Chemical Nomenclature, Formulas, and Equations
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Transcript of Chemical Nomenclature, Formulas, and Equations
Chemical Nomenclature, Formulas, and Equations
2
Formulas and Models
3
A molecular formula shows the exact number of atoms of each element in the smallest unit of a substance.
An empirical formula shows the simplest whole-number ratio of the atoms in a substance.
H2OH2O
molecular empirical
C6H12O6 CH2O
O3 O
N2H4 NH2
2.2
Write the molecular formula of methylamine, a colorless gas used in the production of pharmaceuticals and pesticides, from its ball-and-stick model, shown below.
2.2
Solution
Refer to the labels (also see back end papers).
There are five H atoms, one C atom, and one N atom. Therefore, the molecular formula is CH5N.
However, the standard way of writing the molecular formula for methylamine is CH3NH2 because it shows how the atoms are
joined in the molecule.
2.3
Write the empirical formulas for the following molecules:
(a) biborane (B2H6), which is used in rocket propellants
(b) glucose (C6H12O6), a substance known as blood sugar
(c) nitrous oxide (N2O), a gas that is used as an anesthetic gas (“laughing gas”) and as an aerosol propellant for whipped creams.
2.3
Strategy
Recall that to write the empirical formula, the subscripts in the molecular formula must be converted to the smallest possible whole numbers.
2.3
Solution
(a) There are two boron atoms and six hydrogen atoms in diborane. Dividing the subscripts by 2, we obtain the empirical formula BH3.
(b) In glucose there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. Dividing the subscripts by 6, we obtain the empirical formula CH2O. Note that if we had
divided the subscripts by 3, we would have obtained the formula C2H4O2. Although the ratio of carbon to hydrogen to oxygen atoms in C2H4O2 is the same as that in C6H12O6 (1:2:1), C2H4O2 is not the simplest formula because its
subscripts are not in the smallest whole-number ratio.
2.3
(c) Because the subscripts in N2O are already the smallest
possible whole numbers, the empirical formula for nitrous oxide is the same as its molecular formula.
10
Ionic compounds consist of a combination of cations and anions.• The formula is usually the same as the empirical formula.
• The sum of the charges on the cation(s) and anion(s) in each formula unit must equal zero.
The ionic compound NaCl
11
The most reactive metals (green) and the most reactive nonmetals (blue) combine to form ionic compounds.
12
Formulas of Ionic Compounds
Al2O3
2 x +3 = +6 3 x -2 = -6
Al3+ O2-
CaBr2
1 x +2 = +2 2 x -1 = -2
Ca2+ Br-
Na2CO3
2 x +1 = +2 1 x -2 = -2
Na+ CO32-
2.4
Write the formula of magnesium nitride, containing the Mg2+ and N3− ions.
When magnesium burns in air, it forms both magnesium oxide and magnesium nitride.
2.4
Strategy Our guide for writing formulas for ionic compounds is electrical neutrality; that is, the total charge on the cation(s) must be equal to the total charge on the anion(s).
Because the charges on the Mg2+ and N3− ions are not equal, we know the formula cannot be MgN.
Instead, we write the formula as MgxNy, where x and y are
subscripts to be determined.
2.4
Solution To satisfy electrical neutrality, the following relationship must hold:
(+2)x + (−3)y = 0
Solving, we obtain x/y = 3/2. Setting x = 3 and y = 2, we write
Check The subscripts are reduced to the smallest whole- number ratio of the atoms because the chemical formula of an ionic compound is usually its empirical formula.
16
Chemical Nomenclature
• Ionic Compounds– Often a metal + nonmetal– Anion (nonmetal), add “-ide” to element name
BaCl2 barium chloride
K2O potassium oxide
Mg(OH)2 magnesium hydroxide
KNO3 potassium nitrate
17
• Transition metal ionic compounds– indicate charge on metal with Roman numerals
FeCl2 2 Cl- -2 so Fe is +2 iron(II) chloride
FeCl3 3 Cl- -3 so Fe is +3 iron(III) chloride
Cr2S3 3 S-2 -6 so Cr is +3 (6/2) chromium(III) sulfide
18
19
(a) Fe(NO3)2
(b) Na2HPO4
(c) (NH4)2SO3
2.5
Name the following compounds:
(a) Fe(NO3)2
(b) Na2HPO4
(c) (NH4)2SO3
2.5
Strategy Our reference for the names of cations and anions is Table 2.3.
