Avogadro’s hypothesis states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules. Such being the case, 1 L of oxygen contains the same number of molecules as 1 L of hydrogen (or any other gas). Also, 1 ft3 of helium contains the same number of molecules as 1 ft3 of nitrogen (or any other gas). Of course, these gases must be measured under the same conditions of temperature and pressure.
Avogadro’s hypothesis can be applied to the problem of relative masses of molecules of two gases and, if one is known, the other can be determined as is shown in Example 1.
EXAMPLE 1 At standard conditions, 1 L of oxygen weighs 1.43 g and 1 L of an unknown gas that we know contains carbon and oxygen (CxOy) weighs 1.25 g. By applying Avogadro’s hypothesis, we can state that 1 L of CxOy (S.T.P.) contains the same number of molecules as 1 L of O2 (S.T.P.). This further tells us that a molecule of CxOy must weigh 1.25/1.43 times as much as a molecule of oxygen. According to the Periodic Table, O2 weighs 32 grams per mole, so we can calculate the mass of CxOy.
It appears as though this compound, CxOy, is carbon monoxide (C + O = 12 + 16 = 28 g/mol). By using this technique, we are able to determine atomic mass, especially for the lighter elements. Even crude gas density experiments may be used along with chemical composition data and known atomic masses to establish molecular mass, which leads us to the molecular formula of a gas.
EXAMPLE 2 A hydride of silicon that has the empirical formula SiH3 (approximately 31 g/empirical formula) was found to have an approximate gas density at S.T.P. of 2.9 g/L. By comparison with oxygen, whose molecular mass and density are known, the molecular mass of the hydride is
This is an approximate molecular mass and might be off by as much as 10 percent. Regardless, the value is sufficiently accurate to tell us that the molecular formula is approximately two times the empirical formula, Si2H6, with a molecular mass of 62 g ruling out other multiples of the empirical formula.
If 1 mol of any gas has the same number of molecules, NA, as 1 mol of any other gas (Chapter 2), and if equal numbers of molecules occupy equal volumes at S.T.P. (Avogadro’s hypothesis), then 1 mol of any gas has the same volume at S.T.P. as 1 mol of any other gas. This standard molar volume has the value of 22.414 L.
As a caution to the reader: Of course, Avogadro’s hypotheses and the gas laws assume that gases are all ideal gases. The gases in the real world are not all ideal gases; the molar volume at S.T.P. is most often a little below the 22.414 L indicated here. In the remainder of this chapter, the rounded value 22.4 L/mol will be used for all gases and, if not so identified, all the gases are ideal.
Let us take a look at the combined gas law (Chapter 5) and substitute in standard conditions (S.T.P.) for 1 mol gas. We can use the subscript of zero to specifically indicate standard conditions.
This calculation provides us with the universal gas constant, R. If we were dealing with more than 1 mol of an ideal gas at S.T.P., the volume of the gas would be n times greater. The relationship can be expressed as PV/T = nR or, by rearrangement,
This relationship is the ideal gas law and your knowledge of the law and the value of R is critical. When SI units are used for P and V (pascals and cubic meters), then
must be used.
The mass, in grams, of gas present is given by
or by
Performing the substitution of the above terms provides us with an alternative form of the ideal gas law, which is
and, by further substitution,
For an ideal gas, V is proportional to n when P and T are fixed. This means that the awkward concept of volume percentage (or fraction) discussed in Problem 5.12 can now be replaced with mole percentage or mole fraction. An assumption is that each gas occupies the entire volume of the mixture, but is at its own partial pressure.
A chemical equation represents the relationship of the reactants and products through a numerical relationship expressed by the coefficients associated with the participants. The coefficients can be interpreted as telling us the number of molecules or moles of materials involved; but they also represent the volumes of those participants that are gases, assuming a constant temperature and pressure (T and P). An example of these relationships is as follows:
The interpretation of the relationships of water to the other substances is valid only if the water is also a gas (vapor) under the conditions of temperature and pressure specified. At S.T.P. the water would be condensed into a liquid (or may be solid) and its volume would be very small compared with its volume as a gas and could be ignored. So, at S.T.P. with water as a liquid or solid, the 7 volumes of reactants would only produce 2 volumes of products (water’s volume being negligible).
There are circumstances where weight is an important factor (Example 3), but the calculations involving gases may be in terms of volumes of gases involved. The conversion from volumes of gas to mass is done through the numbers of moles. The methods used in these problem solutions are as in Chapter 4 except that the numbers of moles converted to mass (g, lb, etc.) must be determined from the volume, temperature, and pressure of the gases.
EXAMPLE 3 Carbon dioxide can be removed from the recirculated air aboard a spaceship by passing it over lithium hydroxide.
Calculate the number of grams of LiOH consumed in the above reaction when 100 L of air containing 1.20% CO2 at 29°C and 776 torr are treated by being passed through the LiOH.
