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Previous Next The best candle and soap tips online! TECHNICAL TREATISE ON SOAP AND CANDLES. SOAP ANALYSIS. 377 drachm), and that in saturating the normal acid the volume of alkaline liquor employed was Ty0. Subtract from the volume 6.25 cubic centimetres the volume 0.15, which gives for the oleic acid 6.250.15 =» 6.10 cubic centimetres (1.65 fluidrachms), we have: Oleic acid 6.10 c. c. x 0.9003 grm. = 5.49183 grms. (84.74 grains). Rosin by difference (6.45 grms. 5.49183) 0.95817 " (14.78 " ). Soda 1.2 grm*. X j%\ 0.81600 " (12.59 " ). Water by difference 2.73400 " (42.21 " ). (154.32 10.00000 Soap of oleic acid and rosin The weight of the candleshop.com/cgi-bin/affiliates/clickthru.cgi?id=soforreal">wax will give that of the fatty matter and rosin; the volume x cubic centimetre of the fatty matter the volume 0.15 of the rosin multiplied by 0.9003 weight of a cubic centimetre of oleic acid will give the weight of that volume. By difference the weight of the rosin will be known ; the volume of the alkaline liquor not used, will give the weight of the soda or potash ; lastly, by difference, the weight of the water is obtained. Mixtures of Potash and Soda.In some soaps there is a mixture of potash and soda. The weight of each alkali is known by the following method. Burn 10 grammes (0.35 oz.) of soap. Weigh the ashes and treat them by boiling distilled water: filter, wash the filter with a little warm water, and add the washings to the alkaline solution; then burn the filter, deduct the known weight of its ash from the total weight of the ashes, and by difference we have the weight of the potash and soda mixed in the state of carbonates. Let us suppose that the mixture weighs 3 grammes (46.29 grains). The volume of normal acid necessary to saturate 3 grammes of the mixture is found by a direct experiment. Then the volume of normal acid necessary to saturate 3 grammes of carbonate of potash and 3 grammes of carbonate of soda is found by calculation. The volume of normal acid by which the three grammes of the mixture have been saturated will be intermediate between the volumes which ought to saturate 3 grammes of carbonate of potash, and 3 grammes of carbonate of soda. By a proportional division, we shall have fractions of potash and soda to compose the weight of the mixture examined. Let us suppose that the volume of normal acid used directly for the saturation of the 3 grammes of the mixture is 13 cubic centimetres (3.51 fluidrachms). The volume of normal acid to saturate 3 grammes of carbonate of potash and 3 grammes of carbonate of soda is to be ascertained ; knowing that 10 cubic centimetres (0.33 fluidoz.) of normal acid contain 1.8984 grammes (29.29 grains) of monohydrated sulphuric acid, and that its equivalent is 612.5. For the carbonate of soda, we have: NaO,CO2 662.17 SO3,HO NaO,CO2 8O3,HO 612.5 :: 3 : x = 2.774 grms. (42.80 grains). To know the volume of normal acid which contains 2.774 grammes of monohydrated acid, we have: SO3,HO Vol. SO.t,HO Vol. 1.8984 : 10 c. c. :: 2.774 grms. : x = 14.612 c, c. (3.94 fluidraclims). In the same manner we ascertain the weight of monohy-drated acid and afterwards the volume of normal acid which contains the weight of monohydrated acid necessary to saturate 3 grammes of carbonate of potash. "We have: KO,CO2 SO3,HO KO,CO2 SO.,, HO 863.93 : 612.5 : : 3 : x 2.126 grms. (32.80 grains). SO.,, 110 Vol. SO3,HO 1.8984 grms. : 10 c. c. : : 2.126 grms. : x = 11.198 c. c. (3.02 fldrnw.) of normal acid, which contains 2.126 grammes of monohydrated acid. By experiment 13 cubic centimetres (3.51 fluidrachms) of normal acid have been employed to saturate the alkalies found in the ashes of the calcined soap. It is evident that if the weight of alkali found in the ashes is formed only of carbonate of soda, the volume used should be 14.612 cubic centimetres (3.94 fluidrachms) of normal acid; if, on the contrary, the 3 grammes are only formed of carbonate of potash, the volume used should be 11.198 cubic centimetres (3.02 378 TECHNICAL TREATISE ON SOAP AND CANDLES. SOAP ANALYSIS. 379 fluidrachms) of normal acid. But as the volume of normal acid used has been 13 cubic centimetres, this volume alone indicates a mixture of carbonate of potash and soda. By a proportional division we have the weight of the carbonate of soda proportional to a fraction of the volume 14.612 cubic centimetres of normal acid; we have also the weight of the carbonate of potash proportional to a fraction of the volume 11.198 cubic centimetres of normal acid. To establish this division, we have:? 1. The gain that, the volume 11.198 ought to make to give vol ume 13 cubic centimetres which is 1.802 cubic centimetres (0.49 fluidrachm). 2. The loss that the volume 14.612 cubic centimetres ought to make to give 13 cubic centimetres, which is 1.612 cubic centime tres (0.43 fluidrachm). Which gives: Gain .... 1.802 ? = 3.414 (0.92 fluidrachm). Loss . 1.612 >' To compose the volume 13 cubic centimetres, we take: 1. The ||Y| of the volume 14.612 cubic centimetres, corre sponding to the jfxf of 3 grammes of carbonate of soda. 2. The I f|| of the volume 11.198 cubic centimetres, corre sponding to the \\\\ of 3 grammes of carbonate of potash. We obtain: x= carb. of soda 1.5834 grms. corr. to soda 0.92-58 grm. corr. to vol. of acid 7.7126 c. c. y " " pot. 1.4165 " " pot. 0.9656 " " " " " 5.2873 a;-fj/ = mixture 2.9999 " both 1.8914 for volume found 12.S999 (46.29 grains).. (29.18 grains). (3.51 fldrm.) In this analysis, the following method due to Gay Lussac cannot be well used,'because too much soap would have to be burned so as to operate on 50 grammes of mixed chlorides. Operate as follows: Transform the two carbonates into chlorides and calcine to evaporate the excess of acid; take 50 grammes (1.75 ozs.) of the mixture which is finely powdered, introduce this mixture into a bottle weighing 185 grammes (6.48 ozs.), and containing 200 grammes (7.00 ozs.) of water, stir with a glass Previous Next
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