Chapter 1 - Solutions



CHAPTER 2 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

1) What are the hypotheses on which Dalton’s atomic theory is based?

The hypotheses are:

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.

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.

4) Two different compounds, each containing only phosphorus and chlorine, were decomposed into their constituent elements. One produced 0.2912 g P for every gram of Cl; the other produced 0.1747 g P for every gram of Cl. Show that these results are consistent with the law of multiple proportions.

If we look at the ratio of the amount of phosphorus in each compound we get

0.2912 g P = 1.667 ( 5

0.1747 g P 3

which is the ratio of two small integers.

19) Explain the meaning of each term in the symbol AZX.

A = Mass number = number of protons + number of neutrons

Z = Atomic number = number of protons (usually omitted)

X = Symbol for element

24) Write the appropriate symbol for each of the following isotopes:

a) Z = 11; A = 23 2311Na = 23Na

b) Z = 28, A = 64 6428Ni = 64Ni

c) Z = 50, A = 115 11550Sn = 115Sn

d) Z = 20; A = 42 4220Ca = 42Ca

28) The following radioactive isotopes are used in medicine for imaging organs, studying blood circulation, treating cancer, and so on. Give the number of protons, neutron, and electrons present in each isotope:

The number of protons, Z, is found in the periodic table with the symbol for the element. The atoms are neutral, and so the number of electrons is equal to the number of protons. The mass number, A, is given as a superscript to the left of the symbol for the element. The number of neutrons is A - Z.

a) 198Au 79 protons, 119 neutrons, 79 electrons

b) 47Ca 20 protons, 27 neutrons, 20 electrons

c) 60Co 27 protons, 33 neutrons, 27 electrons

d) 18F 9 protons, 9 neutrons, 9 electrons

e) 125I 53 protons, 72 neutrons, 53 electrons

f) 131I 53 protons, 78 neutrons, 53 electrons

g) 42K 19 protons, 23 neutrons, 19 electrons

h) 43K 19 protons, 24 neutrons, 19 electrons

i) 24Na 11 protons, 13 neutrons, 11 electrons

j) 32P 15 protons, 17 neutrons, 15 electrons

k) 85Sr 38 protons, 47 neutrons, 38 electrons

l) 99Tc 43 protons, 56 neutrons, 43 electrons

30) State two differences between a metal and a nonmetal

Metals are shiny; nonmetals (as solids or liquids) are usually dull colored

Metals are good conductors of heat and electricity; nonmetals are usually poor conductors of heat and electricity

Metals are malleable and ductile; nonmetals are not malleable and are not ductile.

38) Group the following elements in pairs that you would expect to show similar chemical properties: I, Ba, O, Br, S, Ca.

Pairing is based on the elements being in the same group (column) of the periodic table

I and Br Ba and Ca O and S

41) What is the mass (in amu) of a carbon-12 atom? Why is the atomic mass of carbon listed as 12.01 amu in the table on the inside front cover of the book?

The mass of one 12C atom, by definition, is exactly 12. amu. The mass for carbon in the periodic table is the average mass of naturally occurring carbon. It is slightly larger than 12.00 because of the small amount of 13C and 14C that occurs in nature.

44) The atomic masses of 35Cl (75.53 percent) and 37Cl (24.47 percent) are 34.968 amu and 36.956 amu, respectively. Calculate the average atomic mass of chlorine. The percentages in parentheses denote the relative abundances.

Average mass = (0.7553)(34.968 amu) + (0.2447)(36.956 amu) = 35.45 amu

46) The atomic mass of 203Tl and 205Tl are 202.972320 amu and 204.974401 amu, respectively. Calculate the natural abundances of these two isotopes. The average atomic mass of thallium is 204.4 amu.

Let x = fraction of 203Tl. Then (1 - x) = fraction of 205Tl.

So 204.4 = x (202.972320) + (1 - x) (204.974401)

204.4 = 202.972320 x + 204.974401 - 204.974401 x

204.4 = - 2.002081 x + 204.974401

So 204.4 - 204.974401 = - 2.002081 x

- 0.574401 = - 2.002081 x

x = ( - 0.574401) = 0.2869

( - 2.002081)

So natural abundance is 29% 203Tl and 71% 205Tl. (Note: I am giving one more signifi-cant figure to the result than is justified.)