It is important to note that the calculations are tied to the properly written and balanced equation. The equation sets the relationships between participants that we need to complete the solution.
The ideal gas law can be derived solely from theoretical principles by making a few assumptions about the nature of gases and the meaning of temperature. The derivation can be found in any physical chemistry textbook.
The basic assumptions involved are as follows:
1. A gas consists of a very large number of molecules which are in a state of continual random motion. A molecule is extremely small and cannot even be seen. Pressure is due to the force of molecular collisions on a surface, such as the container walls or a diaphragm in a gauge used for measuring pressure.
2. Collisions between molecules or between a molecule and an inert surface are perfectly elastic. This means they bounce off each other with no loss of kinetic energy.
3. No forces are exerted between molecules except through collisions. Therefore, between collisions, a molecule travels in a straight line at constant speed.
4. The average kinetic energy per gas molecule, (
mu2)avg, is independent of the nature of the gas and is directly proportional to the temperature. In this case, the kinetic energy is defined by the standard physical formula in terms of molecule mass, m, and the molecular speed, u. This statement may be taken as a more exact definition of temperature than the qualitative one given in Chapter 1. A list of some common energy units is given in Table 6-1.
Table 6-1 Some Common Energy Units
(a) A mechanical argument based on Assumptions 1, 2, and 3 above shows that for a gas composed of N molecules,
(b) Using Assumption 4, the distribution of speeds over the assemblage of molecules is predicted in mathematical form known as the Maxwell-Boltzmann distribution. Figure 6-1 is the distribution for hydrogen and shows the plots of the fraction of molecules having velocities close to a given value, u, and as a function of u at two different temperatures. The most probable velocity, ump, is the maximum speed for that set of conditions. It is slightly smaller than the average velocity. Another velocity, slight larger than the uavg, is the root-mean-square speed, urms, which is defined as that speed for which a molecule’s kinetic energy would be equal to the average kinetic energy over the entire sample, which is represented by
Fig. 6-1
For the Maxwell-Boltzmann distribution, recalling that raising to the 1/2 power is taking the square root:
(c) From the results of (a) and (b), using N = nNA and NAm = M, we can derive the ideal gas law.
The agreement of this expression with the empirically determined (lab-determined) ideal gas law validates the important definition of temperature in Assumption 4 above.
(d) The frequency of molecular collisions with a particular area of the container wall is predicted by
Imagine a small hole in the container. The rate at which a gas passes through that hole (effuses) into a vacuum is exactly the rate at which the molecules are predicted to collide with a portion of the wall the size of the hole. This means that the equation directly above can be used to describe the effusion rates of two gases at the same pressure and temperature by the ratio
where M is the molar mass. This explains Graham’s law of effusion, which is the experimental finding that the rates of effusion of gases at equal pressures and temperatures are inversely proportional to the square roots (1/2 power) of their densities. This can also be stated in terms of gas densities because the density of a gas is proportional to the molar mass.
(e) The inverse relationship discussed in (d) also applies in the areas of diffusion, thermal conduction, and nonturbulent flow, although the theory is not as exact in these phenomena as with Graham’s law of effusion. A reason for the variation from the theoretical is that the consideration of the nature of intermolecular collisions must be made, which is a process involved in the understanding of some of the nonideal behavior of gases. Problem 6.68 looks at an important process in which one gas penetrates another on the way to equalization of concentrations throughout the mixture. Although the relationship is more accurate for effusion, the assumption will be made in this book that the relative diffusion rates of two gases are inversely proportional to the square roots of their molar masses.
6.1. Determine the approximate molar mass of a gas if 560 cm3 weighs 1.55 g at S.T.P.
Alternate Method
Of course, the second method is faster; but you have to have the data for S.T.P. and converting would definitely slow down the process.
6.2. At 18°C and 765 torr, 1.29 L of gas weighs 2.71 g. Calculate the approximate molar mass of the gas.
This problem is similar to 6.1 and is given to demonstrate a different approach. First, let us convert the gas data to moles.
Using this information (n = 0.0544 mol) and given the sample size (2.71 g), solve the problem.
Note that 
is used as the units of R. A commonly used alternative is L · atm · mol–1 · K–1.
6.3. Determine the volume occupied by 4.0 g of oxygen at S.T.P. (molar mass of O2 is 32).
The volume of 4.0 g O2 at S.T.P. = number of moles in 4.0 g O2 × standard molar volume.
6.4. What volume would 15.0 g Ar occupy at 90°C and 735 torr?
6.5. Compute the approximate density of methane, CH4, at 20°C and 5.00 atm. The molar mass of methane is 16.0.
By rearrangement,
Alternate Method
6.6. Dry air consists of approximately 21% O2, 78% N2, and 1% Ar by moles. (a) What is the average (i.e., “apparent”) molar mass of dry air? (b) Calculate the density of dry air at S.T.P.
