the mass of a 12C atom. The mass of the 12C atom is taken to be exactly 12 u; the mass of the 23Na atom is 22.9898 u. Table 2-1 lists the masses of some nuclides to which reference will be made in this chapter, as well as others.The atomic theory was proposed by John Dalton in 1805. Dalton thought that all atoms of a given element were identical. Chemists in following decades embraced the task of finding the relative masses of atoms of the different elements by precise quantitative chemical analysis. Over a hundred years after Dalton’s proposal was made, investigations with radioactive substances showed that not all atoms of a given element were identical. The Periodic Chart (Table) of the Elements recognizes the differing masses of atoms by providing the average atomic mass for each of the elements. An element can exist in several isotopic forms in which the number of neutrons is different for each isotope; however, all atoms of the same element have the same number of protons, as is discussed directly below.
Every atom has a positively charged nucleus which contains over 99.9 percent of the total mass of the atom. There are numerous particles found in the nucleus, but nuclei may be described by considering only two particles. These particles are the proton and the neutron, collectively known as nucleons. These two nucleons have nearly the same mass (1 atomic mass unit, u, although there is an informal use of amu or AMU). Of these two nucleons, only the proton has an electrical charge and the charge is a positive charge. The size of the proton’s charge may by considered the fundamental unit of charge for atomic and nuclear phenomena, since no smaller charge than this has been discovered in any free particle. The charge of the proton is assigned the value of +1 and all other charges are discussed relative to that charge. Since the neutron has no charge, the charge on the nucleus of an atom is solely due to the number of protons.
The atoms of all isotopes of any specific element have the same number of protons. This number is called the atomic number, Z, and is a characteristic of the element. The nuclei of different isotopes differ in the number of neutrons providing for a different number of nucleons in the nuclei. One way of referring to specific isotopes is to provide the total number of nucleons, A, which is the mass number. Atoms of the different isotopic forms of an element, the nuclides, are distinguished by using the mass number as a superscript to the left of the element’s symbol. So, the nitrogen isotope containing 8 neutrons will have a mass number of 15 and is represented by 15N (or N-15). Working from the other direction, we can determine the number of neutrons in an isotope by subtracting the atomic number from the mass number A – Z = 15 – 7 = 8 neutrons. Further, the charge on the nucleus of a nitrogen atom is +7, which is due to the number of protons (atomic number).
The masses of individual atoms are very small. Even the heaviest atom discovered has a mass less than 5 × 10–25 kg. Since 1 kg is 2.2 lb, the mass referred to is less than 1.10 × 10–24 lb. It is convenient to define a special unit in which the masses of the atoms are expressed without having to use exponents. This unit is called the atomic mass unit, referred to by the symbol u in the literature. It is defined as exactly the mass of a 12C atom. The mass of the 12C atom is taken to be exactly 12 u; the mass of the 23Na atom is 22.9898 u. Table 2-1 lists the masses of some nuclides to which reference will be made in this chapter, as well as others.
Table 2-1 Some Nuclidic Masses (u)
Most chemical reactions do not discriminate significantly among the various isotopes. For example, the percentages of iron atoms which are 54Fe, 56Fe, 57Fe, and 58Fe are 5.8, 91.8, 2.1, and 0.3, respectively, in all iron ores, meteorites, and iron compounds prepared synthetically. For chemical purposes it is of interest to know the average mass of an iron atom in this natural isotopic mixture. These average masses are also tabulated in terms of the unit u and are designated by Ar (E), where E is the symbol for the particular element. The term atomic mass will be used in this book to mean the average atomic mass, and nuclidic mass will be used when referring to one particular isotope of an element. Ar values, which are listed at the end of this book, form the basis for practically all chemical weight calculations. Ar values used to be determined by precise chemical analysis, but nearly all modern values are the weighted average of the nuclidic masses measured by mass spectroscopy, an extremely accurate process.