54) Define molecular formula and empirical formula. What are the similarities and differences between the empirical formula and the molecular formula of a compound?

The molecular formula gives the number of atoms of each type of element per molecule of substance (for those substances that exist as individual molecules). The empirical formula gives the relative number of atoms of each type of element in the compound, reduced to the smallest set of whole numbers.

Both the molecular formula and the empirical formula contain the correct relative number of atoms of each element present in a pure chemical substance. For molecular compounds the molecular formula will be the empirical formula multiplied by some integer value.

64) Write the empirical formulas of the following compounds:

a) Al2Br6 AlBr3 d) K2Cr2O7 K2Cr2O7

b) Na2S2O4 NaSO2 e) H2C2O4 HCO2

c) N2O5 N2O5

66) Write the molecular formula of ethanol. The color codes are black (carbon), red (oxygen), and white (hydrogen).

The molecular formula is C2H6O. Note that an organic chemist would likely write the formula as CH3CH2OH to communicate information about the structure of the molecule.

67) Name the following binary molecular compounds:

a) NCl3 nitrogen trichloride

b) IF7 iodine heptafluoride

c) P4O6 tetraphosphorus hexoxide

d) S2Cl2 disulfur dichloride

Note that we use a prefix to indicate the number of atoms of the first element only if that number is different than one.

72) What is an ionic compound? How is electrical neutrality maintained in an ionic compound?

An ionic compound is a pure chemical substance composed of cations (positive ions) and anions (negative ions). Electrical neutrality is maintained by keeping the positive charge from the cations and the negative charge from the anions equal to one another.

76) Give the number of protons and electrons in each of the following common ions:

a) K+ 19 protons, 18 electrons

b) Mg2+ 12 protons, 10 electrons

c) Fe3+ 26 protons, 23 electrons

d) Br- 35 protons, 36 electrons

e) Mn2+ 25 protons, 23 electrons

f) C4- 6 protons, 10 electrons (this is not actually a common ion)

g) Cu2+ 29 protons, 27 electrons

82) Name the following compounds:

a) KClO potassium hypochlorite

b) Ag2CO3 silver carbonate (I would also accept silver (I) carbonate)

c) HNO2 nitrous acid

d) KMnO4 potassium permanganate

e) CsClO3 cesium chloate

f) KNH4SO4 potassium ammonium sulfate

g) FeO iron (II) oxide

h) Fe2O3 iron (III) oxide

i) TiCl4 titanium (IV) chloride

j) NaH sodium hydride

k) Li3N lithium nitride

l) Na2O sodium oxide

m) Na2O2 sodium peroxide

84) Write formulas for the following compounds:

a) copper (I) cyanide CuCN

b) strontium chlorite Sr(ClO2)2

c) perbromic acid HBrO4

d) hydroiodic acid HI

e) disodium ammonium phosphate Na2NH3PO4

f) potassium dihydrogen phosphate KH2PO4

g) iodine heptafluoride IF7

h) tetraphosphorus decasulfide P4S10

i) mercury (II) oxide HgO

j) mercury (I) iodide Hg2I2

k) selenium hexafluoride SeF6

As discussed n the book, the mercury (I) ion is a polyatomic ion, existing as Hg22+(see 84 j).

96) What is wrong with the name for each of the following compounds:

a) BaCl2 (barium dichloride) We know that barium is a Ba2+ ion (group 2A) and chlorine is a Cl- ion (group 7A), so we do not have to use di- in the second part of the name. The correct name for the compound is barium chloride.

b) Fe2O3 (iron (II) oxide) Oxygen is a O2- ion (group 6A). There are three of these, and so the total charge from the oxygens is -6. For electrical neutrality, the total charge from the two irons must be +6. Therefore the iron ion is the Fe3+ ion, and the correct name for the compound is iron (III) oxide.

c) CsNO2 (cesium nitrate) The nitrate ion is NO3-. So NO2- is the nitrite ion, and the correct name for the compound is cesium nitrite.

d) Mg(HCO3)2 (magnesium (II) bicarbonate) Since magnesium is in group 2A, we know it forms the Mg2+ ion, and so do not have to indicate the charge of the ion in the name. So the correct name for the compound is magnesium bicarbonate (or magnesium hydrogen carbonate).