(a) One mole of air contains 0.21 mol O2, 0.78 mol N2, and 0.01 molAr. By incorporating the molecular masses, we can determine the apparent mass of one mole of dry air.
(b) The mass of 22.4 L of a gas at S.T.P. (assuming dry air) is its molar mass.
6.7. An organic compound had the following analysis: C = 55.8%, H = 7.03%, and O = 37.2%. A 1.500-g sample was vaporized and was found to occupy 530 cm3 at 100°C and 740 torr. What is the molecular formula of the compound?
The approximate molar mass, calculated from the gas density data, is 89 g/mol. The empirical formula, calculated from the percentage composition data, is C2H3O with the empirical formula unit mass of 43.0. The exact molar mass must be (2)(43) = 86.0 g/mol since this is the only multiple of 43.0 (whole-number multiple) reasonably close to the approximate molecular formula of 89 g/mol. The molecule must be the equivalent of 2 empirical formulas: C4H6O.
Alternate Method
The percentage composition of each component, along with the molecular mass of 89 g/mol, can be used to determine the moles of atoms of each element in the compound.
These numbers approximate the numbers of atoms in the molecule with small deviations from values resulting from the approximate nature of the molar mass measurement. The molecular formula, C4H6O2, is obtained without going through the intermediate evaluation of an empirical formula.
6.8. Mercury diffusion pumps may be used in the lab to produce a high vacuum. Cold traps are generally placed between the pump and the system to be evacuated. The traps cause the condensation of mercury vapor, which prevents diffusion back into the system. The maximum pressure of mercury that can exist in the system is the vapor pressure of mercury at the temperature of the cold trap. Calculate the number of mercury vapor molecules per unit of volume in a cold trap maintained at –120°C. The vapor pressure of mercury at this temperature is 10–16 torr.
Molecules/L = (1.0 × 10–20 mol/L)(6.0 × 1023 molecules/mol) = 6 × 103 molecules/L or 6 molecules/mL
In each of the four problems in this section, all gases are measured at the same temperature and pressure.
6.9. (a) What volume of hydrogen will combine with 12 L of chlorine to form hydrogen chloride? (b) What volume of hydrogen chloride will be formed?
We must write a balanced equation for this problem, which is
(a) The equation shows that 1 molecule (or mole) H2 reacts with 1 molecule (or mole) Cl2 to form 2 molecules (or moles) HCl. By Avogadro’s hypothesis, equal numbers of molecules of gases under the same conditions of temperature and pressure occupy equal volumes. Therefore, the equation also indicates that 1 volume of H2 reacts with 1 volume of Cl2 to form 2 volumes of HCl.
(b) Since 12 L of Cl2 are used in the reaction and looking at the logic in part (a), 2 × 12 L = 24 L of HCl are formed.
6.10. (a) What volume of hydrogen will react with 6 ft3 of nitrogen to form ammonia? (b) What volume of ammonia will be produced?
(a) Applying Avogadro’s hypothesis as in 6.9(a), since 1 volume of nitrogen (6 ft3) requires 3 volumes of hydrogen for complete reaction, 3 volumes × 6 ft3/volume = 18 ft3 of hydrogen.
(b) One volume of nitrogen leads to the formation of 2 volumes of ammonia. Therefore, 6 ft3 of nitrogen gas will form 2 × 6 = 12 ft3 of ammonia gas.
6.11. Sixty-four liters of NO are mixed with 40 L of O2 and allowed to react. What is the total volume of gas present after completion of the reaction?
The equation tells us that the volume of oxygen required is only half that of NO (2NO:O2), (64) = 32 L O2 will be required and (40 – 32) O2 = 8L O2 will be in excess. The equation also tells us that the volume of NO2 formed is the same as that of the NO reacted (2NO:2NO2). The final volumes are 0 L NO + 8L O2 + 64 L NO2 = 72 L gas.
6.12. (a) What volume of O2 at S.T.P. is required for the complete combustion of 1 mol of carbon disulfide, CS2? (b) What volumes of CO2 and SO2 are produced (also at S.T.P.)?
(a) Under the conditions of S.T.P., 1 mole of gas takes up 22.4 liters. Since there are 3 moles of oxygen required to use up 1 mole of carbon disulfide,
For the oxygen used:(3 mol gas)(22.4 L/mol gas) = 67.2LO2.
(b) In the same manner as (a),
For the carbon dioxide produced: (1 mol gas)(22.4 L/mol gas) = 22.4LCO2
For the sulfur dioxide produced: (2 mol gas)(22.4 L/mol gas) = 44.8LSO2
6.13. How many liters of oxygen, at standard conditions, can be obtained from the heating of 100 g of potassium chlorate?
Molar Method
Note that the equation tells us that 2 mol KClO3 produce 3 mol O2. As in previous chapters, the symbol n represents the number of moles and M the molecular mass.