Any chemical experiment involves the reaction of enormous numbers of atoms or molecules. The term mole is used to indicate a collection of a large, fixed number of fundamental chemical entities, comparable to the quantity that might be involved in an actual experiment. In fact, the mole is recognized in SI as the unit for one of the dimensionally independent quantities, the amount of substance. The abbreviation for the unit is mol. A mole of atoms of any element is defined as that amount of substance containing the same number of atoms as there are carbon atoms in exactly 12 g of pure 12C. This number is called Avogadro’s number or Avogadro’s constant, NA. The value of this quantity may be related to the value of the u, listed in Table 2-1, as follows:
All units in the expression for NA canceled except for mol, which remained in the denominator and may be expressed as mol–1 (6.0221 × 1023 mol–1). The answer may be interpreted as 6.0221 × 1023 things/mol; of course, in chemistry we are usually referring to atoms or molecules.
Let us look at a mole of atoms of some other element of atomic mass Ar. The average mass of an atom of this element is Ar u and the mass of a mole of such atoms is NA × Ar u, or simply Ar g/mol. In other words, the mass in grams of a mole of atoms of an element is equal to the atomic mass, and Ar may be considered to have the units of g/mol. Therefore, a “mole of gold” is 197.0 g of gold.
Each element has a specific symbol that is different from the symbol for any other element. In a chemical formula, the symbol stands for an atom of an element. Molecular substances are composed of two or more atoms that are tightly bound together. The formula for a molecular substance consists of the symbols for the atoms that are found in that molecule. For instance, the formula for carbon dioxide is CO2. Note the use of the subscript to show that each molecule contains two oxygen atoms in addition to the one carbon atom. Also note that the “1” for the one carbon atom is not written. The molecular mass of CO2 is the sum of the atomic mass of carbon plus twice the atomic mass of oxygen and is expressed in u. As was discussed directly above, the molar mass of CO2 is the mass in grams equal to the molecular mass in u. A “mole of carbon dioxide” is 12.0 u + 2(16.0 u) = 44 u. This result can be expressed as 44 g to indicate one Avogadro’s number, NA, of CO2 molecules. Recall that NA is 6.0221 × 1023 things—molecules in this case.
Many common substances are ionic in nature. This means that the atoms are in the form of charged particles, ions, and are arranged in a potentially huge spatial array that may have no fixed size. In such cases, the formula indicates the relative number of each element present. Table salt is composed of sodium and chloride ions (chlorine ions are called chloride ions) in close association. Although the size of a crystal of table salt is not fixed, the ratio of sodium to chloride ions is 1:1; then, the formula for table salt is expressed as NaCl.
P2O10 is the formula for a compound in which 2 atoms of phosphorus are present for every 10 atoms of oxygen. This formula is called the molecular formula. If the subscripts are the smallest possible ratio of whole numbers, the formula is called an empirical formula—PO5 is the empirical formula for P2O10. P2O10 can also refer to the particular amounts of the components of the compound. One mole of P2O10 contains 2 moles of phosphorus atoms and 10 moles of oxygen atoms. We can calculate the mass of one mol of P2O10 by means of adding up the masses of the components—(2 × 31.0) + (10 × 16.0) = 222 g/mol of P2O10.
The term “atomic weight” had been widely used rather than “atomic mass,” and “molecular weight” rather than “molar mass.” (Many writers used “molecular weight” for “molar mass” even for ionic substances.) Because “weight” is a force rather than a mass, such usage is discouraged. The beginning student, however, must be aware of the old terms because they are certainly found in the literature and may still be used. The term “molar mass” is a particularly welcome change because of its universal applicability, referring to Avogadro’s number of molecules, ions, formula units, or individual atoms (e.g., the molar mass of gold is 197.0 g/mol; the molar mass of hydroxide ion, OH–, is 17.0 g/mol).
2.1. It has been found by mass spectrometric analysis that in nature the relative abundances of the various isotopic atoms of silicon are 92.23% 28Si, 4.67% 29Si, and 3.10% 30Si. Calculate the atomic mass of silicon from this information and from the nuclidic masses.
