1
Chemistry Lecture ’93 B. Rife CHS
Text: Modern Chemistry; Holt, Rinehart & Winston 1993
Chemical Formulas and Chemical Compounds Chapter 7
Homework:
1
Section Reviews (pg 212,216,223,228)
2
Reviewing Concepts:
(1,2,6,8,9,10,11,12,13,14,15,19,20,21,23)
3
Problems (all) (pg 230-231)
4
Chapter/Section Review (Handout)
(pg 229-230)
Exam Date _
7.1 Chemical names and Formulas
7.1A Explain the significance of a chemical formula. Significance of a Chemical Formula
CHEMICAL FORMULA IS A SHORTHAND METHOD OF REPRESENTING THE
COMPOSITION OF A SUBSTANCE BY USING CHEMICAL SYMBOLS AND
NUMERICAL SUBSCRIPTS.
CHEMICAL FORMULAS INDICATE THE RELATIVE NUMBER OF ATOMS FOR EACH
ELEMENT PRESENT IN A COMPOUND.
A COMPOUND IS A SUBSTANCE IN WHICH TWO OR MORE ELEMENTS ARE CHEMICALLY
COMBINED. THE FORMULA H2O REPRESENTS: ONE MOLECULE OF H2O
ONE MOLE OF H2O
ONE MOLAR MASS (18 g) OF H2O
STRUCTURAL FORMULA REPRESENTS THE ARRANGEMENT OF ATOMS IN A
COMPOUND.
7.1B Determine the formula of an ionic compound between any two given ions
( )
MONATOMIC IONS ARE IONS FORMED FROM A SINGLE ATOM
CATIONS (METALS) HAVE LOST ELECTRONS POSITIVE IONS ARE NAMED BY THE
ELEMENT NAME FOLLOWED BY “ION”.
ANIONS (NONMETALS) HAVE GAINED ELECTRONS
NEGATIVE IONS ARE NAMED BY DROPPING THE ENDING OF THE ELEMENT NAME AND
ADDING THE SUFFIX “IDE”
NOTE COMMON MONATOMIC IONS TABLE 7-3
BINARY COMPOUNDS ARE COMPOUNDS COMPOSED OF TWO DIFFERENT ELEMENTS.
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ION CHARGE METHOD OF WRITING FORMULAS
1.
RECOGNIZE POSITIVE AND NEGATIVE IONS NAMED IN COMPOUND.
WRITE THE SYMBOLS FOR THE IONS SIDE BY SIDE, WITH THE POSITIVE ION FOR
MOST METALLIC ELEMENT FIRST.
2.
ADJUST THE NUMBER OF EACH ION SO THAT THE TOTAL POSITIVE AND TOTAL
NEGATIVE CHARGES EQUAL IN MAGNITUDE. CROSSOVER THE CHARGE VALUES
TO GIVE SUBSCRIPTS.
3.
CHECK THE SUBSCRIPTS SO THAT CHARGE IS BALANCED AND WRITE THE
FORMULA CHARGES OF IONS ARE NOT USUALLY INCLUDED IN CHEMICAL
FORMULAS OF IONIC COMPOUNDS.
THE SUBSCRIPT 1 IS UNDERSTOOD WHEN NO SUBSCRIPT IS PRESENT AND IS NOT
WRITTEN
EXAMPLE: MONATOMIC IONS
SODIUM CHLORIDE
MAGNESIUM BROMIDE
ALUMINUM SULFIDE
Na+ & ClMg2+ & Br-Al3+ & S2-
NaCl
MgBr2
Al2S3
Pb2+ & SO42-
PbSO4
MAGNESIUM HYDROXIDE
Mg2+ & OH-
Mg(OH)2
AMMONIUM SULFATE
NH4+ & SO42-
(NH4)2SO4
POLYATOMIC IONS
LEAD(II) SULFATE
A ROMAN NUMERAL USED IN THE NAME OF A METALLIC ION DOES NOT APPEAR IN
THE FORMULA OF A COMPOUND CONTAINING THE ION.
IT IS POSSIBLE TO WRITE THE FORMULA FOR A COMPOUND AND THEN LEARN
THAT SUCH A COMPOUND DOES NOT EXIST.
7.1C Explain the two systems for distinguishing different ionic compounds of the same two
elements.
