Calculations Compounding II

21. Express 0.00022% w/v as PPM. Round to the nearest tenth.

0.00022g/100= x/1,000,000 parts x=2.2 PPM

15. Express 0.04% as a ratio strength.

0.04/100= 1 part/ x x=2500

2. How many grams of NaCl are in 500 ml of 1/2 NS. Round to nearest 100th.

0.45 g/100 ml = x/ 500ml x=2.25 grams

55. How many moles are equivalent to 875 milligrams of aluminum acetate MW = 204..... Round to three decimal places.

0.875/204 = .0043

1. How many Grams of NaCl are in one liter of normal saline?

0.9 g/100 ml = x/ 1000ml X=9 grams

Express 1:4000 as a percentage strength.

1 part/ 4000 parts = x/100 x=0.025%

53. The pharmacist receives an order for 10 ml of tobramycin 1% ophthalmic solution. He has tobramycin 40 mg/ml solution. Tobramycin does not dissociate and has a mw of 467. Find the E value for tobramycin and determine how many milligrams of NaCl are needed to make the solution isotonic?

1. E= 58.5 (1)/ 467(1.8)= .07 2. 0.9/100= X/10 ml = .09 3. 1g/100ml= x/10 = .1 g or 100mg 4. 100 mg X .07= 7 mg 5. 90-7= 93 mg

50. The E-value for ephedrine sulfate is 0.23. How many grams of sodium chloride are needed to compound the following prescription? Round to 3 decimal places.

1. First determine the amount of NaCl to make this Isotonic so..... 0.9/100= x/30= 0.27g 2. Determine amount of sodium chloride represented from ephedrine sulfate. 0.4 g X 0.23 = 0.092 of sodium chloride. 3. Subtract step 2 from step 1 to determine the total amount of NaCl needed to prepare an isotonic prescription. 0.092-0.27= .178 of NaCl are needed to make an isotonic solution.

41. What is the osmolarity, in mOsmol/L of normal saline? Mw=58.5 round to the nearest whole number.

1. How many particles does NaCl dissociate to? 2 2. Calculate the number of grams present in one liter .9/100=x/1000 X=9 3. Solve using the molecular weight mOsmol/L= 9/58.5 X 2 X1000 = 308

43. How many milliosmoles of CaCl2 Mw=111 are represented in 150ml of a 10% w/v calcium chloride solution?

1. How many particles does it dissociate to? 2. Calculate the number of grams present in one liter. 10/100=X/150 X=15 3. 15/111 X 3 X 1000= 405

42. What is the osmolarity in mOsmol/L of D5W? MW 198

1. How many particles does it dissociate to? Answer is 2 2. Calculate the number of grams present in one liter. 5/100= X/1000 X=50 3. Use the molecular weight to solve. 50/198 X 1 X 1000 = 252.5

What is the weight of 750 ml of concentrated acetic acid (SG=1.2)?

1.2= x / 750 ml x=900 g

54. How many moles of anhydrous magnesium sulfate (mw=120.4) are present in 250 grams of the substance?

250/120.4= 2.08

32. A pharmacist has an order for parenteral nutrition that includes 550 ml of D70%. The pharmacist checks the supplies and finds the closest strength he has is D50%. How many ml of D50% will provide an equal energy requirement?

550 ml X 0.70 = x ml X 0.50 x= 770 ml of D50%

Calculate the E value for mannitol MW 182. Round to the nearest tenth

58.5(1)/182(1.8)= 0.18

Calculate the E value for potassium iodide, which dissociates into 2 particles MW 166. Round to two decimal places.

58.5(1.8)/166(1.8)= 0.35

7. If 1,250 grams of a mixture contains 80 grams of a drug, what is the percentage strength (W/W) of the mixture? Round to the nearest tenth.

80 g/ 1250 g = x/100 x= 6.4%

56. How many millimoles of sodium phosphate MW =138 are present in 90 g of substance? Round to the nearest whole number.

90,000/138 = 652.17

Alligation

Alligation is used to obtain a new strength (percentage) that is between two strengths the pharmacist has in stock. It is used when the problem deals with three [ ]s.

