Transcript H 2 SO 4 - cloudfront.net

Chapter 7
Chemical Formulas
and Chemical
Compounds
Significance of Chemical Formula
• Chemical formula shows number and types of
atoms in compound
C8H18
Subscript indicates that
there are 8 carbon atoms
in a molecule of octane.
Subscript indicates that
there are 18 hydrogen atoms
in a molecule of octane.
• Ionic compound made of mixture of + and –
ions held by attraction
• Combination of cation and anion
Al2S3
Al3+ is the positively charged
cation. S2- is the negatively
charged anion
Monoatomic Ions
• Monoatomic ions – ions formed from single
atom
• Group 1 – lose one e-, form +1
• Group 2 – lose two e-, form +2
• Groups 15, 16, 17 gain electrons to form anions
• -1, -2, -3
• Not all main-group elements form ions easily
• C and Si form covalent bonds
• Transition metals can form +1, +2, +3, +4
Naming Monoatomic Ions
• Cations named same as element name
• K+ = potassium
• Mg2+ = magnesium
• Anions named by dropping ending and adding
–ide
• F- = fluoride
• N3- = nitride
Binary Ionic Compounds
• Binary compounds  compounds made of
two different elements
• Total positive and negative charge must be equal
Mg2+
+2
+
Br-1 = +1
+2
+ 2(-1) = 0
Formula = MgBr2
Naming Binary Ionic Compounds
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Nomenclature  naming system
Combine names of cations and anions
Cation name ALWAYS comes first
Then anion name
Al2O3
Aluminum oxide
Practice Problem 1
• Write formulas for the binary ionic
compounds formed between the
following elements:
• a. potassium and iodine
• KI
• b. magnesium and chlorine
• MgCl2
• c. sodium and sulfur
• Na2S
• d. aluminum and sulfur
• Al2S3
• e. aluminum and nitrogen
• AlN
Practice Problem 2
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Name the binary ionic compounds:
a. AgCl
Silver chloride
b. ZnO
Zinc oxide
c. CaBr2
Calcium bromide
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d. SrF2
Strontium fluoride
e. BaO
Barium oxide
f. CaCl2
Calcium chloride
Stock System
• Some elements form two or more cations with
different charges
• Stock system uses Roman numerals to show
ion’s charge
• Roman numeral included in () right after metal
name
Fe2+ Iron (II)
CuCl2
Name of
cation
Roman
numeral
indicating
charge
Copper (II)
Name of anion
chloride
Practice Problem 1
• Write the formula and give the name for
the compounds formed between the
following ions:
• a. Cu2+ and Br−
• CuBr2, copper(II) bromide
• b. Fe2+ and O2−
• FeO, iron(II) oxide
• c. Pb2+ and Cl−
• PbCl2, lead(II) chloride
• d. Hg2+ and S2−
• HgS, mercury(II) sulfide
• e. Sn2+ and F−
• SnF2, tin(II) fluoride
• f. Fe3+ and O2−
• Fe2O3, iron(III) oxide
Polyatomic Ions
• Polyatomic ion  ion containing more than
one atom
• Most are oxyanions  polyatomic ions that
contain oxygen
NO2- NO3• Ion with more oxygen ends in –ate (nitrate)
• Ion with less oxygen ends in –ite (nitrite)
Hypo- and Hyper• One oxyanion family has 4 members
ClOHypochlorite
ClO2-
Chlorite
ClO3-
Chlorate
ClO4-
Perchlorate
• Naming same as binary ionic compounds
• Cation first, anion second
Practice Problem 1
• Write formulas for the following ionic
compounds:
• a. sodium iodide
• NaI
• b. calcium chloride
• CaCl2
• c. potassium sulfide
• K2S
• d. lithium nitrate
• LiNO3
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e. copper(II) sulfate
CuSO4
f. sodium carbonate
Na2CO3
g. calcium nitrite
Ca(NO2)2
h. potassium perchlorate
KClO4
Practice Problem 2
• Give the names for the following
compounds:
• a. Ag2O
• silver oxide
• b. Ca(OH)2
• calcium hydroxide
• c. KClO3
• potassium chlorate
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d. NH4OH
ammonium hydroxide
e. FeCrO4
iron(II) chromate
f. KClO
potassium hypochlorite
Naming Binary Molecular Compounds
• Molecular compounds made of covalently
bonded molecules
• Usually happens between 2 NON-metals
Prefix System
• Uses prefixes to show how many atoms present
Number
Prefix
1
Mono-
2
Di-
3
Tri-
4
Tetra-
5
Penta-
6
Hexa-
7
Hepta-
8
Octa-
9
Nona-
10
Deca-
Examples
• CCl4  carbon tetrachloride
• P4O10  tetraphosphorous decoxide
• There are rules for determining which ion comes
first
Rules
1. The less-electronegative element is given first.
If only 1 atom, no prefix used
2. Second element named by combining (a)
prefix, (b) element root name, (c) –ide
3. The o or a at the end of prefix dropped when
word following beings with vowel
Practice Problem
• Name the following binary molecular
compounds:
• a. SO3
• sulfur trioxide
• b. ICl3
• iodine trichloride
• c. PBr5
• phosphorous pentabromide
Acids
• Most acids either binary acids or oxyacids
• Binary acids  acids that are made of 2
elements
• Usually H and one of the halogens
• Oxyacids  acids that contain H, O, and a 3rd
element (usually nonmetal)
Naming Acids
• Binary acids – hydro(element root)ic acid
• Ex. HCl – hydrochloric acid
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Oxyacids
– with less oxygen, (element root)ous acid
– with more oxygen, (elementroot)ic acid
Ex. HNO2 = nitrous acid
HNO3 = nitric acid
Formula
Name
HF
Hydrofluoric acid
HCl
Hydrochloric acid
HBr
Hydrobromic acid
HI
Hydroiodic acid
H3PO4
Phosphoric acid
HNO2
Nitrous acid
HNO3
Nitric acid
H2SO3
Sulfurous acid
H2SO4
Sulfuric acid
CH3COOH
Acetic acid
HClO
Hypochlorous acid
HClO2
Chlorous acid
HClO3
Chloric acid
HClO4
Perchloric acid
H2CO3
Carbonic acid
Salts
• Salt  an ionic compound made of a cation
and an anion from an acid
• Table salt – NaCl – contains anion from HCl
• Some salts have anions where one or more H
atoms from acid are kept
• Named by adding hydrogen OR prefix bi- to
anion name
HCO3-
hydrogen carbonate ion
bicarbonate ion
Practice Problem 1
• Write formulas for the compounds
formed between the following:
• a. aluminum and bromine
• AlBr3
• b. sodium and oxygen
• Na2O
• c. magnesium and iodine
• MgI2
• d. Pb2+ and O2−
• PbO
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e. Sn2+ and I−
SnI2
f. Fe3+ and S2−
Fe2S3
g. Cu2+ and NO3−
Cu(NO3)2
h. NH4 + and SO42−
(NH4)2SO4
Practice Problem 2
• Name the following compounds using the
Stock system:
• a. NaI
• sodium iodide
• b. MgS
• magnesium sulfide
• c. CaO
• calcium oxide
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d. K2S
potassium sulfide
e. CuBr
copper (I) bromide
f. FeCl2
iron (II) chloride
Practice Problem 3
• Write formulas for each of the following
compounds:
• a. barium sulfide
• BaS
• b. sodium hydroxide
• NaOH
• c. lead(II) nitrate
• Pb(NO3)2
• d. potassium permanganate
• KMnO4
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e. iron(II) sulfate
FeSO4
f. diphosphorus trioxide
P2O3
g. disulfur dichloride
S2Cl2
h. carbon diselenide
CSe2
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i. acetic acid
CH3COOH
j. chloric acid
HClO3
k. sulfurous acid
H2SO3
l. phosphoric acid
H3PO4
The charges on the ions making an ionic compound reflect
the electron distribution of the compound. In order to
indicate the general distribution of electrons among the
bonded atoms in a molecular compound or a polyatomic
ion, oxidation numbers, also called oxidation states,
are assigned to the atoms composing the compound or
ion.
Unlike ionic charges, oxidation numbers do not have an
exact physical meaning. However, oxidation numbers are
useful in naming compounds, in writing formulas, and in
balancing chemical equations. And they are helpful in
studying certain types of chemical reactions.
