141
Summer Chemistry 2008
Silberberg Chapter 2
Dr. Ben Tovrog
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Silberberg Chapter 2 Problem Assignment
These problems are for your practice. They will not be collected or graded. It is essential you develop the ability to work easy and medium problems in a fast and efficient manner. Most have answers given in the back of the text. After you master these I encourage you to complete hard problems.
Easy and Medium: 2.4, 2.11, 2.13, 2.14, 2.18, 2.20, 2.22, 2.24, 2.27, 2.54, 2.58, 2.60, 2.64, 2.66, 2.68, 2.72, 2.74, 2.76, 2.78, 2.82, 2.84, 2.88, 2.90, 2.92, 2.98, 2.100.2.102, 2.104, , 2.126, 2.128, 2.129, 2.139, 2.143 Hard: 2.130, 2.135, 2.145
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Chapter 2: The Components of Matter 2.1 Elements, Compounds, and Mixtures: An Atomic Overview
2.2 The Observations That Led to an Atomic View of Matter
2.3 Dalton's Atomic Theory
2.4 The Observations That Led to the Nuclear Atom Model
2.5 The Atomic Theory Today 2.6 Elements: A First Look at the Periodic Table 2.7 Compounds: Introduction to Bonding 2.8 Compounds: Formulas, Names, and Masses 2.9 Mixtures: Classification and Separation ----------------------------------------------------------
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Definitions for Components of Matter
Element - the simplest type of substance with unique physical and chemical properties. An element consists of only one type of atom. It cannot be broken down into any simpler substances by physical or chemical means.
Molecule - a structure that consists of two or more atoms that are chemically bound together and thus behaves as an independent unit.
Figure 2.1
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Definitions for Components of Matter
Compound - a substance composed of two or more elements which are chemically combined.
Figure 2.1
Mixture - a group of two or more elements and/or compounds that are physically intermingled.
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Law of Mass Conservation:
The total mass of substances does not change during a chemical reaction.
reactant 1
+
reactant 2
product
total mass calcium oxide CaO 56.08g + + + carbon dioxide CO2 44.00g
=
total mass calcium carbonate CaCO3 100.08g
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Figure 2.3
Law of Definite (or Constant) Composition:
No matter the source, a particular compound is composed of the same elements in the same parts (fractions) by mass. Calcium carbonate
Analysis by Mass (grams/20.0g) 8.0 g calcium 2.4 g carbon 9.6 g oxygen 20.0 g
Mass Fraction (parts/1.00 part) 0.40 calcium 0.12 carbon 0.48 oxygen 1.00 part by mass
Percent by Mass (parts/100 parts) 40% calcium 12% carbon 48% oxygen 100% by mass
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Dalton's Atomic Theory
The Postulates
1. All matter consists of atoms. 2. Atoms of one element cannot be converted into atoms of another element. 3. Atoms of an element are identical in mass and other properties and are different from atoms of any other element. 4. Compounds result from the chemical combination of a specific ratio of atoms of different elements.
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Experiments to Determine the Properties of Cathode Rays
OBSERVATION 1. Ray bends in magnetic field. CONCLUSION
2. Ray bends towards positive plate in electric field.
consists of charged particles 3. Ray is identical for any cathode. consists of negative particles
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particles found in all matter
Figure 2.6
Millikan's oil-drop experiment
for measuring an electron's charge.
(1909)
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Millikan used his findings to also calculate the mass of an electron.
determined by J.J. Thomson and others
mass of electron =
mass charge
X
charge
= (-5.686x10-12 kg/C) X (-1.602x10-19C) = 9.109x10-31kg = 9.109x10-28g
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Figure 2.7
Rutherford's -scattering experiment and discovery of the atomic nucleus.
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Figure 2.8
General features of the atom today.
The atom is an electrically neutral, spherical entity composed of a positively charged central nucleus surrounded by one or more negatively charge electrons. The atomic nucleus consists of protons and neutrons.
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Table 2.2
Properties of the Three Key Subatomic Particles Charge Mass
Location Name(Symbol) Relative Absolute(C)* Relative(amu) Absolute(g) in the Atom
Proton (p+) 1+ +1.60218x10-19 1.00727 1.67262x1024
Nucleus
Neutron (n0)
0
0
1.00866
1.67493x10-24 Nucleus Outside Nucleus
Electron (e-)
1-
-1.60218x10-19
0.00054858
9.10939x10-28
* The coulomb (C) is the SI unit of charge.
The atomic mass unit (amu) equals 1.66054x10-24 g.