101) Write the formula for the common ion derived from each of the following:

a) Li Li+ e) Al Al3+

b) S S2- f) Cs Cs+

c) I I- g) Mg Mg2+

d) N N3-

104) Determine the molecular and empirical formulas of the compounds shown here. (black = carbon, white = hydrogen).

Molecular Empirical

a) C2H2 CH

b) C6H6 CH

c) C2H6 CH3

d) C3H8 C3H8

114) Predict the formula and name of the binary compound formed from the following elements:

a) Na and H NaH sodium hydride

b) B and O B2O3 diboron trioxide

c) Na and S Na2S sodium sulfide

d) Al and F AlF3 aluminum fluoride

e) F and O OF2 oxygen difluoride

f) Sr and Cl SrCl2 strontium chloride

I don't believe you can predict the formula for b and e based on what we have discussed so far in class (see Figure 2.14, page 60 of the text). I have used the naming convention for binary molecular compounds for these two substances.

116) Show the location of the a) alkali metals; b) alkaline earth metals; c) halogens; d) noble gases in the given outline of the periodic table. Also draw dividing lines between metals and metalloids and between metalloids and nonmetals.

a) alkali metals are all the elements in the 1A column except the first element (which is hydrogen)

b) alkaline earth metals are all the elements in the 2A column

c) halogens are all the elements in the 7A column

d) noble gases are all the elements in the 8A column

The red diagonal represents the metalloids. The elements to the right of the red diagonal are nonmetals (along with hydrogen). The elements to the left of the red diagonal are metals (except for hydrogen).

117) Fill in the blanks in the table.

|Cation |Anion |Formula |Name |

|Mg2+ |HCO3- |Mg(HCO3)2 |Magnesium bicarbonate |

|Sr2+ |Cl- |SrCl2 |Strontium chloride |

|Fe3+ |NO3- |Fe(NO3)3 |Iron (III) nitrate |

|Mn2+ |ClO3- |Mn(ClO3)2 |Manganese (II) chlorate |

|Sn4+ |Br- |SnBr4 |Tin (IV) bromide |

|Co2+ |PO43- |Co3(PO4)2 |Cobalt (II) phosphate |

|Hg22+ |I- |Hg2I2 |Mercury (I) iodide |

|Cu+ |CO32- |Cu2CO3 |Copper (I) carbonate |

|Li+ |N3- |Li3N |Lithium nitride |

|Al3+ |S2- |Al2S3 |Aluminum sulfide |

The first compound above can also be named magnesium hydrogen carbonate..

124) A cube made of platinum (Pt) has an edge length of 1.000 cm. The density of Pt is 21.45 g/cm3, and the average mass of a single Pt atom is 3.240 x 10-22 g.

a) Calculate the number of Pt atoms in the cube

# atoms = 1.000 cm3 21.45 g 1 atom = 6.620 x 1022 atoms

1 cm3 3.240 x 10-22 g

b) Atoms are spherical in shape. Therefore, the Pt atoms in the cube cannot fill all of the available space. If only 74 percent of the space inside of the cube is taken up by Pt atoms, calculate the radius (in pm = picometers) of a Pt atom. Recall that for a sphere with radius r is V = 4/3 (r3.

Since only 74% of the space inside the cube is occupied the atoms, then the volume per atom is

V (per atom) = (0.74) (1.000 cm3) = 1.118 x 10-23 cm3

6.620 x 1022 atoms

Since V = 4/3 (r3

r = [ 3V/4( ]1/3 = [ 3 (1.118 x 10-23 cm3) / 4( ]1/3 = (2.668 x 10-24 cm3)1/3

= 1.39 x 10-8 cm

In terms of picometers

r = 1.39 x 10-8 cm 1 m 1012 pm = 139. pm

100 cm 1 m

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