Equations and Calculations Flashcards
Clark’s Rule
[weight (lbs) × average adult dose] / 150 = children’s dose
Young’s Rule
[Age (yr) × average adult dose] / [Age (yr) + 12] = children’s dose
Fried’s Rule
[Age (months) × average adult dose] / 150 = infant’s dose
Mg/kg/day
mg dose / kg of patient weight / over 24 hour period
A form of daily dose. Use as conversion factor for determining patient dose based on their weight.
Temperature Conversion
°F = (1.8°C) + 32
°C = (°F - 32) / 1.8
% mark up
[mark up ($) / item cost ($)] × 100%
Selling price
[number of items x item cost ($)] + [number of items × % mark up ($)]
% gross profit
[markup ($) / selling price ($)] × 100%
Net profit
Selling price ($) - [item cost ($) - overhead ($)]
% concentration and dilution
(old vol) × (old %) = (new vol) × (new %)
Allegation Alternate Method
High % ___________ High Parts (Parts of higher % needed for product)
________ Desired % ___________
Low % ____________ Low Parts (Parts of lower % needed for product)
High Parts = Desired % - Low %
Low Parts = High % - Desired %
Total Parts = High Parts + Low Parts
High Volume = [High Parts/Total Parts] x Volume of Final Product = volume of higher % required to make desired volume of desired %
Low Volume = [Low Parts/Total Parts] x Volume of Final Product = volume of lower % required to make desired volume of desired %
Flow rate
(Volume to be infused) / (amount of time being infused)
% (w/w)
for solid drug in solid vehicle. Units in numerator and denominator must match. What those units are is not set.
% (v/v)
for liquid drug in liquid vehicle. Units in numerator and denominator must match. What those units are is not set.
% (w/v)
for solid drug in liquid vehicle. Units of numerator are always grams of solid drug and denominator is always milliliters of liquid vehicle.
x% (w/w) product to ratio
x parts solid drug and 100 parts solid vehicle, or x:100
x% (v/v) product to ratio
x parts liquid drug and 100 parts liquid vehicle, or x:100
x% (w/v) product to ratio
x parts solid drug and 100 parts liquid vehicle, or x:100.
Allegation Medial Method
Amount of A Used x Concentration of A = result of multiplication for A
Amount of B Used x Concentration of B = result of multiplication for B
Amount of C Used x Concentration of C = result of multiplication for C
Sum of Amounts = Amount of A + Amount of B + Amount of C
Sum of Results = Result for A + Result for B + Result for C
Sum of Results / Sum of Amounts = Average Strength of Mixture of A, B, and C (often in percent)
This can be used to check answers for allegation alternate method.
Monovalent Milliequivalents (mEq)
Used for electrolytes. 1 mEq = # mmols of H+ or OH- that reacts with 1 mmol of Ion in molecule.
Monovalent: 1 mEq = Atomic Weight (AW) or Molecular Weight (MW)
NaCl has MW of 58.5, so 1 mEq = 58.5 mg. Dissociates into 1 mEq of Na+ and 1 mEq of Cl-
Divalent Milliequivalents (mEq)
Used for electrolytes. 1 mEq = # mmols of H+ or OH- that reacts with 1 mmol of Ion in molecule.
Divalent: 1 mEq = Atomic Weight (AW) or Molecular Weight (MW) / 2
CaCl2 has MW of 111, so 1 mEq = 111 / 2 = 55.5 mg. Dissociates into 1 mEq of Ca2+ and 2 mEq of Cl-
Trivalent Milliequivalents (mEq)
Used for electrolytes. 1 mEq = # mmols of H+ or OH- that reacts with 1 mmol of Ion in molecule.
Trivalent: 1 mEq = Atomic Weight (AW) or Molecular Weight (MW) / 3
Al(OH)3 has MW of 78, so 1 mEq = 78 / 3 = 26 mg. Dissociates into 1 mEq of Al3+ and 3 mEq of OH-
If solution has 10g of KCl, how many mEq of K+ does it have? AW K+ = 39 and AW of Cl- = 35.5
MW of KCl = [1 x 39] + [1 x 35.5] = 74.5
mEq = 74.5 / 1 (K+ is monovalent) = 74.5 mg
10g KCl x [1000 mg / 1 g] = 10,000 mg KCl
10,000 mg KCl x [1 mEq / 74.5 mg] = 134 mEq of KCl
Solution with 10 g KCl has 134 mEq of K+ and 134 mEq of Cl-
Vial label says “Add 9.2ml diluent to get 10ml of 100 mg per ml solution.” How many ml of recon sol would provide 250mg dose?
[250 mg / 1 dose] x [1 ml / 100 mg] = 2.5ml / 1 dose
Vial label says “Add 9.2ml diluent to get 10ml of 100 mg per ml solution.” If accidently used 11 ml diluent, how many ml of recon sol would provide 250mg dose?
10 ml recon vol x [100 mg / 1 ml] = 1000 mg drug in vial
10 ml recon vol - 9.2 ml diluent = 0.8 ml power volume displacement
11 ml diluent added + 0.8 ml vol disp = 11.8 ml total recon vol
[250 mg / 1 dose] x [11.8 ml / 1000 mg] = 2.95 ml / 1 dose
How much diluent is needed to recon a vial with 5,000,000 units of drug to a concentration of 500,000 units per ml? Drug has volume displacement of 2 ml.
[5,000,000 u / vial] x [1 ml / 500,000 u] = 10 ml total vol / vial
10 ml total vol - 2 ml vol disp = 8 ml diluent needed
