remove the receivers, keeping them inverted (Fig. 11), or the H
32. The Chemical Change is as follows: -
Zinc + hydrogen chloride = zinc chloride + hydrogen.
Zn + 2 HCl = ZnCl2 + 2H.
Complete by adding the weights, and explain. Notice that the
water does not take part in the change; it is added to dissolve
the ZnCl2 formed, and thus keep it from coating the Zn and
preventing further action of the acid. Note also that Zn has
simply changed places with H, one atom of the former having
driven off two atoms of the latter. The H, having nothing to
unite with, is set free as a gas, and collected over water. Of
course Zn must have a stronger chemical affinity for Cl than H
has, or the change could not have taken place. Why one Zn atom
replaces two H atoms will be explained later, asfar as an
explanation is possible. This equation, should be studied
carefully, as a type of all equations. The left-hand member shows
what were taken, i.e. the factors; the right-hand shows what were
obtained, i.e. the products. H2SO4 might have been used instead
of HCl. In that case the reaction, or equation, would have been:
Zinc + hydrogen sulphate = zinc sulphate + hydrogen.
Zn + H2SO4 = ZnSO4 + 2H.
Iron might have been used instead of zinc, in which case the
reactions would have been: -
Iron + hydrogen chloride = iron chloride + hydrogen.
Fe + 2 HCl = FeCl2 + 2 H.
Iron + hydrogen sulphate = iron sulphate + hydrogen.
Fe + H2SO4 = FeSO4 + 2 H.
Write the weights and explain the equations. The latter should be
Experiment 21. - Lift with the left hand a receiver of H, still
inverted, and insert a burning splinter with the right (Fig. 12).
Does the splinter continue to burn? Does the gas burn? If so,
where? Is the light brilliant? Note the color of the flame. Is
there any explosion? Try this experiment with several receivers.
Is the gas a supporter of combustion? i.e. will carbon burn in
it? Is it combustible? i.e. does it burn? If so, it unites with
some part of the air. With what part?34. Collecting H by Upward
Experiment 22. - Pass a d.t. from a H generator to the top of a
receiver or t.t. (Fig. 13). The escaping H being so much lighter
than air will force the latter down. To obtain the gas unmixed
with air, the d.t. should tightly fit a cardboard placed under
the mouth of the receiver. When filled, the receiver can be
removed, inverted as usual, and the gas tested. In this and other
experiments for generating H, a thistle-tube, the end of which
dips under the liquid, can be used for pouring in acid, as in
35. Philosopher's Lamp and Musical Flame.
Experiment 23. - Fit to a cork a piece of glass tubing 10 or 15
cm. long, having the outer end drawn out to a point with a small
opening, and insert it in the H generator. Before igniting the
gas at the end of the tube take the, precaution to collect a t.t.
of it by upward displacement, and bring this in contact with a
flame. If a sharp explosion ensues, air is not wholly expelled
from the generator, and it would be dangerous to light the gas.
When no sound, or very little, follows, light the escaping gas.
The generation of H must not be too rapid, neither should the
t.t. be held under the face, as the cork is liable to be forced
out by the pressure of H. A safety-tube, similar to the thistle-
tube above, will prevent this. This apparatus is called the
"philosopher's lamp." Thrust the flame into a long glass tube 1-
1/2 to 3 cm. in diameter, as shown in Figure 14, and listen for a
36. Product of Burning H in Air.
Experiment 24. - Fill a tube 2 or 3 cm. in diameter with calcium
chloride, CaCl2, and connect one end with a generator of H (Fig.
15). At the other end have a philosopher's lamp-tube.Observing
the usual precautions, light the gas and hold over it a receiver,
till quite a quantity of moisture collects. All water was taken
from the gas by the dryer, CaCl2. What is, therefore, the product
of burning H in air? Complete this equation and explain it: 2H +
O = ? Figure 16 shows a drying apparatus arranged to hold CaCl2.