The atomic mass is the average of the three nuclides, each weighted according to its own relative abundance. The nuclidic masses are given in Table 2-1.
2.2. Naturally occurring carbon consists of two isotopes, 12C and 13C. What are the percentage abundances of the two isotopes in a sample of carbon whose atomic mass is 12.01112?
Let
Then,
and
2.3. Before 1961, a physical atomic mass scale was used whose basis was an assignment of the value 16.00000 to 16O. What would have been the physical atomic mass of 12C on the old scale?
We can use the ratio of the two reference points to determine the older value.
2.4. A 1.5276-g sample of CdCl2 underwent an electrolytic process separating all of the cadmium from the sample. The weight of the metallic cadmium was 0.9367 g. If the atomic mass of chlorine is taken as 35.453, what must be the atomic mass of cadmium from this experiment?
Throughout this book we will specify the amount of a substance in terms of the chemist’s unit, the mole. We will use the symbol n(Symbol or formula) to refer to the number of moles of the substance. Since in most laboratory work mass is determined by weighing, the word “weight” (as in the second sentence in the problem) is commonly used where “mass” would be more precise. Unless it leads to an ambiguity, we will follow common usage and not bother to distinguish between “mass” and “weight.”
We can approach this problem by first calculating the number of moles of Cl atoms in the weighed sample.
From the formula CdCl2 we see that the number of moles of Cd is exactly half the number of moles of Cl.
The atomic mass is the mass per mole.
2.5. In a chemical determination of the atomic mass of vanadium, 2.8934 g of pure VOCl3 was allowed to undergo a set of reactions as a result of which all the chlorine contained in this compound reacted with silver to produce AgCl. The weight of the AgCl was 7.1801 g. Assuming the atomic masses of Ag and Cl are 107.868 and 35.453, what is the experimental value for the atomic mass of vanadium?
This problem is similar to Problem 2.4, except that n(Cl) must be obtained by way of n(AgCl). The three Cl atoms of VOCl3 are converted to 3 formula units of AgCl, the molar mass of which is 143.321 (the sum of 107.868 and 35.453).
From the formula AgCl,
nC(l) = n(AgCl) = 0.050098 mol Cl
Also, from the formula VOCl3,
To find the weight of vanadium in the weighed sample of VOCl3, we must subtract the weights of the chlorine and oxygen. If we designate the mass of any substance or chemical constituent X by m(X), then
m(X) = n(X) × M(X)
where M(X) is the molar mass of X. Note that if X is a single atom, then M(X) is Ar(X). In this problem X is Cl.
m(Cl) = n(Cl) × Ar (Cl) = (0.050098 mol)(35.453 g/mol) = 1.7761 g Cl
The formula VOCl3 tells us that the number of moles of oxygen and vanadium are the same.
m(O) = n(O) × Ar(O) = (0.016699 mol)(15.999 g/mol) = 0.2672 g O
and by difference,
m(V) = m(VOCl3) – m(O) – m(Cl)
m(V) = (2.8934 – 0.2672 – 1.7761)g = 0.8501 g V
and then,
Note that this result differs slightly from the accepted value (50.9415 g/mol). The difference can be ascribed to experimental error in this determination.
2.6. Determine the molar mass of (a) potassium hexachloroiridate(IV), K2IrCl6, and (b) the molar mass of trifluorosilane, SiHF3.
Potassium hexachloroiridate(IV) does not exist as discrete molecules represented by the empirical formula, but trifluorosilane does. The term “molar mass” in either case refers to the mass of NA formula units, which in grams is numerically equal to the sum of all Ar, which appear in the formula (or each element, multiplying its Ar by the number of atoms of that element in the formula).
(a)
(b)
Note that the atomic masses are not all known to the same number of significant figures or to the same number of decimal places in u. In general, the rules for significant figures discussed in Appendix B apply. The value of Ar(Ir) is known to only 0.01 u. Note that in order to express 6 times the atomic mass of Cl to 0.01 u, it was necessary to use the atomic mass to 0.001 u. Similarly, an extra figure was used in the atomic mass for fluorine to give the maximum significance to the last digit in the sum column.
