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gives rise to diffusion. See page 114.

10. We Know Nothing Definite of the Form of Molecules. - In this
book they will always be represented as of the same size, that of
two squares. A molecule is itself composed of atoms, - from two to
several hundred. The size of the atom of most elements we
represent by one square.11. Atoms. - If the gaseous molecules be
of the same size, it is clear that either the atoms themselves
must be condensed, or the spaces between them must be smaller
than before. We suppose the latter to be the case, and that they
do not touch one another, the same thing being true of molecules.
Atoms composing sugar must be crowded nearer together than those
of salt. These atoms are probably in constant motion in the
molecule, as the latter is in the mass. If we regard this square
as a mass of matter, the dots may represent molecules; if we call
it a molecule, the dots may be called atoms, though many
molecules have no more than two or three atoms.

The following experiments illustrate the union of atoms to form
molecules, and of elements to form compounds.

12. Union of Atoms.

Experiment 6. - Mix, on a paper, 5 g. of iron turnings, and the
same bulk of powdered sulphur, and transfer them to an ignition
tube, a tube of hard glass for withstanding high temperatures.
Hold the tube in the flame of a burner till the contents have
become red-hot. After a minute break it by holding it under a jet
of water. Put the contents into an evaporating-dish, and look for
any uncombined iron or sulphur. Both iron and sulphur are
elements. Is this an example of synthesis or of analysis? Why? Is
the chemical union between masses of iron and sulphur, or between
molecules, or between atoms? Is the product a compound, an
element, or a mixture?

Experiment 7. - Try the same experiment, using copper instead of
iron. The full explanation of these experiments is given on page
13.

CHAPTER IV.

ELEMENTS AND BINARIES.

13. About Seventy Different Elements are now recognized, half of
which have been discovered within little more than a century.
These differ from one another in (1) atomic weight, (2) physical
and chemical properties, (3) mode of occurrence, etc. Page 12
contains the most important elements.

The symbol of an element is usually the initial letter or letters
of its Latin name, and stands for one atom of the element. C is
the symbol for carbon, and represents one atom of it. O means one
atom of oxygen.[The symbols of elements will also be used in this
book to stand for an indefinite quantity of them; e.g. O will be
used for oxygen in general as well as for one atom. The text will
readily decide when symbols have a definite meaning, and when
they are used in place of words.] Write, explain, and memorize
the symbols of the elements in heavy type.

14. The Atomic Weight of an element is the weight of its atom
compared with that of hydrogen. H is taken as the standard
because it has the least atomic weight. The atomic weight of O is
16, which means that its atom weighs 16 times as much as the H
atom. Every symbol, then, stands for a definite weight of the
element, i.e. its atomic weight, as well as for its atom.

How much bromine by weight does Br stand for? What do these
symbols mean - As, Na, N, P? If O represents one atom, how much
does O2 or 2 O stand for? How much by weight? Most elements have
two atoms in the molecule. How many molecules in 6 H? 10 N? S8?
I20?

The symbol of a compound is formed by writing in succession the
symbols of the elements of which it is composed. How many atoms
in the following molecules, and how many of each element: C2H60?
HNO3? PbSO4? MgCl2? (Hg2(NO3)2?)

15. The Simplest Compounds are Binaries. - A binary is a substance
composed of two elements; e.g. common salt, which is a compound
of sodium and chlorine. Its symbol is NaCl, its chemical name
sodium chloride. The ending ide is applied to the last name of
binaries. How many parts by weight of Na and of Cl in NaCl? What
is the molecular weight, i.e. the weight of its molecule? Name
KCl. How many atoms in its molecule? Parts by weight of each
element? Molecular weight? Does the symbol stand for more than
one molecule? How many molecules in 4 NaCl? How many atoms of Na
and of Cl? Name these: HCl, NaBr, NaI, KBr, AgCl, AgI, HBr, HI,
HF, HgO, ZnO, ZnS, MgO, CaO. Compute the proportion by weight of
each element in the last three.

A coefficient before the symbol of a compound includes all the
elements of the symbol, and shows the number of molecules. How
many in these: 6 KBr? 3 Sn0? 12 NaCl? How many atoms of each
element in the above?

An exponent, always written below, applies only to the element
after which it is written, and shows the number of atoms. Explain
these: AuCl3, ZnCl2, Hg2Cl2.

Write symbols for four molecules of sodium bromide, one of silver
iodide (always omit coefficient one), eight of potassium bromide,
ten of hydrogen chloride; also for one molecule of each of these:
hydrogen fluoride, potassium iodide, silver chloride.