Alternate Method
The equations shows that 2 mol KClO3, 2(122.6) = 245.2 g, gives 3 mole volumes O2, 3(22.4) = 67.2 L. Then, at S.T.P.,
and
Note that neither method requires calculations involving the mass of oxygen formed.
6.14. What volume of oxygen can be obtained from the decomposition of 100 g KClO3? The conditions of 18°C and 750 torr are held constant in this experiment.
This problem is identical with Problem 6.13, except that the volume of oxygen must be recalculated using conditions that are not standard.
Alternate Method
When we are not concerned with S.T.P., we can calculate the volume directly from moles.
6.15. How many grams of zinc must react with sulfuric acid in order to obtain 500 cm3 (0.5 L) of hydrogen at 20°C and 770 torr?
Let us find the number of moles of hydrogen gas produced by using the ideal gas law.
We can now use dimensional analysis (factor-label method) to arrive at the answer to the problem using the mole ratio (1Zn:1H2) from the balanced equation.
6.16. A natural gas sample contains 84% CH4, 10% C2H6, 3% C3H8, and 3% N2, all determined by percent volume. If a series of catalytic reactions could be used for converting all the carbon atoms of the gas into butadiene, C4H6, with 100% efficiency, how much butadiene could be prepared from 100 g of the natural gas?
This problem starts out as an application of mole volume. If we had 100 moles of the mixture, then we would have 84 mol CH4, 10 mol C2H6, 3 mol C3H8, and 3 mol N2. We can calculate the amount of natural gas in 100 moles of the mixture by utilizing the molecular weights.
The number of moles of carbon in 100 mol mixture is 84(1) + 10(2) + 3(3) + 3(0) = 113 mol C. Since 4 mol C gives us 1 mol C4H6, 54 g, 113 mol C provides us with
Then
1860 g natural gas yields 1530 g C4H6
and
6.17. A reaction for the combustion of SO2 was prepared by opening a stopcock connecting two separate chambers. One of the chambers has a volume of 2.125 L and is filled at 0.750 atm with SO2. The second chamber has the volume of 1.500 L filled at 0.500 atm with O2. The temperature of both is 80°C. (a) What were the mole fraction of SO2 in the mixture, the total pressure, and the partial pressures? (b) If the mixture were to be passed over a catalyst that promoted the formation of SO3 and then was returned to the original two connected vessels, what were the mole fractions in the final mixture, and what was the final total pressure? Assume that the final temperature is 80°C and that the conversion of the SO2 is complete to the extent of the availability of O2.
(a)
Each mole fraction is evaluated by dividing the number of moles of the component by the total number of moles in the mixture. Let X be the symbol for mole fraction.
Note that the mole fraction is dimensionless (no units) and that the sum of the mole fractions is 1.
The total pressure before the reaction can be evaluated by using total volume (2.125 L + 1.500 L = 3.625 L) and the total number of moles, n = 0.0809 mol.
Knowing the total pressure, the calculation of the partial pressure of each substance (X) in the mixture is computed by multiplying the respective mole fractions (X) by the total pressure. This statement can be proven by
then
The sum of the partial pressures, of course, must equal the total pressure.
(b) The chemical reaction for the equation is required for this part of the problem.
2SO2 + O2 → 2SO3
All substances are gases under the experimental conditions. The number of moles of oxygen required is half the number of moles of sulfur dioxide; but this experiment had 0.0259 mol O2, which was less than half of the mol SO2 (0.0550 mol SO2). This means that the oxygen is a limiting reactant and there is an excess of sulfur dioxide. The calculation of the O2 is first, then that of the SO2 used.
The amount of product, SO3, will be the same amount as the SO2 used, according to the balanced equation. Therefore, the amount of SO3 produced will be 0.0518 mol, assuming a 100% efficient reaction. There would be 0.0032 mol SO2 in excess (0.0550 original – 0.0518 used).
The amounts of gas on completion of the reaction (100% efficiency assumed) are 0.0032 mol SO2, 0 mol O2, and 0.0518 mol SO3. On the basis of this information, we calculate the mole fractions of each.
The final total pressure is calculated on the basis of the total number of moles on completion of the reaction.
On the other hand, if we were to notice that the chemical equation tells us that the amount of SO3 produced is the same as the amount of SO2 used, then we can also realize that the final total pressure is the same as the initial partial pressure of SO2. This happens because the only gases present at the end are SO2 (excess amount) and SO3. Further, the amount of SO3 produced is exactly the amount of SO2 used; there is no change in the total amount of gas present before and after the reaction as far as they are concerned. The difference in the final pressure is due to using up the oxygen; this means that the total pressure will be dropped by the amount of oxygen used—all of it.
6.18. (a) Show how the value of R, 8.3145 
in 
can be derived from the value in SI units.
(b) Express R in terms of calories.