Keep in mind that if a metal can form cations of different charges (see Figure 2.10), need to use the Stock system.
2.5
Solution
(a) The nitrate ion (NO3−) bears one negative charge, so the
iron ion must have two positive charges. Because iron forms both Fe+ and Fe2+ ions, we need to use the Stock system and call the compound iron(II) nitrate.
(b) The cation is Na+ and the anion is HPO42− (hydrogen
phosphate). Because sodium only forms one type of ion (Na+), there is no need to use sodium(I) in the name. The compound is sodium hydrogen phosphate.
(c) The cation is NH4+ (ammonium ion) and the anion is SO3
2− (sulfite ion). The compound is ammonium sulfite.
2.6
Write chemical formulas
for the following
compounds:
(a) mercury(I) nitrite
(b) cesium sulfide
(c) calcium phosphate
2.6
Strategy
We refer to Table 2.3 for the formulas of cations and anions.
Recall that the Roman numerals in the Stock system provide useful information about the charges of the cation.
2.6
Solution
(a) The Roman numeral shows that the mercury ion bears a +1 charge. According to Table 2.3, however, the mercury(I) ion is diatomic (that is, ) and the nitrite ion is . Therefore, the formula is Hg2(NO2)2.
(b) Each sulfide ion bears two negative charges, and each cesium ion bears one positive charge (cesium is in Group 1A, as is sodium). Therefore, the formula is Cs2S.
2.6
(c) Each calcium ion (Ca2+) bears two positive charges, and each phosphate ion ( ) bears three negative charges.
To make the sum of the charges equal zero, we must adjust the numbers of cations and anions:
3(+2) + 2(−3) = 0
Thus, the formula is Ca3(PO4)2.
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• Molecular compounds− Nonmetals or nonmetals + metalloids− Common names
− H2O, NH3, CH4
− Element furthest to the left in a period and closest to the bottom of a group on periodic table is placed first in formula
− If more than one compound can be formed from the same elements, use prefixes to indicate number of each kind of atom
− Last element name ends in -ide
28
HI hydrogen iodide
NF3 nitrogen trifluoride
SO2 sulfur dioxide
N2Cl4 dinitrogen tetrachloride
NO2 nitrogen dioxide
N2O dinitrogen monoxide
Molecular Compounds
2.7
Name the following molecular compounds:
(a) SiCl4
(b) P4O10
2.7
Strategy
We refer to Table 2.4 for prefixes.
In (a) there is only one Si atom so we do not use the prefix “mono.”
Solution
(a) Because there are four chlorine atoms present, the compound is silicon tetrachloride.
(b) There are four phosphorus atoms and ten oxygen atoms present, so the compound is tetraphosphorus decoxide. Note that the “a” is omitted in “deca.”
2.8
Write chemical formulas for the following molecular compounds:
(a) carbon disulfide
(b) disilicon hexabromide
2.8
Strategy
Here we need to convert prefixes to numbers of atoms (see Table 2.4).
Because there is no prefix for carbon in (a), it means that there is only one carbon atom present.
Solution
(a) Because there are two sulfur atoms and one carbon atom present, the formula is CS2.
(b) There are two silicon atoms and six bromine atoms present, so the formula is Si2Br6.
33
34
An acid can be defined as a substance that yields hydrogen ions (H+) when dissolved in water.
For example: HCl gas and HCl in water
• Pure substance, hydrogen chloride
• Dissolved in water (H3O+ and Cl−), hydrochloric acid
35
36
An oxoacid is an acid that contains hydrogen, oxygen, and another element.
HNO3 nitric acid
H2CO3 carbonic acid
H3PO4 phosphoric acid
37
Naming Oxoacids and Oxoanions
38
The rules for naming oxoanions, anions of oxoacids, are as follows:1. When all the H ions are removed from the
“-ic” acid, the anion’s name ends with “-ate.” 2. When all the H ions are removed from the
“-ous” acid, the anion’s name ends with “-ite.”3. The names of anions in which one or more
but not all the hydrogen ions have been removed must indicate the number of H ions present.
For example:– H2PO4
- dihydrogen phosphate– HPO4 2- hydrogen phosphate– PO4
3- phosphate
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2.9
Name the following oxoacid and oxoanion:
(a) H2SO3, a very unstable acid formed when SO2(g) reacts with water (sulfurous acid sulfite +2H+)
(b) H2AsO4−, once used to control ticks and lice on livestock
(dihydrogen arsenate)
(c) SeO32−, used to manufacture colorless glass. (selenite)
H3AsO4 is arsenic acid, and H2SeO4 is selenic acid.