Assigning Oxidation Numbers
• Shared electrons assumed to belong to moreelectronegative atom in each bond
Rules
1. Atoms in a pure element have Ox# of zero.
1. Na, O2, P4 = 0
2. More-electronegative element in binary
molecular compound assigned the number
equal to negative charge it has as ion. Less
electronegative assigned equal to positive
charge it has as ion.
3. Fluorine always is -1 because it is most
electronegative.
4. Oxygen has -2 in almost all compounds, except
Peroxides (H2O2, O = -1)
Compounds with halogens (O = +)
5. H = +1 in all compounds with more
electronegative element, H = -1 in compounds
with metals
6. Sum of Ox# of all atoms in neutral compound =
zero
7. Sum of Ox# of all atoms in polyatomic ion =
charge of ion
8. Ox# of ions in ionic compounds = its charge
Sample Problem 1
• Assign Ox# to each atom in UF6.
• Start by placing known Ox#s above appropriate
elements.
• From rules, we know F is always -1
• Multiply known Ox#s by appropriate number of
atoms and place totals underneath matching
elements.
• There are 6 F atoms, 6 x -1 = -6
• UF6 is molecular
• According to guidelines, sum of Ox#s must be
zero
• Total positive Ox#s must be +6
• Divide total positive Ox#s by number of atoms
• +6 ÷ 1 = +6
H2SO4
+1 -2
H 2 S O4
+2 -8
+1 +6 -2
H 2 S O4
+2 +6 -8
Practice Problem
• Assign oxidation numbers to each atom in
the following compounds or ions:
• a. HCl
• +1, -1
• b. CF4
• +4, -1
• c. PCl3
• +3, -1
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d. SO2
+4, -2
e. HNO3
+1, +5, -2
f. KH
+1, -1
g. P4O10
+5, -2
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h. HClO3
+1, +5, -2
i. N2O5
+5, -2
j. GeCl2
+2, -1
Stock System
• Remember there were two ways to name
covalent compounds
▫ Stock system
▫ Prefix system
• Some nonmetals can have more than one
oxidation number
• Listed in Table A-15 in handout
Group 14
Carbon
-4, +2, +4
Group 15
Nitrogen
Phosphorous
-3, +3, +5
-3, +3, +5
Group 16
Sulfur
-2, +4, +6
Group 17
Chlorine
Bromine
Iodine
-1, +1, +3, +5, +7
-1, +1, +3, +5, +7
-1, +1, +3, +5, +7
• Can use Roman numerals to show oxidation
number
Prefix System
Stock System
PCl3
Phosphorous
trichloride
Phosphorous(IV)
chloride
PCl5
Phosphorous
pentachloride
Phosphorous(V)
chloride
N2O
Dinitrogen oxide
Nitrogen(I) oxide
NO
Nitrogen monoxide
Nitrogen(II) oxide
PbO2
Lead dioxide
Lead(IV) oxide
Mo2O3
Dimolybdenum
trioxide
Molybdenum(III)
oxide
Practice Problem 1
• Assign oxidation numbers to each atom in
the following compounds or ions:
• a. HF
• +1, -1
• b. CI4
• +4, -1
• c. H2O
• +1, -2
• d. PI3
• +3, -1
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e. CS2
+4, -2
f. Na2O2
+1, -1
g. H2CO3
+1, +4, -2
h. NO2 −
+3, -2
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i. SO42−
+6, -2
j. ClO2 −
+3, -2
k. IO3 −
+5, -2
Practice Problem 2
• Name each of the following binary
molecular compounds according to the
Stock system:
• a. CI4
• carbon(IV) iodide
• b. SO3
• sulfur(VI) oxide
• c. As2S3
• arsenic(III) sulfide
• d. NCl3
• nitrogen(III) chloride
As you have seen, a chemical formula indicates the
elements as well as the relative number of atoms or ions
of each element present in a compound. Chemical
formulas also allow chemists to calculate a number of
characteristic values for a given compound. In this
section, you will learn how to use chemical formulas to
calculate the formula mass, the molar mass, and the
percentage composition by mass of a compound.
