Fundamentals of Medicinal and Pharmaceutical Chemistry FUNCHEM.4: Moles, Chemical Equations and Reactions. Composition of Physiological Solutions

FUNCHEM.4: Moles, Chemical Equations and Reactions. Composition of Physiological Solutions

Learning Outcomes

Upon completion of this module, students should be able to:

Define 'molecule', 'molecular formula', 'molecular mass', 'mole', and 'Avogadro’s number'.
Solve calculations involving moles.
Demonstrate the method of balancing chemical equations.
Define 'solution' and relate the human body to solutions in terms of blood plasma, extra- and intracellular body fluids.
Apply medical applications related to solutions.
Recall and calculate percentage solutions (w/v %, w/w %, mg %), parts per million (ppm), and molar concentrations.

What is an Atom?

An atom is the basic unit of an element that can enter into chemical combination.
It is the smallest part of an element that can undergo a chemical reaction.
An atom is composed of electrons, protons, and neutrons, along with other subatomic particles.

What are Molecules?

A molecule is an aggregate of at least two atoms in a definite arrangement, held together by chemical forces (also known as chemical bonds).
Diatomic molecule: Contains only two atoms, which can be of the same or different elements.
- Examples: $H2, N2, O2, Br2, HCl, CO$
Polyatomic molecule: Contains more than two atoms.
- Examples: $O3, H2O, NH3, CH4$
A molecule may contain atoms of the same element (e.g., $H2$ ) or atoms of two or more different elements (e.g., $H2O$ ).

Molecular Formula

The molecular formula shows the exact number of atoms of each element present in a molecule.
Example: Glucose has the molecular formula $C6H{12}O_6$ , indicating that each molecule contains $6$ carbon atoms, $12$ hydrogen atoms, and $6$ oxygen atoms.

Atomic Mass

Atomic mass is the mass of an atom expressed in atomic mass units (amu).
Definition of amu: One atomic mass unit (amu) is precisely defined as a mass exactly equal to one-twelfth ( $1/12$ ) the mass of one carbon-12 atom.
Average atomic mass: This is the weighted average of the atomic masses of all the naturally occurring isotopes of an element. This value is typically found on the periodic table.
- Example: Carbon ( $C$ ) has an atomic number of $6$ and an average atomic mass of $12.01$ amu.

Molecular Mass (or Molecular Weight)

The molecular mass is the sum of the atomic masses (in amu) of all the atoms in a molecule.
Example: For Sulfur Dioxide ( $SO_2$ )
- Atomic mass of Sulfur ( $S$ ) = $32.07$ amu
- Atomic mass of Oxygen ( $O$ ) = $16.00$ amu
- Molecular mass of $SO_2 = 1 imes (32.07 ext{ amu}) + 2 imes (16.00 ext{ amu}) = 32.07 + 32.00 = 64.07$ amu.
Relationship between Molecular Mass and Molar Mass: For any molecule, the numerical value of its molecular mass (in amu) is equal to its molar mass (in grams/mole).
- Thus, $1$ molecule of $SO_2 = 64.07$ amu.
- And $1$ mole of $SO2 = 64.07$ g of $SO2$ .

Chemical Reactions

A chemical reaction is a process in which a substance (or several substances) is transformed into one or more new and different substances.
A chemical equation is a symbolic representation that uses chemical symbols and formulas to illustrate what occurs during a chemical reaction, showing the reactants and products.

Balancing Chemical Equations: Glucose Synthesis Example

The chemical equation for glucose synthesis (photosynthesis):
- Unbalanced: $CO2 + H2O ightarrow C6H{12}O6 + O2$
- To be useful for quantitative analysis, chemical equations must be balanced, ensuring the number of atoms of each element is the same on both the reactant and product sides.
- Balanced: $6CO2 + 6H2O ext{ (sunlight)} ightarrow C6H{12}O6 + 6O2$

Measuring Quantities: The Need for the Mole

Individual atoms and molecules are extraordinarily small, making it impractical to measure them directly in grams. For instance, $6$ molecules of $CO_2$ weigh approximately $4.78 imes 10^{-22}$ g.
To work with realistic quantities in chemistry, we need a way to relate the macroscopic mass of a substance to the microscopic number of atoms or molecules it contains.

