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Homework Problems

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Homework Problems

## Homework Problems

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1. Homework Problems Chapter 2 Homework Problems: 1, 4, 19, 24, 28 (give the number of protons, neutrons, and electrons), 30, 38, 41, 44, 46, 54, 64, 66, 67, 72, 76, 82, 84, 96, 101, 104, 114, 116, 117, 124

2. CHAPTER 2 Atoms, Molecules, and Ions

3. Early Theories of Matter The ancient Greeks discussed two possibilities for the essential property of matter. Matter is continuous (no “particles” of matter) - Plato, Aristotle, and a majority of Greek philosophers. Matter is discrete (composed of particles) - Democritus, Leucippus, and a small number of Greek philosophers. Does this process have an end? yes – particles of matter exist no – matter is continuous

4. General Properties of Chemical Systems As scientists began studying chemical systems they discovered several general properties of such systems. 1) Conservation of Mass. The total mass in a closed system remains constant, even if chemical reactions occur. 2) Law of Definite Proportions. All samples of a particular pure chemical substance contain the same relative amounts of each element making up the substance. Examples: methane 74.9 % C, 25.1 % H water 88.8 % O, 11.2 % H copper (II) sulfate 39.8 % Cu, 20.1 % S, 40.1 % O

5. 3) Law of Multiple Proportions. When two elements can combine to form several different chemical compounds, the ratio of the amount of the second element combining with a fixed amount of the first element will be the ratio of small whole numbers. Example: There are two common compounds of carbon and oxygen carbon monoxide 1.000 g of C reacts with 1.332 g of O carbon dioxide 1.000 g of C reacts with 2.664 g of O g O in carbon dioxide = 2.664 g = 2.000 2 g O in carbon monoxide 1.332 g 1 It is easier to calculate the ratios with the larger number on top, but that is not required.

6. Example: The following pure chemical substances can be formed out of the elements nitrogen and oxygen nitrogen monoxide 1.000 g of N reacts with 1.142 g of O nitrogen dioxide 1.000 g of N reacts with 2.285 g of O nitrous oxide 1.000g of N reacts with 0.5711 g of O Do these substances demonstrate the law of multiple proportions?

7. nitrogen monoxide 1.000 g of N reacts with 1.142 g of O nitrogen dioxide 1.000 g of N reacts with 2.285 g of O nitrous oxide 1.000g of N reacts with 0.5711 g of O g O in nitrogen monoxide = 1.142 g = 2.000  2 g O in nitrous oxide 0.5711g 1 g O in nitrogen dioxide = 2.285 g = 4.001 4 g O in nitrous oxide 0.5711 g 1 g O in nitrogen dioxide = 2.285 g = 2.001 2 g O in nitrogen monoxide 1.142 g 1 So yes, these data are consistent with the law of multiple proportions.

8. Dalton’s Atomic Theory A comprehensive theory that accounted for the above obser-vations was proposed by John Dalton, an English chemist, in 1808. There were three parts to the theory. 1) Elements are composed of particles, called atoms. a) All atoms of the same element are identical in size, mass, and chemical properties. b) Atoms of different elements differ in their size, mass, and chemical properties.

9. Dalton’s Atomic Theory (continued) 2) Chemical compounds are composed of atoms of more than one element. a) In any particular pure chemical compound the same kinds of atoms are present in the same relative numbers. 3) Chemical reactions can rearrange atoms, but atoms cannot be created, destroyed, or converted from atoms of one element to atoms of a different element. We now know that some of the hypotheses in Dalton’s atomic theory are not completely correct; however, the theory represents a good starting point in understanding the composition of matter.

10. Consequences of Dalton’s Atomic Theory Dalton’s theory can be used to explain the observations cited above. 1) Conservation of mass. Explained by (1) and (3). 2) Law of definite proportion. Explained by (1) and (2). Example: methane = CH4 Chemical formula - A list of the elements making up a compound, giving the number of atoms of each element per molecule or per formula unit of the compound.

11. 3) Law of multiple proportions. Explained by (1) and (2) . carbon monoxide (CO) carbon dioxide (CO2) 1 C atom: 1 O atom 1 C atom: 2 O atom So for a given amount of carbon, carbon dioxide will have twice as many oxygen atoms (and therefore twice the mass of oxygen) as carbon monoxide.

12. Atomic Structure In Dalton’s atomic theory the smallest particles (atoms) could not be further broken down. However, a series of experiments, beginning in the mid-19th century, demonstrated that atoms themselves were composed of smaller particles.