(a)
(b)
6.19. Calculate the root-mean-square velocity of H2 at 0°C.
6.20. Calculate the relative rates of effusion of H2 and CO2 through a fine pinhole.
6.21. Let us assume that octane (C8H18) burns explosively (gasoline certainly can) and that rubbing alcohol (C3H7OH) does, too. Which will explode more violently on the basis of gas expansion on complete burning?
Ans. Octane will produce much more gas than the alcohol (34:14 on the mole basis), producing a more violent explosion.
6.22. If 200 cm3 of a gas at S.T.P. weighs 0.268 g, what is its molar mass?
Ans. 30.0 g/mol
6.23. A design of a tank for holding nitrous oxide, N2O, for automobile racing requires the choice of a material that will support the stress of 1500 g stored in a volume of 7.5 liters. The temperature is not expected to exceed 125°C. Calculate the stress (atm) to which the tank may be exposed.
Ans. 218 atm
6.24. What is the volume of 16 g of nitrous oxide, N2O, at S.T.P.?
Ans. 8.1 L
6.25. What volume will 1.216 g of SO2 gas occupy at 18°C and 755 torr?
Ans. 456 cm3
6.26. A chemist had to determine the molar mass of a compound in the liquid state, but was concerned that it would decompose if heated. So he used a syringe to inject 0.436 grams of the liquid sample into a 5.00-L flask with argon at 17°C and connected an open-tube manometer. The liquid vaporized completely and the mercury in the manometer went from 16.7 mm to 52.4 mm. What is the molar mass of the compound?
Ans. 44.2 g/mol
6.27. Calculate the weight of 2.65 L of carbon dioxide gas at 34°C and 1.047 atm.
Ans. 4.84 g
6.28. Determine the density of H2S gas at 27°C and 2.00 atm.
Ans. 2.77 g/L
6.29. What is the mass of 1 mole of a gas whose density at 40°C and 785 torr is 1.286 kg/m3?
Ans. 32.0 g/mol
6.30. A large piece of magnesium, 20 kg, is accidentally dropped into a vat containing 500 liters of concentrated HCl solution. What will be the volume of hydrogen gas released (1 atm, 32°C)?
Ans. 2.1 × 104 LH2
6.31. Phosgene, COCl2, inhalation is suspected as a cause of death. Gas from the lungs was dried, the components separated and identified. A small sample of a puzzling component was placed in a 25-mL container which changed in weight from 18.6600 g to 18.7613 g. The temperature was stable at 24°C and the final pressure was 1 atm. Determine the molecular mass.
Ans. 99 u—this is close enough to phosgene’s mass of 98.9164 u to be a candidate.
6.32. A collapsed balloon and its load weighs 216 kg. Calculate how many cubic meters of hydrogen gas is required so that the balloon is sufficiently inflated to launch it from a mountain top at –12°C and at a pressure of 628 torr. The density of air on the mountain is 1.11 g/L.
Ans. 210 m3
6.33. One of the methods for estimating the temperature of the center of the sun is based on the ideal gas law. What is the calculated temperature if the sun’s center is assumed to consist of gases whose average molar mass is 2.0, the density is 1.4 × 103 kg/m3, and the pressure is 1.3 × 109 atm?
Ans. 2.3 × 107 K
6.34. A 100-cm3 experimental chamber was sealed off at a pressure of 1.2 × 10–5 torr and 27°C. What is the number of gas molecules remaining in the tube?
Ans. 3.9 × 1013 molecules
6.35. Photochemical smog contains NO2, the gas responsible for the “brown blanket” over larger cities. This gas can produce a dimer (two identical units chemically bonded) by the reaction, 2NO2 → N2O4. If the reaction of 750 g NO2 goes all the way to the right, what will be the pressure in a 10-L chamber at 42°C?
Ans. 21 atm
6.36. One of the important considerations of hydrogen as automotive fuel is its compactness. Compare the number of hydrogen atoms per cubic meter available in (a) hydrogen gas under a pressure of 14.0 MPa at 300 K; (b) liquid hydrogen at 20 K at a density of 70.0 kg/m3; (c) the solid compound DyCo3H5, which has a density of 8200 kg/m3 at 300 K, from which all of the hydrogen can be made available for combustion.
Ans. (a)0.68 × 1028 atoms/m3; (b)4.2 × 1028 atoms/m3; (c)7.2 × 1028 atoms/m3
6.37. A 28-L tank designed for storing compressed gas can hold no more than 15 atm pressure. (a) What number of moles of helium can that tank hold at standard temperature (0°C)? (b) What number of moles of helium can the tank hold at 140°F (60°C, the temperature of a car on a hot summer day)?
Ans. (a) 19 mol He at 0°C; (b) 12 mol He at 60°C
6.38. A plastic bottle is designed to hold 1 liter of carbonated beverage, but pressure is a problem. The air space above the beverage holds 100 mL volume and the bottle can support 7.5 atm of pressure. How much CO2 can be in that space at room temperature (25°C) in moles and grams?
