Compounds & Molecules NaCl, salt Buckyball, C 60 Ethanol, C 2 H 6 O Molecule: The smallest identifiable unit that retains the physical and chemical properties of the pure substances.

Compounds & Molecules COMPOUND is a combination of 2 or more elements in definite ratios by mass.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).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.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 2Formula for glycine is C 2 H 5 NO 2 (description of the composition) (description of the composition) In one molecule there areIn 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 HFormula for glycine is NH 2 CH 2 CO 2 H (composition and functional groups) (composition and functional groups) In one molecule there areIn one molecule there are –1 NH 2 (amine group) –1 CH 2 group –1 CO 2 H group

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

WRITING FORMULAS condensed formulaCan also write glycine as the condensed formula H 2 NCH 2 COOH showing functional groups (atom ordering and connectivity) structural formulaor in the form of a structural formula showing how atoms are attached to each other (bond orders)

MOLECULAR MODELING Ball & stick Space-filling Drawing of glycine An even higher level of structural detail

Resources for Molecular Modeling Modeling softwareModeling 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) = 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) = g O TOTAL = molar mass = g/mol

Formula = Molar mass = TylenolTylenol C 8 H 9 NO 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 = g/mol (b) Calc. moles of alcohol

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? = 2.78 x molecules We know there are mol of C 2 H 6 O.

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? = 5.57 x C atoms There are 2.78 x molecules. Each molecule contains 2 C atoms. Therefore, the number of C atoms is

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.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 chargeTaking 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.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 e- F + e- --> F - oxidation reduction

oxidation reduction

PREDICTING ION CHARGES In general metals (Mg) lose electrons ---> cationsmetals (Mg) lose electrons ---> cations nonmetals (F) gain electrons ---> anionsnonmetals (F) gain electrons ---> anions

Charges on Common Ions By losing or gaining e-, atom has same number of electrons as nearest Group 8A atom. Cation charge=group # Anion charge=group #-8

Predicting Charges on Monatomic Ions

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

NONMETALSNONMETALS NONMETAL + n e > X n- where n = 8 - Group no. C 4-,carbide N 3-, nitride O 2-, oxide S 2-, sulfide F -, fluoride Cl -, chloride Group 7AGroup 6A Group 4A Group 5A 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 #

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

PO 4 3- phosphate ion CH 3 CO 2 - acetate ion Polyatomic Ions

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

IONIC COMPOUNDS NH 4 + Cl - ammonium chloride, NH 4 Cl

Some Ionic Compounds Mg 2+ + NO > Mg(NO 3 ) 2 magnesium nitrate magnesium nitrate Fe 2+ + PO > Fe 3 (PO 4 ) 2 iron(II) phosphate iron(II) phosphate calcium fluoride Ca F - ---> CaF 2

Properties of Ionic Compounds Forming NaCl from Na and Cl 2 A metal atom can transfer an electron to a nonmetal.A metal atom can transfer an electron to a nonmetal. The resulting cation and anion are attracted to each other by electrostatic forces.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 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 o C

ELEMENTS THAT EXIST AS MOLECULES See SCREEN 3.2 on the CD-ROM Allotropes of C

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 CH 4 methane CO 2 Carbon dioxide BCl 3 boron trichloride

Naming Molecular Compounds CH 4 methane BCl 3 boron trichloride CO 2 Carbon dioxide All are formed from two or more nonmetals. 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) Structure of NO 2 Chemistry of NO, nitrogen monoxide

Percent Composition Consider NO 2, Molar mass = ? What is the weight percent of N and of O? What are the weight percentages of N and O in NO?

How to Determine a Formula? Mass spectrometer

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

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 : 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.Because it contains only B and H, it must contain 18.90% H. In g of the compound there are g of B and g of H.In g of the compound there are g of B and g of H. Calculate the number of moles of each constitutent.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 g of sample. 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. 1 atom B + 3 atoms H --> 1 molecule BH 3 or 1 mol B atoms + 3 mol H atoms ---> 1 mol BH 3 molecules 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?

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 Take the ratio of moles of B and H. Always divide by the smaller number. A compound of B and H is 81.10% B. What is its empirical formula?

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 is one example of this class of compounds. B2H6B2H6

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 to find the MOLAR MASS. Here experiment gives 53.3 g/mol Compare with the mass of B 2 H 5 = g/unit = g/unit Find the ratio of these masses. Molecular formula = B 4 H 10

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.Reaction of Sn and I 2 is done using excess Sn. Mass of Sn in the beginning = gMass of Sn in the beginning = g Mass of iodine (I 2 ) used = gMass of iodine (I 2 ) used = g Mass of Sn remaining = gMass of Sn remaining = g

Find the mass of Sn that combined with g I 2. Mass of Sn initially = g Mass of Sn recovered = g Mass of Sn used = g Find moles of Sn used: Tin and Iodine Compound

Now find the number of moles of I 2 that combined with 3.83 x mol Sn. Mass of I 2 used was g. How many mol of iodine atoms ? = x mol I atoms

Tin and Iodine Compound Now find the ratio of number of moles of moles of I and Sn that combined. Empirical formula is SnI 4