Concentration Calculations in Pharmaceutical Solutions

1

Molar Mass Calculations

a) Calculate the molar mass of urea (NH₂)₂CO.
b) Calculate the molar mass of nitroglycerin CH₂ONO₂CHONO₂CH₂ONO₂.
c) Calculate the molar mass of antiseptic nitrofural.

Part A: Urea (NH₂)₂CO

Identify atomic composition:

2 Nitrogen (N) atoms
4 Hydrogen (H) atoms
1 Carbon (C) atom
1 Oxygen (O) atom

Find atomic masses:

N = 14.01 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol

Calculate molar mass:

\(M = 2(14.01) + 4(1.01) + 12.01 + 16.00\)

\(M = 28.02 + 4.04 + 12.01 + 16.00 = 60.07 \text{g/mol}\)

Part B: Nitroglycerin (C₃H₅N₃O₉)

Identify atomic composition:

3 Carbon (C) atoms
5 Hydrogen (H) atoms
3 Nitrogen (N) atoms
9 Oxygen (O) atoms

Calculate molar mass:

\(M = 3(12.01) + 5(1.01) + 3(14.01) + 9(16.00)\)

\(M = 36.03 + 5.05 + 42.03 + 144.00 = 227.11 \text{g/mol}\)

Part C: Nitrofural (C₈H₆N₄O₇)

Identify atomic composition:

8 Carbon (C) atoms
6 Hydrogen (H) atoms
4 Nitrogen (N) atoms
7 Oxygen (O) atoms

Calculate molar mass:

\(M = 8(12.01) + 6(1.01) + 4(14.01) + 7(16.00)\)

\(M = 96.08 + 6.06 + 56.04 + 112.00 = 270.18 \text{g/mol}\)

Molar mass calculations are fundamental for preparing pharmaceutical solutions with precise concentrations. Urea is used in topical dermatological products, nitroglycerin for angina treatment, and nitrofural as an antiseptic.

a) Urea: 60.07 g/mol
b) Nitroglycerin: 227.11 g/mol
c) Nitrofural: 270.18 g/mol

2

Ascorbic Acid Solution Volume

What volume of solution with a molar concentration of 0.01 mol/L can be prepared from 1.76 g of ascorbic acid?

Step-by-Step Solution:

Calculate molar mass of ascorbic acid (C₆H₈O₆):

\(M = 6(12.01) + 8(1.01) + 6(16.00) = 176.14 \text{g/mol}\)

Calculate moles of ascorbic acid:

\(n = \frac{\text{mass}}{\text{molar mass}} = \frac{1.76}{176.14} = 0.0100 \text{mol}\)

Use molarity formula to find volume:

\(M = \frac{n}{V}\)

Rearrange to solve for volume:

\(V = \frac{n}{M} = \frac{0.0100 \text{mol}}{0.01 \text{mol/L}} = 1.00 \text{L}\)

Convert to appropriate units:

\(1.00 \text{L} = 1000 \text{mL}\)

Ascorbic acid (vitamin C) is commonly used in pharmaceutical formulations. This calculation demonstrates how to prepare solutions with specific molar concentrations, which is essential for creating standardized pharmaceutical preparations.

1.00 L (1000 mL) of 0.01 mol/L solution can be prepared

3

Boric Acid Solution Preparation

Calculate the mass of boric acid required to prepare one 10 ml bottle of 3% solution (solution density is 1.01 g/ml).

Step-by-Step Solution:

Understand 3% mass percent concentration:

3% = 3 g boric acid per 100 g solution

Calculate mass of 10 mL solution:

\(\text{mass} = \text{volume} \times \text{density} = 10 \text{mL} \times 1.01 \frac{\text{g}}{\text{mL}} = 10.1 \text{g}\)

Calculate mass of boric acid needed:

\(\text{mass of boric acid} = 3\% \times 10.1 \text{g} = 0.03 \times 10.1 = 0.303 \text{g}\)

Convert to milligrams (for pharmaceutical precision):

\(0.303 \text{g} = 303 \text{mg}\)

Boric acid solutions are commonly used as antiseptics for eye washes. The 3% concentration is a standard formulation that provides antimicrobial activity without causing significant irritation to sensitive tissues.

