An Introduction to Chemical Science Part 15

Vapor density is very easily determined from the formula by the method given above. But in practice the formula is obtained from the vapor density, and hence the method there given has to be reversed.

173. Vapor Density of Oxygen.--Suppose we were to obtain the vapor density of O. We should carefully seal and weigh a given volume, say a liter, at a noted temperature and barometric pressure, which are reducedto 0 degrees and 760 mm, and compare it with the weight of the same volume of H. This has been done repeatedly, and O has been found to weigh 16 times as much as H, volume for volume, or, more exactly, 15.96+. Now a liter of each gas has the same number of molecules, therefore the O molecule weighs 16 times the H molecule. The half-molecule of each has the same proportion, and the vapor density of O is 16. Atomic weight is obtained in a very different way.

PROBLEMS.

(1) A liter of Cl is found to weigh 3.195 g. Compute its vapor density, and explain fully.

(2) A liter of Hg vapor, under standard conditions, weighs 9 g.

Find its vapor density, and explain.

The vapor density of only a few elements has been satisfactorily determined. See page 12. Some cannot be vaporized; others can be, but only under conditions which prevent weighing them. The vapor density of very many compounds also is unknown.

(3) A liter of CO2 weighs 1.98 g. Find the vapor density, and from that the molecular weight, remembering that the latter is twice the former. See whether it corresponds to that obtained from the formula, CO2. This is,in fact, the way a formula is ascertained, if the atomic weights of its elements are known.

(4) A liter of a compound gas weighs 2.88 g. a.n.a.lysis shows that its weight is half S and half O. As the atomic weight of S is 32, and that of O is 16, what is the symbol for the gas?

Solution. Its molecular weight is 64, i.e. (2.88=0.09) X 2, of which 32 is S and 32 O. The atomic weight of S is 32, hence there is one atom of S, while of O there are two atoms. The formula is SO2.

(5) A liter of a compound gas, which is found to contain 1 C and 3 O by weight, weighs 1.26 g. What is its formula? Atomic weights are taken from page 12. Prove your answer.

(6) A liter of a compound of N and O weighs 1.98 g. The N is 7/11; and the O 4/11. What is the gas?

(7) A compound of N and H gas weighs 0.765 g. to the liter. The N is 14/17 of the whole, the H 3/17. What gas is it? CHAPTER x.x.xV.

ATOMIC WEIGHT.

174. Definition.--We have seen that the molecular weight of a compound, as well as of most elements, is obtained from the vapor density by doubling the latter. It remains to explain how atomic weights are obtained. The term is rather misleading. The atomic weight of an element is its least combining weight, the smallest portion that enters into chemical union, which is, of course, the weight of an atom.

175. Atomic Weight of Oxygen.--Suppose we wish to find the atomic weight of oxygen. We must find the smallest proportion by weight in which it occurs in any compound. This can only be done by a.n.a.lyzing all the compounds of O that can be vaporized. As ill.u.s.trative of these compounds take the six following:--

Wt. of other Names. V. d. Mol. Wt. Wt. of O. Elem. Symbol.

Carbon monoxide... 14 2816 12 ?

Carbon dioxide.... 22 4432 12 ?

Hydrogen monoxide... 9 1816 2 ?

Nitrogen monoxide... 22 4416 28 ?

Nitrogen trioxide... 38 7648 28 ?

Nitrogen pentoxide... 54 10880 28 ?

176. Molecular Symbols.--From the vapor density of the gases-- column 2--we obtain their molecular weight-- column 3. To find the proportion of O, it must be separated by chemical means from its compounds and separately weighed. These relative weights are given in column 4. Now the smallest weight of O which unites in any case is its atomic weight. If any compound of O should in future be found in which its combining weight is 8 or 4, that would be called its atomic weight. By dividing the numbers in column 4, wt. of O, by 16, the atomic weight of O, we obtain the number of O atoms in the molecule. Subtracting the weights of O from the molecular weights, we have the parts of the other elements, column 5, and dividing these by the atomic weight of the respective elements, we have the number of atoms of those elements, these last, combined with the number of O atoms, give the symbol. In this way complete the last column.

Show how to get the atomic weight of Cl from these compounds, arranging them in tabular form, and completing as above: HCl, KCl, NaCl, ZnCl2, MgCl2; the atomic weight of N in these: N2O, NO, NH3.

177. Molecular and Atomic Volumes.--We thus see that vapor density and atomic weight are obtained in two quite different ways. In the case of elements the two are usually identical, i.e.

with the few whose vapor density is known; but this is not always true, and it leads to interesting conclusions regarding atomic volume. In O both vapor density and atomic weight are 16. This gives 2 atoms of O to the molecule, i.e. the molecular weight / the atomic weight. The size of an O atom is therefore half the gaseous molecule, and is represented by one square. S has a vapor density and an atomic weight of 32 each. Compute the number of atoms in the molecule. Compute for I, in which the two are identical, 127. P has an atomic weight of 31, while its vapor density is 62. Its molecule must consist of 4 atoms, each half the size of the H atom, The vapor density of As is 150, the atomic weight 75. Compute the number of atoms in its molecule, and represent their relative size. Hg has an atomic weight of 200, a vapor density of 100. Compute as before, and compare the results with those on page 12. Ozone has an atomic weight of 16, a vapor density 24. Compute.

Chapter x.x.xVI.

DIFFUSION AND CONDENSATION OF GASES.