( )
NOMENCLATURE - IS A TERM THAT REFERS TO METHODS OF NAMING CHEMICAL
COMPOUNDS.
THE “OLD SYSTEM” DIFFERENT NAMES ARE NEEDED FOR POSITIVE IONS (TRANSITION
METALS) OF TWO DIFFERENT CHARGES.
THE SUFFIX -OUS IS GIVEN TO THE ION WITH THE LOWER CHARGE
THE SUFFIX -IC IS GIVEN TO THE ION WITH THE HIGHER CHARGE
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THE STOCK SYSTEM
COMPOUNDS OF METALS THAT FORM MORE THAN ONE ION (HAVE MORE THAN ONE
OXIDATION STATE) ARE NAMED BY INCLUDING THE A ROMAN NUMERAL
REPRESENTING THE CHARGE.
IN BOTH SYSTEMS MONATOMIC ANIONS ARE NAMED WITH THE ROOT NAME OF THE
ANION AND THE THE SUFFIX -IDE.
A POLYATOMIC ION IS A GROUP OF ATOMS BONDED TOGETHER COVALENTLY
(SHARED) BUT POSSESSING AN OVERALL CHARGE.
NOTE COMMON POLYATOMIC IONS TABLE 7-4
OXYANIONS - ARE POLYATOMIC IONS THAT CONTAIN OXYGEN
THE ION WITH THE LARGER NUMBER OF OXYGEN ATOMS IS NAMED WITH
THE -ATE SUFFIX.
(NITRATE NO3, SULFATE SO4)
THE ION WITH THE SMALLER NUMBER OF OXYGEN ATOMS IS NAMED WITH
THE -ITE SUFFIX.
(NITRITE NO2, SULFITE SO3)
7.1D Name an ionic compound, given its formula
( )
PSEUDOBINARY COMPOUNDS - CONTAIN MORE THAN TWO ELEMENTS. IN THESE
COMPOUNDS ONE OR MORE OF THE IONS CONSIST OF MORE THAN ONE ELEMENT BUT
BEHAVE AS IF THEY WERE SIMPLE IONS.
THE MOST COMMON EXAMPLES OF SUCH ANIONS ARE THE HYDROXIDE ION OHAND THE CYANIDE ION, CN-.
THE AMMONIUM ION NH4+, IS THE MOST COMMON CATION THAT BEHAVES LIKE A
SIMPLE METAL CATION.
IN NAMING COMPOUNDS CONTAINING POLYATOMIC IONS, THE NAME OF THE
POLYATOMIC ION IS NOT CHANGED.
Fe2O3
HCN
NaOH
NH4NO3
IRON (III) OXIDE
HYDROGEN CYANIDE
SODIUM HYDROXIDE
AMMONIUM NITRATE
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7.1E Using prefixes, name a binary molecular compound from its formula.
7.1F Write the formula of a binary molecular compound, given its name.
( )
( )
THE “OLD SYSTEM” OF NAMING BINARY MOLECULAR COMPOUNDS
THE GENERAL RULE IS TO NAME THE LESS ELECTRONEGATIVE ELEMENT FIRST AND
THE MORE ELECTRONEGATIVE ELEMENT SECOND.
THE FIRST WORD OF A BINARY MOLECULAR COMPOUND IS MADE UP OF
(A) THE PREFIX OF THE NUMBER OF ATOMS OF THE FIRST ELEMENT AND
(B) THE NAME OF THE FIRST ELEMENT.
IF ONLY ONE ATOM OF THE FIRST ELEMENT APPEARS IN THE FORMULA THEN NO
PREFIX (MON) IS USED.
THE SECOND WORD OF A BINARY MOLECULAR COMPOUND IS MADE UP OF
(A) THE PREFIX OF THE NUMBER OF ATOMS OF THE SECOND ELEMENT AND
(B) THE ROOT OF THE NAME OF THE SECOND ELEMENT AND
(C) THE SUFFIX -IDE WHICH MEANS THAT ONLY THE TWO ELEMENTS NAMED
ARE PRESENT.