0.303 g (303 mg) of boric acid is required

4

Nitroglycerin Dosage Calculation

Calculate how many nitroglycerin tablets containing 0.1 mg of active substance can replace 8 ml of solution containing 1 mg/ml of nitroglycerin.

Step-by-Step Solution:

Calculate total nitroglycerin in solution:

\(\text{Total nitroglycerin} = 8 \text{mL} \times 1 \frac{\text{mg}}{\text{mL}} = 8 \text{mg}\)

Determine number of tablets needed:

\(\text{Number of tablets} = \frac{\text{Total nitroglycerin needed}}{\text{Nitroglycerin per tablet}} = \frac{8 \text{mg}}{0.1 \text{mg/tablet}} = 80 \text{tablets}\)

Verify the calculation:

\(80 \text{tablets} \times 0.1 \frac{\text{mg}}{\text{tablet}} = 8 \text{mg}\)

This matches the amount in the solution.

Nitroglycerin is used for acute angina pectoris. This calculation demonstrates the importance of precise dosage conversion between different pharmaceutical forms, which is critical for patient safety and therapeutic effectiveness.

80 nitroglycerin tablets are needed

5

Saline Solution Preparation

How to prepare 200 ml of physiological saline solution (0.15 mol/L NaCl) from a one-molar NaCl solution?

Step-by-Step Solution:

Use dilution formula:

\(C_1V_1 = C_2V_2\)

Where:

C₁ = 1 mol/L (concentration of stock solution)
V₁ = ? (volume of stock solution needed)
C₂ = 0.15 mol/L (desired concentration)
V₂ = 200 mL = 0.2 L (desired volume)

Solve for V₁:

\(V_1 = \frac{C_2V_2}{C_1} = \frac{0.15 \text{mol/L} \times 0.2 \text{L}}{1 \text{mol/L}} = 0.03 \text{L}\)

Convert to milliliters:

\(0.03 \text{L} = 30 \text{mL}\)

Determine volume of water to add:

\(\text{Water volume} = 200 \text{mL} - 30 \text{mL} = 170 \text{mL}\)

Physiological saline (0.9% NaCl) is isotonic with blood and commonly used for wound irrigation. This calculation shows how to properly dilute concentrated stock solutions to prepare standard pharmaceutical solutions.

Mix 30 mL of 1M NaCl with 170 mL of distilled water

6

Nitrofural Solution Preparation

How to prepare 100 ml of a 0.02% nitrofural solution from Furacillin powder (1 packet = 10 mg, solution density = 1 g/ml)?

Step-by-Step Solution:

Understand 0.02% concentration:

0.02% = 0.02 g nitrofural per 100 g solution

Since density = 1 g/ml, 100 g solution = 100 ml solution

Calculate mass of nitrofural needed for 100 ml:

\(\text{Mass} = 0.02\% \times 100 \text{g} = 0.0002 \times 100 = 0.02 \text{g} = 20 \text{mg}\)

Determine number of Furacillin packets needed:

\(\text{Number of packets} = \frac{20 \text{mg}}{10 \text{mg/packet}} = 2 \text{packets}\)

Prepare the solution:

Dissolve 2 packets (20 mg) of Furacillin powder in distilled water to make 100 ml of solution.

Nitrofural (Furacillin) is used for treating skin burns. This calculation demonstrates how to prepare a specific percentage solution from dry powder, which is a common pharmaceutical preparation method for topical antiseptics.

Dissolve 2 packets (20 mg) of Furacillin powder in distilled water to make 100 ml of solution

7

Physiological Saline for Dehydration

Calculate the volume, molarity and normality of physiological saline solution (0.9% NaCl) for a 70 kg patient (density = 1.003 g/cm³).