H2SeO3 is selenous acid
2.9Strategy We refer to Figure 2.14 and Table 2.6 for the conventions used in naming oxoacids and oxoanions.
Solution
(a) We start with our reference acid, sulfuric acid (H2SO4). Because H2SO3 has one fewer O atom, it is called sulfurous acid.
(b) Because H3AsO4 is arsenic acid, the AsO43− is named arsenate.
The H2AsO4− anion is formed by adding two H+ ions to AsO4
3−, so H2AsO4
− is called dihydrogen arsenate.
(c) The parent acid is H2SeO3. Because the acid has one fewer O atom than selenic acid (H2SeO4), it is called selenous acid. Therefore, the SeO3
2− anion derived from H2SeO3 is called selenite.
42
A base can be defined as a substance that yields hydroxide ions (OH-) when dissolved in water.
NaOH sodium hydroxide
KOH potassium hydroxide
Ba(OH)2 barium hydroxide
43
Hydrates are compounds that have a specific number of water molecules attached to them.
BaCl2•2H2O
LiCl•H2O
MgSO4•7H2O
Sr(NO3)2 •4H2O
barium chloride dihydrate
lithium chloride monohydrate
magnesium sulfate heptahydrate
strontium nitrate tetrahydrate
CuSO4•5H2O CuSO4
44
Organic chemistry is the branch of chemistry that deals with carbon compounds.
C
H
H
H OH C
H
H
H NH2 C
H
H
H C OH
O
methanol methylamine acetic acid
Functional Groups:
45
46
Chemical Equations
• Symbolic representation of a chemical reaction that shows:
1. reactants on left side of reaction2. products on right side of equation3. relative amounts of each using stoichiometric
coefficients
47
Chemical Equations
• Attempt to show on paper what is happening at the laboratory and molecular levels.
48
Chemical Equations
• Law of Conservation of Matter – There is no detectable change in quantity of matter in an
ordinary chemical reaction.– Balanced chemical equations must always include the
same number of each kind of atom on both sides of the equation.
– This law was determined by Antoine Lavoisier.
49
Chemical Equations
• Look at the information an equation provides:
Fe O + 3 CO 2 Fe + 3 CO2 3 2
Reactants Yields Products
50
Chemical Equations
• Look at the information an equation provides:
Fe O + 3 CO 2 Fe + 3 CO2 3 2
Reactants Yields Products
1 formula unit
3 molecules
2 atoms 3 molecules
51
Chemical Equations
• Look at the information an equation provides:
Fe O + 3 CO 2 Fe + 3 CO2 3 2
Reactants Yields Products
1 formula unit
3 molecules
2 atoms 3 molecules
1 mole 3 moles 2 moles 3 moles
52
How to “Read” Chemical Equations
2 Mg + O2 2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
53
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2 CO2 + H2O
2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
2C2H6 NOT C4H12
54
Balancing Chemical Equations
3. Start by balancing those elements that appear in only one reactant and one product.
C2H6 + O2 CO2 + H2O start with C or H but not O
2 carbonon left
1 carbonon right
multiply CO2 by 2
C2H6 + O2 2CO2 + H2O
6 hydrogenon left
2 hydrogenon right
multiply H2O by 3
C2H6 + O2 2CO2 + 3H2O
55
Balancing Chemical Equations
4. Balance those elements that appear in two or more reactants or products.
2 oxygenon left
4 oxygen(2x2)
C2H6 + O2 2CO2 + 3H2O
+ 3 oxygen(3x1)
multiply O2 by 72
= 7 oxygenon right
C2H6 + O2 2CO2 + 3H2O72
remove fractionmultiply both sides by 2
2C2H6 + 7O2 4CO2 + 6H2O
56
Balancing Chemical Equations
5. Check to make sure that you have the same number of each type of atom on both sides of the equation.
2C2H6 + 7O2 4CO2 + 6H2O
Reactants Products
4 C12 H14 O
4 C12 H14 O
4 C (2 x 2) 4 C
12 H (2 x 6) 12 H (6 x 2)
14 O (7 x 2) 14 O (4 x 2 + 6)
3.12
When aluminum metal is exposed to air, a protective layer of aluminum oxide (Al2O3) forms on its surface. This layer prevents further reaction between aluminum and oxygen, and it is the reason that aluminum beverage cans do not corrode. [In the case of iron, the rust, or iron(III) oxide, that forms is too porous to protect the iron metal underneath, so rusting continues.]