In all the above cases the elements have united atom for atom.
Some elements will not so unite. In CaCl2 how many atoms of each
element? Parts by weight of each? Give molecular weight. Is the
size of the molecule thereby changed? Name these, give the number
of atoms of each element in the molecule, and the proportion by
weight, also their molecular weights: AuCl3, ZnCl2, MnCl2, Na2O,
K2S, H3P, H4C.

Principal Elements.
Name. Sym. At. Wt. Valence. Vap.D. At.Vol. Mol.Vol. State.
Aluminium Al 27. II, IV ... ... ... Solid
Antimony Sb 120. III, V. ... ... ... "
Arsenic As 75. III, V 150. "
Barium Ba 137. II ... ... ... "
Bismuth Bi 210. III, V ... ... ... "
Boron B 11. III ... ... ... "
Bromine Br 80. I, (V) 80. Liquid
Cadmium Cd 112. II 56. Solid
Calcium Ca 40. II ... ... ... "
Carbon C 12. (II), IV ... ... ... "
Chlorine Cl 35.5 I, (V) 35.5 Gas
Chromium Cr 52. (II),IV,VI ... ... ... Solid
Cobalt Co 59. II, IV ... ... ... Gas
Copper Cu 63. I, II ... ... ... "
Fluorine F 19. I, (V) ... ... ... Gas
Gold Au 196. (I), III ... ... ... Solid
Hydrogen H 1. I 1. Gas
Iodine I 127. I, (V) 127. ... ... Solid
Iron Fe 56. II,IV,(VI) ... ... ... "
Lead Pb 206. II, IV ... ... ... "
Lithium Li 7. I ... ... ... "
Magnesium Mg 24. II ... ... ... "
Manganese Mn 55. II, IV, VI ... ... ... "
Mercury Hg 200. I, II 100. Liquid
Nickel Ni 59. II, IV ... ... ... Solid
Nitrogen N 14. (I),III,V 14. Gas
Oxygen O 16. II 16. "
Phosphorus P 31. (I),III, V 62. Solid
Platinum Pt 197. (II), IV ... ... ... "
Potassium K 39. I ... ... ... "
Silicon Si 28. IV ... ... ... "
Silver Ag 108. I ... ... ... "
Sodium Na 23. I ... ... ... "
Strontium Sr 87. II ... ... ... "
Sulphur S 32. II,IV,(VI) 32(96) "
Tin Sn 118. II, IV ... ... ... "
Zinc Zn 65. II 32.5 "

If more than one atom of an element enters into the composition
of a binary, a prefix is often used to denote the number. SO2 is
called sulphur dioxide, to distinguish it from SO3, sulphur
trioxide. Name these: CO2, SiO2, MnO2. The prefixes are: mono or
proto, one; di or bi, two; tri or ter, three; tetra, four; pente,
five; hex, six; etc. Diarsenic pentoxide is written, As2O5.
Symbolize these: carbon protoxide, diphosphorus pentoxide,
diphosphorus trioxide, iron disulphide, iron protosulphide. Often
only the prefix of the last name is used.

16. An Oxide is a Compound of Oxygen and Some Other Element, as
HgO. What is a chloride? Define sulphide, phosphide, arsenide,
carbide, bromide, iodide, fluoride.

In Experiment 6, where S and Fe united, the symbol of the product
was FeS. Name it. How many parts by weight of each element? What
is its molecular weight? To produce FeS a chemical union took
place between each atom of the Fe and of the S. We may express
this reaction, i.e. chemical action, by an equation: -


Iron + Sulphur = Iron Sulphide
Or, using symbols Fe + S = FeS
Using atomic weights, 56 32 = 88.


These equations are explained by saying that 56 parts by weight
of iron unite chemically with 32 parts by weight of sulphur to
produce 88 parts by weight of iron sulphide. This, then,
indicates the proportion of each element which combines, and
which should be taken for the experiment. If 56 g. of Fe be used,
32 g. of S should be taken. If we use more than 56 parts of Fe
with 32 of S, will it all combine? If more than 32 of S with 56
of Fe? There is found to be a definite quantity of each element
in every chemical compound. Symbols would have no meaning if this
were not so.

Write and explain the equation for the experiment with copper and
sulphur, using names, symbols, and weights, as above.

CHAPTER V.

MANIPULATION.