Step-by-Step Solution:

Calculate volume of saline needed:

\(\text{Volume} = 20 \frac{\text{mL}}{\text{kg}} \times 70 \text{kg} = 1400 \text{mL}\)

Calculate mass of NaCl in solution:

\(\text{Mass of NaCl} = 0.9\% \times \text{mass of solution}\)

Mass of solution = volume × density = 1400 mL × 1.003 g/mL = 1404.2 g

\(\text{Mass of NaCl} = 0.009 \times 1404.2 = 12.64 \text{g}\)

Calculate molarity:

Molar mass of NaCl = 58.44 g/mol

\(\text{Moles of NaCl} = \frac{12.64}{58.44} = 0.2163 \text{mol}\)

\(\text{Molarity} = \frac{0.2163 \text{mol}}{1.4 \text{L}} = 0.155 \text{mol/L}\)

Calculate normality:

For NaCl, normality = molarity since it dissociates into 2 ions (Na⁺ and Cl⁻)

\(\text{Normality} = 0.155 \text{N}\)

Physiological saline (0.9% NaCl) is isotonic with blood and commonly used for IV rehydration. The molarity and normality calculations help ensure the solution has the correct ionic strength for therapeutic use without causing hemolysis.

Volume: 1400 mL
Molarity: 0.155 mol/L
Normality: 0.155 N

8

Isotonic Glucose Solution

Calculate the mass percent, molar and normal concentration of an isotonic glucose solution containing 100 g of C₆H₁₂O₆ per 2 liters (density = 1.0 g/ml).

Step-by-Step Solution:

Calculate mass percent:

Mass of solution = 2 L × 1000 mL/L × 1.0 g/mL = 2000 g

\(\text{Mass percent} = \frac{100 \text{g}}{2000 \text{g}} \times 100\% = 5.0\%\)

Calculate molar concentration:

Molar mass of glucose = 180.16 g/mol

\(\text{Moles of glucose} = \frac{100}{180.16} = 0.555 \text{mol}\)

\(\text{Molarity} = \frac{0.555 \text{mol}}{2 \text{L}} = 0.278 \text{mol/L}\)

Calculate normal concentration:

Glucose is a non-electrolyte (doesn't dissociate), so normality = molarity

\(\text{Normality} = 0.278 \text{N}\)

Verify isotonicity:

Isotonic glucose solution is typically 5% (mass percent), which matches our calculation.

Isotonic glucose solutions are used for detoxification and as a source of energy. The 5% concentration is isotonic with blood, preventing cell damage during intravenous administration.

Mass percent: 5.0%
Molar concentration: 0.278 mol/L
Normal concentration: 0.278 N

9

Hemoglobin Content Analysis

Determine whether 2.5 mmol/L hemoglobin content corresponds to the norm if the lower limit of the norm for women is 139 g/L (Mr(Hb) = 67000 g/mol).

Step-by-Step Solution:

Convert hemoglobin concentration to g/L:

2.5 mmol/L = 0.0025 mol/L

\(\text{Concentration (g/L)} = \text{molar concentration} \times \text{molar mass}\)

\(= 0.0025 \frac{\text{mol}}{\text{L}} \times 67000 \frac{\text{g}}{\text{mol}} = 167.5 \frac{\text{g}}{\text{L}}\)

Compare with normal range:

Lower limit of normal = 139 g/L

Calculated value = 167.5 g/L

Determine if within normal range:

167.5 g/L > 139 g/L, so the value is above the lower limit of normal

Hemoglobin concentration is a critical blood parameter. This conversion between molar and mass concentrations is essential for interpreting laboratory results in clinical practice and determining if values fall within normal ranges.

Yes, 2.5 mmol/L (167.5 g/L) is above the lower limit of normal (139 g/L)