The empirical formula is the formula for a compound that is expressed in the lowest ratio that can be calculated (refer to Chapter 2). Often, a substance must be analyzed to gather information leading to its identity. Various processes can be used to determine the composition of a sample, and an effective way of expressing these data is in the form of weight. Weights can be converted to moles and expressing a formula is the next logical step. The empirical formula is not necessarily the actual molecular formula; however, the empirical formula does contain important information.
Consider a compound that analyzes 17.09 percent magnesium, 37.93 percent aluminum, and 44.98 percent oxygen. (Unless stated to the contrary, percentages are weight percentages, i.e., number of grams of element per 100 g of compound.) Table 3-1 gives a systematic scheme for handling provided data.
The numbers in column (4) represent the numbers of moles of atoms of each of the components of the compound found in the 100-g sample. These three numbers state the ratio of the components of the compound—0.703:1.406:2.812. We could write the compound as Mg0.703Al1.406O2.812 except, of course, the numbers used must be whole numbers. Suppose we were to divide all three amounts of moles by the smallest number; such a manipulation (5) preserves the ratio since all three are divided by the same number, and the division does result in a whole-number ratio. This final ratio can be used to correctly write the empirical formula, MgAl2O4.
The existence of a formula for a compound implies that fixed relationships exist between the weights of any two elements in a specific compound or between the weight of any element and the weight of the compound as a whole. These relationships can best be seen by writing the formula in a vertical form, as illustrated in Table 3-2 for the compound Al2O3.
The sum of the entries in column (4) for the elements equals the molar mass of the compound. The entries in column (5) represent the fractional content of the various elements in the compound. These numbers are really dimensionless (g/g cancel out) and are the same in any unit of mass that may be used in similar calculations. This means that we can express the amounts of the elements in 1 ton of the compound in tons of elements, or 1 pound of the compound in terms of pounds of elements.
The percentage of aluminum in Al2O3 is the number of parts by weight of Al in 100 parts by weight of Al2O3. It follows that the percentage is expressed by a number 100 times as great as the fraction. Then, the percentages of aluminum and oxygen are 52.9 percent and 47.1 percent, respectively. The sum of the constituent percentages of any compound must equal 100 percent.
Sometimes it is desirable to state the composition of a substance with respect to a particular element so contained. For example, the aluminum content of glass may be expressed in terms of Al2O3, even though there is no aluminum oxide in the glass formulation. Then, a sample of glass that is 1.3 percent Al2O3 contains sufficient aluminum so that, if all the aluminum in a 100-g sample of the glass were converted to Al2O3, the weight of Al2O3 would be 1.3 g. In many cases oxide notations are the result of historical errors in the assignment of chemical structures to complex substances. Whatever the origin, it is a straightforward procedure to convert the data in such a form to direct elementary composition, or vice versa, by the use of a quantitative factor such as that found in column (5) of Table 3-2. The ratio of aluminum to aluminum oxide can be expressed as
and is called a quantitative factor. These factors may be used as special conversion factors in numerical problems, such as was found in Chapter 1, Problem 1.19.
Stoichiometry is the series of calculations on the basis of formulas and chemical equations and will be covered in Chapter 4. The use of conversion factors is common even when the relative proportions are not fixed by a chemical formula. Consider a silver alloy used for jewelry production. (Alloys are mixtures of metals and, as mixtures, may be produced in differing ratios of the metals.) A particular alloy contains 86 percent silver. Factors based on this composition, such as
may be used as conversion factors in all problems involving alloys of this particular composition, and are referred to as nonstoichiometric factors.
The molecular mass of a compound is calculated by adding up the atomic masses of the constituent elements. Those masses are the average atomic masses, which are the weighted masses of the various isotopic forms of the elements involved. The nuclidic molecular mass may be defined for a molecule made up of particular nuclides by adding nuclidic atomic masses in the same way that the usual molecular mass is computed from the atomic masses.
The mass spectrometer is an instrument that is capable of separating particles of different isotopic composition and measuring their individual relative masses. The mass spectrometer also will pull a compound apart one atom at a time, producing fragments that can be detected by their specific masses. The distinction of various fragments and the level of precision in providing masses can supply information from which the exact molecular formula can be deduced without resorting to a quantitative chemical composition analysis.
EXAMPLE 1 Consider the three gases CO, C2H4, and N2. Since 12C, 16O, 14N, and 1H dominate over all other isotopes, the mass spectrometer will reveal the presence of a particle of approximate mass 28 in all three cases. If the measurements are made with great precision, the three gases can easily be distinguished on the basis of their nuclidic masses, which are calculated below.
EXAMPLE 2 Find the formula of an organic compound whose dominant nuclidic species was found to have a precise molecular mass of 44.025. It is known that no elements other than C, H, O, and N are present.
The number of carbon atoms in the molecule, n(C), must be at least 1, otherwise the compound would not be organic. n(C)cannot be greater than 3, because 4 carbon atoms would contribute 48 to the total mass number given of the molecule, 44. Similar constraints limit the number of oxygen or nitrogen atoms per molecule. The possible combinations of carbon, oxygen, and nitrogen consistent with the limiting mass are listed in column (1) of Table 3-3.