[Fig. 15][Fig. 16]
37. Explosiveness of H.
Experiment 25. - Fill a soda-water bottle of thick glass with
water, invert it in a pneumatic trough, and collect not over 1/4
full of H. Now remove the bottle, still inverted, letting air in
to fill the other 3/4. Mix the air and H by covering the mouth of
the bottle with the hand, and shaking well; then hold the mouth
of the bottle, slightly inclined, in a flame. Explain the
explosion which follows. If 3/4 was air, what part was O? What
use did the N serve? Note any danger in exploding H mixed with
pure O. What proportions of O and H by volume would be most
dangerously explosive? What proportion by weight?
By the rapid union of the two elements, the high temperature
suddenly expanded the gaseous product, which immediately
contracted; both expansion and contraction produced the noise of
38. Pure H Is a Gas without Color, Odor, or Taste.
- It is the lightest of the elements, 14 1/2 times as light
asair. It occurs uncombined in coal-mines, and some other places,
but the readiness with which it unites with other elements,
particularly O, prevents its accumulation in large quantities. It
constitutes two-thirds of the volume of the gases resulting from
the decomposition of water, and one-ninth of the weight. Compute
the latter from its symbol. It is a constituent of plants and
animals, and some rocks. Considering the volume of the ocean, the
total amount of H is large. It can be separated from H2O by
electrolysis, or by C, as in the manufacture of water gas.
When burned with O it forms H2O. Pure O and H when burning give
great heat, but little light. The oxy-hydrogen blow-pipe (Fig.
17) is a device for producing the highest temperatures of
combustion. It has O in the inner tube and H in the outer. Why
would it not be better the other way? These unite at the end, and
are burned, giving great heat. A piece of lime put into the flame
gives the brilliant Drummond or calcium light.
Chapter IX. UNION BY WEIGHT.
39. In the Equation -
Zn + 2 HCl = ZnCl2 + 2 H
65 + 73 = 136 + 2
65 parts by weight of Zn are required to liberate 2 parts by weight of
H; or, by using 65 g Zn with 73 g HCl, we obtain 2 g H. If twice as
much Zn (130 g) were used, 4 g H could be obtained, with, of course,
twice as much HCl. With 260 g. Zn, how much H could be liberated?
A proportion may be made as follows: -
Zn given : Zn required :: H given : H required.
65 : 260 :: 2 : x.
[footnote: Given, as here used, means the weight called for by the
equation; required means that called for by the question.]
Solving, we have 8 g H.
How much H is obtainable by using 5 g Zn, as in the experiment?
To avoid error in solving similar problems, the best plan is as
Zn + 2HCl = ZnCl2 + 2 H | 65:5::2:x
65 2 | 65 x = 10
5 x | x = 10/65 = 2/13 Ans. 2/13 g.
The equation should first be written; next, the atomic or molecular
weights which you wish to use, and only those, to avoid confusion;
then, on the third line, the quantity of the substance to be used, with
underneath the substance wanted. The example above will best
how this. This plan will prevent the possibility of error. The proportion
will then be: -
a given : a required :: b given : b required.
How much Zn is required to produce 30 g. H?
Zn + 2HCl = ZnCl2 + 2H | 2:30::65:x
65 2 | 2x = 1950
x 30 | x = 975 Ans. 975 g. Zn.
(1) How much Zn is necessary for 14 g. H?
(2) How many pounds of Zn are necessary for 3 pounds of H?
(3) How many grams of H from 17 g. of Zn?
(4) How many tons of H from 1/2 ton of Zn?
Suppose we wish to find how much chlorhydric acid - pure gas -
will give 12 g. H. The question involves only HCl and H. Arrange
as follows: -
Zn + 2HCl = ZnCl2 + 2 H | H giv. : H req. :: HCl giv. : HCl req.
73 2 | 2 : 12 :: 73 x
x 12 | 2x=876 x=438
Ans. 438 g. HCl.
(1) How much HCl is needed to produce 100 g. H?