13. Radioactivity In 1895, Antoine Becquerel discovered that some substances (such as radium and uranium) spontaneously emit “radiation”, a process called radioactivity. Three types of radioactivity were found: alpha () radiation: positively charged particles, now known to be He2+ nuclei (2 protons + 2 neutrons) beta () radiation: negatively charged particles, now known to be electrons gamma () radiation: uncharged, now known to be high energy photons (particles) of light

14. Electrons J. J. Thompson (1897) found that when a high voltage was applied across two electrodes at low pressure a beam of particles moved from the negative to the positive electrode. The particles, named electrons, were negatively charged and the same regardless of the gas between the electrodes or the metal used in the electrodes. The charge and mass of an electron were determined experimentally by Millikan (1909).

15. The “Plum Pudding” Model Since atoms are electrically neutral, the negative charge of the electrons in an atom had to be balanced by a positive charge. Thompson suggested that most of the space within an atom consisted of a positively charged substance, with electrons embedded within, the “plum pudding” model.

16. Rutherford and the Nuclear Atom To test Thompson’s plum pudding model, Ernest Rutherford (1909) carried out an experiment where a beam of positively charged particles (alpha particles) were directed at a thin sheet of gold metal.

17. The results of this experiment were inconsistent with the plum pudding model. Rutherford proposed a new model, called the nuclear model of the atom, that did account for the experimental results.

18. Subatomic Particles particlechargemass Coulombs elementary kg amu proton (p+) + 1.60 x 10-19 +1 1.673 x 10-27 1 neutron (n) 0 0 1.675 x 10-27 1 electron (e-) - 1.60 x 10-19 - 1 9.11 x 10-31  0 ________ 1 amu = 1.6605 x 10-27 kg mp/me = 1836.

19. Atomic Structure nucleus electron charge cloud 1) The protons and neutrons of the atom are found in a small region in the center of the atom, called the nucleus. This region contains most of the mass of the atom, and all of the positive charge. 2) Electrons in the atom form a diffuse cloud of negative charge centered on the nucleus and occupying most of the volume of the atom. 3) The size of the charge for the proton and electron is the same. The charge for the proton is positive, and the charge for the electron is negative. Neutrons have no charge.

20. 4) The type of element for an atom is determined by the number of protons in the atomic nucleus. Element (new definition) - An element is a pure chemical substance composed of atoms, each of which has the same number of protons in the nucleus. Hydrogen - one proton per atom Helium - two protons per atom Lithium - three protons per atom . . . . Similarly, we can now define a compound (new definition) as a pure chemical substance composed of two or more different kinds of atoms.

21. The Periodic Table

22. Atomic Number and Mass Number 1) The atomic number (Z) is equal to the number of protons in the atom. 2) Since atoms are electrically neutral, the number of electrons in an atom is also equal to Z, the atomic number. 3) The mass number (A) is equal to the number of protons + neutrons in the atom. a) Because protons and neutrons have a mass of approximately 1 (in amu) and electrons have a mass of approximately 0 (in amu) the mass number is equal to the approximate mass of the atom in amu. b) Based on the above, the number of neutrons in an atom is equal to A - Z. So for an atom: # protons = Z # electrons = Z # neutrons = A - Z

23. Notation For Atoms We use the following general notation to represent isotopes of atoms. mass number atomic number symbol for element Since we can use the symbol for the element and the periodic table to determine Z, the atomic number, we often omit Z in giving the symbol for the atom. Example: 3416S = 34S We can omit the subscript because all sulfur atoms contain 16 protons.

24. Isotopes The atomic number determines the number of protons and electrons in an atom. This does not place any restrictions on the number of neutrons in the atom. It is possible for atoms of the same element to have different numbers of neutrons. These different types of atoms are called isotopes. Isotopes of Hydrogen normal hydrogen deuterium tritium 1H 2H 3H Note that to a very good approximation isotopes of a particular element are chemically identical to one another.

25. Example of Notation for Isotopes As an example of using the above notation, consider the follow-ing naturally occurring isotopes of oxygen (Z = 8). protons neutrons electrons mass number symbol 8 8 8 16 16O 8 9 8 17 17O 8 10 8 18 18O Example: How many protons, neutrons, and electrons are there in one atom of 56Fe? What is the approximate mass of one atom of 56Fe in amu and in kg?

