oxoanions)

HNO 3 nitric acid NO 3 nitrate ion Prefix per- and suffix –ate: largest # Suffix -ate : greater # of oxygen atoms Suffix -ite : smaller # of oxygen atoms Prefix hypo- and suffix –ite: smallest #

Polyatomic Ions

SO 4 2 sulfate ion SO 3 2 sulfite ion

Polyatomic Ions

NO 3 nitrate ion NO 2 nitrite ion

Polyatomic Ions

CO 3 2 carbonate ion HCO 3 bicarbonate ion hydrogen carbonate

Polyatomic Ions

PO 4 3 phosphate ion CH 3 CO 2 acetate ion

COMPOUNDS FORMED FROM IONS

CATION ANION + ---> COMPOUND Na + + Cl --> NaCl A neutral compd. requires equal number of + and - charges.

IONIC COMPOUNDS

NH 4 + Cl ammonium chlor ide , NH 4 Cl

Some Ionic Compounds

Ca 2+ + 2 F ---> CaF 2 Mg 2+ + NO 3 ----> Mg(NO 3 ) 2 magnesium nitrate calcium fluor ide Fe 2+ + PO 4 3 ----> Fe 3 (PO 4 ) 2 iron(II) phosphate

Properties of Ionic Compounds

Forming NaCl from Na and Cl 2

• •

A metal atom can transfer an electron to a nonmetal.

The resulting cation and anion are attracted to each other by

electrostatic forces

.

Electrostatic Forces

The oppositely charged ions in ionic compounds are attracted to one another by ELECTROSTATIC FORCES .

These forces are governed by COULOMB’S LAW .

Electrostatic Forces

COULOMB’S LAW

Force of attraction = (charge on +)(charge on -) (distance between ions) 2 As ion charge increases, the attractive force _______________.

As the distance between ions increases, the attractive force ________________.

This idea is important and will come up many times in future discussions!

Importance of Coulomb’s Law

NaCl, Na + and Cl , m.p. 804 o C MgO, Mg 2+ and O 2 m.p. 2800 o C

ELEMENTS THAT EXIST AS MOLECULES

Allotropes of C

See SCREEN 3.2 on the CD-ROM

Screen 3.2

ELEMENTS THAT EXIST AS DIATOMIC MOLECULES (gases)

ELEMENTS THAT EXIST AS POLYATOMIC MOLECULES White P 4 and polymeric red phosphorus S 8 sulfur molecules

Molecular Compounds

Compounds without Ions CO 2 Carbon dioxide CH 4 methane BCl 3 boron trichloride

CH 4 methane

Naming Molecular Compounds

CO 2 Carbon dioxide All are formed from two or more nonmetals. BCl 3 boron trichloride Ionic compounds generally involve a metal and nonmetal (NaCl)

Empirical & Molecular Formulas

A pure compound always consists of the same elements combined in the same proportions by weight.

Therefore, we can express molecular composition as PERCENT BY WEIGHT Ethanol, C 2 H 6 O 52.13% C 13.15% H 34.72% O

Percent Composition

Consider some of the family of nitrogen oxygen compounds: NO 2 , nitrogen dioxide and closely related, NO, nitrogen monoxide (or nitric oxide) Chemistry of NO, nitrogen monoxide Structure of NO 2

Percent Composition

Consider NO 2 , Molar mass = ?

What is the weight percent of N and of O?

14.0 g N Wt. % N = 46.0 g NO 2 • 100% = 30.4 % Wt. % O



2 (16 .0 g O per mole ) 46 .0 g x 100 %



69 .6% What are the weight percentages of N and O in NO?

Determining Formulas

In

chemical analysis

we determine the % by weight of each element in a given amount of pure compound and derive the EMPIRICAL or SIMPLEST formula.

PROBLEM formula?

: A compound of B and H is 81.10% B. What is its empirical

A compound of B and H is 81.10% B. What is its empirical formula?

•

Because it contains only B and H, it must contain 18.90% H.

•

In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H.

•

Calculate the number of moles of each constitutent.

A compound of B and H is 81.10% B. What is its empirical formula?

Calculate the number of moles of each element in 100.0 g of sample.

81.10 g B • 1 mol 10.81 g = 7.502 mol B 18.90 g H • 1 mol 1.008 g = 18.75 mol H

A compound of B and H is 81.10% B. What is its empirical formula?

Now, recognize that atoms combine in the ratio of small whole numbers Find the ratio of moles of elements in the compound.

A compound of B and H is 81.10% B. What is its empirical formula?

Take the ratio of moles of B and H. Always divide by the smaller number.

18.75 mol H 7.502 mol B = 2.499 mol H 1.000 mol B = 2.5 mol H 1.0 mol B But we need a whole number ratio. 2.5 mol H/1.0 mol B = 5 mol H to 2 mol B EMPIRICAL FORMULA = B 2 H 5

A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5 . What is its

molecular formula

?

Is the molecular formula B 2 H 5 , B 4 H 10 , B 6 H 15 , B 8 H 20 , etc.? B 2 H 6 B 2 H 6 is one example of this class of compounds.

A compound of B and H is 81.10% B. Its empirical formula is B 2 H 5 . What is its molecular formula ?

We need to do an EXPERIMENT the MOLAR MASS.

to find Here experiment gives 53.3 g/mol Compare with the mass of B 2 H 5 = 26.66 g/unit Find the ratio of these masses.

53.3 g/mol 26.66 g/unit of B 2 H 5 = 2 units of B 2 H 5 1 mol Molecular formula = B 4 H 10

How to Determine the molar mass?

Mass spectrometer

Mass Spectrum of Ethanol Mass Spectrum of Ethanol (from the NIST site) CH 2 O + 31 CH 3 CH 2 O + 45 CH 3 CH 2 OH + 46

Determine the formula of a compound of Sn and I using the following data.

• • • •

Reaction of Sn and I 2 is done using excess Sn.

Mass of Sn in the beginning = 1.056 g Mass of iodine (I 2 ) used = 1.947 g Mass of Sn remaining = 0.601 g

Tin and Iodine Compound

Find the mass of Sn that combined with 1.947 g I 2 .

Mass of Sn initially = 1.056 g Mass of Sn recovered = 0.601 g Mass of Sn used = 0.455 g Find moles of Sn used: 0.455 g Sn • 1 mol 118.7 g = 3.83 x 10 -3 mol Sn

Tin and Iodine Compound

Now find the number of moles of I 2 combined with 3.83 x 10 -3 that mol Sn. Mass of I 2 used was 1.947 g.

1.947 g I 2 • 1 mol 253.81 g = 7.671 x 10 -3 mol I 2 How many moles of iodine atoms ? 7.671 x 10 -3 mol I 2



2 mol I atoms 1 mol I 2



= 1.534 x 10 -2 mol I atoms

Tin and Iodine Compound

Now find the ratio of number of moles of moles of I and Sn that combined.

1.534 x 10 -2 mol I 3.83 x 10 -3 mol Sn 4.01 mol I = 1.00 mol Sn

Empirical formula is

SnI 4