The Mole (mol): A Unit to Count Numbers of Particles

The mole (mol) is the SI unit for the amount of substance. It is defined as the amount of a substance that contains as many elementary entities (atoms, molecules, ions, etc.) as there are atoms in exactly $12.00$ grams of the isotope carbon-12 (carbon-12 is used as the reference standard).
Avogadro’s number ( $N_A$ ): This fundamental constant represents the number of elementary entities in one mole of any substance.
- $1 ext{ mol} = N_A = 6.0221415 imes 10^{23}$
Analogies for understanding the mole:
- A 'dozen' always means $12$ items (e.g., $1$ dozen eggs = $12$ eggs, $1$ dozen chickens = $12$ chickens).
- Similarly, $1$ mole of any substance always means $6.022 imes 10^{23}$ particles of that substance.
  - $1$ mole of atoms = $6.022 imes 10^{23}$ atoms.
  - $1$ mole of molecules = $6.022 imes 10^{23}$ molecules.

Molar Mass

Molar mass is defined as the mass in grams of $1$ mole of molecules or $1$ mole of atoms of a substance.
The unit for molar mass is grams per mole (g/mol).
For atoms: The molar mass (in g/mol) is numerically equal to the atomic mass (in amu).
For molecules: The molar mass (in g/mol) is numerically equal to the molecular mass (in amu).

Mole Calculations - Exercises and Examples

Exercise with Carbon (C, molar mass = 12.01 g/mol):
- $1$ mole of C has a mass of $12.01$ g and contains $6.022 imes 10^{23}$ atoms of carbon.
- $2$ moles of C has a mass of $24.02$ g and contains $1.2044 imes 10^{24}$ atoms of carbon.
- $0.5$ mole of C has a mass of $6.005$ g and contains $3.011 imes 10^{23}$ atoms of carbon.
Key Formulas for Mole Calculations:
- $ext{No. of moles (mol)} = rac{ ext{mass (g)}}{ ext{Molar mass (g/mol)}}$
- $ext{Mass (g)} = ext{No. of moles (mol)} imes ext{Molar mass (g/mol)}$
- $ext{Number of particles} = ext{No. of moles (mol)} imes N_A$
Summary of Conversions:
- To convert between mass and moles: Multiply or divide by molar mass.
- To convert between moles and number of particles: Multiply or divide by Avogadro's number ( $6.022 imes 10^{23}$ ).
Example 1: Calculating mass from moles (Zinc)
- Problem: How many grams of Zinc (Zn) are in $0.356$ mole of Zn? (Atomic mass of Zn = $65.38$ amu, so molar mass = $65.38$ g/mol)
- Solution: $ext{mass} = ext{No. of moles} imes ext{molar mass} = 0.356 ext{ mol} imes 65.38 ext{ g/mol} = 23.3 ext{ g}$
Example 2: Calculating moles and mass from number of atoms (Gold)
- Problem: A sample of gold (Au) contains $3 imes 10^{22}$ atoms. How many moles is this, and what is the mass of the sample? (Atomic mass of Au = $196.97$ amu, so molar mass = $196.97$ g/mol)
- Solution:
  - $ext{No. of moles} = rac{ ext{Number of atoms}}{N_A} = rac{3 imes 10^{22}}{6.022 imes 10^{23} ext{ atoms/mol}} hickapprox 0.05 ext{ mol}$
  - $ext{Mass} = ext{No. of moles} imes ext{Molar mass} = 0.05 ext{ mol} imes 196.97 ext{ g/mol} = 9.85 ext{ g}$
Example 3: Calculating molecules from mass (Glucose)
- Problem: How many glucose molecules are in $30$ g of glucose? (Molecular mass of glucose ( $C6H{12}O_6$ ) = $180.16$ amu, so molar mass = $180.16$ g/mol)
- Solution:
 - $ext{No. of moles} = rac{ ext{mass}}{ ext{Molar mass}} = rac{30 ext{ g}}{180.16 ext{ g/mol}} hickapprox 0.166 ext{ mol}$
 - $ext{No. of glucose molecules} = 0.166 ext{ mol} imes 6.022 imes 10^{23} ext{ molecules/mol} hickapprox 1.00 imes 10^{23} ext{ molecules}$