Write a balanced equation for the formation of Al2O3.
An atomic scale image of aluminum oxide.
3.12
Strategy Remember that the formula of an element or compound cannot be changed when balancing a chemical equation. The equation is balanced by placing the appropriate coefficients in front of the formulas. Follow the procedure described on p. 92.
Solution The unbalanced equation is
In a balanced equation, the number and types of atoms on each side of the equation must be the same. We see that there is one Al atom on the reactants side and there are two Al atoms on the product side.
3.12
We can balance the Al atoms by placing a coefficient of 2 in front of Al on the reactants side.
There are two O atoms on the reactants side, and three O atoms on the product side of the equation. We can balance the
O atoms by placing a coefficient of in front of O2 on the reactants side.
This is a balanced equation. However, equations are normally balanced with the smallest set of whole-number coefficients.
3.12
Multiplying both sides of the equation by 2 gives whole-number
coefficients.
or
Check For an equation to be balanced, the number and types of atoms on each side of the equation must be the same. The final tally is
The equation is balanced. Also, the coefficients are reduced to the simplest set of whole numbers.
61
Combination Reactions
• Combination reactions occur when two or more substances combine to form a compound.
• There are three basic types of combination reactions.
1. Two elements react to form a new compound2. An element and a compound react to form one new
compound3. Two compounds react to form one compound
62
Combination Reactions
1. Element + Element CompoundA. Metal + Nonmetal Binary Ionic Compound
sg2s NaCl 2ClNa 2
63
Combination Reactions
1. Element + Element CompoundA. Metal + Nonmetal Binary Ionic Compound
sg2s MgO 2OMg 2
64
Combination Reactions
1. Element + Element CompoundA. Metal + Nonmetal Binary Ionic Compound
s32s AlBr 2 Br3Al 2
65
Combination Reactions
1. Element + Element CompoundB. Nonmetal + Nonmetal Covalent Binary
Compound
s104g2s4 O PO 5P
66
Combination Reactions
1. Element + Element CompoundB. Nonmetal + Nonmetal Covalent Binary
Compound
3g2s4 PCl4 Cl 6P
67
Combination Reactions
1. Element + Element CompoundB. Nonmetal + Nonmetal Covalent Binary
Compound• Can control which product is made with the reaction
conditions.
chlorine limitedin
AsCl 2 Cl 3As 2 s3g2s
chlorine excessin
AsCl 2 Cl 5As 2 s5g2s
68
Combination Reactions
1. Element + Element CompoundB. Nonmetal + Nonmetal Covalent Binary
Compound• Can control which product is made with the reaction
conditions.
fluorine limitedin
SeF F 2Se s4g2s
fluorine excessin
SeF F 3Se g6g2s
69
Combination Reactions
2. Compound + Element Compound
s5g2s3 AsClClAsCl
g6g2s4 SFFSF
70
Combination Reactions
The reaction of oxygen with oxides of nonmetals is an example of this type of combination reaction.
g3 &catalyst
g2g2 SO 2OSO 2
g2g2g CO 2OCO 2
104264 OPO 2OP
71
Combination Reactions
3. Compound + Compound Compound– gaseous ammonia and hydrogen chloride
– lithium oxide and sulfur dioxide
s4gg3 ClNH HClNH
4232 SOLi SOOLi
72
Decomposition Reactions
• Decomposition reactions occur when one compound decomposes to form:
1. Two elements2. One or more elements and one or more
compounds3. Two or more compounds
73
Decomposition Reactions
1. Compound Element + Element– decomposition of dinitrogen oxide
• decomposition of calcium chloride g2g2g2 ON 2ON 2
g2yelectricit
2 ClCaCaCl
2sh
s BAg 2AgBr 2 r
decomposition of silver halides
74
Decomposition Reactions
2. Compound One Element + Compound(s)
– decomposition of hydrogen peroxide
g22or Mn or Feνh
aq22 OO H2O H23
75
Decomposition Reactions
3. Compound Compound + Compound– decomposition of ammonium hydrogen
carbonate
g2g2g3s34 COO HNHHCONH
76
Displacement Reactions
• Displacement reactions occur when one element displaces another element from a compound.– These are redox reactions in which the more
active metal displaces the less active metal of hydrogen from a compound in aqueous solution.
– Activity series is given in Table 4-14.