Column (2) lists the mass numbers of the carbon, nitrogen, oxygen skeletons. Column (3) lists the number of hydrogen atoms needed to bring the mass number of the molecule to 44. Column (4) lists the maximum number of H atoms consistent with rules for molecular structure as discussed in Chapters 9 and 15. One such role is that n(H, max) is equal to twice the number of carbon atoms plus the number of nitrogen atoms plus 2. Column (5) lists the allowed formulas consistent with the total mass number and with all the assumptions and rules. Note that all skeletons for which the number in column (3) (the mass shortage to be made up by hydrogen) exceeds the number in column (4) (the amount of hydrogen allowable for the skeleton by the rules of oxidations numbers) are rejected. Column (6) tabulates the nuclidic molecular masses for the allowed formulas, computed from the nuclidic masses in Table 2-1. When the computed molecular masses are compared with the experimental value, 44.025, it is seen that C2OH4 is the only allowable formula that fits the data within the claimed precision; therefore, this must be the formula of the substance.
3.1. Derive the empirical formula of a hydrocarbon that on analysis gave the following percentage composition: C = 85.63% and H = 14.37%.
The tabular solution, based on 100 g of compound, is as follows:
where E = element; m(E) = mass of the element per 100 g of compound; Ar (E) = atomic mass of the element; n(E) = amount of element per 100 g of compound, expressed in moles of atoms.
The procedure of dividing n(E) by n(C) is equivalent to finding the number of atoms of each element for each atom of carbon. The ratio of H to C atoms is 2:1. This means that the empirical formula is CH2 and that the molecular formula will be a multiple of CH2; that is, if we had the information to determine the molecular formula, which we don’t.
The empirical formula CH2 is not a stable substance. It is necessary to determine the molar mass to determine the molecular formula. If this hydrocarbon were a gas or an easily volatilized liquid, its molar mass could be determined from the density of the gas, as shown in Chapter 5. Supposing such a determination yields a molar mass of about 55 g/mol, what is the molecular formula?
Since the mass of the empirical formula, CH2, is 14 u, we can divide the given molar mass to determine the number of empirical formula units we need to achieve the molar mass of 55. EF in the calculation represents the empirical formula.
The calculation tells us there are essentially 4 empirical formula units per molecular formula, which then must be C4H8. As a verification, butene is a compound with the formula C4H8.
3.2. The analysis of a compound indicates a composition that is 26.57% K, 35.36% Cr, and 38.07% O. Derive the empirical formula of the compound.
We can set up the information and solution in the form of the following table:
In contrast to the previous example, the numbers in column (5) are not all integers. The ratio of the numbers of atoms of the two elements must be the ratio of small whole numbers, in order to satisfy one of the postulates of Dalton’s atomic theory. Allowing for experimental errors and any uncertainty from calculations, we see that the entry of oxygen in column (5), 3.499, is essentially 3.500 when we allow for error. If we round off to 3.5 we can come to a whole number by multiplying by 2 (to get rid of the 0.5). Of course, we must multiply the remaining elements in the ratio to preserve the relationship. When we do so, we arrive at a 2:2:7 ratio shown in column (6), arriving at K2Cr2O7 as the formula of the compound.
3.3. A 15.00-g sample of a hydrated salt, Na2SO4 · xH2O, was found to contain 7.05 g of water. Determine the empirical formula of the salt.
Hydrates are compounds containing water molecules loosely bound to the other components. H2O may usually be removed intact by heating and then may be replaced by wetting or possibly by absorbing moisture from the air. The Na2SO4 and H2O groups may be considered as units of which the compound is made and their formula unit masses are used in place of atomic masses. This problem is different from the previous problems because percent compositions were not provided; instead, we are to work with a mass of the hydrated salt and are given the mass of the anhydrous (without the water) salt. We can set up the tabular solution using these data.
Column (5) is determined in this problem, as in the previous problem, by dividing both numbers by the smallest (0.0559) which preserves the mole ratio of 0.0559:0.391 and leads to the whole-number ratio required to write a chemical formula. Column (5) contains 6.99, a number that is so close to a whole number that the difference can be taken for experimental error. The mole ratio of Na2SO4 to H2O is 1 to 7, providing us with an empirical formula of Na2SO4· 7H2O.
3.4. A 2.500-g sample of uranium was heated in air. The resulting oxide weighed 2.949 g. Determine the empirical formula of the oxide.
The oxide contains 2.500 g uranium and, by subtraction (2.949 g uranium oxide – 2.500 g uranium), 0.449 g oxygen. Division of the weights of uranium and oxygen by their respective atomic masses tells us that there are 0.01050 mol U and 0.02806 mol O. Dividing both by the smallest number of moles (0.01050) yields the ratio of 1 mol U:2.672 mol O. Rounding off here would be a mistake, as is discussed directly below. Since 2.672 is close to 2
, multiplying by 3 would give us a number that is either a whole number or, hopefully, close to one. The result is nearly a whole-number relationship, 3.00 mol U:8.02 mol O. Considering the possibility of both human error in the analysis and the error involved in calculation, the compound formula is U3O8.