2.7. How many (a) grams of H2S, (b) moles of H and of S, (c) grams of H and of S, (d) molecules of H2S, (e) atoms of H and of S, are contained in 0.400 mol H2S?
The atomic masses involved are H, 1.008; of S, 32.066. The molecular mass of H2Sis2(1.008)+32.066 = 34.08.
Note that it is not necessary to express the molecular mass to 0.001 u, even though the atomic masses are known to this significance. Since the limiting factor in this problem is n(H2S), known to one part in 400, the value 34.08 (expressed to one part in over 3000) for the molecular mass is more than adequate. This a time-saving device; if you had used the complete atomic masses, your answer would be the same.
(a) Number of grams of compound = (number of moles) × (mass of 1 mole)
Number of grams of H2S = (0.400 mol)(34.08 g/mol) = 13.63 g H2S
(b) One mole of H2S contains 2 moles of H and 1 mole of S. Then 0.400 mol H2S contains
and 0.400 mol S (half as much as H).
(c) Number of grams of element = (number of moles) × (mass of 1 mole)
Number of grams of H = (0.800 mol)(1.008 g/mol) = 0.806 g H
Number of grams of S = (0.400 mol)(32.066 g/mol) = 12.83 g S
(d) Number of molecules = (number of moles) × (number of molecules in 1 mole)
= (0.400 mol)(6.02 × 1023 molecules/mol) = 2.41 × 1023 molecules
(e) Number of atoms of element = (numbers of moles) × (number of atoms per mole)
Number of atoms of H = (0.800 mol)(6.02 × 1023 atoms/mol) = 4.82 × 1023 atoms H
Number of atoms of S = (0.400 mol)(6.02 × 1023 atoms/mol) = 2.41 × 1023 atoms S
2.8. How many moles of atoms are contained in (a) 10.02 g calcium, (b) 92.91 g phosphorus? (c) How many moles of molecular phosphorus are contained in 92.91 g phosphorus if the formula of the molecule is P4? (d) How many atoms are contained in 92.91 g phosphorus? (e) How many molecules are contained in 92.91 g phosphorus?
Atomic masses of Ca and P are 40.08 and 30.974; when expressed in grams, we have one mole of each.
(a) 
(c) Molar mass of P4 is (4)(30.974) = 123.90. Then
(d) Number of atoms P = (3.000 mol)(6.022 × 1023 atoms/mol) = 1.807 × 1024 atoms P
(e) Number of molecules of P4 = (0.7500 mol)(6.022 × 1023 molecules/mol)
= 4.517 × 1023 molecules P4
2.9. How many moles are represented by (a) 6.35 g of CO2, (b) 9.11 g of SiO2, (c) 15.02 g of Ca(NO3)2?
Refer to the periodic table for the appropriate atomic masses. The molecular masses are calculated using the atomic masses.
(a) The amount of CO2 = 6.35 g × (1 mol/44.01 g) = 0.1443 mol CO2.
(b) Amount of SiO2 = 9.11 g × (1 mol/60.09 g) = 0.1516 mol SiO2.
(c) Amount of Ca(NO3)2 = 15.02 g × (1 mol/164.10 g) = 0.0915 mol Ca(NO3)2.
The result (a) is a measure of the number of CO2 molecules (CO2 is normally a gas in which the CO2 molecules are separated from each other and have individual physical identities). SiO2 on the other hand is a complicated crystalline solid (quartz), in which each silicon is surrounded by more than two oxygens and each oxygen by more than one silicon. Because of these factors, there is no physically distinct cluster of one silicon with two oxygens. The result of (b) represents a count of the number of SiO2 formula units. The Ca(NO3)2 discussed in (c) is an ionic crystal of no specific size, the given sample containing 0.0915 moles of calcium ions and twice that number of moles of nitrate ions.