77
Displacement Reactions
1. [More Active Metal + Salt of Less Active Metal] [Less Active Metal + Salt of More Active Metal]
– molecular equation
(s)aq3(s) aq3 Ag CuNO Cu +AgNO
79
Displacement Reactions
2. [Active Metal + Nonoxidizing Acid] [Hydrogen + Salt of Acid]
– Common method for preparing hydrogen in the laboratory.– HNO3 is an oxidizing acid.
• Molecular equation
g2aq342aq42(s) H 3 + )(SOAl SO3H + Al 2
80
Displacement Reactions
• Total ionic equationYou do it!
• Net ionic equationYou do it!
g2-2aq4
3aq
-2aq4aq(s) H 3 + SO 3 + Al 2 SO 3+H 6 + Al 2
g23aqaq(s) H 3 +Al 2 H 6 + Al 2
81
Displacement Reactions
• The following metals are active enough to displace hydrogen– K, Ca, Na, Mg, Al, Zn, Fe, Sn, & Pb
• Notice how the reaction changes with an oxidizing acid.– Reaction of Cu with HNO3.
• H2 is no longer produced.
82
Displacement Reactions
3. [Active Nonmetal + Salt of Less Active Nonmetal] [Less Active Nonmetal + Salt of More Active
Nonmetal]
• Molecular equation
(aq)s2aqg2 NaCl 2 I NaI 2 + Cl Total ionic equation
You do it!
-aqaqs2
-aqaqg2 Cl 2 +Na 2 I I 2 + Na 2 +Cl
84
Metathesis Reactions
• Metathesis reactions occur when two ionic aqueous solutions are mixed and the ions switch partners.
AX + BY AY + BX• Metathesis reactions remove ions from solution in
two ways:1. form predominantly unionized molecules like H2O2. form an insoluble solid
• Ion removal is the driving force of metathesis reactions.
85
Metathesis Reactions
1. Acid-Base (neutralization) Reactions– Formation of the nonelectrolyte H2O – acid + base salt + water
86
Metathesis Reactions
• Molecular equation
)(2 (aq)(aq)(aq) OH + KBr KOH + HBr Total ionic equation
You do it!
)(2-aqaq
-aqaq
-aqaq OH + Br+KOH+K+Br+H
Net ionic equation
You do it!
)(2-aqaq OH OH +H
87
Metathesis Reactions
• Molecular equation
)(2aq)(23(aq)3(aq)2 OH 2 + )Ca(NOHNO 2 + Ca(OH) Total ionic equation
You do it!
)(2-
aq32aq
-aq3aq
-aq
2aq OH 2 +NO 2+ CaNO 2+ H 2+OH 2+Ca
Net ionic equation
You do it!
)(2aq-aq
)(2aq-aq
OH H+OH
betteror
OH 2 H 2+OH 2
88
Metathesis Reactions
2. Precipitation reactions are metathesis reactions in which an insoluble compound is formed.
– The solid precipitates out of the solution much like rain or snow precipitates out of the air.
89
Metathesis Reactions
• Precipitation Reactions • Molecular equation
(s)3)aq(3aq)(32(aq)23 CaCO +KNO 2 COK + )Ca(NO Total ionic reaction
You do it!
s3-
aq3aq
-2aq3aq
-aq3
2aq
CaCO NO 2K 2
COK 2 NO 2 Ca
91
Metathesis Reactions
• Molecular equation
2(s)43)aq(aq)(43(aq)2 POCa +NaCl 6 PONa 2 + CaCl 3
Total ionic reactionYou do it!
s243-1aq
1aq
-3aq4
1aq
-1aq
2aq
POCa +Cl 6 Na 6
PO2 Na 6 + Cl 6 Ca 3
93
Metathesis Reactions
• Molecular equation
g22)aq(aq)(32(aq) SO O H+NaCl 2 SONa + HCl2
Total ionic reactionYou do it!
g22-1aq
1aq
-2aq3
1aq
-1aq
1aq
SO OH +Cl 2Na 2
SO Na 2 + Cl 2H 2
• LiBF4 + H2O H3BO3 + HF + LiF
• Al(OH)3 + HCl AlCl3 + H2O
• C4H9SO + O2 CO2 + SO2 + H2O
• Cd + NiO2 + H2O Cd(OH)2 + Ni(OH)2
• SiO2 + C + Cl2 CO + SiCl4
• Br2 + H2O HBr + HBrO3
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