Emphasis must be placed on the importance of carrying out the computations to as many significant figures as the analytical precision requires; the information given in the problem is in 4 significant digits. If the numbers in the ratio, 1:2.67, are rounded off at this point, that would result in a ratio of 1:3 and a formula of UO3. Although this formula is close (notice U3O9), it is not correct. Further, if the numbers in the ratio 1:2.67 had been multiplied by 2 to give 2:5.34 and these numbers had been rounded off to 2:5 (U2O5), the wrong formula would have also been advanced.
3.5. A 1.367-g sample of an organic compound was combusted in a stream of dry oxygen to yield 3.002 g CO2 and 1.640 g H2O. If the original compound contained only carbon, hydrogen, and oxygen, what is its empirical formula?
Since the relationship within a chemical formula is a small whole-number relationship of moles of elements to each other, the first step in the solution is to determine the number of grams of each of the elements. This step must isolate the desired element from the compound produced by the burning, which can be performed by multiplying by the fraction of the compound that is the element.
The amount of oxygen in the organic compound is obtained by a simple subtraction of the carbon and hydrogen from the sample mass (1.367 – 0.819 – 0.184 = 0.364 g O).
Since the numbers of moles are necessary to write the chemical formula, we must perform a set of conversions from grams to moles.
Dividing each of the results by the smallest value (0.0228) lead us to the ratio of 3:8:1 and the chemical formula, C3H8O.
3.6. A strip of copper weighing 3.178 g is strongly heated in a stream of oxygen until it is all converted to 3.978 g of the black copper-oxygen compound. What is the percentage composition of copper and oxygen in this compound?
Note that the total of the two percentages is 100%, which forms a check for the calculations.
3.7. (a) Determine the percentages of iron in FeCO3, Fe2O3, and Fe3O4. (b) How many kilograms of iron could be obtained from 2.000 kg of Fe2O3?
(a) Molar mass of FeCO3 is 115.86; of Fe2O3 is 159.69; of Fe3O4 is 231.54. The key is to calculate the fraction of iron found in each compound, then convert each fraction to percent composition.
(b) From (a), the weight of Fe in 2.00 kg Fe2O3 is 0.06994 × 2.000 kg = 1.399 kg Fe.
3.8. Given the formula K2CO3, determine the percent composition of each of the elements in potassium carbonate.
One mole of K2CO3 contains
The sum of the percentages is 100%, our check on our work.
3.9. CaO can be isolated from limestone, CaCO3, by means of driving off CO2 during heating. (a) Calculate the percentage of CaO in CaCO3. (b) How many pounds of CaO can be obtained from 1 ton of limestone that is 97.0% CaCO3? (1 ton = 2000 lb)
(a) There is 1 mole of CaO per mole of CaCO3. We can write a quantitative factor (conversion factor) and apply to determine the fraction, then the percentage composition, of CaO in CaCO3.
(b) We must first calculate the weight of CaCO3 in one ton of the limestone, then the weight of the CaO.
3.10. How much 58.0% sulfuric acid solution is needed to provide 150 g of H2SO4?
Let w = mass (weight) of sulfuric acid solution. Also, notice that there are 58.0 grams of pure acid per 100 grams of solution (58.0% H2SO4 solution).
A second solution avoids the algebra in the solution by using the concept of the conversion factor and setting it up so that units cancel properly.
3.11. How much calcium is in the amount of Ca(NO3)2 that contains 20.0 g nitrogen?
It is not necessary to find the weight of calcium nitrate containing 20.0 g N. The relationship between the calcium and the nitrogen can be found directly from the formula. There are 2 atoms of nitrogen for each atom of calcium. This relationship can also be expressed in terms of moles: 2 mol N:1 mol Ca.
3.12. (a) How much sulfuric acid, H2SO4, could be produced from 500 kg of sulfur? (b) How many kilograms of Glauber’s salt, Na2SO4 · 10H2O, could be obtained from 1000 kg of H2SO4?
(a) The formula of sulfuric acid indicates that 1 mol S (32.07 g S) will give 1 mol H2SO4 (98.08 g H2SO4). We assume a 100% efficient reaction. Then, since the ratio of any two elements involved in the formula can be expressed in a ratio of the mass units (g/mol), we can use a conversion factor containing the information for both the sulfuric acid and the sulfur.
(b) 1 mol H2SO4 (98.08 g/mol) will give 1 mol Na2SO4 · 10H2O (322.2 g/mol), since each substance contains one sulfate (SO4) group per formula unit. Then
3.13. How many tons of Ca3(PO4)2 must be treated with carbon and sand in an electric furnace to make 1 ton of phosphorus? Assume complete conversion of the phosphorus.
