This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1893 Excerpt: ...of the oxides which are isomorphous, are then added together. Thus: SiO, = 36.02 H-00 = 0.6103, Al, O, = 7.53-h 102 = 0.0738, Pe, O, = 22.18 4-160 = 0.1380, CaO = 31.80-4-56 = 0.5678, MgO = 1.95-i-40 = 0.0487. Adding respectively the values for Al, O, and for Fe, O, and the values for CaO and for MgO, there results: SiO, = 0.6103 = 3, (Al, O, .Fe, O, ) = 0.2118 = 1, (CaO.MgO) = 0.6165 = 3; which result is the same as that previously obtained. The formula of the mineral as derived from these data can be expressed as: 3(CaMgO), (Al, Fe.O, ),3SiO, or as: (CaMg), (Al, Fe, )Si, O1, . In accordance with the custom of mineralogists, who prefer to represent the formulae of such minerals even more concisely, the letter R can be used to denote metallic radicals, and Roman numerals placed above the same, can be made to indicate the valence of the radicals: The above formula would thus be written: II VI IV R, (R, )Si, Oi, a shorter, and an equally correct expression of the composition of this mineral. A formula of this kind can easily be changed to a formula n in the old dualistic system, by converting R into RO and VI (R, ) into R, O, . 'It would then read: 3RO, R, O, Si, O, . Comparing now the quantities of oxygen in combination with the several bases and with the silicon, it will be seen that they bear to one another the relation 3:3:6; that is, as 1: 1: 2. This ratio, in this instance 1: 1: 2, is termed the oxygen ratio, and it will be noticed that it is the same as the ratio of the total valence, the so-called atomic ratio, of each radical in the formula: For, II X 3 = 6; VI X 1 = 6; IV X 3 = 12. and 6: 6: 12, is the same ratio as 1: 1: 2. If it be required to calculate the mineralogical formula of a silicate from its percentage composition, the first step to be taken wil