(2) How much H in 10 g. HCl?
(3) How much ZnCl2 is formed by using 50 g. HCl? The question
is now between HCl and ZnCl2.
Zn + 2HC1 = ZnCl2 + 2H
73 136 | Arrange the proportion, and solve.
Suppose we have generated H by using H2S04: the equation is
Zn + H2S04 = ZnSO4 + 2 H. There is the same relation as before
between the quantities of Zn and of H, but the H2S04 and ZnS04 are
How much H2SO4 is needed to generate 12 g. H?
Zn + H2SO4 = ZnS04 + 2 H
98 2 | Make the proportion, and solve
(1) How much H in 200 g. H2S04?
(2) How much ZnS04 is produced from 200 g. H2S04?
(3) How much H2S04 is needed for 7 1/2 g H?
(4) How much Zn will 40 g. H2SO4 combine with?
(5) How much Fe will 40 g. H2SO4 combine with?
(6) How much H can be obtained by using 75 g Fe?
These principles apply to all reactions. Suppose, for example, we
wish to get l0 g. of O: how much KClO3 will it be necessary to use?
The reaction is: -
KClO3 = KCl + O3 | 48 : 10 :: 122.5 : x
122.5 48 |
x 10 | Ans. 25.5+ g. KClO3.
The pupil should be required to make up problems of his own,
using various reactions, and to solve them.
Examine graphite, anthracite coal, bituminous coal, cannel coal,
wood, gas carbon, coke.
40. Preparation of C.
Experiment 26. - Hold a porcelain dish or a plate in the flame of
a candle, or of a Bunsen burner with the openings at the bottom
closed. After a minute examine the deposit. It is carbon, i.e.
lamp- black or soot, which is a constituent of gas, or of the
candle. Open the valve at the base of the Bunsen burner, and hold
the deposit in the flame. Does the C gradually disappear? If so,
it has been burned to CO2. C + 2 O = CO2. Is C a combustible
Experiment 27. - Ignite a splinter, and observe the combustion and
the smoke, if any. Try to collect some C in the same way as
With plenty of O and high enough temperature, all the C is burned
to CO2, whether in gas, candle, or wood. CO2 is an invisible gas.
The porcelain, when held in the flame, cools the C below the
point at which it burns, called the kindling-point, and hence it
is deposited. The greater part of smoke is unburned carbon.
Experiment 28. - Hold an inverted dry t.t. or receiver over the
flame of a burning candle, and look for any moisture (H2O). What
two elements are shown by these experiments to exist in the
candle? The same two are found in wood and in gas. Experiment
29. - Put into a small Hessian crucible (Fig. 18) some pieces of
wood 2 or 3 cm long, cover with sand, and heat the crucible
strongly. When smoking stops, cool the crucible, remove the
contents, and examine the charcoal. The gases have been driven
off from the wood, and the greater part of what is left is C.
Experiment 30. - Put 1 g. of sugar into a porcelain crucible, and
heat till the sugar is black. C is left. See Experiment 5. Remove
the C with a strong solution of sodium hydrate (page 208).
41. Allotropic Forms. - Carbon is peculiar in that it occurs in at
least three allotropic, i.e. different, forms, all having
different properties. These are diamond, graphite, and amorphous
- not crystalline - carbon. The latter includes charcoal, lamp-
black, bone-black, gas carbon, coke, and mineral coal. All these
forms of C have one property in common; they burn in O at a high
temperature, forming CO2. This proves that each is the element C,
though it is often mixed with some impurities.
Allotropy, or allotropism, is the quality which an element often
has of appearing under various forms, with different properties.
The forms of C are a good illustration.
42. Diamond is the purest C; but even this in burning leaves a
little ash, showing that it is not quite pure. It is a rare
mineral, found in India, South Africa, and Brazil, and is the
hardest and most highly refractive to light of all minerals.