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Atomic Symbols, Isotopes, Numbers
A
X Z
The Symbol of the Atom or Isotope
X = Atomic symbol of the element A = mass number; A = Z + N Z = atomic number
(the number of protons in the nucleus)
N = number of neutrons in the nucleus
Isotope = atoms of an element with the same number of protons, but a different number of neutrons
Figure 2.9
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Sample Problem 2.2
Determining the Number of Subatomic Particles in the Isotopes of an Element
PROBLEM: Silicon(Si) is essential to the computer industry as a major component of semiconductor chips. It has three naturally occurring isoltopes: 28Si, 29Si, and 30Si. Determine the number of protons, neutrons, and electrons in each silicon isotope.
PLAN:
We have to use the atomic number and atomic masses.
SOLUTION: The atomic number of silicon is 14. Therefore
28Si 29Si 30Si
has 14p+, 14e- and 14n0 (28-14) has 14p+, 14e- and 15n0 (29-14) has 14p+, 14e- and 16n0 (30-14)
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Follow-up Problem 2.2
How many protons, neutrons and electrons are in: 11 5 Q
41 20 X 131 53 Y
a.
a. a.
What symbols do Q, X and Y represent?
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More Isotope Problems
How many electrons, protons, and neutrons are found in each of the following?
24
Mg
23
Na +
35
Cl
35
Cl-
When an atom becomes an ion subatomic particle undergoes a change in number?
- Is that particle gained or lost?
14 (e.g. 7 N
14 becomes 7 N 3- which
- How is the magnitude of the charge on an ion determined?
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Sample Problem 2.3
Calculating the Atomic Mass of an Element
PROBLEM: Silver(Ag: Z = 47) has 46 known isotopes, but only two occur naturally, 107Ag and 109Ag. Given the following mass spectrometric data, calculate the atomic mass of Ag:
Isotope
107Ag
Mass(amu)
106.90509
Abundance(%)
51.84
109Ag
108.90476
48.16
PLAN:
We have to find the weighted average of the isotopic masses, so we multiply each isotopic mass by its fractional abundance and then sum those isotopic portions.
SOLUTION:
mass portion from 107Ag = 106.90509amu x 0.5184 = 55.42amu mass portion from 109Ag = 108.90476amu x 0.4816 = 52.45amu atomic mass of Ag = 55.42amu + 52.45amu = 107.87amu
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Follow-up Problem 2.3
Boron (B, Z = 5) has two naturally occurring isotopes. Find the percentage abundances of 10B and 11B given the atomic mass of B = 10.81 amu, and the isotopic mass of 10B is 10.0129 amu and 11B is 11.0093 amu.
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Figure 2.12
The formation of an ionic compound.
Transferring electrons from the atoms of one element to those of another results in an ionic compound. Positive ions are called cations (e.g. Na+)
Negative ions are called anions (e.g. Cl-)
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Predicting Ion Charges
Metals Strong tendency to lose electrons to become cations.
Main Group: Cation charge = Group Number (e.g. Na +, Ca2+, Al3+) Transition: Cation charge can vary (typically +1, +2, +3)
Nonmetals Strong tendency to gain electrons and become anions.
Anion charge = Group # - 8 (or 18) (e.g. Cl-, O2-)
The driving force in forming cations and anions is to achieve the same number of electrons as the Nobel gas in the same period
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Table 2.3 Charge
Common Monoatomic Ions Common ions are in blue.
Cations Formula
H+ Name hydrogen Charge Anions Formula HName hydride
Li+
lithium
sodium
F-
fluoride
chloride
+1
Na+
-1
Cl-
K+
Cs+ Ag+
potassium
cesium silver
BrI-
bromide
iodide
Mg2+
Ca2+
magnesium
calcium strontium barium zinc
O2-
oxide sulfide
+2
Sr2+ Ba2+ Zn2+
-2
S2-
Cd2+
cadmium
aluminum
+3
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Al3+
-3
N3-
nitride
Figure 2.14
The relationship between ions formed and the nearest noble gas.
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Sample Problem 2.4
Predicting the Ion and Element Forms
PROBLEM: What monatomic ions do the following elements form?
(a) Iodine (Z = 53) PLAN: (b) Calcium (Z = 20) (c) Aluminum (Z = 13)
Use Z to find the element. Find its relationship to the nearest noble gas. Elements occurring before the noble gas gain electrons and elements following lose electrons.
SOLUTION:
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Types of Chemical Formulas
A chemical is formula comprised of element symbols and numerical subscripts that show the type and number of each atom present in the smallest unit of the substance.
