(41) In the following prescription what mercurial salt does the patient get, and how much at a dose?

M.

Sig

Teaspoonful in water after meals.

Discussion.

Mercuric chloride with potassium iodide forms the red mercuric iodide, which, in the presence of an excess of the potassium iodide, forms the potassio-mercuric iodide. The molecule of mercuric chloride is HgCl2 with a molecular weight of about 270. The molecule of red mercuric iodide is Hgl2 with a molecular weight of about 450, so for each 270 parts of the chloride used 450 parts of the iodide are formed. If 1/24 of a grain of the chloride is ordered for each dose (2 grains in 48 doses), the resulting iodide of each dose is found by the following:

1/24: x:: 270:450 = about 1/14 grain at each dose of red mercuric iodide in the form of potassio-mercuric iodide.

(42) What does the patient get in the following prescription, and how much sodium bicarbonate is required to neutralize the acid?

M.

Sig

Tablespoonful in water each evening at 4, 6, 8 and 10.

Discussion.

Sodium salicylate and carbon dioxide are formed so the patient gets a freshly formed sodium salicylate in water charged with carbon dioxide.

Salicylic acid had the chemical formula HC7H5O3 which has the molecular weight of about 138. Sodium bicarbonate has a formula NaHCO3, which has a molecular weight of about 84. So for each 138 grains of the acid 84 grains of the sodium salt will be required. The prescription calls for 240 grains of the acid, so the amount of bicarbonate is found as follows:

240: x:: 138: 84 or 146 grains of the sodium bicarbonate. It is usually considered better for therapeutic reasons to prescribe an excess of the bicarbonate, say 4 drachms for this prescription.

(43) Demonstrate why the compound tincture of cinchona cannot be used in the treatment of malaria.

Discussion.

This preparation represents 10 per cent. of cinchona, and cinchona contains about 5 per cent. of alkaloids (about 4 per cent. of quinine), so a teaspoonful (60 minims) of the tincture represents 6 grains of cinchona, about 1/20 of which is active (1/20 of 6 is 6/20, or about 1/3 grain). If it contains about 1/3 grain of alkaloid to 1 fluidrachm, it would require 3 fluidrachms to yield 1 grain, or 90 fluidrachms (nearly a pint), to give a day's treatment. The alcohol, tannic acid, bad taste, etc., would be intolerable.

(44) Write a prescription for quinine for malaria in a child 6 years old, based on 30 grains per day for an adult.

Discussion.

A child 6 years old should receive one-third the adult dose (Young's rule); this would be 10 grains per day. Following the custom of dividing into about five daily doses, it would give 2 grains per dose. A child this age can seldom swallow a capsule and would not take a bitter solution. In making a tasteless mixture a nearly insoluble salt should be used. The sulphate is the most desirable common salt. Two grains of this can be conveniently disguised in a teaspoonful of syrup. The usual three (or more) days intensive treatment should be ordered.

Name and age of patient.

Date,

M.

Sig

Teaspoonful each morning at 7, 8, 9, 10 and 11. (Shake-label.)

Signature.

(45) What would be the daily amount of quinine for the first few days of the treatment of malaria in an adult weighing 75 pounds if the average adult dose was 30 grains?

Discussion.

An"average dose"means the dose for an average man. In using drugs for systemic effect by absorption into the blood, it is usually intended to secure a certain percentage concentration-so much of the drug to so much blood. The probable amount of blood is based largely on the size of patient, so a patient about half the average size would usually receive half the average dose, which would give about 15 grains of quinine for this patient.

(46) One doctor gives 30 grains quinine sulphate, another gives 30 grains quinine hydrobromide. Demonstrate which gives the most of the alkaloid.

Discussion.

Quinine sulphate has the formula (C20H24O2N2)2H2SO4 + 7H2O, which has the molecular weight 872, the quinine 2(C20H24O2N2) being 648 and the sulphate radical and water (H22SO4 + 7H2O) being 224. Quinine hydrobromide has the formula C20H24O2N2HBr + H2O, which has the molecular weight 423, the quinine being 324 and the hydrobromide radical and water (HBr + H2O) being 99. So quinine sulphate is 74 per cent. quinine (648/872), while quinine hydrobromide is 76 per cent. quinine (324/423), 74 per cent. of 30 is 22;2 (grains of quinine in 30 grains of quinine sulphate), 76 per cent. of 30 is 22.8 (grains of quinine in 30 grains of quinine hydrobromide).

(47) Demonstrate which official quinine salt contains the larger percentage of quinine.

Discussion.

The official salts, with the formulae and molecular weights of their acid radicals and water of crystallization, are as follows: Bisulphate (H2SO4+(7H2O)=224), dihydrochloride (2HC1 = 72), hydrobromide (HBr + H2O = 98), hydrochloride (HC1 + 2H2O = 72), sulphate (H2SO4-H-7H2O = 224), salicylate (HC7H5O3 + H20 - 156), the radical in quinine and urea hydrochloride (HC1CO(NH2)2HC1 + (5H2O)H20 = 222 ). Quinine sulphate contains 2 quinine radicals to each acid radical, so the 224 should be divided by 2. Quinine tannate is not constant in composition, but as it only contains from 30 to 35 per cent. of quinine, it is out of consideration. As the quinine radical (C20H24O2N2) is constant in all, the salt having the lightest radical naturally has the largest percentage of quinine alkaloid. This shows the hydrochloride and the dihydrochloride to contain the largest percentage of the alkaloid.