Boron is harder. [Footnote: B, not occurring free, is not a
mineral.] When heated in the electric arc, at very high
temperatures, diamond swells and turns black. 43. Graphite, or
Plumbago, is One of the Softest Minerals. - It is black and
infusible, and oxidizes only at very high temperatures, higher
than the diamond. It contains from 95 to 98 per cent C. Graphite
is found in the oldest rock formations, in the United States and
Siberia. It is artificially formed in the iron furnace. Graphite
is employed for crucibles where great heat is required, for a
lubricant, for making metal castings, and, mixed with clay, for
lead-pencils. It is often called black-lead.
44. Amorphous Carbon comprises the following varieties.
Charcoal is made by heating wood, for a long time, out of contact
with the air. The volatile gases are thus driven off from the
wood; what is left is C, and a small quantity of mineral matter
which remains as ash when the coal is burned.
45. Lamp-black is prepared as in Experiment 26, or by igniting
turpentine (C1OH16), naphtha, and various oils, and collecting
the C of the smoke. It is used for making printers' ink, India
ink, etc. A very pure variety is obtained from natural gas.
Bone-black, or animal charcoal, is obtained by distilling bones,
i.e. by heating them in retorts into which no air is admitted.
The C is the charred residue.
Gas Carbon is formed in the retorts of the gas-house. See page
182. It is used to some extent in electrical work.
46. Coke is the residue left after distilling soft coal. It is
tolerably pure carbon, with some ash and a little volatile
matter. It burns without flame. 47. Mineral Coal is fossilized
wood or other vegetable matter. Millions of years ago trees and
other vegetation covered the earth as they do to-day. In certain
places they slowly sank, together with the land, into the
interior of the earth, were covered with sand, rock, and water,
and heated from the earth's interior. A slow distillation took
place, which drove off some of the gases, and converted vegetable
matter into coal. All the coal dug from the earth represents
vegetable life of a former period. Millions of years were
required for the transformation; but the same change is in
progress now, where peat beds are forming from turf.
Coal is found in all countries, the largest beds being in the
United States. From the nature of its formation, coal varies much
Anthracite, or hard coal, is purest in carbon, some varieties
having from 90 to 95 per cent. This represents most complete
distillation in the earth; i.e. the gases have mostly been driven
off. It is much used in New England.
48. Bituminous, or soft coal, crocks the hands, and burns rapidly
with much flame and smoke. The greater part of the coal in the
earth is bituminous. It represents incomplete distillation.
Hence, by artificially distilling it, illuminating gas is made.
See page 180. It is far less pure C than anthracite.
49. Cannel Coal is a variety of bituminous coal which can be
ignited like a candle. This is because so many of the gases are
still left, and it shows cannel to be less pure C than bituminous
50. Lignite, Peat, Turf, etc., are still less pure varieties of
C. Construct a table of the naturally occurring forms of this
element, in the order of their purity. Carbon forms the basis of
all vegetable and animal life; it is found in many rocks, mineral
oils, asphaltum, natural gas, and in the air as CO2.
51. C a Reducing Agent.
Experiment 31. - Put into a small ignition-tube a mixture of 4 or
5 g. of powdered copper oxide (CuO), with half its bulk of
powdered charcoal. Heat strongly for ten or fifteen minutes.
Examine the contents for metallic copper. With which element of
CuO has C united? The reaction may be written: Cu0 + C = CO + Cu.
Complete and explain.
A Reducing, or Deoxidizing, Agent is a substance which takes away
oxygen from a compound. C is the most common and important
reducing agent, being used for this purpose in smelting iron and
other ores, making water-gas, etc.
An Oxidizing Agent is a substance that gives up its O to a
reducing agent. What oxidizing agent in the above experiment?
52. C a Decolorizer.
Experiment 32. - Put 3 or 4 g. of bone-black into a receiver, and
add 10 or 15 cc.of cochineal solution. Shake this thoroughly,
covering the bottle with the hand. Then pour the whole on a
filter paper, and examine the filtrate. If all the color is not
removed, filter again. What property of C is shown by this
experiment? Any other coloring solution may be tried.