26. Example: How many protons, neutrons, and electrons are there in one atom of 56Fe? What is the approximate mass of one atom of 56Fe in amu and in kg? # protons = Z = 26 # neutrons = A – Z = 56 – 26 = 30 # electrons = Z = 26 approximate mass (amu) = A = 56 approximate mass (kg) = 56 amu 1.6605 x 10-27 kg 1 amu = 9.30 x 10-26 kg Note that the actual mass of one atom of 56Fe is 55.934939 amu.

27. Atomic Mass Units (amu) The mass of a single atom of an element, expressed in SI units, is an extremely small number. For example, the mass of a single atom of 16O is 2.6560 x 10-26 kg. For convenience, we often express values for atomic mass in terms of atomic mass units (amu). Atomic mass units are defined as follows 12.00 amu = mass of one atom of 12C (exact) From this we get 1. amu = 1.6605 x 10-27 kg (approximate) The mass of any other atom (or particle) is found relative to the ratio of its mass to the mass of a 12C atom, which can be measured experimentally. Mass of particle (amu) = mass particle • (12.00 amu) mass 12C atom

28. Example: A mass spectrometer is a device for determining values for mass for atoms and molecules. In a particular experiment, the ratio (mass M/mass 12C) is measured and found to be equal to 7.337. What is the mass of the molecule M (in amu)?

29. Example: A mass spectrometer is a device for determining values for mass for atoms and molecules. In a particular experiment, the ratio (mass M/mass 12C) is measured and found to be equal to 7.337. What is the mass of the molecule M (in amu)? mass M = 7.337 mass 12C atom Mass M = 7.337 (mass 12C atom) = 7.337 (12.00 amu) = 88.04 amu Note that because of the way we define atomic mass units, the only isotope whose mass is exactly equal to its mass number is 12C. isotope mass (amu) 1H 1.007825 12C 12.000000... (exact) 238U 238.0508

30. Atomic Mass in the Periodic Table Because different isotopes of an element have different masses, the question arises as to which mass should be given in the periodic table. For short lived radioactive elements the mass number of the most stable isotope of the element is listed. Element Z A technetium (Tc) 43 98 radon (Rn) 86 222 plutonium (Pu) 94 244

31. Average Atomic Mass For naturally occurring elements, the value for mass given in the periodic table is the average atomic mass, based on the natural abundance of the isotopes that is observed. In general, we find the average atomic mass as follows: Mave = f1 M1 + f2 M2 + f3 M3 + … = i=1n fi Mi where f1, f2,...are the fractions of each isotope observed in nature M1, M2,…are the corresponding masses for each isotope (in amu) Note the following f1 + f2 + f3 + …= 1 fx = % X 100 %

32. Non-chemical Example A person has a box of sandwiches. Half of the sandwiches are 6.0 ounces, and half of the sandwiches are 10.0 ounces. What is the average weight of a sandwich? Average weight = (0.50)(6.0 oz) + (0.50)(10.0 oz) = 8.0 ounces We use the same procedure in finding the average mass of an atom. We multiply the fraction of each isotope by the mass of that isotope, and then add the results to find the average mass.

33. Chemical Example There are three naturally occurring isotopes of the element magnesium. Based on the information below, find the average atomic mass of a magnesium atom. Isotope percent f M(amu) 24Mg 78.70 % 23.98504 25Mg 10.03% 24.98584 26Mg 11.17% 25.98259

34. Chemical Example There are three naturally occurring isotopes of the element magnesium. Based on the information below, find the atomic mass of a magnesium atom. Isotope percent f M(amu) 24Mg 78.70 % 0.7870 23.98504 25Mg 10.03% 0.1003 24.98584 26Mg 11.17% 0.1117 25.98259 So Mave = (0.7870)(23.98504 amu) + (0.1003)(24.98584 amu) + (0.1117)(25.98259 amu) = 24.30 amu, the value given in the periodic table.

35. Periodic Table The periodic table is an arrangement of the chemical elements based on similarities in their physical and chemical properties The periodic table contains a large amount of useful information about the chemical elements. Organization There are several ways in which the elements in the periodic table may be classified. Rows = Periods Columns = Groups This is the more important classification. Elements in the same group usually have similar physical and chemical properties.

36. Simplified Periodic Table 1A 2A 3A 4A 5A 6A 7A 8A

37. 1A 2A 3A 4A 5A 6A 7A 8A You are responsible for knowing the names/symbols for elements 1-57, 72-86, and 92.