Ans. 0.030 mol or 1.3 g CO2
6.39. An empty 1.5-ft3 steel gas tank and its valve weighed 125 lb. The tank was filled with oxygen to 2000 lbf/in2 absolute at 25°C. (a) What percent of the total weight of the full tank was O2? Assume the ideal gas law applies. (Note: 1 atm = 14.7 lb/in2.)
For parts (b), (c), and (d), note that pure oxygen gas is not necessarily the most compact source of oxygen for a confined fuel system because of the weight of the cylinder necessary to confine the gas. Other compact sources are hydrogen peroxide and lithium peroxide. The oxygen-yielding reactions are
(b) Rate 65% by weight solution of H2O2 in H2O, and (c) rate pure Li2O2 in terms of % of total weight which is “available” oxygen. Neglect the weights of the containers. (d) Compare with (a).
Ans. (a) 12%; (b) 31%; (c) 35%; (d) Both (b) and (c) are more efficient than (a) on a weight % basis.
6.40. A 0.331-g sample of hydrated ammonium carbonate was placed in an evacuated 30.0 cm3 pyrolysis tube rated at 45.0 atm. It was heated at 250°C until complete decomposition by
(a) Was there any liquid water in the tube? (b) Knowing that the vapor pressure of water at 250°C is 39.2 atm, did the tube explode?
Ans. (a) No, P(water) = 8.30 atm; (b) No, P(total) = 20.7 atm
6.41. An iron meteorite was analyzed for its isotopic argon constant. The amount of 36Ar was 0.200 mm3 (S.T.P.) per kilogram of meteorite. If each 36Ar atom had been formed by a single cosmic event, how many such events must there have been per kilogram of meteorite?
Ans. 5.4 × 1015
6.42. Three volatile compounds of a certain element have gaseous densities calculated back to S.T.P. as follows: 6.75, 9.56, and 10.08 kg/m3. The three compounds contain 96.0%, 33.9%, and 96.4% of the element in question, respectively. What is the element’s most probable atomic mass?
Ans. 72.6, although the data do not exclude 72.6/n, where n is a positive integer.
6.43. During short spaceflights, chemical absorbers can be used to remove space travelers’ exhaled CO2. Li2O is very efficient in terms of absorbing capacity per unit weight. What is the absorption efficiency of pure Li2O as stated in liters CO2 (S.T.P.) per kilogram Li2CO3 reacting as below?
Li2O + CO2 → Li2CO3
Ans. 752 L/kg
6.44. A mixture of methane, CH4, and ethane, C2H6, exerted a pressure of 2.33 atm when confined within a reaction vessel at a fixed pressure and temperature. It was mixed with an excess of oxygen and burned completely to CO2 and H2O. After removing all the H2O and excess oxygen, the CO2 was returned to the vessel where its pressure was 3.02 atm, when measured at the same pressure and temperature as the original mixture. Calculate the mole fractions of the gases in the original mixture.
Ans. X(CH4) = 0.704, X(C2H6) = 0.296
6.45. Exactly 500 cm3 of a gas at S.T.P. weighs 0.581 g. The gas composition is C = 92.24% and H = 7.76%. Derive the molecular formula of the gas.
Ans. C2H2
6.46. A hydrocarbon has the composition of 82.66% carbon and 17.34% hydrogen. The density of the vapor is 0.2308 g/L at 30°C and 75 torr. Determine its molar mass and its molecular formula.
Ans. 58.2 g/mol, C4H10
6.47. How many grams of O2 are contained in 10.5 L of oxygen measured over water at 25°C and 740 torr? The vapor pressure of water at 25°C is 24 torr.
Ans. 12.9 g
6.48. An empty flask open to the air weighed 24.173 g. The flask was filled with the vapor of an organic liquid and was sealed off at 100°C. At room temperature, the flask then weighed 25.002 g. The flask was then opened and filled with water at room temperature, after which it weighed 176 g. The barometric reading was 725 mm Hg. All weighings were taken at room temperature, which was 25°C. What is the molar mass of the organic vapor? Allow for the buoyancy of the air in the weighing of the sealed-off flask using 1.18 g/L for the density of air at 25°C and 1 atm.
Ans. 213 g/mol
6.49. A student decides that the proverbial “lead balloon” can be built and proceeds to the design stage. He assumes a spherical shape of radius r and a skin weight of 5.00 g/cm2. The balloon is to be filled with helium at 25°C and 711 torr. At these conditions the density of air is 1.10 g/L. For a sphere of radius r, the surface (skin) area is 4πr2 and the volume is 4πr3/3. Calculate how large the radius must be to get the balloon off the ground and the final weight.