The formula for calcium phosphate tells us that 2 mol P (2 × 30.974 g P = 61.95 g P) is contained in 1 mol Ca3(PO4)2 (310.2 g/mol). Then changing grams to tons in the weight ratio, we obtain
3.14. A 5.82-g silver coin is dissolved in nitric acid. When sodium chloride is added to the solution, all the silver is precipitated as AgCl. The AgCl precipitate weighs 7.20 g. Determine the percentage of silver in the coin.
and, since the 5.82-g coin contains 5.42 g Ag,
3.15. A sample of impure sulfide ore contains 42.34% Zn. Find the percentage of pure ZnS in the sample.
The formula ZnS shows that 1 mole of ZnS contains 1 mole Zn, providing us with the conversion factor
Consider 100.0 g of sample; it contains 42.34 g Zn. Then, applying the conversion factor,
3.16. Fertilizers are compounds or mixtures usually used to supply potassium, nitrogen, and phosphorus to soil. If a bag of fertilizer consists of almost pure KNO3 (potassium nitrate), what percentages of those three important elements should appear on the label?
Let us take a look at one mole pure KNO3, which contains
3.17. (a) A Pennsylvania bituminous coal is analyzed as follows: Exactly 2.500 g is weighed into a fused silica crucible. After drying for 1 hour at 110°C, the moisture-free residue weighs 2.415 g. The crucible is then covered with a vented lid and strongly heated until no volatile matter remains. The residual coke button weighs 1.528 g. The crucible is further heated, but without the cover, until all specks of carbon have disappeared, resulting in a final ash that weighs 0.245 g. What is the proximate analysis of this coal, i.e., the percents of moisture, volatile combustible matter (VCM), fixed carbon (FC), and ash?
Similarly, the other percentages are calculated to be 35.5% VCM, 51.3% FR, and 9.8% ash.
(b) On the “dry basis” a sample of coal analyzes as follows: VCM is 21.06%; fixed carbon is 71.80%; and ash is 7.14%. If the moisture present in the coal is 2.49%, what is the analysis on the “wet basis”?
If we were to consider a 100-g sample of the coal, then the percents without the water present (100 – 2.49 = 97.5 g of dry sample) can be used and the solution becomes
When we add these percentages and include the 2.5% for moisture, the sum is 100%.
3.18. A certain fertilizer, “A,” contains 38.7% K, 13.9% N, and no P. Another, “B,” contains 12.2% N, 26.9% P, and no K. (a) What will be the percentages of K, N, and P in a fertilizer made by blending equal weights of A and B? (b) The manufacturer wishes to market a mixture of A and B in which the two elements K and P are present in equal proportions. What proportions of A and B should be used?
(a) If we choose to mix 100 g of each, we arrive at the composition of 200 g of the mixture from which we can calculate the percentages requested.
(b) We can consider 100 g of the mixture letting c = g of A; then 100 – c = g of B. Setting %K = %P in the mixture,
0.387c = 0.269(100 – c)
Solving this, we find that c = 41.0 g of A, and (100 – c) = 59.0 g of B.
3.19. When the Bayer process is used for recovering aluminum from siliceous ores, some aluminum is always lost because of the formation of an unworkable “mud” having the average formula 3Na2O · 3Al2O3 · 5SiO2 · 5H2O. Since aluminum and sodium ions are always in excess in the solution from which this precipitate is formed, the precipitation of the silicon in the “mud” is complete. A certain ore contained 13% (by weight) kaolin (Al2O3 · 2SiO2 · 2H2O) and 87% gibbsite (Al2O3 · 3H2O). What percent of the total aluminum in this ore is recoverable in the Bayer process?
If we were to obtain 100 g of the ore, it would contain 13 g kaolin and 87 g gibbsite. We can determine the amounts of aluminum by the following procedures:
Kaolin has equal numbers of Al and Si atoms, and 13 g kaolin contains 2.7 g Al. The mud takes 6 Al atoms for 5 Si atoms or 6 Al atoms lost for every 5 Si atoms in the kaolin. This means that the precipitation of all the Si from 13 g kaolin involves the loss of (
) (2.7 g) = 3.2 g Al.
3.20. A clay was partially dried and was found to contain 50% silica and 7% water. The original clay contained 12% water. What is the percentage of silica in the original sample?
We have to make the assumption that only water was lost in the drying process. The original and the partially dried clays have the compositions as indicated directly below.
The ratio of silica to the other dry constituents must be the same in both clays; then
Solving, p = 47. This means that there was 47% silica in the original clay.
3.21. Bronze is an alloy of copper and tin. A 0.6554-g sample of a certain bronze was reacted with nitric acid and the tin removed. After appropriate treatment of the solution, titration with sodium thiosulfate revealed that it contained 8.351 millimoles of copper. Calculate the percentages of copper and tin in this bronze.
3.22. A nugget of gold and quartz weighs 100 g and has a density of 6.4 g/cm3. The density of gold is 19.3 g/cm3 and that of quartz is 2.65 g/cm3. Determine the weight of gold in the nugget.
If we let w be the weight of gold in the nugget, the amount of quartz is 100 g – w.
Solving for w, we find that the nugget contains 68 g of gold.
3.23. A purified cytochrome protein isolated from a bacterial preparation was found to contain 0.376% iron. What can be deduced about the molar mass of the protein?