PREFIXES: (TABLE 7-5)
MON = 1
DI = 2
TRI = 3
ROOTS:
H HYDRO
III
B BOR
IV
C CARB
Si SILIC
TETRA = 4
V
N NITR
P PHOSPH
As ARSEN
PENTA = 5
VI
O OX
S SULF
Se SELEN
VII
F FLUOR
Cl CHLOR
Br BROM
EXAMPLES:
CO
CARBON MONOXIDE
CO2
CARBON DIOXIDE
SbCl3 ANTIMONY TRICHLORIDE
CCl4
CARBON TETRACHLORIDE
As2S5 DIARSENIC PENTASULFIDE
7.1G List the names and formulas of the common laboratory acids.
( )
SOME MOLECULAR COMPOUNDS CONTAINING H ATOMS DISSOCIATE IN WATER TO
FORM H+ IONS. THESE ARE CALLED ACIDS.
BINARY ACIDS - CONTAIN HYDROGEN AND ONE OF THE HALOGENS
IF THE ANION DOES NOT CONTAIN OXYGEN, THE ACID IS NAMED WITH THE PREFIX
HYDRO- AND THE SUFFIX -IC.
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HCl (g)
HYDROGEN CHLORIDE
HF (g)
HYDROGEN FLUORIDE
HCl (aq)
HYDROCHLORIC ACID
HF (aq)
HYDROFLUORIC ACID
OXYACIDS - CONTAIN HYDROGEN, OXYGEN, AND A THIRD ELEMENT
WHEN THE ANION CONTAINS OXYGEN, THE ACID NAME FORMED IS FROM AN
ANION WITH -ATE SUFFIX IS THE ROOT NAME AND THE SUFFIX -IC (acids)
FROM AN ANION WITH -ITE SUFFIX IS THE ROOT NAME AND
THE SUFFIX -OUS (acids)
ClO3- CHLORATE ION
ClO2- CHLORITE ION
HClO3 CHLORIC ACID
HClO2 CHLOROUS ACID
AN IONIC COMPOUND COMPOSED OF A CATION (FROM A BASE) AND THE ANION
FROM AN ACID IS REFERRED TO AS A SALT.
SOME SALTS CONTAIN ANIONS IN WHICH ONE OR MORE HYDROGEN ATOMS FROM THE
ACID ARE RETAINED SUCH ANIONS ARE NAMED BY ADDING THE WORD
HYDROGEN OR THE PREFIX BIHCO3HYDROGEN CARBONATE
BICARBONATE
7.2 Oxidation Numbers
OXIDATION NUMBERS REPRESENT APPARENT CHARGE ON AN ATOM OR ATOMS AND
ARE USED TO WRITE BALANCED CHEMICAL FORMULAS.
7.2A List the rules for assigning oxidation numbers
( )
1. AN UNCOMBINED ELEMENT (MOLECULAR) HAS AN OXIDATION NUMBER OF ZERO.
2. A MONATOMIC ION HAS AN OXIDATION NUMBER EQUAL TO THE CHARGE OF THE
ION.
3. FLUORINE ALWAYS HAS AN OXIDATION NUMBER OF -1.
4. OXYGEN HAS AN OXIDATION NUMBER OF -2 IN MOST COMPOUNDS EXCEPT IN
PEROXIDES SUCH AS H2O2 WHERE IT IS -1.
5. HYDROGEN HAS AN OXIDATION NUMBER OF +1 IN ALL COMPOUNDS WITH MORE
ELECTRONEGATIVE ELEMENTS, AND AN OXIDATION NUMBER OF -1 IN
COMPOUNDS WITH METALS KNOWN AS HYDRIDES.
6. THE MORE ELECTRONEGATIVE ELEMENT IN A BINARY COMPOUND IS ASSIGNED
THE NUMBER EQUAL TO THE CHARGE IT WOULD HAVE IF IT WERE AN ION.
7. THE ALGEBRAIC SUM OF THE OXIDATION NUMBERS OF ALL ATOMS IN A NEUTRAL
COMPOUND IS ZERO.
8. THE ALGEBRAIC SUM OF THE OXIDATION NUMBERS OF ALL ATOMS IN A
POLYATOMIC ION IS EQUAL TO THE CHARGE OF THE ION.
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7.2B Give the oxidation number for each element in the formula of a chemical compound
( )
+1 -2 +1
NaOH
+2 +5 -2
Cu(NO3)2
+1 +7 -2
KMnO4
7.2C Name binary compounds, using oxidation numbers and the Stock system.