Ans. r = 158 m, weight = 1.82 × 107 kg, including the skin and the helium
6.50. A 50-cm3 sample of a hydrogen-oxygen mixture was placed in a gas buret at 18°C and confined at ambient pressure. A spark was passed through the sample so that the formation of water could go to completion. The resulting pure gas had a volume of 10 cm3 at ambient pressure. What was the initial mole fraction of hydrogen in the mixture (a)if the residual gas after sparking was hydrogen? (b) if the residual gas was oxygen?
Ans. (a) 0.73; (b) 0.53
6.51. How much water vapor is contained in a cubic room 4.0 m along an edge if the relative humidity is 50% and the temperature is 27°C? The vapor pressure of water at 27°C is 26.7 torr. The relative humidity expresses the partial pressure of water as a percentage of the water vapor pressure.
Ans. 0.82 kg
6.52. A batch of wet clothes in a dryer contains 0.983 kg water. Assuming the air leaves the dryer saturated with water vapor at 48°C and 738 torr total pressure, calculate the volume of dry air at 24°C and 738 torr required to dry the clothes. The vapor pressure of water at 48°C is 83.7 torr.
Ans. 1.070 × 104 L
6.53. Ethane gas, C2H6, burns in air as indicated by the equation 2C2H6 + 7O2 → 4CO2 + 6H2O. Determine the number of (a) moles CO2 and of H2O forms when 1 mol of C2H6 burned, (b) liters of O2 required to burn 1 L of C2H6, (c) liters of CO2 formed when 25 L of C2H6 are burned, (d) liters (S.T.P.) of CO2 formed when 1 mole C2H6 is burned, (e) moles CO2 formed when 25 L (S.T.P.) of C2H6 are burned, and (f) grams of CO2 formed when 25 L (S.T.P.) of C2H6 are burned.
Ans. (a) 2 mol CO2, 3 mol H2O; (b) 3.5 L; (c) 50L; (d) 44.8 L; (e) 2.23 mol; (f) 98.2 g
6.54. Pure nitrogen can be prepared by the decomposition in aqueous solution of ammonium nitrite.
If 56.0 mL of N2 were collected over water at 42°C and a total pressure of 778 torr, what mass of NH4NO2 must have been decomposed? The vapor pressure water at 42°C is 61 torr.
Ans. 0.131 g
6.55. CO2 can be collected over water; however, since CO2 is very soluble in water and reacts with it, the water must be saturated with CO2. A volume of 372 mL gas are collected over CO2-saturated water. The conditions are 1.001 atm and 23.40°C. The CO2 is obtained by the reaction below. What mass of CaCO3 decomposed? (Vapor pressure of water at 23.40°C is 21.58mmHg.)
Ans. 1.50 g CaCO3
6.56. Lithium reacts with hydrogen to produce the hydride, LiH. Sometimes the product is contaminated with unreacted lithium metal. The extent of the contamination can be measured by measuring the amount of hydrogen gas generated by reacting a sample with water.
A 0.205-g sample of contaminated LiH yielded 561 mL of gas measured over water at 22°C and a total pressure of 731 torr. Calculate the percent by weight of lithium metal in the sample. The vapor pressure of water at 22°C is 20 torr.
Ans. 37%
6.57. A mixture of 5.00 g of water, 5.00 g of methanol (CH3OH), and 5.00 g of ethanol (C2H5OH) was put into a large sealed vessel that had been previously evacuated. The vessel was heated until all traces of liquid had evaporated. If the total pressure was 2.57 atm, what was the partial pressure of the ethanol?
Ans. 0.514 atm
6.58. Fifty grams of aluminum are to be treated with a 10% excess of H2SO4.
(a) What volume of concentrated sulfuric acid (density of 1.80 g/cm3) containing 96.5% H2SO4 by weight must be taken? (b) What volume of hydrogen would be collected over water at 20°C and 785 torr? The vapor pressure of water at 20°C is 17.5 torr.
Ans. (a) 173 cm3; (b) 66.2 L
6.59. A 0.750-g sample of solid benzoic acid, C7H6O2, was placed in a 0.500-L pressurized reaction vessel filled with O2 at 10.0 atm pressure and 25°C. The benzoic acid was burned completely to water and CO2. What were the final mole fractions of CO2 and H2O vapor in the resulting gas mixture brought to the initial temperature? The vapor pressure of water at 25°C is 23.8 torr. Neglect both the volume occupied by nongaseous substances and the solubility of CO2 in water. The water pressure in the gas phase cannot exceed the vapor pressure of water, so most of the water is condensed to the liquid.
Ans. X(CO2) = 0.213, X(H2O) = 0.0032
6.60. Two gases in adjoining vessels were brought into contact by opening a stopcock between them. One vessel measured 0.250 L and contained NO at 800 torr and 220 K; the other measured 0.100 L and contained O2 at 600 torr and 220 K. The reaction to form N2O4 as a solid exhausted the limiting reactant completely. (a) Neglecting the vapor pressure of N2O4, what was the pressure and composition of the gas remaining at 220 K after completion of the reaction? (b) What weight of N2O4 was formed?