The decolorizing power of charcoal is an important
characteristic. Animal charcoal is used in large quantities for
decolorizing sugar. The coloring matter is taken out mechanically
by the C, there being no chemical action. 53. C a Disinfectant.
Experiment 33. - Repeat the previous experiment, adding a solution
of H2S3 i.e. hydrogen sulphide, in water, instead of cochineal
solution. See page 120. Note whether the bad odor is removed. If
Charcoal has the property of absorbing large quantities of many
gases. Ill-smelling and noxious gases are condensed in the pores
of the C; O is taken in at the same time from the air, and these
gases are there oxidized and rendered odorless and harmless. For
this reason charcoal is much used in hospitals and sick-rooms, as
a disinfectant. This property of condensing O, as well as other
gases, is shown in the experiment below.
54. C an Absorber of Gases and a Retainer of Heat.
Experiment 34. - Put a piece of phosphorus of the size of a pea,
and well dried, on a thick paper. Cover it well with bone-black,
and look for combustion after a while. O has been condensed from
the air, absorbed by the C, and thus communicated to the P. Burn
all the P at last.
55. The Symbols NaCl and MgCl2 differ in two ways. - What are
they? Let us see why the atom of Mg unites with two Cl atoms,
while that of Na takes but one. If the atoms of two elements
attract each other, there must be either a general attraction all
over their surfaces, or else some one or more points of
attraction. Suppose the latter to be true, each atom must have
one or more poles or bonds of attraction, like the poles of a
magnet. Different elements differ in their number of bonds. Na
has one, which may be written graphically Na-; Cl has one, -Cl.
When Na unites with Cl, the bonds of each element balance, as
follows: Na-Cl. The element Mg, however, has two such bonds, as
Mg= or -Mg-. When Mg unites with Cl, in order to balance, or
saturate, the bonds, it is evident that two atoms of Cl must be
used, as Cl-Mg-Cl, or MgCl2.
A compound or an element, in order to exist, must have no free
bonds. In organic chemistry the exceptions to this rule are very
numerous, and, in fact, we do not know that atoms have bonds at
all; but we can best explain the phenomena by supposing them, and
for a general statement we may say that there must be no free
bonds. In binaries the bonds of each element must balance.
56. The Valence, Quantivalence, of an Element is its Combining
Power Measured by Bonds. - H, having the least number of bonds,
one, is taken as the unit. Valence has always to be taken into
account in writing the symbol of a compound. It is often written
above and after the elements [i.e. written like an exponent], as
An element having a valence of one is a monad; of two, a dyad;
three, a triad; four, tetrad; five, pentad; six, hexad, etc. It
is also said to be monovalent, di- or bivalent, etc. This theory
of bonds shows why an atom cannot exist alone. It would have free
or unused bonds, and hence must combine with its fellow to form a
molecule, in case of an element as well as in that of a compound.
This is illustrated by these graphic symbols in which there are
no free bonds: H-H, O=O, N[3-bond symbol]N, C[4-bond symbol]C. A
graphic symbol shows apparent molecular structure.
After all, how do we know that there are twice as many Cl atoms
in the chloride of magnesium as in that of sodium? The compounds
have been analyzed over and over again, and have been found to
correspond to the symbols MgCl2 and NaCl. This will be better
understood after studying the chapter on atomic weights. In
writing the symbol for the union of H with O, if we take an atom
of each, the bonds do not balance, H-=O, the former having one;
the latter, two. Evidently two atoms of H are needed, as H-O-H,
= O , or H2O. In the union of Zn and O, each has two bonds;
hence they unite atom with atom, Zn = O, or ZnO.