An empirical formula indicates the relative number of atoms of each element in the compound. It is the simplest type of formula. The empirical formula for hydrogen peroxide is HO. A molecular formula shows the actual number of atoms of each element in a molecule of the compound. The molecular formula for hydrogen peroxide is H 2O2.
A structural formula shows the number of atoms and the bonds between them, that is, the relative placement and connections of atoms in the molecule.
The structural formula for hydrogen peroxide is H-O-O-H.
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Figure 2.19
Some common monatomic ions of the elements.
Can you see any patterns?
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Naming binary ionic compounds The name of the cation is written first, followed by that of the anion.
The name of the cation is the same as the name of the metal. Many metal names end in -ium. The name of the anion takes the root of the nonmetal name and adds the suffix -ide. Calcium and bromine form calcium bromide.
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Sample Problem 2.5
Naming Binary Ionic Compounds
PROBLEM: Name the ionic compound formed from the following pairs of elements: (a) magnesium and nitrogen (b) iodine and cadmium
(c) strontium and fluorine PLAN: (d) sulfur and cesium
Use the periodic table to decide which element is the metal and which the nonmetal. The metal (cation) is named first and we use the -ide suffix on the nonmetal name root.
SOLUTION:
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Sample Problem 2.6
Determining Formulas of Binary Ionic Compounds
PROBLEM: Write empirical formulas for the compounds named in Sample Problem 2.5.
PLAN: Compounds are neutral. We find the smallest number of each ion which will produce a neutral formula. Use subscripts to the right of the element symbol.
SOLUTION:
(a) Mg2+ and N3-; three Mg2+(6+) and two N3-(6-); Mg3N2 (b) Cd2+ and I-; one Cd2+(2+) and two I-(2-); CdI2
(c) Sr2+ and F-; one Sr2+(2+) and two F-(2-); SrF2 (d) Cs+ and S2-; two Cs+(2+) and one S2- (2-); Cs2S
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Table 2.4 (partial) Element Copper
Metals With Several Oxidation States
Systematic Name copper(I) copper(II) cobalt(II) Common Name
Ion Formula Cu+1 Cu+2 Co+2
cuprous
cupric
Cobalt
Co+3
Fe+2
cobalt (III)
iron(II) ferrous
Iron Manganese Tin
Fe+3
Mn+2 Mn+3 Sn+2 Sn+4
iron(III)
manganese(II) manganese(III) tin(II) tin(IV)
ferric
stannous stannic
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Sample Problem 2.7
Determining Names and Formulas of Ionic Compounds of Elements That Form More Than One Ion
PROBLEM: Give the systematic names for the formulas or the formulas for the names of the following compounds: (a) tin(II) fluoride (c) ferric oxide PLAN: (b) CrI3
(d) CoS
Compounds are neutral. We find the smallest number of each ion which will produce a neutral formula. Use subscripts to the right of the element symbol. (a) Tin (II) is Sn2+; fluoride is F-; so the formula is SnF2.
SOLUTION:
(b) The anion I is iodide(I-); 3I- means that Cr(chromium) is +3. CrI3 is chromium(III) iodide (c) Ferric is a common name for Fe3+; oxide is O2-, therefore the formula is Fe2O3. (d) Co is cobalt; the anion S is sulfide(2-); the compound is cobalt (II) sulfide.
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Table 2.5 (partial) Some Common Polyatomic Ions
Formula Name Formula Name hydronium
Cations
NH4+ ammonium H3O+
Common Anions
CH3COOCNacetate cyanide
CO3-2
CrO4-2 Cr2O7-2 O2-2
carbonate
chromate dichromate oxide sulfate phosphate
OHClO3-
hydroxide chlorate nitrite
nitrate
NO2NO3MnO4-
SO4-2
PO4-3
permanganate
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Figure 2.20
Naming oxoanions Root Suffixes Examples ClO4perchlorate
Prefixes No. of O atoms
per
root root
root
ate ate
ite ite
ClO3ClO2ClO-
chlorate chlorite
hypochlorite
hypo
Table 2.6
root
Numerical Prefixes for Hydrates and Binary Covalent Compounds Number
1 2
Prefix
mono di
Number
4 5
Prefix
tetra penta
Number
8 9
Prefix
octa nona
3
tri
6
7
hexa
hepta
10
deca
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Sample Problem 2.8
Determining Names and Formulas of Ionic Compounds Containing Polyatomic Ions
PROBLEM: Give the systematic names or the formula or the formulas for the names of the following compounds: (a) Fe(ClO4)2 PLAN: (b) sodium sulfite (c) Ba(OH)2 8H2O
Note that polyatomic ions have an overall charge so when writing a formula with more than one polyatomic unit, place the ion in a set of parentheses.