The percent iron content is rather small at 0.376 g per 100 g of sample. Further, the information in the problem implies that each molecule must contain a minimum of one atom of iron. If it does contain only one atom of iron (55.8 u), then the molar mass, M, is given by
This means that, if the protein molecule contained n number of atoms of Fe, the molar mass would be 14,800n u (14,800n g/mol) cytochrome.
This method of calculation is useful for the determination of the minimum molar mass of a macromolecular (large molecule) substance when an analysis can be done for only one of the minor components. Frequently, the approximate molar mass can be determined by a physical method, such as osmotic pressure or sedimentation rate.
3.24. A purified pepsin isolated from a bovine preparation was subject to an amino acid analysis of its hydrolytic products. The amino acid present in the smallest amount was lysine, C6H14N2O2, and the amount of lysine recovered was found to be 0.43 g per 100 g protein. What is the minimum molar mass of the protein?
Proteins do not contain free amino acids, but they do contain chemically linked forms of amino acids, which on degradative hydrolysis can be reconverted to the free amino acid form. The molar mass lysine is 146, and we let M be the minimum molar mass of the protein. As in Problem 3.23, the protein molecule must be at least heavy enough to contain one lysine residue.
Solving for M, we find that the pepsin has a molecular mass of 34,000 u.
3.25. Unsaturated polyesters produced from maleic acid (C4H4O4) and ethylene glycol (C2H6O2) are widely used (with styrene and fiberglass) to make reinforced resin structures. A 5.00-g portion of a batch of polyester was dissolved and treated with 0.00420 moles of sodium hydroxide (NaOH), an amount just sufficient to neutralize all of the acid “end-groups” present. Since there are two end-groups in each molecule, what is the average molar mass of the polyester?
3.26. An organic compound was prepared containing at least one and no more than two sulfur atoms per molecule. The compound had no nitrogen, but oxygen could have been present. The mass-spectrometrically determined molecular mass of the predominant nuclidic species was 110.020. (a) What are the allowable molecular formulas consistent with the mass number 110 and with the facts about the elementary composition? (b) What is the molecular formula of the compound?
(a) The skeleton, minus the hydrogen present in the molecule, would be made up of the elements C, O, and S. The number of possible skeletons can be reduced by the following considerations: (i) The maximum number of carbon atoms is 6, since the mass number of carbons plus 1 sulfur would be 116, which is too heavy to conform to the problem. (ii) The maximum number of hydrogen atoms is 2n(C) + 2 = 14 by the rules of organic molecular composition. (iii) The skeleton (C, O, S) must contribute between 96 and 110 u to the mass number. A rather short list, Table 3-4, will now suffice.
(b) Of the six formulas consistent with the known mass number, only C6SH6 is consistent with the precise molecular mass.
3.27. What are the empirical formulas for the chlorides of vanadium containing 58.0%, 67.8%, and 73.6% chlorine?
Ans. VCl2, VCl3, VCl4
3.28. A compound contains 21.6% sodium, 33.3% chlorine, 45.1% oxygen. Record the empirical formula of the compound (atomic masses: Na = 23.0, Cl = 35.5, and O = 16).
Ans. NaClO3
3.29. When 1.010 g of zinc vapor is burned in air, 1.257 g of the oxide is produced. What is the empirical formula of the oxide?
Ans. ZnO
3.30. A compound has the following percentage composition: H = 2.24%, C = 26.69%, O = 71.07%, and a molar mass of 90. Derive its molecular formula.
Ans. H2C2O4
3.31. Determine the simplest formula of a compound that has the following composition: Cr = 26.52%, S = 24.52%, and O = 48.96%.
Ans. Cr2S3O12 or Cr2(SO4)3
3.32. A 3.245-g sample of titanium chloride was reduced with sodium to metallic titanium. After the resultant sodium chloride was washed out, the residual metal was dried and found to weigh 0.819 g. What is the empirical formula of the original compound of titanium?
Ans. TiCl4
3.33. Magnesium can be used to produce strong and light-weight alloys with aluminum. (a) Calculate the percent magnesium available from MgSO4 · 7H2O, Epsom salts, and (b) from Mg3(PO4)2.
Ans. (a) 7.1% Mg; (b) 27.7% Mg
3.34. Calculate the formula of the compound, which exists as a hydrate, containing 44.6% ytterbium and 27.5% chlorine.
Ans. YbCl3 · 6H2O
3.35. An organic compound was found to contain 47.37% carbon and 10.59% hydrogen. The balance was presumed to be oxygen. What is the empirical formula of the compound?
Ans. C3H8O2
3.36. Derive the empirical formulas of the minerals that have the following compositions: (a) ZnSO4 = 56.14%, H2O = 43.86%; (b) MgO = 27.16%, SiO2 = 60.70%, H2O = 12.14%; (c) Na = 12.10%, Al = 14.19%, Si = 22.14%; O = 42.09%, H2O = 9.48%.
Ans. (a) ZnSO4 · 7H2O; (b) 2MgO · 3SiO2 · 2H2O; (c)Na2Al2Si3O10 · 2H2O
3.37. A borane (composed of boron and hydrogen) analyzed 88.45% boron. What is the empirical formula?
Ans. B5H7
3.38. What is the empirical formula of a catalyst that can be used in the polymerization of butadiene if its composition is 23.3% Co, 25.3% Mo, and 51.4% Cl.