Write the grapbic and the common symbols for the union of H^I and
Cl^I; of K^I and Br^I; Ag^I and O^II; Na^I and S^II; H^I and
P^III. Study valences. It will be seen that some elements have a
variable quantivalence. Sn has either 2 or 4; P has 3 or 5. It
usually varies by two for a given element, as though a pair of
bonds sometimes saturated each other;. e.g. =Sn=, a quantivalence
of 4, and |Sn=, a quantivalence of 2. There are, therefore, two
oxides of tin, SnO and SnO2, or Sn=O and O=Sn=O. Write symbols
for the two chlorides of tin; two oxides of P; two oxides of
The chlorides of iron are FeCl2 and Fe2Cl6. In the latter, it
might be supposed that the quantivalence of Fe is 3, but the
graphic symbol shows it to be 4. It is called a pseudo-triad, or
false triad. Cr and Al are also pseudo-triads.
Cl Cl | | Cl - Fe - Fe - Cl | | Cl Cl
Write formulae for two oxides of iron; the oxide of Al.
57. A Radical is a Group of Elements which has no separate
existence, but enters into combination like a single atom; e.g.
(NO3) in the compounds HNO3 or KNO3; (SO4) in H2SO4. In HNO3 the
radical has a valence of 1, to balance that of H, H-NO3). In
H2SO4, what is the valence of (SO4)? Give it in each of these
radicals, noting first that of the first element: K(NO3),
Na2(SO4), Na2(CO3), K(ClO3), H3(PO4), Ca3(PO4)2, Na4(SiO4).
Suppose we wish to know the symbol for calcium phosphate. Ca and
PO4 are the two parts. In H3(PO4) the radical is a triad, to
balance H3. Ca is a dyad, Ca==(P04). The least common multiple of
the bonds (2 and 3) is 6, which, divided by 2 (no. Ca bonds),
gives 3 (no. Ca atoms to be taken). 6 / 3 (no. (PO4) bonds) gives
2 (no. PO4 radicals to be taken). Hence the symbol Ca3(P04)2.
Verify this by writing graphically.
Write symbols for the union of Mg and (SO4), Na and (PO4), Zn and
(NO3), K and (NO3), K and (SO4), Mg and (PO4), Fe and (SO4) (both
valences of Fe), Fe and (NO3), taking the valences of the
radicals from HNO3, H2SO4, H3PO4.
ELECTRO-CHEMICAL RELATION OF ELEMENTS.
58. Examine untarnished pieces of iron, silver, nickel, lead,
etc.; also quartz, resin, silk, wood, paper. Notice that from the
first four light is reflected in a different way from that of the
others. This property of reflecting light is known as luster.
Metals have a metallic luster which is peculiar to themselves;
and this, for the present, may be regarded as their chief
characteristic. Are they at the positive or negative end of the
list? See page 43. How is it with the non-metals? This
arrangement has a significance in chemistry which we must now
examine. The three appended experiments show how one metal can be
withdrawn from solution by a second, this second by a third, the
third by a fourth, and so on. For expedition, three pupils can
work together for the three following experiments, each doing
one, and examining the results of the others.
59. Deposition of Silver.
Experiment 35. - Put a ten-cent Ag coin into an evaporating-dish,
and pour over it a mixture of 5 cc. HNO3 and 10 cc. H2O. Warm
till all, or nearly all, the Ag dissolves. Remove the lamp. 3 Ag
+ 4 HNO3 = 3 AgNO3 + 2 H2O + NO. Then add 10 cc. H2O, and at once
put in a short piece of Cu wire, or a cent. Leave till quite a
deposit appears, then pour off the liquid, wash the deposit
thoroughly, and remove it from the coin. See whether the metal
resembles Ag. 2 AgNO3 + Cu =?60. Deposition of Copper.
Experiment 36. - Dissolve a cent or some Cu turnings in dilute
HNO3, as in Experiment 35, and dilute the solution. 3 Cu + 8 HN09
- 3 Cu (NOA+4 H2O+2 NO.)
Then put in a clean strip of Pb, and set aside as before,
examining the deposit finally. Cu(NO3), + Pb - ?