atmosphere. Barometer height, 754'8 mm.

Volume of

globe, 250 c.c. Residual air, 245 c.c. at 10'3 mm. and 17.2° C. Excess of weight of globe, o 8765 gram. Weight of air displaced o‘3020 gram.

AVOGADRO'S HYPOTHESIS; DENSITY AND MOLECULAR WEIGHT; VALENCY, EQUIVALENTS AND ATOMIC WEIGHTS.

According to Avogadro's Hypothesis equal volumes of gases, at the same temperature and pressure, contain the same number of molecules.

Hence it follows that the molecular weights of bodies in the state of gas or vapour are in the same ratio as their specific gravities or vapour densities: e.g. if H sp. gr. of gaseous HCl. = 18'18.

=

I, the Assuming the molecular weight of H to be 2, the molecular weight of HCl is 36°36.

The molecular weight of a gaseous element or compound may be thus defined: The molecular weight of a gaseous element or compound is a number which expresses how many times greater than two unit masses of hydrogen is the mass of the specified element or compound which occupies (under the same conditions of temperature and pressure) the same volume as is occupied by these two unit masses of hydrogen. The maximum atomic weight of an element is a number which expresses how many times greater is the smallest mass of that element which combines with other elements to form a compound gaseous molecule, than the smallest mass of hydrogen which combines with other elements to produce a compound gaseous molecule, such smallest mass of hydrogen being taken as unity. (Watts's Dict. I. p. 341.)

That property of an atom which determines the number of other atoms with which it can combine to form a compound molecule is termed its valency or atomic value. Valency is usually expressed as the number of hydrogen atoms, or atoms chemically equivalent to hydrogen, which one atom of the specified element can combine with or replace when in the same state as in the compound considered.

D

The equivalent of an element is the number expressing that weight of the element which will combine with or replace the unit weight of hydrogen.

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=

Let E the equivalent of any element, A its atomic weight, and V = its valency in the compound used for determining the equivalent; the relations between these quantities are expressed by the equation A = E . V.

The numerical value of the molecular weight of any substance is obtained by the summation of the atomic weights of the component elements each multiplied by the number denoting how many of such atoms are contained in the molecule, e.g. the molecular weight of alcohol, C2H6O, is found thus (12 × 2) + (1 × 6) + 16 = 24 +6 +16= 46.

QUESTIONS.

=

= I'791,

=

86. If the density of sulphuretted hydrogen ammonia = O'597, nitrous oxide 1520, marsh gas 0555, chloroform 420, and stannic chloride = 9°20, as compared with air, what are the molecular weights of these bodies taking the molecular weight of hydrogen as 2?

87. Determine the weight of one litre of the following simple gases and vapours at o° and 760 mm., on the supposition that they can all exist as perfect gases at the standard temperature and pressure :—

88. Calculate the volume, at the standard temperature and pressure, of a kilogram of each of the following gases

and vapours :—

Carbon monoxide.

Hydrogen sulphide.

Marsh gas.

Water.

Ethylene.
Oxysulphide of carbon.
Bromine.

Hydrochloric acid.

89. What weight of each of the specified substances is represented by the appended formulæ, taking one gram as the unit of weight?

Lead chloride, PbCl,; Silicon fluoride, SiF;

Ferric chloride, FeCl3; Methyl bromide, CH3Br.

=

90. Find the density (air 1) of :-Carbon monoxide, CO; Carbon bisulphide, CS,; Sulphur trioxide, SO; Boron trifluoride, BF,; Phosphorus pentafluoride, PF¿.

=

=

=

=

91. If the electro-chemical equivalent of: Oxygen 7'98, Silver 107.66, Chlorine 35 37, Antimony 39.86, and Copper 3159, and the hydrogen-replacing power of these elements in the compounds used in finding the equivalents be for one atom of the element taken, Oxygen 2, Silver 1, Chlorine 1, Antimony 3, and Copper 2 atoms; what are the atomic weights of these elements?

92. From the data given deduce the probable atomic weight of oxygen:

93. Find the most probable atomic weight of carbon from the appended experimental results :

94. Find the molecular weights of the following bodies :

95. Select the most probable atomic weight for phosphorus as indicated by the given experimental results :—

DEDUCTION OF THE EMPIRICAL FORMULA OF A BODY FROM ITS PERCENTAGE COMPOSITION; FORMULÆ OF MINERALS.

In order to calculate the empirical formula of a compound, divide the percentage amount of each constituent by its corresponding atomic weight: divide each of the quotients so obtained by the lowest number, and reduce them to their simplest ratios.

EXAMPLES.

1. A body on analysis yielded the following percentage composition :

The simplest relation between the carbon and oxygen is at once seen to be as I to 2, since

2272745454 :: I: 2.

Hence the formula of the compound is CO2

2. A compound of hydrogen and nitrogen was found to possess the following percentage composition. Calculate its formula.

3. A compound of iron and oxygen possesses the following percentage composition. Calculate its formula.

At. weight of Fe = 560. At. weight of O = 16.