37. A pharmacist is asked to prepare 80 grams of a 12.5% ichthammol ointment with 16% and 12% ichthammol ointment that she has on stock. How many grams of the 16% and the 12% ointment are required?

Higher (16%)- - - - - -Label- - - - - - - -0.5 parts of 16% \ / \ / \ / 12.5% Desired / \ / \ / \ Lower 12%- - - - - - - -label- - - - - - - -3.5 parts of 12% 1. Divide the total weight (80 g) by the number of parts to get the weight per part. 80 g/ 4 parts= 20 grams per part 2. Take the amount per part (20 g) and multiply it by the parts from each of the concentrations (From high to low) 0.5 parts of 16% X 20g/part= 10g of the 16% ichthammol ointment 3.5 parts of the 12% X 20g/part= 70 g of the 12% ichthammol ointment. Once together there will be 80 grams of the 12.5% ointment

38. A Pharmacist is asked to prepare 1 gallon of tincture containing 5.5% iodine. The pharmacy had 3% iodine tincture and an 8.5% iodine tincture in stock. How many milliliters of the 3% and 8.5% iodine tincture should be used? Use 1 gallon = 3,785 ml

Higher (8.5%)- - - - - -Label- - - - - - - -2.5 parts of 8.5% \ / \ / \ / 5.5% Desired / \ / \ / \ Lower 3%- - - - - - - -label- - - - - - - -3.0 parts of 3% 1. Divide the total weight (3,785 ml) by the number of parts to get the weight per part. 3785 ml/ 5.5 parts= 688.2 per part 2. Take the amount per part (688.2) and multiply it by the parts from each of the concentrations (From high to low) 2.5 parts of 8.5% X 688.2/part= 1721 of the 8.5% 3.0 parts of the 3% X 688.2/part= 2064 of the 3% . Once together there will be 3785 ml of the 5.5%

Moles and Millimoles

Mol is the molecular weight of a substance in grams or g/m. A mmol is 1/1,000 of the molecular weight in grams or 1/1,000 of a mole. For monovalent species, the numeric value of the millliequivalent and millimole are identical.

Percentage Strength

Number/ratio as fraction of 100 Expressed as -%w/v g/100ml -%v/v ml/100ml -%w/w g/100g

Dilution and Concentration

Q1 x C1 = Q2 x C2

What is the specific gravity of 150 ml glycerin weighing 165 grams? Round to the nearest tenth.

SG= 165 grams glycerin/150 ml water =1.1; it is heavier than water

milliequivalents

The amount in mg of a solute equal to 1/1000 of its gram equivalent weight, taking into account the valence of the ions. Quantity of particles is important but so is the electrical charge.

Osmolarity

The measure of the total number of particles or solutes per liter of solution. To be defined as osmoles/liters or more commonly as mOsmol/L. Solutes can be either ionic such as NaCl which dissociates into two solutes in solution that being Na+ and Cl- or non ionic such as glucose and urea which do not dissociate.

Specific Gravity

The ratio of the density of a substance to the density of water. Water has a SG of 1. 1 g water= 1 ml water -Substances with an SG less than one are lighter than water and substances with an SG greater than one are heavier than water. SG= Weight of substance (G)/ Weight equal volume of water(G) simplified to SG=g/ml

Isotonicity

equal concentrations outside and inside a cell, which causes no change in cell volume. E= 58.5(i)/ (MW of Drug) X (1.8) Once E-value is determined then... 1. Calculate the total amount of NaCl needed to make the final product/prescription isotonic by multiplying 0.9% NS. 2. Calculate the amount of NaCl represented by the drug. To do this, multiply the total drug amount in milligrams or grams by the "E-Value) 3. Subtract step 2 from step 1 to determine the total amount of NaCl needed to prepare an isotonic prescription.

Osmolarity formula

mOsmol/L = Wt of substance (g/L)/MW (g/mole) X # of particles X 1000 Steps 1. Add up the number of particles into which the compound dissociates. 2. Calculate the number of grams of the compound present in 1L. 3. Use the molecular weight to solve the problem.

mols and mmols formula

mols=g/MW mmols = mg/MW