Equations and Calculations

Question 1

Clark’s Rule

Answer 1

[weight (lbs) × average adult dose] / 150 = children’s dose

Question 2

Young’s Rule

Answer 2

[Age (yr) × average adult dose] / [Age (yr) + 12] = children’s dose

Question 3

Fried’s Rule

Answer 3

[Age (months) × average adult dose] / 150 = infant’s dose

Question 4

Mg/kg/day

Answer 4

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.

Question 5

Temperature Conversion

Answer 5

°F = (1.8°C) + 32
°C = (°F - 32) / 1.8

Question 6

% mark up

Answer 6

[mark up ($) / item cost ($)] × 100%

Question 7

Selling price

Answer 7

[number of items x item cost ($)] + [number of items × % mark up ($)]

Question 8

% gross profit

Answer 8

[markup ($) / selling price ($)] × 100%

Question 9

Net profit

Answer 9

Selling price ($) - [item cost ($) - overhead ($)]

Question 10

% concentration and dilution

Answer 10

(old vol) × (old %) = (new vol) × (new %)

Question 11

Allegation Alternate Method

Answer 11

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 %

Question 12

Flow rate

Answer 12

(Volume to be infused) / (amount of time being infused)

Question 13

% (w/w)

Answer 13

for solid drug in solid vehicle. Units in numerator and denominator must match. What those units are is not set.

Question 14

% (v/v)

Answer 14

for liquid drug in liquid vehicle. Units in numerator and denominator must match. What those units are is not set.

Question 15

% (w/v)

Answer 15

for solid drug in liquid vehicle. Units of numerator are always grams of solid drug and denominator is always milliliters of liquid vehicle.

Question 16

x% (w/w) product to ratio

Answer 16

x parts solid drug and 100 parts solid vehicle, or x:100

Question 17

x% (v/v) product to ratio

Answer 17

x parts liquid drug and 100 parts liquid vehicle, or x:100

Question 18

x% (w/v) product to ratio

Answer 18

x parts solid drug and 100 parts liquid vehicle, or x:100.

Question 19

Allegation Medial Method

Answer 19

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.

Question 20

Monovalent Milliequivalents (mEq)

Answer 20

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-

Question 21

Divalent Milliequivalents (mEq)

Answer 21

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-

Question 22

Trivalent Milliequivalents (mEq)

Answer 22

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-

Question 23

If solution has 10g of KCl, how many mEq of K+ does it have? AW K+ = 39 and AW of Cl- = 35.5

Answer 23

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-

Question 24

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?

Answer 24

[250 mg / 1 dose] x [1 ml / 100 mg] = 2.5ml / 1 dose

Question 25

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?

Answer 25

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

Question 26

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.

Answer 26

[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