Ans. (a) 229 torr NO; (b) 0.402 g N2O4
6.61. The industrial production of ammonia by use of natural gas feedstock can be represented by the following simplified set of reactions:
By assuming (i) that only the above reactions take place, plus the chemical absorption of CO2, (ii) that natural gas consists of only CH4, (iii) that air consists of 0.80 mole fraction N2 and 0.20 mole fraction O2, and (iv) that the ratio of conversion of CH4 by processes (1) and (2) is controlled through admitting oxygen for reaction (2) by adding just enough air so that the mole ratio of N2 to H2 is exactly 1:3, consider the overall efficiency of a process in which 1200 m3 (S.T.P.) of natural gas yields 1.00 metric tons of NH3. (a) How many moles of NH3 would be formed from each mole of natural gas if there were to be the complete conversion of natural gas subject to the stated assumptions? (b) What percentage of the maximum yield calculated in (a) is the actual yield?
Ans. (a) 2.29 mol NH3/mol CH4; (b) 48% yield
6.62. Calculate the ratio of (a) urms to ump and (b) uavg to ump.
Ans. (a) 1.22; (b) 1.13
6.63. At what temperature do N2 molecules have the same average speed as He atoms at 300 K?
Ans. 2100 K
6.64. At what temperature would the most probable speed of CO molecules be twice that at 0°C?
Ans. 819°C
6.65. Two gases were introduced into a 3-meter glass tube, NH3 from the right and an unknown from the left. When the two met, a white smoke appeared and deposited inside the tubing a distance of 2.1 from the right end. (a) What is the molecular mass of the unknown? (b) Make a proposal as to the unknown’s formula.
Ans. (a) 36 g/mol; (b) The mass is very close to that of HCl, 36.46
6.66. Two toxic gases are placed in an argon-filled tube. The relative rates for the travel of the tube are 0.80 for #1 and 1.0 for #2. The first gas is known to be hydrogen bromide, HBr, and the second is suspected to be diborane, B2H6, cyanogen chloride, ClCN, or cyanogen, (CN)2. Which is the second gas most likely to be?
Ans. Cyanogen, (CN)2, is the closest with cyanogen chloride, ClCN, at 0.87:1.
6.67. What is the ratio of the average molecular kinetic energy of UF6 to that of He, both at 300 K?
Ans. 1.000
6.68. What is the kinetic energy of a mole of CO2 at 400 K (a) in kJ and (b) in kcal?
Ans. (a) 4.99 kJ; (b) 1.192 kcal (Cal)
6.69. Uranium isotopes have been separated by taking advantage of the different rates of diffusion of the two isotopic forms of UF6. One form contains uranium of atomic mass 238 and the other of atomic mass 235. What are the relative rates of diffusion of these two molecules if Graham’s law applies?
Ans. UF6 with 235U is faster by a factor of 1.004.
6.70. A gas sample is suspected to be HCN, H2S, or CO. The sample is mixed with N2; the mixture is placed at one end of a tube containing only argon. It is determined that nitrogen arrives at the other end of the tube 1.1 times faster than the unknown. Which is the most likely formula for the unknown gas?
Ans. H2S
6.71. The pressure of a vessel that contained pure oxygen dropped from 2000 torr to 1500 torr in 47 minutes as the oxygen leaked through a small hole into a vacuum. When the same vessel was filled with another gas, the pressure dropped from 2000 torr to 1500 torr in 74 minutes. What is the molar mass of the second gas?
Ans. 79 g/mol
6.72. A large cylinder of helium filled at 2000 lbf/in2 has a small thin orifice through which helium escaped into an evacuated space at the rate of 3.4 millimoles per hour. How long would it take for 10 millimoles of CO to leak through a similar orifice if the CO were confined at the same pressure?
Ans. 7.8 hours
6.73. Calculate the average speed of a nitrogen molecule in air at room temperature (25°C). Hint: If you use SI units for R and for the molar mass, u will come out in m/s.
Ans. 475 m/s
6.74. How does the average speed of a nitrogen molecule in air compare to that of an oxygen molecule in air? (Air is approximately 20% oxygen and 80% nitrogen by moles.)
Ans. The N2 molecules move at 1.07 times the speed of the O2 molecules (7% faster).
6.75. You learned that in a mixture each gas exerted a partial pressure in proportion to its mole fraction. Yet, in a box of air the average N2 molecule, because of its greater speed, hits the walls more frequently than the average O2 molecule. This suggests that nitrogen might contribute more than 80% of the pressure. Explain the apparent discrepancy.
Ans. Although the average O2 molecule collides with the wall less frequently (by a factor of 1/1.07), it carries a momentum (mu) which is 1.07 times greater than that of the average N2 molecule (see Problem 6.74). The force of an impact depends on momentum.