Ans. Co3Mo2Cl11
3.39. A 1.500-g sample of a compound containing only C, H, and O was burned completely. The only combustion products were 1.738 g CO2 and 0.711 g H2O. What is the empirical formula of the compound?
Ans. C2H4O3
3.40. Elementary analysis showed that an organic compound contained C, H, N, and O as its only elementary constituents. A 1.279-g sample was burned completely, as a result of which 1.60 g of CO2 and 0.77 g of H2O were obtained. A separately weighed 1.625-g sample contained 0.216 g nitrogen. What is the empirical formula of the compound?
Ans. C3H7O3N
3.41. Plaster of Paris is supplied as a white powder which can be mixed with water for producing impressions of footprints, tire tracks, and the like. Plaster of Paris is a hydrated calcium sulfate, CaSO4, containing 6.20% H2O. What is the formula for Plaster of Paris?
Ans. 2CaSO4 · H2O or CaSO4 · 
H2O
3.42. A hydrocarbon containing 92.3% C and 7.74% H was found to have a molar mass of approximately 79. What is the molecular formula?
Ans. C6H6
3.43. An alloy with a low melting point is produced from 10.6 lb bismuth, 6.4 lb lead, and 3.0 lb tin. (a) What is the percentage composition of the alloy? (b) How much of each metal is required to make 70.0 g of alloy? (c) What weight of the alloy can be made from 4.2 lb tin?
Ans. (a) 53% Bi, 32% Pb, 15% Sn; (b) 37.1g Bi, 22.4 g Pb, 10.5 g Sn; (c)28lb
3.44. Some golden flakes were collected from a fatal puncture wound. The sample of 0.0022 g was collected and turned out to be 0.0019 g copper and the remainder zinc. A brass letter opener was found nearby and was 87% copper and 13% zinc. Are the flakes likely to be from the letter opener?
Ans. The percentage of copper, 86%, is close and appears to be within the intrinsic error of measurement and calculation, but may not be conclusive.
3.45. Calculate the percentage of copper in each of the following minerals: (a) cuprite, Cu2O; copper pyrite, CuFeS2; malachite, CuCO3 · Cu(OH)2. (b) How many grams of cuprite will give 500 kg of copper?
Ans. (a) 88.82%, 34.63%, 57.48%; (b) 563 kg
3.46. Titanium is used for its strength and light weight; some alloys are 50% the weight of steel with no sacrifice of strength. Titanium is isolated from rutile, TiO2, and other ores. If the rutile ore is 11.48% titanium(IV) oxide by weight, how much ore must be processed to obtain 900 kg (a little less than a metric ton) of metallic titanium (assume pure)?
Ans. 13,000 kg ore (13 metric tons)
3.47. What is the nitrogen content (fertilizer rating, usually expressed in the percentage composition of nitrogen) of NH4NO3? of (NH4)2SO4?ofNH3?
Ans. 35.0% N, 21.2% N, 82.3% N
3.48. Determine the percentage composition of (a) silver chromate, Ag2CrO4;(b) calcium pyrophosphate, Ca2P2O7.
Ans. (a) 65.03% Ag, 15.67% Cr, 19.29% O; (b) 31.54% Ca, 24.38% P, 44.08% O
3.49. Find the percentage of arsenic in a polymer having the empirical formula C2H8AsB.
Ans. 63.6% As
3.50. A famous case of poisoning occurred in the orange-growing region of Florida (1988, soft drink tampering). The analysis of 5.000 g suspected toxin revealed the chemical composition of 4.049 g thallium, 0.318 g sulfur, and 0.639 g oxygen. Provide the chemical formula and name.
Ans. Tl2SO4, thallium(I) sulfate (Use: rat and ant poison; banned in 1975)
3.51. The lethal dose of thallium can be as little as 14 mg Tl/kg body weight; thallium accumulates in the body. (a) What mass (g) of the thallium compound in Problem 3.50 would be a single fatal dose for a 220 lb (100 kg) man? (b)If the compound were to be introduced into the meals of this man at the rate of 50 mg/day total over three meals/day, how long would it take (days) to build up the fatal dose of thallium assuming no excretion of the metal?
Ans. (a) 1.73 g Tl2SO4; (b) 35 days
3.52. Determine the compound that contains more arsenic: Na3AsO4, As2O3, or As2S3.
Ans. Approximate percents As: 36% in Na3AsO4, 75% in As2O3, and 70% in As2S3
3.53. The specifications for a transistor material called for one boron atom in 1010 silicon atoms. What would be the boron content of 1 kg of this material?
Ans. 4 × 10–11 kg B
3.54. The purest form of carbon is prepared by decomposing pure sugar, C12H22O11 (driving off the contained H2O from the carbohydrate). What is the maximum number of grams of carbon that could be obtained from 500 g sugar?
Ans. 211 g C
3.55. The empirical formula for vinyl plastic (a polymer of vinyl chloride, PVC) used in pipe that can be used for building sprinkler systems is CH2CHCl. (a) What is the percentage of chlorine in this plastic, and, as a contrast, (b) calculate the percentage of chlorine in table salt, NaCl?
