10. Calculations II: Compounding Flashcards
Common IV Fluids: Normal Saline content
0.9% NaCl in water
Common IV Fluids: 1/2 NS content
0.45% NaCl in water
Common IV Fluids: 1/4 NS content
0.225% NaCl in water
Common IV Fluids: D5W content
5% dextrose in water
Common IV Fluids: D20W
20% dextrose in water
Common IV Fluids: D5NS
5% dextrose and 0.9% NaCl in water
Common IV Fluids: D51/2NS
5% dextrose and 0.45% NaCl in water
Rx “Prepare a 3% w/w coal tar preparation; qs with petrolatum to 150g.”
How many grams of petrolatum is required to compound this prescription?
(0.03*150g) = 4.5g coal tar
150g - 4.5g = 145.5 g petrolatum
Percent weight-in-volume (%w/v) units
g per 100 mL
Percent volume-in-volume (%v/v) units
mL per 100mL
Percent weight-in-weight (%w/w) units
g per 100 g
Express 0.04% as a ratio strength
0.04 / 100 = 1 part / x
x = 100 / 0.04 = 2500
1:2500
50mg of drug in 50mL of solution
Express conc as a ratio strength
50mg = 0.05g
0.05g/50mL = 0.1g / 100mL = 1 part / x
x = 100/0.1 = 1000
1:1000
Ratio strength to percentage strength math
% strength = 100 / ratio strength
Percent strength to ratio strength
Ratio strength = 100 / % strength
Express 0.00022% (w/v) as part per million (PPM)
0.00022g/100mL = x parts / 1,000,000
X = 2.2 PPM
Water has a specific gravity of ___
1; 1g water = 1mL water
____ is the ratio of the density of a substance to the density of water
Specific gravity
Substances with a specific gravity <1 are (lighter/heavier) than water
lighter
Substances with a specific gravity >1 are (lighter/heavier) than water
heavier
What is the specific gravity of 150mL of glycerin weighing 165g?
SG = 165g/150mL = 1.1
Specific Gravity (SG) formula
SG = g/mL
Q1C1 formula
Q1 x C1 = Q2 x C2
Q1 = old quantity
C1 = old conc
Q2 = new quantity
C2 = new conc
Alligation example question
Prepare 80g of a 12.5% ichthammol ointment with 16% and 12% ichthammol ointments. How many grams of each?
16% - 12.5% = 3.5 parts of lower conc (12%)
12% - 12.5% = 0.5 parts of higher conc (16%)
80g/ 4 parts = 20g per part
0.5 parts of 16% = 10g of 16%
3.5 parts of 12% = 70g of 12%
Alligation example question
Prepare 1% morphine sulfate oral solution. 120mL bottle of morphine sulfate labeled 20mg/5mL and a 240mL bottle of morphine sulfate labeled 100mg/5mL
How much of 100mg/5mL should be added to the 20mg/5mL bottle?
20mg/5mL = 0.4%
100mg/5mL = 2%
2%-1% = 1 part of lower conc (0.4%)
0.4-1% = 0.6 part of lower conc (2%)
120mL of 20mg/5mL bottle (0.4%) = 1 part
0.6 part of 120mL = 72mL of higher conc (2%)
___ is the measure of total # of particles (or solutes) per L of solution
Osmolarity (Osmol/L or mOsmol/L)
Osmolarity will always need to be normalized to a volume of ___
1L
mOsmol/L Formula
mOsmol/L = (Wt of substance (g/L)) / (MW (g/mol)) * (# of particles) * 1000
1. Add up the # of particles when compound dissociates
2. Calculate $ of grams of compound present in 1 L
3. Use MW to solve problem
Note: milliosomole calculations do not normalize to 1 liter
Dissociation factor (i) based on number of dissociated ions
i = 1 + 0.8 (for each additional dissociated ion)
Example
4 dissociated ions = 1 + (0.8*3) // i = 3.4
Sodium chloride equivalent or “E value” formula
E = (58.5)(i) / (MW of drug)(1.8)
Calculate E value for mannitol (MW=182g/mol)
58.5(1) / 182(1.8) = 0.18
T/F: For monovalent species, the numeric value of the milliequivalent and millimole are identical
True
mEQ formula
mEq = mg*valence / MW