17. To Break Glass Tubing.

Experiment 8. - Lay the tubing on a flat surface, and draw a sharp
three-cornered file two or three times at right angles across it
where it is to be broken, till a scratch is made. Take the tube
in the hands, having the two thumbs nearly opposite the scratch,
and the fingers on the other side. Press outward quickly with the
thumbs, and at the same time pull the hands strongly apart, and
the tubing should break squarely at the scratch.

To break large tubing, or cut off bottles, lamp chimneys, etc.,
first make a scratch as before; then heat the handle of a file,
or a blunt iron - in a blast-lamp flame by preference - till it is
red-hot, and at once press it against the scratch till the glass
begins to crack. The fracture can be led in any direction by
keeping the iron just in front of it. Re-heat the iron as often
as necessary.

18. To Make Ignition-Tubes.

Experiment 9. - Hold the glass tubing between the thumb and
forefinger of each hand, resting it against the second finger.
Heat it in the upper flame, slowly at first, then strongly, but
heat only a very small portion in length, and keep it in constant
rotation with the right hand. Hold it steadily, and avoid
twisting it as the glass softens. The yielding is detected by the
yellow flame above the glass and by an uneven pressure on the
hands. Pull it a little as it yields, then heat a part just at
one side of the most softened portion. Rotate constantly without
twisting, and soon it can be separated into two closed tubes. No
thread should be attached; but if there be one, it can be broken
off and the end welded. The bottom can be made more symmetrical
by heating it red-hot, then blowing, gradually, into the open
end, this being inserted in the mouth. The parts should be
annealed by holding above the flame for a short time, to cool
slowly.

For hard glass - Bohemian - or large tubes, the blast-lamp or
blowpipe is needed. In the blast-lamp air is forced out with
illuminating gas. This gives a high degree of heat. Bulbs can be
made in the same way as ignition-tubes, and thistle-tubes are
made by blowing out the end of a heated bulb, and rounding it
with charcoal.

19. To Bend Glass Tubing.

Experiment 10. - Hold the tube in the upper flame. Rotate it so as
to heat all parts equally, and let the flame spread over 3 or 4
cm. in length. When the glass begins to yield, without removing
from the flame slowly bend it as desired. Avoid twisting, and be
sure to have all parts in the same plane; also avoid bending too
quickly, if you would have a well-rounded joint. Anneal each bend
as made. Heated glass of any kind should never be brought in
contact with a cool body. For making O, H, etc., a glass tube -
delivery-tube - 50 cm. long should have three bends, as in Figure
6. The pupil should first experiment with short pieces of glass,
10 or 15 cm. long. An ordinary gas flame is the best for bending
glass.

20. To Cut Glass.

Experiment 11. - Lay the glass plate on a flat surface, and draw a
steel glass-cutter - revolving wheel - over it, holding this
against a ruler for a guide, and pressing down hard enough to
scratch the glass. Then break it by holding between the thumb and
fingers, having the thumbs on the side opposite to the scratch,
and pressing them outward while bending the ends of the glass
inward. The break will follow the scratch.

Holes can be bored through glass and bottles with a broken end of
a round file kept wet with a solution of camphor in oil of
turpentine.

21. To Perforate Corks.

Experiment 12. - First make a small hole in the cork with the
pointed handle of a round - rat-tail - file. Have the hole
perpendicular to the surface of the cork. This can be done by
holding the cork in the left hand and pressing against the larger
surface, or upper part, of the cork, with the file in the right
hand. Only a mere opening is made in this way, which must be
enlarged by the other end of the file. A second or third file of
larger size may be employed, according to the size of the hole to
be made, which must be a little smaller than the tube it is to
receive, and perfectly round.

CHAPTER VI.

OXYGEN.

22. To Obtain Oxygen.

Experiment 13. - Take 5 g. of crystals of potassium chlorate
(KClO3) and, without pulverizing, mix with the same weight of
pure powdered manganese dioxide (MnO2). Put the mixture into a
t.t., and insert a d.t. - delivery-tube - having the cork fit
tightly. Hang it on a r.s. - ring-stand, - as in Figure 7, having
the other end of the d.t.

(Fig 7.)

under the shelf, in a pneumatic trough, filled with water just
above the shelf. Fill three or more receivers - wide-mouthed
bottles - with water, cover the mouth of each with a glass plate,
invert it with its mouth under water, and put it on the shelf of
the trough, removing the plate. No air should be in the bottles.
Have the end of the d.t. so that the gas will rise through the
orifice. Hold a lighted lamp in the hand, and bring the flame
against the mixture in the t.t. Keep

the lamp slightly in motion, with the hand, so as not to break
the t.t. by over-heating in one place. Heat the mixture strongly,
if necessary. The upper part of the t.t. is filled with air:
allow this to escape for a few seconds; then move a receiver over
the orifice, and fill it with gas. As soon as the lamp is taken
away, remove the d.t. from the water. The gas contracts, on
cooling, and if not removed, water will be drawn over, and the
t.t. will be broken. Let the t.t. hang on the r.s. till cool.