Ans. (a) 56.7%; (b) Table salt is 60.7% chlorine.
3.56. A compound is known to be composed of 40.002% carbon, 8.063% hydrogen, and 53.285% oxygen. The mass spectrophotometer indicates a molecular mass about 121 u. (a) What is the empirical formula? (b) What is the likely molecular formula?
Ans. (a)CH2O; (b)C4H8O4, a carbohydrate, as is glucose, C6H12O6
3.57. What weight of CuO will be required to furnish 200 kg of copper?
Ans. 250 kg CuO
3.58. Ordinary table salt, NaCl, can be electrolyzed in the molten state to produce sodium and chlorine. Electrolysis of an aqueous solution produces sodium hydroxide (NaOH), hydrogen, and chlorine. The latter two products may be combined to form hydrogen chloride (HCl). How many pounds of metallic sodium and of liquid chlorine can be obtained from 1 ton of salt? Alternately, how many pounds of NaOH and how many pounds of hydrogen chlorine can be obtained from that sample?
Ans. 787 lb Na, 1213 lb liquid Cl2, 1370 lb NaOH, 1248 lb HCl
3.59. Compute the amount of zinc in a metric ton of ore containing 60.0% zincite, ZnO.
Ans. 482 kg Zn
3.60. How much phosphorus is contained in 5.00 g of the compound CaCO3 · 3Ca3(PO4)2? How much P2O5?
Ans. 0.902 g P, 2.07 g P2O5
3.61. A 10.00-g sample of crude ore contains 2.80 g HgS. What is the percentage of mercury in the ore?
Ans. 24.1% Hg
3.62. A procedure for analyzing oxalic acid content of a solution involves the formation of the insoluble complex Mo4O3(C2O4)3 · 12H2O. (a) How many grams of this complex would form per gram of oxalic acid, H2C2O4, if 1 mol of the complex results from the reaction with 3 mol of oxalic acid? (b) How many grams of molybdenum are contained in the complex formed by the reaction of 1 g of oxalic acid?
Ans. (a) 3.38 g complex; (b) 1.42 g Mo
3.63. An agricultural insecticide contains 18% arsenic. Express this as the percentage As2O5.
Ans. 28% As2O5
3.64. During autopsy, a small amount of a white powder was found inside the victim’s mouth. The analysis indicates that the molecule mass is in the range of 210 u and that the composition is 33.18% sodium, 74.92% arsenic, and the remainder oxygen. What are the formula and name of the compound?
Ans. Na3AsO4, sodium arsenate
3.65. Express the potassium content of a fertilizer in percent potassium if it is 6.8% K2O.
Ans. 5.6% potassium
3.66. A typical analysis of Pyrex™ glass showed 12.9% B2O3, 2.2% Al2O3, 3.8% Na2O, 0.4% K2O, and the balance was SiO2. What is the ratio of silicon to boron atoms in the glass?
Ans. 3.6
3.67. A piece of plumber’s solder weighing 3.00 g was dissolved in dilute nitric acid, then treated with dilute sulfuric acid. This precipitated the lead as PbSO4, which after washing and drying weighed 2.93 g. The solution was then neutralized to precipitate stannic acid, which was decomposed by heating, yielding 1.27 g SnO2. What is the analysis of the solder as percent lead and percent tin?
Ans. 66.7% Pb, 33.3% Sn
3.68. Two sources for copper can be purchased in pure form—CuSO4 and Cu(NO3)2. Suppose you can purchase either at $28/kg. (a) Which is the better deal when considered as a source for copper? (b) Suppose you had to purchase for 10,000 kg (22 tons) copper, but purchased the wrong copper source. How much money did you overspend?
Ans. (a) Cu(NO3)2; (b) $88,340 overspent (Does this come out of your pay?)
3.69. Determine the weight of sulfur required to make 1 metric ton of H2SO4, sulfuric acid.
Ans. 327 kg S
3.70. A sample of impure cuprite, Cu2O, contains 66.6% copper. What is the percentage of pure Cu2O in the sample?
Ans. 75.0% Cu2O
3.71. A cold cream sample weighing 8.41 g lost 5.83 g of moisture on heating to 110°C. The residue on extracting with water and drying lost 1.27 g of water-soluble glycerol (glycerin). The balance was oil. Calculate the composition of this cream.
Ans. 69.3% moisture, 15.1% glycerol, 15.6% oil
3.72. A household cement gave the following analytical data: A 28.5-g sample, on dilution with acetone, yielded a residue of 4.6 g aluminum powder. The filtrate, on evaporation of the acetone and solvent, yielded 3.2 g of plasticized nitrocellulose, which contained 0.8 g of benzene-soluble plasticizer. Determine the composition of this cement.
Ans. 16.2% Al, 72.6% solvent, 2.8% plasticizer, 8.4% nitrocellulose
3.73. A sample of coal contains 2.4% water. After drying, the moisture-free residue contains 71.0% carbon. Determine the percent carbon on the “wet basis.”