( )
THE STOCK SYSTEM IS AN ALTERNATIVE TO THE PREFIX SYSTEM FOR NAMING
BINARY MOLECULAR COMPOUNDS
N2 O
LAUGHING GAS
DINITROGEN MONOXIDE
NITROGEN (I) OXIDE
7.3 Using Chemical Formulas
7.3A Calculate the formula mass or molar mass of any given compound.
( )
FORMULA MASS - IS THE SUM OF THE ATOMIC MASSES OF THE ATOMS IN A FORMULA
UNIT (IONIC COMPOUND)
MOLECULAR MASS - THE MASS OF ONE MOLECULE OF A MOLECULAR SUBSTANCE
(COVALENT), IN ATOMIC MASS UNITS.
MOLECULAR MASS IS THE SUM OF THE ATOMIC MASSES OF THE ATOMS IN THE
MOLECULE.
MOLECULAR WEIGHT OF CANE SUGAR C12H22O11
12 x C = 12 x 12 amu =
144 amu
22 x H = 22 x 1 amu =
22 amu
11 x O = 11 x 16 amu =
176 amu
__________________________________
MOLECULAR WEIGHT =
342 amu
Ca3(PO4)2 = 3(40.08) + 2(30.97) + 2(4)(16.0) = 310.2 u
7.3B Use molar mass to convert between mass in grams and amount in moles of a chemical
compound ( )
MOLES X MOLAR MASS (g) = MASS IN GRAMS MOLES
MASS IN GRAMS X
MOLES
= MOLES MOLAR MASS
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7.3C Give the number of molecules, formula units, or ions in a given molar amount of a chemical
compound.( )
6Ba3(AsO4)2
STATE THE NUMBER OF MOLECULES IN THE ABOVE FORMULA.
STATE THE NUMBER OF IONS IN THE ABOVE FORMULA.
STATE THE NUMBER OF ATOMS IN THE ABOVE FORMULA.
7.3D Calculate the percent composition of a given chemical compound.
( )
THE PERCENT COMPOSITION OF A COMPOUND IS THE PERCENT BY MASS OF EACH
ELEMENT IN THE COMPOUND. THE PERCENT IS THE SAME, NO MATTER WHAT THE SIZE
OF THE SAMPLE (LAW OF CONSTANT OR DEFINITE COMPOSITION.
PERCENTAGE COMPOSITION MAY BE USED IN THE IDENTIFICATION OF SOME
COMPOUNDS.
ATOMIC WT. x # OF ATOMS OF THE ELEMENT
MOLAR MASS OF THE COMPOUND
x 100% COMPOUND =
= % ELEMENT
TOTAL PERCENT OF ALL ELEMENTS IN A COMPOUND EQUALS 100% (CHECK)
EXAMPLE:
HYDRATE - A SOLID COMPOUND CONTAINING A DEFINITE PERCENTAGE OF
BOUND WATER.
SODIUM CARBONATE DECAHYDRATE Na2CO3 . 10H2O
FORMULA WEIGHT Na2CO3 . 10H2O
2 x Na = 2 x 23 =
46 amu
1 x C = 1 x 12 =
12
3 x O = 3 x 16 =
48
20 x H = 20 x 1 =
20
10 x O = 10 x 16 =
160
__________________________
FORMULA WEIGHT = 286 amu
PERCENT OF Na2
46 Na2
286 Na2CO3 . 10H2O X 100% Na2CO3 . 10H2O = 16.8% Na
PERCENT OF H2O
180 H2O
286 Na2CO3 . 10H2O X 100% Na2CO3 . 10H2O = 62.9% H2O
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7.4 Determining Chemical Formulas
7.4A Define simplest formula, and explain how the term applies to ionic and molecular
compounds.
( )
EMPIRICAL (SIMPLEST) FORMULA - THE SMALLEST WHOLE-NUMBER RATIO OF
ATOMS PRESENT IN A COMPOUND. (IONIC COMPOUNDS)
LAW OF CONSTANT COMPOSITION - DIFFERENT SAMPLES OF ANY PURE COMPOUND
CONTAIN THE SAME ELEMENTS IN THE SAME PROPORTIONS BY MASS.
7.4B Find a simplest formula from either percent or mass composition.
( )
FROM PERCENTAGE COMPOSITION
A.
CONVERT PERCENT INTO GRAMS.
ONE HUNDRED PERCENT EQUALS ONE HUNDRED GRAMS.
B.