Formula Masses
• Like atoms, a molecule, formula unit, or ion has
an average mass (amu)
• Simply add amu of all atoms in molecule
H2O
Amu H = 2 x 1.01 amu
Amu O = 1 x 16.00 amu
Amu H2O = 18.02 amu
• Mass of water molecule can be named molecular
mass
• Mass of 1 NaCl is not molecular mass b/c it’s not
a molecule, it’s an ionic compound
• Formula mass  sum of the average atomic
masses of all the atoms represented in its
formula
Practice Problem
• Find the formula mass of each of the
following:
• a. H2SO4
• b. Ca(NO3)2
• c. PO43−
• d. MgCl2
• a. H2SO4
98.08 amu
• b. Ca(NO3)2
164.10 amu
• c. PO43−
94.97 amu
• d. MgCl2
95.21 amu
Molar Masses
• Molar mass is the same are formula mass, but
units are in grams/mole (g/mol)
Molar Mass H2O = ?
H = 2 x 1.01 g/mol
O = 1 x 16.00 g/mol
H2O = 18.02 g/mol
Practice Problem
• Find the molar mass of each of the
compounds
• a. Al2S3
• b. NaNO3
• c. Ba(OH)2
• a. Al2S3
150.17 g/mol
• b. NaNO3
85.00 g/mol
• c. Ba(OH)2
171.35 g/mol
Molar Mass as Conversion Factor
• g/mol can be used as a conversion factor to
change from moles to mass
• What is the mass in grams of 2.50 mol of
oyxgen gas?
Practice Problem 1
• Ibuprofen, C13H18O2, is the active
ingredient in many nonprescription pain
relievers. Its molar mass is 206.29 g/mol.
• a. If the tablets in a bottle contain a total
of 33 g of ibuprofen, how many moles of
ibuprofen are in the bottle?
• b. How many molecules of ibuprofen are
in the bottle?
• c. What is the total mass in grams of
carbon in 33 g of ibuprofen?
• a. 0.16 mol C13H18O2
• b. 9.6 x 1022 molecules C13H18O2
• c. 25 g C
Practice Problem 2
• How many moles of compound are there
in the following?
• a. 6.60 g (NH4)2SO4
• b. 4.5 kg Ca(OH)2
• a. 0.0500 mol
• b. 61 mol
Practice Problem 3
• How many molecules are there in the
following?
• a. 25.0 g H2SO4
• b. 125 g of sugar, C12H22O11
• a. 1.53 × 1023 molecules
• b. 2.20 × 1023 molecules
Percent Composition
• Percentage composition  percentage by
mass of each element in a compound
• Divide mass of element in sample of compound
by total mass of sample, multiply by 100
• Mass percentage of element in compound same
regardless of sample’s size
• Simpler way to calculate….
• Determine how many grams of element are
present in 1 mole of compound
• Then divide this value by molar mass of
compound, multiply by 100
Sample Problem
• Find the percentage composition of
copper(I) sulfide, Cu2S.
1. Analyze
• Given:
• formula, Cu2S
• Unknown:
• percentage composition of Cu2S
2. Plan
• formula → molar mass → mass percentage of
each element
• The molar mass of the compound must be found
• Then the mass of each element present in one
mole of the compound is used to calculate the
mass percentage of each element.
3. Compute
Molar mass:
Cu = 2 x 63.55 g
S = 1 x 32.07 g
Cu = 127.1 g Cu
S = 32.07 g S
159.2 g/mol Cu2S
Sample Problem 2
• As some salts crystallize from a water
solution, they bind water molecules in
their crystal structure. Sodium carbonate
forms such a hydrate, in which 10 water
molecules are present for every formula
unit of sodium carbonate. Find the mass
percentage of water in sodium carbonate
decahydrate, Na2CO3•10H2O, which has a
molar mass of 286.14 g/mol.
1. Analyze
• Given:
• chemical formula, Na2CO3•10H2O
• molar mass of Na2CO3•10H2O
• Unknown:
• mass percentage of H2O
2. Plan
• chemical formula → mass H2O per mole of
Na2CO3•10H2O → % water
• The mass of water per mole of sodium carbonate
decahydrate must first be found.
