time, and is naturally difficult for scientists to appreciate. It is therefore considered advisable now to place all the facts as succintly as possible before scientists (before continuing the demonstration of the structure of each element separately) in order that they may have an opportunity of seeing the certainty of the truth of the matter. There is no hypothesis, guesswork or speculation in any part of the paper. The whole is merely an arrangement of experimental facts in such a manner as to demonstrate the truths which underlie them. The truths have been discovered, not hypothesised; for it is evidently impossible for any one to have guessed them. Table I. is intended to be read in the following manner, taking No. 1 as an example: The atomic weight of Sb is equal to the sum of the atomic weights of Ag and C. 21.29, 13.29 and 8.0 are relative volumes of Sb, Ag and C respectively; and the first is equal to the sum of the other two. No. 74 gives a molecule containing 21.29 (Sb), and so on. RELATIVE VOLUMES AND PART-VOLUMES OF THE ELEMENTS. (1) 119(Sb) = 107(Ag) + 12(C) = (13) 75(As) = 63 (Cu) + 12(C) Part.vols. Refer to Nos. Vols. 21.29 = 13.29 + 8.0 74, 75, 87. 90, 75, 85. 99, 101, 87. 15.70 + 8.0 76, 77, 87. 78, 79, 87. 17.76 = 8.719.05 80, 81, 82. 92. 14.54 = 13.83 0.71 19.59 = 91, 84, 104, 105, 85, 106. 107, 81, 108. 109, 110, 82, 111. 112, 113, 111 114, 88, 111. 2(7.13) + 11.31 115, 110, 84. 69.69 =4(11.85) + 22.29 116, 87, 117. 118, 110. here it is shown to be a volume of Ag when it is a part of another element. It is necessary to notice that there are two coincidences in each one of the 21 cases of Table I. E.g., in the case of Sb, its at.wt. is equal to the sum of the at. wts. of Ag and Taking No. 4, Table I., as an example, because it is the first one mentioned in which the two partvolumes are each of the most general form, i.e., having one figure before the decimal and two after it; the mathematical probability that, after noticing that the at.wt. 51(V) equals 24(Mg) + 27(Al), and after obtaining the exact relative volume 17.76 of V, this latter value would be capable of being split up into two parts which have been shown previously to be rel. vols. of Mg and Al, lies between may be the case the figures for the partvolumes would have to appear as 871905 or 905871, i.e., two particular arrangements of figures out of 531441 possible cases, where there are no zeros, or out of 1,000,000 when zeros are included; i.e., one in 250,000 (at least). This is equivalent to the statement that if there were 250,000 cases to be investigated, only one of them would accidently be as found; but the actual position is that 21 out of about 200 are of this nature, i.e., = TABLE II. about 1 in 10. But other considerations in Tables II-VIII., and the following paper, bring nearly the whole of the 200 into the law. The three volumes 17.76, 8.71 and 9.05 were obtained exactly as they stand several years before this law was discovered, and each one of these volumes has been shewn to be exactly correct for several reasons in preceding papers published by the Chemical News. Similar remarks apply to each one of the twenty other cases in Table I. In Table II., Ti has the volume 5.02 fourteen times. This is a volume of Ti only when forming part of another element. In Table III., Na has the volume 1.9 twenty-four times. This is a volume of Na only when forming part of another element. In Table IV., Na has the volume 9.54 eight times, H has the volume 0.97 eight times, and H, the volume 1.54 six times. In Table V., 7.49(K) occurs five times. In Table VI., 3.04 (S) occurs twice, 3.2(Si) four times, 3.74(Al) three times, and 3.91 (Ca) six times. In Table VII. 5.23(Ca) occurs twice, 3(Zr) four times, 5.02 (Zr) also four times. In Table VIII. 6.54(Na) occurs four times, 4.05 (Al) three times. Each volume here mentioned as occuring in Tables II.-VIII. is only known as such when forming part of the volume of an element, except 0.97(H). THE PART-VOLUME 5.02 OF TITANIUM. Vols. Part-vols. 35.7 = 25.66 + 2(5.02) 24.58 = 14.54 + 2(5.02) 29.00 = 14.54+9.44+5.02 30.68 = 25.66 + 5.02 19.56 = 14.54 + 5.02 15.10 = 10.08 + 5.02 20.55 = 17.55 = Atomic Weights. (21) 136(Ba) = 40(Ca) + 96(Ti2) (22) 136(Ba) = 40(Ca) 96(Ti2) + (23) 136(Ba) = 40(Ca) 96(Ti2) + (24) 88(Sr) 40 (Ca) + 48 (Ti) (25) 88(Sr) 40 (Ca) + 48 (Ti) (26) 104(Pd) = 56(Fe) + 48 (Ti) (27) 128(Te) = 32(S) + 96 (Ti2) (28) 128(Te) = 32(S) + 96 Ti2) (29) 138(Ce) = 63 (Cu)+27 (Al)+48 (Ti) = TABLE III. 10.51 + 2(5.02) 126, 74. 7.51 + 2(5.02) 129, 130. 128, 110,,82. 21.2 = 7.13+9.05+5.02 (58) TABLE VII. THE PART-VOLUMES 5.28 (Ca), 3(Zr) AND 5.02(Zr). Atomic Weights. = = 63(Cu) 23 (Na) + 40 (Ca) (59) 63(Cu) 23 (Na) + 40 (Ca) (60) 209 (Bi) = 90(Zr) +119 (Sb) (61) 118(Sn) = 90(Zr) + 28 (Si) 28(Si) (62) 118(Sn) = 90Zr) + 28 (Si) (63) 118(Sn) = 90(Zr + 28(Si) (64) 194 (Pt) = 90(Zr) + 104 (Pd) (65) 141 (Pr) = 90 (Zr) + 24(Mg) + 27(Al) 90 (Zr) + 24(Mg + 27(Al) = 90(Zr) + 48(Mg2) |