.MOLLCI FORAIL LAS. After analysis has shown the percentage composition of a substance, the next step is to determine its molecular formula. For this purpose the percentages of the constituent elements must, first of all, be ex pressed in terms of their atomic weights. Let, for example, an analysis of pure acetic acid give the following results: carbon, 39.9 per cent.: hydrogen, 6.7 per cent.; and oxygen, 53.4 per cent. Since the atomic weights of the three elements are, respectively, 12, 1, and 16, the analytical results 39.9 + 6.7 + 53.4 are writ ten in the form (3,33 X 12) + (6.7 X 1) + (3.34 X IL), or, using the symbols of the ele ments to denote their atomic weights, in the form, 3.33C + 6.7H + 3.340. Allowing for the errors of analysis. it is therefore evident that for every atom of carbon. acetic acid contains 2 atoms of hydrogen and 1 atom of oxygen—a relation expressed by any one of the formulas, Clip.
etc. According to the first of these formulas, the molecular weight of acetic acid would be 30 (i.e. 12 X 1 + 1 X 2 + 16 X ; according to the seemid it would be 60 (i.e. 12 X 2 + I X 4 + 16 X 2) according to the third it would be 90, etc. Now, according to Avo gadro's rule, the molecular weight of a com pound is twice as great as the density of its vapor (compared with hydrogen). Therefore, in crder to fix the molecular weight of acetic acid, its vapor-density must be determined; and this may be done by one of the methods described in the article NoLEctmEs—MotEc•L.An WEIGHTS. The vapor-density being found to be about 30, the molecular weight is taken to be GO, and hence the formula is accepted as representing a molecule of acetic acid.