Calculos Quimicos En Ingles

Concentration Expressions

D5w = Dextrose 5%

1 DL = means 1 Deciliter = 1/10 of the Liter = 100 ml

1 PPM = 1 gm /1,000,000 ml

5% W/V = 5 gm /100 ml

5% W/W = 5 gm /100g

1:50 W/V = 1 gm /50 ml

1:50 W/W= 1 gm/50 gm

1:50 V/V= 1 ml/50 ml

1 part in 500 part Phenol in Water = 1 gm / 500 ml (called Ratio strength)

1 teaspoonful = 5 ml
1 dessertspoonful =10 ml
1tablespoonful =15 ml

1 teacupful =120 ml
1 measuring cup =glassful = 240 ml

1 Kg = 2.2 Lb
1 Gallon = 3784 ml
1 Fz (Fluid Ounce) = 29.57 ml
1 Pint = 473 ml
1 Inch = 2.54 cm

1 Gallon = 4 quarters / 8 pints / 16 Fluid Ounce

9 C = 5F-160
Ex. Convert 212 C to F (9x 212 = 5 F -160, F= 212

Molarity = Moles of solute /Liters of Solution
Molality = Moles of solute/Kg of Solvent
Normality =Weight of solute (gms) /gm equivalent weight x volume (liters)

N.B. Normality + Molarity changes with Temp.
Molality doesn’t change by temp





Dilution from Stock
I think it’s better to name the quantity required by Volume required since it’s in mL

Also the expression Amount/Unit dose is simply the Final Concentration mg/ml

This type of equations has the expression (When Xml is diluted to Y ml a final conc. is ….) Or (when diluted X in Y using only numbers ex. Diluted 1 in 10)
The X is a Low volume while the Y is the higher volume

We use this expression to get the Dilution Factor and the needed outcome is usually the Weight of the stock solution.

Ex.1 How many grams of Terbutaline Sulphate would be required to prepare a 100 ml solution such that when 1 ml isdiluted to 5 ml, a final concentration of 500 mcg/ml of Terbutaline Sulphate would be produced


X mg/100 ml x 1ml /5ml = 0.5 mg/1 ml x = 250 mg Terbutaline Sulphate

Ex.2 what is the amount of magnesium chloride (Mgcl2.6 H20 ; MW= 203.3) required to prepare 100 ml of a solution such that 10 ml is diluted to 1 liter yields a solution containing 0.3 Meq Mg+2 /ml

Given:

Volume required= 100 ml, Dilution Factor = 10/1000,
Final Conc. = 0.3 Meq Mg+2/ml (has to be mg/ml)

Solution

Mgcl2.6 H20 ( Mg+2 + 2 Cl- + 6 H20
1 millimole Mgcl2.6 H20 ( 2 Meq Mg+2 = Mgcl2.6 H20 m.wt (203.3 mg)
0.3 Meq Mg+2 = X, X = 30.495 mg

Amount of Drug /required Volume x Dilution Factor = Final Conc.
Amount of Drug/ 100 x 10/1000 = 30.495 mg/ 1 mlAmount of Drug /10,000 = 30.495 / 1 , Amount of Drug = 304950 mg = 304.95 gm

Ex. A Hospital Pharmacy Technician adds by a syringe 20 ml of a concentrated sterile 2% w/v dye solution to a 250 ml commercial bag of saline what’s the concentration of the dye in Final solution ?

2% w/v = 2 Gm 100 ml
X Gm 20 ml, X = 40/100 = 0.4 gm

The 4 gm in the 20 ml have an additional 250saline = 0.4 gm 270 ml (Don’t use 250)
Y 100 ml

Y = 400/270 = 0.15% W/V
An important note when solving this don’t do 0.4 gm 20 mlY gm 100 ml, Y = 2% (wrong)

This is because The 0.4 gm Dye is the Total amount in 20 ml while if we use
0.4 Gm /20 ml we will get the amount of Dye in only 1 ml

Ex. Codeine Phosphate 200 mg
Dimetab Elixir qs 120 ml

Sig: 1 tsp t.i.d. P.C. and H.S.
What’s the amount of daily Codeine consumed

1 tea spoonful 3 times daily post cibum + Hora Somni5 ml x 3 after Meals + 5 ml at bed time = 20 ml

Take care not to be mixed up Tsp (tea spoonful) with Tbsp (table spoonful)
200 mg 120 ml
X mg 20 ml, X = 200x 20 /120 = 33 mg / dy , 8.25 mg / dose

EX.How many mg of the codeine base in each dose of the cough product used in question (mwt codeine = 299, mwt Codeine phosphate = 406)

Codeine phosphate ( Codeine base + Po4 2-
1...