Ans. 69.3%
3.74. A certain breakfast food contains 0.637% salt (NaCl). Express this as milligrams of sodium per 60.0 gram serving.
Ans. 150 mg sodium
3.75. A liter flask contains a mixture of two liquids (A and B) of specific gravity 1.4. The specific gravity is the density of the sample relative to that of water. The specific gravity of liquid A is 0.8 and that of liquid B is 1.8. What volume of each must have been put into the flask? Assume the volumes are additive with no volume changes on mixing.
Ans. 400 mL of A and 600 mL of B
3.76. When a zinc sulfide ore, ZnS, is roasted, all the sulfur is released to the atmosphere as SO2. If a maximum of 0.060 mg of SO2 is permitted per cubic meter of air, (a) how many cubic meters of air will be needed for safe disposal of the effluent from the roasting of 1.00 metric tons of zinc sulfide, and (b) how large an area would be covered by such a volume of air if it were 1.00 km high?
Ans. (a) 1.10 × 1010m3; (b) 1.10 × 107 m2 (about 4.2 square miles!)
3.77. A taconite ore consisted of 35.0% Fe3O4 and the balance siliceous impurities. How many tons of the ore must be processed in order to recover a ton of metallic iron (a) if there is 100% recovery, and (b) if there is only 75% recovery?
Ans. (a) 3.94 tons; (b) 5.25 tons
3.78. A typical formulation for a cationic asphalt emulsion calls for 0.5% tallow amine emulsifier and 70% asphalt, the rest consisting of water and water-soluble ingredients. How much asphalt can be emulsified per pound of emulsifier?
Ans. 140 lb
3.79. Uranium hexafluoride, UF6, is used in the gaseous diffusion process for separating uranium isotopes since not all uranium isotopes can be undergo chain reaction, a requirement for use in reactors and nuclear weapons. How many kilograms of elementary uranium can be converted to UF6 per kilogram of combined fluorine?
Ans. 2.09 kg
3.80. One of the earliest methods for determining the molar mass of proteins was based on chemical analysis. A hemoglobin preparation produced from red blood cells separated from the blood by the use of a centrifuge was found to contain 0.335% iron. (a) If the hemoglobin molecule contains 1 atom of iron, what is its molar mass? (b) If it contains 4 atoms of iron, what is its molar mass?
Ans. (a) 16,700 u; (b) 66,700 u
3.81. A polymeric substance, tetrafluoroethylene, can be represented by the formula (C2F4)n, where n is a large number. The material was prepared by polymerizing C2F4 in the presence of a sulfur-bearing catalyst that served as a nucleus upon which the polymer grew. The final product was found to contain 0.012% S. What is the value of n if each polymeric molecule contains (a) 1 sulfur atom, (b) 2 sulfur atoms? In either case, assume that the catalyst contributes a negligible amount to the total mass of the polymer.
Ans. (a) 2700, (b) 5300
3.82. A peroxidase enzyme isolated from human red blood cells was found to contain 0.29% selenium. What is the minimum molecular mass of the enzyme?
Ans. 27,000 u
3.83. Nitroglycerin is an explosive and is also used to produce other explosives, such as dynamite. “Nitro” is 18.5% nitrogen. Calculate the molecular mass of nitro.
Ans. 227.1 u
3.84. A sample of polystyrene prepared by heating styrene with tribromobenzoyl peroxide in the absence of air has the formula Br3C6H3(C8H8)n. The number n varies with the conditions of preparation. One sample of polystyrene prepared in this manner was found to contain 10.46% bromine. What is the value of n?
Ans. 19
In all the following problems the nuclidic molecular mass reported is that of the species containing the most prevalent nuclide of each of its elements.
3.85. An alkaloid was extracted from the seed of a plant and purified. The molecule was known to contain 1 atom of nitrogen, no more than 4 atoms of oxygen, and no other elements besides carbon and hydrogen. The mass-spectrometrically determined nuclidic molecular mass was found to be 297.138. (a) How many molecular formulas are consistent with mass number 297 and with the other known facts except the precise molecular weight? (b) What is the probable molecular formula?
Ans. (a) 17; (b) C18O3NH19
3.86. An organic ester (an organic salt) was decomposed inside a mass spectrometer. An ionic decomposition product had the nuclidic molecular mass 117.090. What is the molecular formula of this product if it is known in advance that the only possible constituent elements are C, O, and H, and that no more than 4 oxygen atoms are present in the molecule?
Ans. C6 O2H13
3.87. An intermediate in the synthesis of a naturally occurring alkaloid had a mass-spectrometrically determined nuclidic molecular mass of 205.147. The compound is known to have no more than 1 nitrogen atom and no more than 2 oxygen atoms per molecule. (a) What is the most probable molecular formula of the compound? (b) What must the precision of the measurement be to exclude the next most probable formula?
Ans. (a) C13ONH19 (nuclidic molecular mass is 205.147). (b) The next closest molecular mass is 205.159 for C14OH21. The range of uncertainty in the experimental value should not exceed half the difference between 205.147 and 205.159, i.e., it should be less than 0.006, or about 1 part in 35,000.