• This value is then divided by the mass of one
mole of Na2CO3•10H2O.
3. Compute
• 1 mol Na2CO3•10H2O contains 10 mol of H2O
4. Evaluate
• Checking shows that the arithmetic is correct
and that units cancel as desired.
• When a new substance is synthesized or is
discovered, it is analyzed to show its percentage
composition
• From this data, the empirical formula is then
determined
• An empirical formula consists of the symbols
for the elements combined in a compound, with
subscripts showing the smallest wholenumber mole ratio of the different atoms in
the compound
• For an ionic compound, the formula unit is
usually the compound’s empirical formula
• For a molecular compound, however, the
empirical formula does not necessarily indicate
the actual numbers of atoms present in each
molecule
• For example, the empirical formula of the gas
diborane is BH3, but the molecular formula is
B2H6
• In this case, the number of atoms given by the
molecular formula corresponds to the empirical
ratio multiplied by two.
Calculation of Empirical Formulas
• Determine empirical formula from percent
composition
• Change percent composition to grams
▫ Ex. 29.9% H …. 29.9 g H
• Use molar mass to change grams to mole
• Divide all moles by smallest mole amount
• Result is the subscript for the empirical formula
Example
• Percent composition of diborane is 78.1% B and
21.9% H
• In 100.0 g sample of diborane, there is 78.1 g B
and 21.9 g H
• Mass composition of each element converted to
composition in moles by dividing by molar mass
• 7.22 mol B to 21.7 mol H
• Not a ratio of smallest whole numbers
• Divide each number of mol by smallest number
in ratio
Rounding, empirical formula is BH3
Sample Problem 1
• Quantitative analysis shows that a
compound contains 32.38% sodium,
22.65% sulfur, and 44.99% oxygen. Find
the empirical formula of this compound.
1. Analyze
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Given: percentage composition:
32.38% Na,
22.65% S, and
44.99% O
• Unknown: empirical formula
2. Plan
percentage composition
mass
moles
smallest whole-number mole ratio of atoms
3. Compute
• Mass composition (mass of each element in
100.0 g sample):
▫ 32.38 g Na,
▫ 22.65 g S,
▫ 44.99 g O
• Divide each mole by smallest mole in ratio
Na2SO4
4. Evaluate
• Calculating the percentage composition of the
compound based on the empirical formula
determined in the problem reveals a percentage
composition of 32.37% Na, 22.58% S, and
45.05% O
• These values equal 100%
Practice Problem 1
• Analysis of a 10.150 g sample of a
compound known to contain only
phosphorus and oxygen indicates a
phosphorus content of 4.433 g. What is
the empirical formula of this compound?
• P2O5
Practice Problem 2
• A compound is found to contain 63.52% iron
and 36.48% sulfur. Find its empirical formula.
• FeS
Practice Problem 3
• Find the empirical formula of a compound found
to contain 26.56% potassium, 35.41%
chromium, and the remainder oxygen.
• K2Cr2O7
Practice Problem 4
• Analysis of 20.0 g of a compound containing
only calcium and bromine indicates that 4.00 g
of calcium are present. What is the empirical
formula of the compound formed?
• CaBr2
Calculating Molecular Formulas
• Empirical formula contains smallest whole
number ratio
• Molecular formula is actual formula for
compound
• Empirical  CH
• Molecular  C2H4 (ethene), C3H6 (cyclopropane)
• Relationship between empirical and molecular:
x(empirical formula) = molecular formula
• x is whole-number multiple
• Must know molecular mass
• Ex. Experiment shows molar mass of diborane
(BH3) is 27.67 g/mol
• Empirical formula mass = 13.84 g/mol
Practice Problem 1
• The empirical formula of a compound of
phosphorus and oxygen was found to be
P2O5. Experimentation shows that the
molar mass of this compound is 283.89
g/mol. What is the compound’s molecular
formula?
• P4O10
Practice Problem 2
• Determine the molecular formula of the
compound with an empirical formula of
CH and a formula mass of 78.110 g/mol.
• C6H6
Practice Problem 3
• A sample of a compound with a formula
mass of 34.00 g/mol is found to consist of
0.44 g H and 6.92 g O. Find its molecular
formula.
• H2O2