Fundamentals of Medicinal and Pharmaceutical Chemistry PDF

Fundamentals of Medicinal and Pharmaceutical Chemistry FUNCHEM.4 Moles, Chemical Equations and Reactions. Composition of Physiological Solutions D r. D a r r e n G r i ffi t h  General Chemistry - The Essential...

Fundamentals of Medicinal and Pharmaceutical Chemistry FUNCHEM.4 Moles, Chemical Equations and Reactions. Composition of Physiological Solutions D r. D a r r e n G r i ffi t h  General Chemistry - The Essential Concepts by Chang and Goldsby 7e  Section 2.2 - The structure of atom  Section 2.5 - Molecules and Ions  Section 2.6 - Chemical Formulas Recommended  Section 3.1 - Atomic mass  Section 3.2 - Avogadro’s Number and the Reading Molar Mass  Section 3.3 - Molecular Mass  Section 3.7- Chemical Reactions and Chemical  Equations  Section 4.1- General Properties of Aqueous Solutions  Section 4.5 - Concentration of Solutions FUNCHEM.5 Moles, Chemical Equations and Reactions. Composition of Physiological 2 Solutions FUNCHEM.4 LEARNING OUTCOMES Define ‘molecule’, molecular formula’, ‘molecular mass’, ‘mole’, ‘Avogadro’s number’. Solve calculations involving moles. Demonstrate method of balancing chemical equations. Define ‘solution’ and relate human body to solutions in terms of blood plasma, extra- and intracellular body fluids. Apply medical applications to solutions. Recall and calculate percentage solutions (w/v %, w/w %, mg %), parts per million (ppm) and molar concentrations. 3 What is an atom? An atom is the basic unit of an element that can enter into chemical combination An atom is made up of electrons, protons, and neutrons as well as other particles. An atom is the smallest part of an element that can undergo a chemical reaction. 4 What are molecules? A molecule is an aggregate of at least two atoms in a definite arrangement held together by chemical forces (also called chemical bonds). H2 H 2O NH3 CH4 A molecule may contain atoms of the same element or atoms of two or more elements A diatomic molecule contains only two atoms: H2, N2, O2, Br2, HCl, CO A polyatomic molecule contains more than two atoms: O3, H2O, NH3, CH4 5 Molecular Formula Molecular Formula shows the exact number of atoms of each element in a molecule e.g. Glucose: C6H12O6 6xC + 12 x H + 6xO 6 Atomic Mass Atomic mass is the mass of the atom in atomic mass units (amu). One amu is defined as a mass exactly equal to one-twelfth the mass of one carbon-12 atom. The average atomic mass is the weighted average of all of the naturally occurring isotopes of the element. 6 Atomic number C 12.01 Atomic mass 7 Molecular mass (or molecular weight) The sum of the atomic masses (in amu) in a molecule. 1S 32.07 amu 2O + 2 x 16.00 amu SO2 SO2 64.07 amu For any molecule molecular mass (amu) = molar mass (grams/mol) 1 molecule SO2 = 64.07 amu 1 mole SO2 = 64.07 g SO2 8 Chemical Reactions A process in which a substance ( or substances) is changed into one or more new substances. A chemical equation uses chemical symbols to show what happens during a chemical reaction 9 ‘Recipe’ to make glucose sunlight CO2 + H2O C6H12O6 + O2 Glucose Balanced sunlight 6CO2 + 6H2O C6H12O6 + 6O2 Glucose 10 How do we measure quantities? sunlight 6CO2 + 6H2O C6H12O6 + 6O2 Glucose 6 molecules of CO2 weighs 4.78 x 10-22g on the balance 0.000000000000000000000478 g We need to relate mass of a substance to the number of atoms/molecules present 11 The Mole (mol): A unit to count numbers of particles Dozen = 12 Pair = 2 The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12C 1 mol = NA = 6.0221415 x 1023 Avogadro’s number (NA) 12 Avogadro’s Number 1 dozen eggs = 12 eggs 1 dozen chickens = 12 chickens 1 mole of oranges = 6.022 x 10 23 oranges 1 mole of atoms = 6.022 x 10 23 atoms 1 mole of molecules = 6.022 x 10 23 molecules 13 