With glass plates take out the receivers, leaving them covered,
mouth upward (Fig. 8), with little or no water inside. When cool,
the t.t. may be cleaned with water, by covering its mouth with
the thumb or hand, and shaking it vigorously.

What elements, and how many, in KClO3? In Mn02? It is evident
that each of these compounds contains O. Why, then, could we not
have taken either separately, instead of mixing the two? This
could have been done at a sufficiently high temperature. Mu02
requires a much higher temperature for dissociation, i.e.
separation into its elements, than KClO3, while a mixture of the
two causes O to come off from KClO3 at a lower temperature than
if alone. It is not known that Mn02 suffers any change.

Each molecule of potassium chlorate undergoes the following
change: -


Potassium Chlorate = Potassium Chloride + Oxygen
KClO3 = KCl + 3 O.


Is this analysis or synthesis? Complete the equation, by using
weights, and explain it. Notice whether the right- hand member of
the equation has the same number of atoms as the left. Has
anything been lost or gained? What element has heat separated?
Does the experiment show whether O is very soluble in water? How
many grams of O are obtainable from 122.58 g. KCIO3? PROPERTIES.

23. Combustion of Carbon.

OXYGEN Experiment 14. - Examine the gas in one of the receivers.
Put a lighted splinter into the receiver, sliding along the glass
cover. Remove it, blow it out, and put in again while glowing. Is
it re-kindled? Repeat till it will no longer burn. Is the gas a
supporter of combustion? How did the combustion compare with that
in air? Is it probable that air is pure O? Why did the flame at
last go out? Has the O been destroyed, or chemically united with
something else?

Wood is in part C. CO2 is formed by the combustion; name it. The
equation is C + 2O = CO2. Affix the names and weights. Is CO2 a
supporter of combustion? Note that when C is burned with plenty
of O, CO2 is always formed, and that no matter how great the
conflagration, the union is atom by atom. Combustion, as here
shown, is only a rapid union of O with some other substance, as C
or H.

24. Combustion of Sulphur.

Experiment 15. - Hollow out one end of a piece of electric-light
pencil, or of crayon, 3 cm. long, and attach it to a Cu wire
(Fig. 9). Put into this a piece of S as large as a pea, ignite it
by holding in the flame, and then hold it in a receiver of O.
Note the color and brightness of the flame, and compare with the
same in the air. Also note the color and odor of the product. The
new gas is SO2. Name it, and write the equation for its
production from S and O. How do you almost daily perform a
similar experiment? Is the product a supporter of combustion?

25. Combustion of Phosphorus.

Experiment 16. - With forceps, which should always be used in
handling this element, put a bit of P, half as large as the S
above,into the crayon, called a deflagrating-spoon. Heat another
wire, touch it to the P, and at once lower the latter into a
receiver of O. Notice the combustion, the color of the flame and
of the product. After removing, be sure to burn every bit of P by
holding it in a flame, as it is liable to take fire if left. The
product of the combustion is a union of what two elements? Is it
an oxide? Its symbol is P2O5. Write the equation, using symbols,
names, and weights. Towards the close of the experiment, when the
O is nearly all combined, P2O3 is formed, as it is also when P
oxidizes at a low temperature. Name it and write the equation.

26. Combustion of Iron.

Experiment 17. - Take in the forceps a piece of iron picture-cord
wire 6 or 8cm long, hold one end in the flame for an instant,
then dip it into some S. Enough S will adhere to be set on fire
by holding it in the flame again. Then at once dip it into a
receiver of O with a little water in the bottom. The iron will
burn with scintillations. Is this analysis or synthesis? What
elements combine? A watch-spring, heated to take out the temper,
may be used, but picture-wire is better.

The product is Fe3O4. Write the equation. How much Fe by weight
in the formula? How much O? What per cent by weight of Fe in the
compound? Multiply the fractional part by 100. What per cent of
0? Whatper cent of C0 .is C? O2? Find the percentage composition
of SO2. P2O5.

From the last five experiments what do you infer of the tendency
of O to unite with other elements?

