Fig.
1In this paper, Dumas sought to make a comparison of salts of ammonia with similar derivatives of the combination, C2H2 (Fig. 1). While the authors did not refer to C2H2 as a radical, later, Berzelius named the species the etherin radical. The paper contains some twenty compounds whose compositions are defined "structurally" as a combination of a base, an acid, and/or water, as shown from left to right in the three columns below (Fig. 1). But first, a few words about the conventions that were used. The dot above one "H" in the last column indicates an attachment to oxygen; thus, water. The number of atoms present in a structure are written as superscripts. The acid, H6C4O3, expresses acetic acid less the water in the last column. From our perspective, this entity would be an anhydride. Dumas and his contemporaries were perfectly capable of determining the percent composition of organic compounds. Where they were in disagreement was on the atomic weights of the elements. That is, does a particular compound have six carbons of mass 12 or twelve carbons of mass six? Today we recognize acetic acid as having the composition C2H4O2, where H=1, C=12, and O=16. If we constitute acetic acid as described by Dumas, we obtain H6C4O3 + H2O = C4H8O4. Obviously, this formula is double the true molecular formula but both have the same percentage of each element. To form ammonium acetate with this double formula acid, two molecules of ammonia [NH3 = AzH3] are required. Ammonium acetate using Dumas's formula is C4H14N2O4; halved it is what we use today, C2H7NO2. From this analysis we conclude that H=1, C=12, N=14, and O=16.
Fig.
1Dumas uses four volumes of the "etherin radical" in place of ammonia to satisfy the empirical formula of éther acétique, ethyl acetate [C4H8O2 today]. The formula of ethyl acetate according to Dumas is C12H16O4, or halved, C6H8O2. While it is clear that the number of hydrogens and oxygens are the same in both the Dumas and the modern formula, it is also clear that the number of carbons is not. What is going on here? For the etherin radical, Dumas is using H=1, C=6! If we reconstruct his formula for ethyl acetate taking into consideration the two different carbon masses, we have C(6)8C(12)4H16O4. If this formula is halved, C(6)4C(12)2H8O2 results. When this formula is converted to all carbons of mass=12, C4H8O2, the modern formula is produced. Dumas has used C=6 for the base etherin, and C=12 for the acid. After all, etherin is nothing more than ethylene [olefiant gas], because 2 x C(6)2H2 = 2 x C(12)H2= C2H4. It is unfair for us today, blessed with 20/20 hindsight, to be too critical. After all he was trying to develop a theory of the constitution of organic compounds. For what was known with certainty in 1828, Dalton's imperfect atomic theory notwithstanding, there may have been two types of carbon. Are there not three isotopes of carbon?
Thirteen years before this paper appeared, Gay-Lussac demonstrated that the vapor density of ether was greater than that of alcohol. From his own studies in this area, Dumas formulated ether as the half hydrate of ethylene while alcohol was the monohydrate of ethylene Fig. 2). As in the example above, C= 6 in the base etherin. Likewise, these data can lead to the conclusion, either algebraically or experimentally, that alcohol is a hydrate of ether. You may visualize this process as the O-H bond of water and alcohol adding across ethylene to form respectively, alcohol and ether. Remember, in 1828 there was no concept of bonds or valence but only one of affinity. The concept of a hydrate was closer to the truth as the names of the four compounds would imply.
Fig.
2Dumas applied the etherin theory to more complex substances such as sucre de raisins [glucose, C6H12O6] and sucre de cannes [sucrose, C12H22O11] (Fig. 3). For these sugars, the acid "radical" is CO2. Summation gives C(6)12H12O6 which is correct for the modern composition of a hexose such as glucose, C(12)6H12O6. Here all carbons have C=6! Sucrose is formed from glucose and fructose (same formula as glucose) less a molecule of water. Therefore, based upon Dumas's formula for glucose, it should be doubled less one molecule of water or:
The formula presented for sucrose in the chart is C(6)12H10O5, which is one hexose less one water. Either this entry was an error in transcription or an error in analysis.
Fig.
3copyright FEZiegler2001