2.10. Three common gaseous compounds of nitrogen and oxygen of different elementary composition are known, (A) laughing gas containing 63.65% nitrogen, (B) a colorless gas containing 46.68% nitrogen, and (C) a brown, toxic gas containing 30.45% nitrogen. Show how these data illustrate the law of multiple proportions.
According to the law of multiple proportions, the relative amounts of an element combining with some fixed amount of a second element in a series of compounds are in ratios of small whole numbers.
Since percent means “parts per hundred” we can assume 100 g. Then, on the basis of 100 g of each compound, we tabulate below the mass of N, the mass of O (obtained by the difference from 100), and the mass of N per gram of O.
The relative amounts are not affected if all three amounts are set up in the form of a ratio and then divided by the smallest of the relative amounts.
The relative amounts are indeed the ratios of small whole numbers—4.000 : 2.000 : 1.000—within the precision of the analyses.
The law of multiple proportions was an important contribution to the credibility of Dalton’s atomic theory. It was discovered before relative atomic masses were well known (note that Ar values were not involved in the calculation above). However, it follows logically that all atoms of the same element have the same mass (which is unchangeable) and that compounds contain elements in the relative proportions of simple whole numbers.
2.11. Naturally occurring argon consists of three isotopes, the atoms of which occur in the following abundances: 0.34% 36Ar, 0.07% 38Ar, and 99.59% 40Ar. Calculate the atomic mass of argon from these data and from data in Table 2-1.
Ans. 39.948
2.12. Naturally occurring boron consists of 80.22% 11B (nuclidic mass = 11.009) and 19.78% of another isotope. To account for the atomic mass, 10.810, what must be the nuclidic mass of the other isotope?
Ans. 10.01
2.13. 35Cl and 37Cl are the only naturally occurring chlorine isotopes. What percentage distribution accounts for the atomic mass, 35.4527?
Ans. 24.23% 37Cl
2.14. Gallium is important, used in high-temperature thermometers, and has two naturally occurring isotopes.69Ga comprises 60.1% and 71Ga is the other 30.9%. What is the average atomic mass for Ga?
Ans. 69.723
2.15. To account for nitrogen’s atomic mass of 14.00674, what must be the ratio of 15N to 14N atoms in natural nitrogen? Ignore the small amount of 16N.
Ans. 0.00369
2.16. At one time there was a chemical atomic mass scale based on the assignment of the value 16.0000 to naturally occurring oxygen. What would have been the atomic mass, on such a table, of silver, if current information had been available? The atomic masses of oxygen and silver on the present table are 15.9994 and 107.8682.
Ans. 107.872
2.17. The nuclidic mass of 90Sr had been determined on the old physical scale (16O = 16.0000) as 89.936. Calculate the mass of 90Sr to the atomic mass scale on which 16O is 15.9949.
Ans. 89.907
2.18. In a chemical atomic mass determination, the tin content of 3.7692 g SnCl4 was found to be 1.7170 g. If the atomic mass of chlorine is taken as 35.453, what is the value for the atomic mass of the tin isolated during this experiment?
Ans. 118.65
2.19. A 12.5843-g sample of ZrBr4 was dissolved and, after several chemical steps, all of the combined bromine was precipitated as AgBr. The silver content of the AgBr was found to be 13.2160 g. Assume the atomic masses of silver and bromine to be 107.868 and 79.904. What value was obtained for the atomic mass of Zr from this experiment?
Ans. 91.23
2.20. The atomic mass of sulfur was determined by decomposing 6.2984 g of Na2CO3 with sulfuric acid. The weight of Na2SO4 formed was 8.4380 g. In this reaction, all sodium in the starting material (Na2CO3) appears in the product (Na2SO4). Calculate the atomic mass of sulfur from this experiment.
Ans. 32.017
2.21. Although there is only one naturally occurring isotope of iodine, 127I, the atomic mass is given as 126.9045. Explain.