Reading Chemical Equations

Consider the balanced equation for glucose synthesis: $6CO2(g) + 6H2O(l) ext{ (sunlight)} ightarrow C6H{12}O6(s) + 6O2(g)$
State Symbols:
- (s) = solid
- (l) = liquid
- (g) = gas
- (aq) = dissolved in water (aqueous solution)
Components:
- Reactants: Substances on the left side of the arrow (e.g., $CO2, H2O$ ).
- Products: Substances on the right side of the arrow (e.g., $C6H{12}O6, O2$ ).
Numbers in Equations:
- Prefix numbers (coefficients): Refer to the whole molecule or formula unit (e.g., $6CO_2$ means $6$ molecules of carbon dioxide).
- Subscript numbers: Refer only to the atom immediately preceding them (e.g., $CO_2$ means one carbon atom and two oxygen atoms per molecule).
Interpreting Coefficients (Stoichiometry):
- For the reaction: $2Mg + O_2 ightarrow 2MgO$
 - This means $2$ atoms of Mg react with $1$ molecule of $O_2$ to produce $2$ formula units of MgO.
 - Crucially, this also means $2$ moles of Mg react with $1$ mole of $O_2$ to produce $2$ moles of MgO.
 - It translates to gram quantities based on molar masses: $48.6$ grams of Mg react with $32.0$ grams of $O2$ to produce $80.6$ grams of MgO. It does not mean $2$ grams Mg react with $1$ gram $O2$ to make $2$ g MgO.

Balancing Chemical Equations - Method

Place the coefficient of $1$ in front of the most complicated molecule on either the reactant or product side.
Balance each element sequentially, leaving oxygen and hydrogen for last.
Ensure the smallest whole number coefficients are used to balance the equation.

Example: Balancing the combustion of ethanol:
- Unbalanced: $C2H5OH(l) + O2(g) ightarrow H2O(l) + CO_2(g)$
- Step 1: Put $1$ in front of $C2H5OH$ (most complicated):
 - $1C2H5OH(l) + O2(g) ightarrow H2O(l) + CO_2(g)$
- Step 2: Balance C atoms (2 on left, need 2 on right):
 - $1C2H5OH(l) + O2(g) ightarrow H2O(l) + 2CO_2(g)$
- Step 3: Balance H atoms (6 on left, $2 imes H2O$ implies need $3H2O$ ):
 - $1C2H5OH(l) + O2(g) ightarrow 3H2O(l) + 2CO_2(g)$
- Step 4: Balance O atoms (Left: $1$ from $C2H5OH + 2$ from $O2 = 3 + X$ for $O2$ ; Right: $3$ from $3H2O + 4$ from $2CO2 = 7$ ). We need $7$ oxygen atoms on the left. We have $1$ from ethanol. So we need $6$ more, which will come from $3O_2$ .
 - Balanced: $C2H5OH(l) + 3O2(g) ightarrow 3H2O(l) + 2CO_2(g)$

Moles in Chemical Reactions: Calculating Amounts of Reactants and Products (Stoichiometry)

To calculate the amounts of reactants consumed or products formed in a chemical reaction:

Write a balanced chemical equation for the reaction.
Convert the given quantities of the reactant (in grams) into moles using its molar mass.
Use the mole ratio derived from the balanced equation to calculate the number of moles of the desired product formed or reactant consumed.
Convert the calculated moles of the product or reactant back into grams using its molar mass.