SOLUTION:
(a) ClO4- is perchlorate; iron must have a 2+ charge. This is iron(II) perchlorate. (b) The anion sulfite is SO32- therefore you need 2 sodiums per sulfite. The formula is Na2SO3. (c) Hydroxide is OH- and barium is a 2+ ion. When water is included in the formula, we use the term "hydrate" and a prefix which indicates the number of waters. So it is barium hydroxide octahydrate.
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Sample Problem 2.9
Recognizing Incorrect Names and Fromulas of Ionic Compounds
PROBLEM: Something is wrong with the second part of each statement. Provide the correct name or formula. (a) Ba(C2H3O2)2 is called barium diacetate.
(b) Sodium sulfide has the formula (Na)2SO3.
(c) Iron(II) sulfate has the formula Fe2(SO4)3. (d) Cesium carbonate has the formula Cs2(CO3).
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Naming Acids Compounds that start with H
1) Binary acids solutions form when certain gaseous compounds dissolve in water. For example, when gaseous hydrogen chloride(HCl) dissolves in water, it forms a solution called hydrochloric acid. Prefix hydro- + anion nonmetal root + suffix -ic + the word acid - hydrochloric acid
2) Oxoacid names are similar to those of the oxoanions, except for two suffix changes: Anion "-ate" suffix becomes an "-ic" suffix in the acid. Anion "-ite" suffix becomes an "-ous" suffix in the acid. The oxoanion prefixes "hypo-" and "per-" are retained. Thus, BrO4is perbromate, and HBrO4 is perbromic acid; IO2- is iodite, and HIO2 is iodous acid.
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Sample Problem 2.10
Determining Names and Formulas of Anions and Acids
PROBLEM: Name the following anions and give the names and formulas of the acids derived from them: (a) Br (b) IO3 (c) CN (d) SO4 2(e) NO2 -
SOLUTION: (a) The anion is bromide; the acid is hydrobromic acid, HBr.
(b) The anion is iodate; the acid is iodic acid, HIO3.
(c) The anion is cyanide; the acid is hydrocyanic acid, HCN.
(d) The anion is sulfate; the acid is sulfuric acid, H 2SO4.
(e) The anion is nitrite; the acid is nitrous acid, HNO 2.
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Sample Problem 2.11
Determining Names and Formulas of Binary Covalent Compounds
(a) What is the formula of carbon disulfide? (b) What is the name of PCl5?
(c) Give the name and formula of the compound whose molecules each consist of two N atoms and four O atoms.
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Sample Problem 2.12
Recognizing Incorrect Names and Formulas of Binary Covalent Compounds
PROBLEM:
Explain what is wrong with the name of formula in the second part of each statement and correct it: (a) SF4 is monosulfur pentafluoride.
(b) Dichlorine heptaoxide is Cl2O6.
(c) N2O3 is dinitrotrioxide. SOLUTION: (a) The prefix mono- is not needed for one atom; the prefix for four is tetra-. So the name is sulfur tetrafluoride. (b) Hepta- means 7; the formula should be Cl2O7. (c) The first element is given its elemental name so this is dinitrogen trioxide.
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Naming Roadmap for Ionic and Molecular Substances
Binary Compound? NO YES YES Metal Present? NO
Covalent compound naming, pp. 68-69
Polyatomic Ion? NO
Impossible
Does cation form >1 ion? YES
NO
Use element name for cation pp 63-64
YES
Same procedure as binary ionic compounds, pp 6667
Use Roman Numeral after cation metal name, p. 65)
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Sample Problem 2.13
Calculating the Molecular Mass of a Compound
PROBLEM:
Using the data in the periodic table, calculate the molecular (or formula) mass of the following compounds: (a) tetraphosphorous trisulfide (b) ammonium nitrate
PLAN:
Write the formula and then multiply the number of atoms(in the subscript) by the respective atomic masses. Add the masses for the compound.
SOLUTION:
(a) P4S3
(b) NH4NO3
molecular = (2xatomic mass of N) mass + (4xatomic mass of H) + (3xatomic mass of O) = (2x14.01amu)+ (4x1.008amu) + (3x16.00amu) = 80.05amu
molecular = (4xatomic mass of P) mass + (3xatomic mass of S)
= (4x30.97amu) + (3x32.07amu)
= 220.09amu
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