Ans. The atomic masses indicated on the Periodic Table of the Elements are averages, but they are calculated relative to the mass of 12C. The mass number for iodine’s naturally occurring isotope is 127, which is a total of the number of protons and neutrons, not true masses.
2.22. Determine the molecular mass (or formula unit mass) to 0.01 u for (a) LiOH, (b) H2SO4, (c) O2, (d) S8, (e) Ca3(PO4)2, (f) Fe4[Fe(CN)6]3.
Ans. (a) 23.95; (b) 98.08; (c) 32.00; (d) 256.53; (e) 310.18; (f) 859.28
2.23. How many grams of each of the constituent elements are contained in one mole of (a) CH4, (b) Fe2O3, (c) Ca3P2? How many atoms of each element are contained in the same amount of compound?
Ans. (a) 12.01 g C, 4.032 g H 6.02 × 1023 atoms C, 2.41 × 1024 atoms H
(b) 111.69 g Fe, 48.00 g O 1.204 × 1024 atoms Fe, 1.81 × 1024 atoms O
(c) 120.23 g Ca, 61.95 g P 1.81 × 1024 atoms Ca, 1.204 × 1024 atoms P
2.24. One of the commercial bullets that can be fired from a 38 special revolver weighs 156 grains (1 lb = 2000 grains). Assuming the bullet is made from only lead, (a) how many moles of lead are required for each bullet? (b) What number of atoms are present in a bullet?
Ans. (a) 0.17 mol Pb; (b) 1.03 × 1023 atoms
2.25. ACO2 cartridge is used to power a rotary tool for smoothing surfaces; it holds 8 g CO2. (a) How many moles of CO2 are stored in the cartridge? (b) How many molecules of CO2 are there in the cartridge?
Ans. (a) 0.18 mol CO2; (b) 1.1 × 1023 molecules CO2
2.26. Calculate the number of grams in a mole of each of the following common substances: (a) calcite, CaCO3; (b) quartz, SiO2; (c) cane sugar, C12H22O11; (d) gypsum, CaSO4 · 2H2O; (e) white lead, Pb(OH)2 · 2PbCO3.
Ans. (a) 100.09 g; (b) 60.09 g; (c) 342.3 g; (d) 172.2 g; (e) 775.7 g
2.27. What is the average mass in kilograms of (a) a helium atom; (b) a fluorine atom; (c) a neptunium atom?
Ans. (a) 6.65 × 10–27 kg; (b) 3.15 × 10–26 kg; (c) 3.94 × 10–25 kg
2.28. What is the mass of one molecule of (a) CH3OH; (b) C60H122; (c) C1200H2000O1000?
Ans. (a) 5.32 × 10–26 kg; (b) 1.40 × 10–24 kg; (c) 5.38 × 10–23 kg
2.29. How many moles of atoms are contained in (a) 32.7 g Zn; (b) 7.09 g Cl; (c) 95.4 g Cu; (d) 4.31 g Fe; (e) 0.378 g S?
Ans. (a) 0.500 mol; (b) 0.200 mol; (c) 1.50 mol; (d) 0.0772 mol; (e) 0.0118 mol
2.30. Two bottles, one labeled potassium cyanide and the other sodium cyanide, were found hidden behind a water heater. They each contained 125 g substance. (a) Which bottle contains more molecules? (b) How many moles are present from (a)? (c) How many more molecules are present in the bottle from (a) than the other bottle?
Ans. (a) NaCN; (b) 2.55 moles NaCN; (c) 3.8 × 1023 molecules NaCN
2.31. How many moles are represented by (a) 24.5 g H2SO4, (b) 4.00 g O2?
Ans. (a) 0.250 mol; (b) 0.125 mol
2.32. A sample of a metal is composed of 4.25 moles molybdenum and 1.63 moles of titanium. Express the ratio of the two metals in terms of (a) atoms and (b) masses.
Ans. (a) 425 atoms Mo to 163 atoms Ti; (b) 407.7 g Mo to 78.04 g Ti
2.33. (a) How many moles of Cd and of N are contained in 132.4 g of Cd(NO3)2 · 4H2O? (b) How many molecules of water of hydration are in this same amount?