Example: Glucose Degradation
- Overall equation for the degradation of glucose ( $C6H{12}O6$ ) to carbon dioxide ( $CO2$ ) and water ( $H2O$ ) (Note: The provided slides imply the balanced equation for this specific problem as $C6H{12}O6 ightarrow 6CO2 + 6H2O$ for calculating purposes, though biological degradation typically includes oxygen as a reactant).
- Problem: If $856$ g of $C6H{12}O6$ is consumed, what is the mass of $CO2$ produced?
- Given Molar Masses: Glucose ( $C6H{12}O6$ ) = $180.2$ g/mol; Carbon Dioxide ( $CO2$ ) = $44.01$ g/mol.
- Step 1: Balanced Chemical Equation (as per the problem's context):
 - $C6H{12}O6 ightarrow 6CO2 + 6H_2O$
- Step 2: Convert grams of $C6H{12}O_6$ to moles:
 - $ext{Moles of } C6H{12}O6 = rac{ ext{mass (g)}}{ ext{Molar mass (g/mol)}} = rac{856 ext{ g}}{180.2 ext{ g/mol}} hickapprox 4.75 ext{ mol } C6H{12}O6$
- Step 3: Use mole ratio to find moles of $CO_2$ .
 - From the balanced equation, the mole ratio between $C6H{12}O6$ and $CO2$ is $1 ext{ mol } C6H{12}O6 : 6 ext{ mol } CO2$ .
 - $ext{Moles of } CO2 = 4.75 ext{ mol } C6H{12}O6 imes rac{6 ext{ mol } CO2}{1 ext{ mol } C6H{12}O6} = 28.5 ext{ mol } CO_2$
- Step 4: Convert moles of $CO2$ to grams of $CO2$ .
 - $ext{Mass of } CO2 = ext{Moles of } CO2 imes ext{Molar mass of } CO_2 = 28.5 ext{ mol} imes 44.01 ext{ g/mol} hickapprox 1254.3 ext{ g}$

What is a Solution?

A solution is a homogeneous mixture of two or more substances.
Solutions can exist in any state of matter (solid, liquid, or gas).
Solute: The substance that is being dissolved (present in the smaller amount).
Solvent: The substance that does the dissolving (present in the larger amount).
Example: In a saline solution, salt (NaCl) is the solute, and water is the solvent.

The Human Body as a Solution

The human body is approximately $60%$ water, effectively making it a large, complex solution.
The total body water, approximately $40$ litres, is distributed among various fluid compartments:
- $3$ litres of blood plasma
- $25$ litres of intracellular fluid (fluid within cells)
- $12$ litres of extracellular fluid (fluid outside cells, including interstitial fluid)

Concentration of Solutions

The concentration of a solution quantifies the amount of solute present in a given amount of solvent or a given amount of solution.
For accuracy in medical and scientific contexts, exact concentrations are often required.
Dilute solution: Contains a relatively small amount of solute in a large quantity of solution.
Concentrated solution: Contains a relatively large amount of solute in a given quantity of solution.

Quantitative Concentration Methods

Quantitative methods are essential for clinical reports, medicines, intravenous drips, and oral rehydration packs.

