A treatise on pharmacy for students and pharmacists / by Charles Caspari, Jr.

  • Caspari, Charles, 1850-1917.
Date:
1895
Catalogue details

Licence: Public Domain Mark

Credit: A treatise on pharmacy for students and pharmacists / by Charles Caspari, Jr. Source: Wellcome Collection.

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    Answer Proof. 1.5X10 = 15 pounds of 3 % bark. 15 X 3 = 45 1.5X10 = 15 3.5 a 15 X 3.5 = 52.5 1.0 X 10 = 10 a ^ u 10 X 4 = 40 1.0X10 = 10 u g a u 10 X 6 = 60 3 5X 10 = 35 6.5 11 35 X 6.5 = 227.5 85 425.0 If a definite quantity of a mixture is required, the quantity of each simple may be ascertained by multiplying its proportional by the ratia which the total quantity required bears to the sum of the propor- tionals of all the simples. Example: How many grammes of powdered opium of 9, 12, 15, 16, and 17 per cent, morphia strength must be taken to produce 250 grammes of a mixture containing 14 per cent, of morphia. 14 9 12- 15— 16 17 3 + 2 5 X 13.89 = 69.45 1X13.89 = 13.89 2X 13.89 = 27.78 5 X 13.89 = 69.45 5 X 13.89 = 69.45 18 the sum of proportionals 250.02 250 18 = 13.89, ratio of required quantity to the sum of the proportionals. Answer: 69.45 Gm. each of 9 per cent., 16 per cent., and 17 per cent, opium, 13.89 Gm. of 12 per cent, opium, and 27.78 Gm. of 15 per cent, opium. The foregoing rule can also be applied to a mixture of liquids of different specific gravities. Example: A pharmacist is called upon to prepare 500 Cc. of a mixture of alcohol, spec. grav. 0.820; glycerin, spec. grav. 1.25; simple syrup, spec. grav. 1.817; and water; the mixture to contain 15 per cent, by volume of glycerin, and to have the same specific grav^ity as water. How will he proceed ? The 500 Cc. of mixture must weigh 500 Gm. in order to have the same specific gravity as water; 15 per cent, of 500 is 75, and 75 Cc. of glycerin of 1.25 specific gravity will weigh 93.75 Gm.; this leaves 425 Cc. as the volume of the alcohol, syrup, and water mix- ture, the weight of which must be 500 — 93.75 = 406.25 Gm. Such a mixture would have the specific gravity 0.9559, for 406.25-^425 = 0.9559, and the necessary quantity of each ingredient may be ascertained by alligation, thus 0.0441 0.9559 425 0.820z 1.000- 1.317- 0.3611 = 0.4052 proportional of alcohol. 0.1359 water. 0.1359 simple syrup. 0.6770 sum of the proportionals. 0.6770 = 627.8, ratio of required quantity to the sum of the proportionals.. 627.8 X 0.4052 = 254.384 Cc of alcohol. 627.8 X 1 359 = 85.318 Cc of water. 627.8 X 1.359 = 85.318 Cc of syrup.