Ans. (a) 0.429 mol Cd and 0.858 mol N; (b) 1.033 × 1024 molecules H2O
2.34. How many moles of Fe and of S are contained in (a) 1 mol of FeS2 (pyrite); (b) 1 kg of FeS2? (c) How many kilograms of S are contained in exactly 1 kg of FeS2?
Ans. (a) 1 mol Fe, 2 mol S; (b) 8.33 mol Fe, 16.7 mol S; (c) 0.535 kg S
2.35. A certain public water supply contained 0.10 ppb (parts per billion) of chloroform, CHCl3. How many molecules of CHCl3 would be contained in a 0.05-mL drop of this water?
Ans. 2.5 × 1010
2.36. Iridium has an extremely high density, 22.65 g/cm3. How many (a) grams Ir, (b) moles Ir, and (c) atoms of Ir are in a cube 2 cm to the side?
Ans. (a) 181.2 g Ir; (b) 0.94 mol Ir; (c) 5.7 × 1023 atoms Ir
2.37. The threshold after which death occurs is 2500 nanograms cyanide per milliliter of blood. Assuming the average blood volume of 5.6 L for an average-sized person, (a) what mass in grams potassium cyanide, KCN, will provide the fatal dose? (b) The density is 1.5 g/cm3 KCN; how large would this sample be in cm3?(c) How many moles KCN are there? (d) How many molecules are present?
Ans. (a) 0.014 g KCN; (b) 0.021 cm3 (a few small crystals); (c) 2.6 × 10–4mol KCN; (d) 1.6 × 1020 molecules KCN
2.38. An alloy named 45 Permalloy contains 54.7% Fe, 45% Ni, and 0.3% Mn by mass. (a) Express the content of a 0.685-g sample in terms of moles each metal. (b) If the source of the information had expressed the percentage composition of Permalloy in moles, rather than mass, would the 45 (percent nickel) in the name still be correct? Explain.
Ans. (a) 6.7 × 10–3 mol Fe; 5.3 × 10–3 mol Ni; 3.7 × 10–5 mol Mn; (b) No, since the masses and moles are different numbers for each of the components, the percentages would also be different in moles from that given in mass.
2.39. A 0.01-g sample of crude gunpowder was collected from the site of a pipe bomb detonation. Analysis told us the sample was 20% sulfur by mass. The estimate of the amount of gunpowder used was 0.350 kg (less than ¾ lb). Calculate (a) the mass (g) of sulfur obtained to produce the bomb, (b) mol S, and (c) number of atoms S.
Ans. (a) 70 g S; (b) 2.1 mol S; (c) 1.3 × 1024 atoms S
2.40. Verify the law of multiple proportions for an element, X, which forms oxides having percentages of X equal to 77.4%, 63.2%, 69.6%, and 72.0%. If the compound with 77.4% X is XO, what element is X, and what are the other compounds?
Ans. The relative amounts of X combining with the fixed amount of oxygen are 2, 1, 
, and 
. The relative amounts of oxygen combining with the fixed amount of X are 1, 2, , and . Since Ar (X) = 54.8, Xis Mn. The other compounds have the empirical formulas MnO2, Mn2O3, and Mn3O4.
2.41. Rust is a variable mixture of a number of iron compounds. A sample of rust has been separated into the various compounds which have been analyzed. There are two sets of data for compounds composed of iron, oxygen, and hydrogen: (1) is 52.12% iron, 45.04% oxygen, and 2.84% hydrogen; (2) is 62.20% iron, 35.73% oxygen, and 2.25% hydrogen. Identify the compounds; how do these data relate to the law of multiple proportions?
Ans. (1) is Fe(OH)3 and (2) is Fe(OH)2. The percentages tell us that there are definitely two compounds. The reason the condition exists is that Fe can have two oxidation numbers, +3 and +2, producing two compounds with a small number ratio of the components.