Percentage Concentration Methods:
- Weight/Volume Percent ( $ext{w/v} %$ ):
  - $ext{w/v} % = rac{ ext{grams of solute}}{100 ext{ mL of solution}} imes 100$
  - Example: A $5%$ w/v saline solution means $5$ grams of NaCl are dissolved in enough water to make a final volume of $100$ mL of solution. This is prepared by weighing $5$ g NaCl, adding it to a volumetric flask, and adding water until the bottom of the meniscus reaches the $100$ mL mark.
- Weight/Weight Percent ( $ext{w/w} %$ ):
  - $ext{w/w} % = rac{ ext{grams of solute}}{100 ext{ grams of solution}} imes 100$
  - Example: A $5%$ w/w saline solution means $5$ grams of NaCl are mixed with enough water to achieve a total mass of $100$ grams of solution. This is prepared by weighing $5$ g NaCl in a beaker and adding water until the total mass is $100$ g.
- Comparison for Aqueous Solutions: For aqueous solutions, where the density of water is approximately $1$ g/mL ( $1$ mL of water = $1$ g of water), the w/v % is nearly equivalent to the w/w %.
Medical Application: Intravenous (IV) Therapy
- Definition: IV therapy involves administering substances directly into a patient's vein.
- Reasons for IV Therapy:
  - When a patient cannot safely swallow (e.g., in a coma or under anesthetic).
  - To deliver medications that would be destroyed by gastric juices or are poorly absorbed by the gastrointestinal tract.
  - To rapidly increase the concentration of medication (e.g., morphine) or electrolytes in the bloodstream.
  - When a patient cannot drink enough fluids to compensate for significant fluid loss (e.g., severe burns, severe diarrhea, hemorrhage).
- Isotonicity: IV drips are typically isotonic with blood plasma, meaning they have the same osmotic pressure as blood plasma to prevent cell damage.
- Example: Treating severe dehydration:
  - Problem with pure water: Injecting pure water directly into the bloodstream in cases of severe dehydration would cause red blood cells (RBCs) to swell and burst (a process called haemolysis) due to osmosis, as water moves into the cells to equalize solute concentration.
  - Solution: $5%$ w/v glucose solution: This solution is nearly isotonic with blood plasma. When administered intravenously, the glucose is metabolized by the body's cells, leaving the water behind to rehydrate the body without causing haemolysis.
Milligram Percent ( $ext{mg}%$ ):
- $ext{mg} % = rac{ ext{milligrams of solute}}{100 ext{ mL of solution}} imes 100$
- This unit is commonly used to express clinical levels, such as Blood Urea Nitrogen (BUN) levels.
- Urea is a waste product from the metabolism of amino acids (from protein) that is excreted by the kidneys into urine.
- Normal BUN levels: Typically range from $7$ to $21$ mg of urea per $100$ mL of blood (i.e., $7-21$ mg%).
- Example: Infant dehydration and BUN: An infant suffering from dehydration might have a blood urea level of $32$ mg%, meaning $32$ mg of urea in $100$ mL of blood. This is equivalent to $0.032$ g of urea in $100$ mL of blood, or $0.032% ext{ w/v}$ .
- Elevated BUN (Azotemia): A BUN level above the normal range can indicate:
  - Impaired renal function (kidney problems).
  - Dehydration (insufficient fluid volume to excrete waste products effectively).
  - Excessive protein intake or increased protein catabolism (breakdown).
Parts Per Million (PPM):
- $ext{ppm} = rac{ ext{milligrams of solute}}{1 ext{ Litre of solution}}$
- PPM is used to describe extremely dilute solutions.
- It is often employed to express the concentration of trace substances, such as toxic metals found in drinking water.
- Some compounds can be toxic to humans even at concentrations as low as $1$ ppm.

Molarity (M)

Molarity (M) is a direct measure of the concentration of a solution, indicating the number of moles of a solute per litre of solution.
$M = rac{ ext{moles of solute}}{ ext{liters of solution}}$
The units for molarity are mol $L^{-1}$ .

Conclusion

This module covered fundamental concepts essential for understanding medicinal and pharmaceutical chemistry:

Moles and Avogadro's Number as tools for quantifying matter.
The method for balancing chemical equations to represent chemical reactions accurately.
The nature of physiological solutions and their relevance to the human body.
Various percentage composition methods ( $ext{w/v} %, ext{w/w} %, ext{milligram percent}$ ) and Parts Per Million (ppm) for expressing concentration.
Molarity (moles per litre) as a standard scientific unit for solution concentration.
The practical and medical applications of these concepts, including IV therapy and understanding blood component levels.

For more information, contact Dr. Darren Griffith at dgriffith@rcsi.com.