38. Major Groups in the Periodic Table 1A 2A 3A 4A 5A 6A 7A 8A

39. Metals, Nonmetals, and Metalloids Metals: Usually solid at room temperature (exceptions Cs, Fr, Hg) Shiny metallic luster Good conductors of electricity and heat Malleable (can be hammered into thin sheets) Ductile (can be drawn into thin wires) Nonmetals: Can be solid, liquid, or gas at room temperature Dull colored (as solids) Poor conductors of electricity and heat Not malleable, not ductile Metalloids (semimetals): Intermediate between metals and nonmetals

40. Metals, Nonmetals, Metalloids in the Periodic Table 1A 2A 3A 4A 5A 6A 7A 8A

41. Examples of Elements (as found in nature) nickel germanium sulfur (metal) (metalloid) (nonmetal)

42. Formation of Ions Ions are charged particles. Ions can be formed from an atom by either adding electrons (to form an anion) or removing electrons (to form a cation). Ions cannot be formed by changing the number of protons in the atom. cationsanions particle Z # electrons particle Z # electrons Na 11 11 Cl 17 17 Na+ 11 10 Cl- 17 18 Ca 20 20 S 16 16 Ca2+ 20 18 S2- 16 18 Note that the charge of an ion is indicated by a superscript to the right of the symbol for the ion. Metals usually form cations, while nonmetals usually form anions.

43. Just as we can predict the number of protons, neutrons, and electrons from the symbol for an atom, we can do the same thing for cations and anions formed from atoms. We do this using the atomic number (Z) the mass number (A) and the charge of the ion. Example: How many protons, neutrons and electrons are there for a 31P3- ion and a 25Mg2+ ion?

44. Example: How many protons, neutrons and electrons are there for a 31P3- ion and a 25Mg2+ ion? 31P3- # protons = Z = 15 # neutrons = A - Z = 31 - 15 = 16 Charge is 3-, so there are 3 more electrons than protons, and so the number of electrons = 18. 25Mg2+ # protons = Z = 12 # neutrons = A - Z = 25 - 12 = 13 Charge is 2+, so there are two more protons than electrons, and so the number of electrons is 10.

45. Ion Charges For Main Group Elements Main group elements tend to form ions by adding or removing electrons so that the number of electrons remaining in the ion is equal to the number of electrons in one atom of the nearest noble gas. cations (metals) group 1A (Li, Na, K, Rb, Cs) form 1+ ions group 2A (Mg, Ca, Sr, Ba) form 2+ ions group 3A (Al) form 3+ ions anions (nonmetals) group 5A (N, P) form 3- ions group 6A (O, S, Se, Te) form 2- ions group 7A (F, Cl, Br, I) form 1- ions

46. Transition Metal Ions Transition metals form cations. Most transition metals, as well as a few main group metals like tin (Sn) and lead (Pb), can form ions with several different charges, but a few transition metals, such as silver, usually form only one type of cation (Ag+ for silver, Cd2+ for cadmium, Zn2+ for zinc). Example: Iron (Fe) Fe2+, Fe3+ Copper (Cu) Cu+, Cu2+ Chromium (Cr) Cr3+, Cr6+ It is generally not easy to predict which cations a transition metal will form.

47. Chemical Formula The chemical formula for a substance provides information concerning the composition of the substance. We can divide substances into two general types. 1) Substances that exist as collections of molecules. In this case the chemical formula indicates the number of atoms of each elements present per molecule. water phosphorus pentachloride nitrous acid (H2O) (PCl5) (HNO2)

48. For organic molecules, the chemical formula is often given in a way that indicates how the molecule is put together. ethyl alcohol dimethyl ether acetone CH3CH2OH = C2H6O CH3OCH3 = C2H6O CH3COCH3 = C3H6O Notice that this longer notation makes it possible to distinguish among different forms (isomers) of organic molecules.

49. 2) Substances that exist as collections of atoms or ions in the form of a crystal structure, a regular arrangement of the particles making up the substance. For these substances, the formula that is given is usually the empirical formula. An empirical formula gives the relative number of atoms of each element making up the compound, reduced to the smallest set of whole number coefficients. Ca2+ F- sodium chloride polonium calcium fluoride NaCl Po CaF2

50. Empirical Formula for Molecular Compounds The empirical formula for a substance gives the relative number of atoms of each element making up a pure chemical substance, reduced to the smallest set of integer values. For substances that exist as molecules, the molecular formula must be an integer multiple of the empirical formula. SubstanceChemical formulaEmpirical formula water H2O H2O hydrogen peroxide H2O2 HO benzene C6H6 CH dichloroethane C2H4Cl2 CH2Cl acetic acid CH3COOH CH2O