27. Oxygen is a Gas without Color, Odor, or Taste.

It is chemically a very active element; that is, it unites with
almost everything. Fluorine is the only element with which it
will not combine. When oxygen combines with a single element,
what is the compound called? We have found that O makes up a
certain portion of the air; later, we shall see how large the
proportion is. Its tendency to combine with almost everything is
a reason for the decay, rust, and oxidation of so many
substances, and for conflagrations, great and small. New
compounds are thusformed, of which O constitutes one factor.
Water, H2O, is only a chemical union of O and H. Iron rust, Fe2O3
and H2O, is composed of O, Fe, and water. The burning of wood or
of coal gives rise to carbon dioxide, CO2, and water. Decay of
animal and vegetable matter is hastened by this all-pervading
element. O forms a portion of all animal and vegetable matter, of
almost all rocks and minerals, and of water. It is the most
abundant of all elements, and makes up from one-half to two-
thirds of the earth's surface. Compute the proportion of it, by
weight, in water, H2O. It is the union of O in the air with C and
H in our blood that keeps up the heat of the body and supports
life. See page 81.

There are many ways of preparing this element besides the one
given above. It may be obtained from water (Experiment 38) and
from many other compounds, e.g. by heating mercury oxide,
HgO.

CHAPTER VII.

NITROGEN.

28. Separation.

Experiment 18. - Fasten a piece of electric-light pencil, or of
crayon, to a wire, as in Experiment 15, and bend the wire so it
will reach half-way to the bottom of a receiver. Using forceps,
put into the crayon a small piece of phosphorus. Pass the wire up
through the orifice in the shelf of a p.t. (pneumatic trough),
having water at least l cm. above the shelf. Heat another wire,
touch it to the P, and quickly invert an empty receiver over the
P, having the mouth under water, so as to admit no air (Fig. 10).
Let the P burn as long as it will, then remove the wire and the
crayon, letting in no air. Note the color of the product, and
leave till it is tolerably clear, then remove the receiver with a
glass plate, leaving the water in the bottom.

Do the fumes resemble those of Experiment 16? Does it seem likely
(Fig 10.) that part of the air is O? Why a part only? Find what
proportion of the receiver is filled with water by measuring the
water with a graduate; then fill it with water and measure that;
compute the percentage which the former is of the latter. What
proportion of the air, then, is O? What was the only means of
escape for the P2O6, and P2O2 formed? These products are solids.
Are they soluble in water? Compute the percentage composition,
always by weight, of P2O2 and P2O5.

The gas left in the receiver is evidently not O. Experiment 19
will prove this conclusively, and show the properties of the new
gas.

29. Properties.

Experiment 19. - When the white cloud has disappeared, slide the
plate along, and insert a burning stick; try one that still
glows.

See whether the P and S on the end of a match will burn. Is the
gas a supporter of combustion? Since it does not unite with C,
S, or P, is it an active or a passive element? Compare it with
O. Air is about 14 1/2 times as heavy as H. Which is heavier, air
or N? See page 12. Air or O?

Write out the chief properties, physical and chemical, of N, as
found in this experiment.

30. Inactivity of N. - N will scarcely unite chemically except on
being set free from compounds. It has, however, an intense
affinity for boron, and will even go through a carbon crucible to
unite with it. It is not combined with O in the air; but the two
form a mixture (page 86), of which N makes up four-fifths, its
use being to dilute the O. What would be the effect, in case of a
fire, if air were pure O? What effect on the human system?

Growing plants need a great deal of N, but they are incapable of
making use of that in the air, on account of the chemical
inactivity of the element. Their supply comes from compounds in
earth, water, and air. By reason of its inertness N is very
easily set free from its compounds. For this reason it is a
constituent of most explosives, as gunpowder, nitro-glycerine,
dynamite, etc. These solids, by heat or concussion, are suddenly
changed to gases, which thereby occupy much more space, causing
an explosion.

Nitrogen exists in many compounds, such as the nitrates; but the
great source of it all is the atmosphere. See page 85.

CHAPTER VIII.

HYDROGEN.

31. Preparation.

Experiment 20. - Prepare apparatus as for making O. Be sure that
the cork perfectly fits both d.t. and t.t., or the H will escape.
Cover 5 g. granulated Zn, in the t.t., with 10 cc. H2O, and add 5
cc. chlorhydric acid, HCl. Adjust as for O (Fig. 7), except that
no heat is to be applied. If the action is not brisk enough, add
more HCl. Collect several receivers of the gas over water, adding
small quantities of HCl when necessary. Observe the black
floating residuum; it is carbon, lead, etc. With a glass plate


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