178. Diffusion of Gases.--Oxygen is 16 times as heavy as H. If the two gases were mixed, without combining, in a confined s.p.a.ce, it might be supposed that O would settle to the bottom and H rise to the top. This would, in fact, take place at first, but only for an instant, for all gases tend to diffuse or become intimately mixed. The lighter the gas the more quickly it diffuses.

179. Law of Diffusion of Gases.--The diffusibility of gases varies inversely as the square roots of their vapor densities.

Compare the diffusibility of H with that of O. dif. H:dif. O:: sqrt(16): sqrt(1), or dif: H: dif. O:: 4: 1.

That is to say, if H and O be set free from separate receivers in a room, the H will become intermingled with the atmosphere four times as quickly as the O. Compare the diffusibility of O and N; of Cl and H. Take the atomic weights of these, since they are the same as the vapor densities. In case of a compound gas, half the molecular weight must be taken for the vapor density; e.g. dif.

N20: dif. O.:: sqrt(16): sqrt(22).

180. Cause.--Diffusion is due to molecular motion; the lighter the gas the more rapid the vibration of its molecules. Compare the diffusibility of CO2 and that of Cl; of HCl and SO2; of HF and I.

181. Liquefaction and Solidification of Gases.--Water boils at 100 degrees, under standard pressure, though evaporating at all temperatures; it vaporizes at a lower point if the pressure be less, as on a mountain, and at a higher temperature if the pressure be greater, as at points below the sea level. Alcohol boils at 78 degrees, standard pressure, and every liquid has a point of temperature and pressure above which it must pa.s.s into the gaseous state. Likewise every gas has a critical temperature above which it cannot be liquefied at any pressure.

This condition was not recognized formerly, and before 1877, O, H, N, C4, CO, NO, etc., had not been liquefied, though put under a pressure of more than 2,000 atmospheres. They were called permanent gases. In 1877 Cailletet and Pictet liquefied and solidified these and others. The lowest temperature, about -225 degrees, was produced by suddenly releasing the pressure from solid N to 4mm, which caused it rapidly to evaporate.

Evaporation, especially under diminished pressure, always lowers the temperature by withdrawing heat.

These low degrees are indicated by a H thermometer, or if too low for that, by a "thermo-electric couple" of copper and German silver.

The pupil can easily liquefy SO, by pa.s.sing it through a U-tube which is surrounded by a mixture of ice and salt in a large receiver. At the meeting of the American a.s.sociation for the Advancement of Science in 1887, a solid brick of CO2 was seen and handled by the members, Liquid H is steel blue.

A few results obtained under a pressure of one atmosphere are:-- Boiling Points: C2H4--102 degrees; CH4--184 degrees; O--181 degrees; N --194 degrees; CO--190 degrees; NO--154 degrees; Air-- 191 degrees.

Solidifying Points: Cl -102 degrees; HCl -115 degrees; Ether -129 degrees; Alcohol -130 degrees.

Chapter x.x.xVII.

SULPHUR.

Examine brimstone, flowers of sulphur, pyrite, chalcopyrite, sphalerite, galenite, gypsum, barite.

182. Separation.

Experiment 103.--To a solution of 2 g. of sodium sulphide,, Na2S2 in 10 cc. H2O add 3 or 4cc. HCl, and look for a ppt. Filter, and examine the residue. It is lac sulphur, or milk of sulphur.

183. Crystals from Fusion.

Experiment 104.--In a beaker of 25 or 50 cc. capacity put 20 g.

brimstone. Place this over a flame with asbestos paper interposed, and melt it slowly. Note the color of the liquid, then let it cool, watching for crystals. When partly solidified pour the liquid portion into an evapo- rating-dish of water, and observe the crystals of S forming in the beaker (Fig. 42). The hard ma.s.s may be separated from the gla.s.s by a little HNO3 and a thin knife-blade, or by CS2.

184. Allotropy.

Experiment 105.--Place in a t.t. 15g of brimstone, then heat slowly till it melts. Notice the thin amber-colored liquid. The temperature is now a little above 100 degrees. As the heat increases, notice that it grows darker till it becomes black and so viscid that it cannot be poured out. It is now above 200 degrees. Still heat, and observe that it changes to a slightly lighter color, and is again a thin liquid. At this time it is above 300 degrees. Now pour a little into an evaporating dish containing water. Examine this, noticing that it can be stretched like rubber. Leave it in the water till it becomes hard. Continue heating thebrimstone in the t.t. till it boils at about 450 degrees, and note the color of the escaping vapor. Just above this point it takes fire. Cool the t.t., holding it in the light meantime, and look for a sublimate of S on the sides.

185. Solution.

Experiment 106.--Place in an evaporating-dish a gram of powdered brimstone, and add 5cc, CS2, carbon disulphide. Stir, and see whether S is dissolved. Put this in a draft of air, and note the evaporation of the liquid CS2, and the deposit of S crystals.

These crystals are different in form from those resulting from cooling from fusion.

186. Theory of Allotropy.--The last three experiments well ill.u.s.trate allotropy. We found S to crystallize in two different ways. Substances can crystallize in seven different systems, and usually a given substance is found in one of these systems only; e.g. galena is invariably cubical. An element having two such forms is said to be dimorphous. If it crystallizes in three systems, it is trimorphous. A crystal has a definite arrangement of its molecules. If without crystalline form, a substance is called amorphous. An ill.u.s.tration of amorphism was S after it had been poured into water. Thus S has at least three allotropic forms, and the gradations between these probably represent others. Allotropy seems to be due to varied molecular structure.