CONVERT GRAMS OF ELEMENT INTO MOLES OF ELEMENT.
DIVIDE BY GRAM-ATOMIC WEIGHT (MOLAR MASS)
C.
REDUCE RATIO OF MOLES OF ELEMENTS
DIVIDE EACH NUMBER BY THE SMALLEST, MULTIPLY THE RESULTING
NUMBERS BY THE SMALLEST WHOLE NUMBER TO REDUCE FRACTIONS.
EXAMPLE:
ANALYSIS OF A SAMPLE OF A PURE COMPOUND REVEALS
THAT IT CONTAINS 50.1% SULFUR AND 49.9% OXYGEN BY MASS.
WHAT IS THE SIMPLEST FORMULA?
A: 50.1% S = 50.1g S AND 49.9% O = 49.9g O
B: 50.1g S (1 MOL S / 32.1g S) = 1.56 MOL S
49.9g O (1 MOL O / 16.0g O) = 3.12 MOL O
C: 1.56/1.56 = 1 S AND 3.12/1.56 = 2 O
= SO2
FROM RELATIVE MASS DATA BEGIN WITH STEP B
7.4C Explain the relationship between the simplest formula and the molecular formula of a given
compound. ( )
A MOLECULAR FORMULA IS A WHOLE NUMBER MULTIPLE OF AN EMPIRICAL
FORMULA.
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MOLECULAR (TRUE) FORMULA - FORMULA THAT INDICATES THE ACTUAL NUMBER
OF ATOMS PRESENT IN A MOLECULE OF A MOLECULAR SUBSTANCE (COVALENT
COMPOUNDS)
7.4D Find a molecular formula from a simplest formula( )
IT IS NOT POSSIBLE TO DECIDE WHICH IS THE TRUE FORMULA UNLESS THE
MOLECULAR WEIGHT OF THE SUBSTANCE HAS BEEN DETERMINED.
MOLECULAR FORMULA IS A WHOLE-NUMBER MULTIPLE (x) OF THE EMPIRICAL
FORMULA.
(EMPIRICAL FORMULA)x = MOLECULAR FORMULA
MOLECULAR WEIGHT IS A WHOLE-NUMBER MULTIPLE (x) OF THE EMPIRICAL
FORMULA WEIGHT
(EMPIRICAL WEIGHT)x = MOLECULAR WEIGHT
EXAMPLE
IN A SAMPLE OF GLUCOSE WE FOUND 40.0% C, 6.72% H, 53.3% O, AND THE MOLECULAR
WEIGHT TO BE 180. DETERMINE THE SIMPLEST FORMULA AND THE MOLECULAR
FORMULA OF GLUCOSE.
SIMPLEST (EMPIRICAL) FORMULA
A & B 40.0 g C (1 mol C / 12.01 g C) = 3.33 mol C
6.72 g H (1 mol H / 1.008 g H) = 6.66 mol H
53.3 g O (1 mol O / 16.0 g O) = 3.33 mol O
C:
3.33/3.33 = 1 C
6.66/3.33 = 2 H
3.33/3.33 = 1 O
THUS THE EMPIRICAL FORMULA = CH2O
MOLECULAR FORMULA
FORMULA WEIGHT OF CH2O IS 30.02
180 / 30.02 = 6
THUS THE MOLECULAR FORMULA IS (CH2O)6 = C6H12O6
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HYDRATES
HYDRATES ARE CRYSTALS, WHICH CONTAIN A SPECIFIC RATIO OF WATER
MOLECULES TO THE COMPOUND.
FINDING THE FORMULA OF A HYDRATE
GIVEN:
0.391 g Li2SiF
0.0903 g H2O
MOLECULAR MASS Li2SiF6
2(6.94) + 28.1 + 6(19.0) = 156 g Li2SiF6
0.391 g Li2SiF6 (1 mol Li2SiF6 /156 g Li2SiF6)
= 0.00251 mol Li2SiF6
MOLECULAR MASS H2O
2(1.0) + 16.0 = 18.0 g H2O
0.0903 g H2O (1 mol H2O /18.0 g H2O) = 0.00502 mol H2O
MOLECULAR RATIO:
0.00251 mol Li2SiF6 / 0.00251 mol = 1
0.00502 mol H2O / 0.00251 mol = 2
THUS Li2SiF6 . 2H2O