Molar Mass Molar mass is the mass in grams of 1 mole of molecules or 1 mole of atoms The unit for molar mass is g/mol 14 1 mole of atoms contains 6.022 x 10 23 atoms and has a mass = atomic mass 1 mole of molecules contains 6.022 x 10 23 molecules and has a mass = molecular mass 15 A bit of exercise …… 1 mole of C has a mass of 12.01g _____ and contains 6.022 x 1023 atoms of carbon ___________ 2 moles of C has a mass of 24.02g _____ and contains 1.21 x 1024 atoms of carbon __________ 6.005g and contains 0.5 mole of C has a mass of _____ 3.012 x 1023 atoms of carbon ___________ 16  Example: Zinc (Zn) is a silvery metal that is used in making brass (with copper) and in plating iron to prevent corrosion. How many grams of Zn are in 0.356 mole of Zn? No. of moles = mass (g) Molar mass (g/mol) Zinc mass = No. of moles x molar mass mass = 0.356 x 65.38 Ans: 23.3 g 17 Moles – summary slide Moles relates mass of a substance to the number of atoms/molecules present Mass Mass No of moles = mass/ molar mass Mass = No of moles x molar mass Moles Moles X 6.022 x 1023 ÷ 6.022 x 1023 Number Number *Remember: Regarding Atoms: Molar mass = atomic mass but units are changed from a.m.u to g/mol Problem: A sample of gold contains 3 x 1022 atoms. How many moles is this and what is the mass of the sample if gold has an atomic mass of 196.97 a.m.u? Answer: 6 x 1023 molecules = 1 mole No. of moles = 3 x 1022/ 6.022 x 1023 = 0.05 mol No. of moles = Mass (g) Molar mass (g/mol) Mass (g) = no. of moles x Molar mass (g/mol) Mass (g) = 0.05 mol x 196.97 g/mol Mass = 9.85 g Problem: How many glucose molecules are in 30 g of glucose given glucose has a molecular mass of 180.16 a.m.u? Answer: No. of moles = mass/ molar mass No. of moles = 30g/180.16 g/mol = 0.166 moles No. of glucose molecules = 0.166 moles x 6.022 x 1023 No. of glucose molecules = 1.00 x 1023 Chemical Equations REACTANTS PRODUCTS sunlight 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2 (g) Glucose Prefix numbers refer to the whole molecule Subscript numbers refer to previous atom only Symbols (s), (l) and (g) indicate the state of the product or reactant (aq) = dissolved in water 21 How to “Read” Chemical Equations 2 Mg + O2 2 MgO 2 atoms Mg + 1 molecule O2 makes 2 formula units MgO 2 moles Mg + 1 mole O2 makes 2 moles MgO 48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO NOT 2 grams Mg + 1 gram O2 makes 2 g MgO 22 Balancing Chemical Equations Make sure you have the same number of atoms on the reactants side as on the product side, if not: 1. Put 1 in front of the most complicated molecule on the reactant side 2. Balance each element, leave oxygen to last 3. Use the smallest whole number to balance the equation 23 Examples C2H5OH(l) + O2(g) H2O(l) + CO2 (g) 2C 6H 3O 1 C 2H 3O C2H5OH(l) + 3 O2(g) 3 H2O(l) + 2 CO2 (g) 2C 6H 7O 2C 6H 7O 24 Calculate the Amounts of Reactants and Products 1. Write balanced chemical equation 2. Convert quantities of reactant A (in grams) into moles 3. Use the mole ratio in the balanced equation to calculate the number of moles of product B formed. 4. Convert moles of product to grams of product 25 Example: A general overall equation for degradation of glucose (C6H12O6) to carbon dioxide (CO2) and water (H2O): If 856 g of C6H12O6 is consumed by a person over a certain period, what is the C6H12O6 mass of CO2 produced? No. of moles = mass (g) Molar mass (g/mol) 26  mass mass ?? moles moles Step 1: The balanced equation is given in the problem. Step 2: To convert grams of C6H12O6 to moles of C6H12O6 Moles = mass/ molar mass = 856/180.2 = 4.75 mol C6H12O6 Step 3: From the mole ratio, we see that 1 mol C6H12O6 : 6 mol CO2. Therefore, the number of moles of CO2 formed is 4.75 x 6 = 28.5 mol 27  mass mass ?? moles moles Step 4: Finally, the number of grams (mass) of CO2 formed is given by: Moles = mass/molar mass (rearrange) Mass = moles x molar mass (of CO2) Mass = 28.5 x 44.01 = 1,254.3 g 28 What is a solution….. Solutions are homogeneous mixture of two or more substances, can be formed in any state of matter; that is they may be solid, liquid, or gas Solute Solvent Solid/ liquid/ liquid/ gas gas Present in larger amount Solute Solvent Saline solution Solute – the substance being dissolved Solvent – the substance “doing” the dissolving 29  The human body is made up of ~ 60% water.. ….hence the body is a big solution Approximately 40 Litres: 3 litres of blood plasma 25 litres of intracellular fluid 12 litres of extracellular fluid 30 What is a solution - Concentration The concentration of a solution is the amount of solute present in a given amount of solvent, or a given amount of solution A dilute solution contains relatively little solute in a large quantity of solution A concentrated solution contains a relatively large amount of solute in a given quantity of solution For the sake of accuracy sometimes we need to know exact concentrations 31 What is a solution – quantitative concentrations Quantitative Percentage Concentration Molar Concentration (Molarity) Clinical reports Medicines Intravenous drips Oral Rehydration packs 32 Percentage composition Weight/Volume percent (w/v %) Weight/Weight percent (w/w %) Medication and healthcare products 33 Quantitative Methods: % Concentration of solution 1. WEIGHT/VOLUME % = g OF SOLUTE X 100 100ml OF SOLN Example: 5% w/v saline solution Weigh 5 g NaCl Carefully add the NaCl to the volumetric flask Add water until the until the bottom of the meniscus touches the 100 ml graduation line 34 Quantitative Methods: % Concentration of solution 2. WEIGHT/WEIGHT = g OF SOLUTE X 100 100g OF SOLN Example: 5% w/w saline solution Weigh 5 g NaCl in beaker Add enough water to make a total mass of 100g 100.00 g 35 Water 1ml = 1g d = mass/volume 100 ml = 100 g Weight/Volume = Weight/Weight for aqueous solutions 36 Intravenous Drip Intravenous (IV) therapy is the giving of substances directly into a vein Can’t Swallow safely (coma, under anaesthetic) Require medications which are destroyed by gastric juices or are poorly absorbed by the gastrointestinal tract Must rapidly increase the concentration of medication (e.g. morphine) or electrolyte Can’t drink enough to keep up with loss of fluids (major burns, severe diarrhoea, haemorrhage) I.V. drip is usually isotonic with blood plasma 37 Example: Severe dehydration Inject water directly, RBC will swell and burst (Haemolysis) Haemolysis 5% w/v glucose is nearly isotonic with blood plasma As the glucose is metabolised the water remains to rehydrate the body 38 Quantitative Methods : % Concentration of solution 3. MILLIGRAM PERCENT = mg OF SOLUTE X 100 100ml OF SOLN *1mg = 0.001g Blood Urea Nitrogen levels (BUN) level is measured in milligram percent Protein Amino Urea metabolised acids (kidneys) Urine 39  Normal BUN = 7 - 21mg urea / 100mL blood = 7 - 21 mg% Elevated BUN level (azotemia) due to ….. Impaired renal function Dehydration (lack of fluid volume to excrete waste products) Excessive protein intake or protein catabolism An Infant suffering dehydration might have a blood urea level 32 mg% i.e. 32 mg of urea per 100 mL of blood or 0.032 g of urea in 100 mL of blood = 0.032 % w/v 40 Quantitative Methods: Concentration of solution 4. PARTS PER MILLION (PPM) = mg OF SOLUTE 1 Litre OF SOLN Used to describe extremely dilute solutions e.g. the concentration of toxic metals present in drinking water Some compounds are toxic to humans at 1 ppm! 41 Molarity (M) Molarity indicates the number of moles of a solute per liter of solution.. moles of solute M = molarity = liters of solution M (units) = mol L-1 42 In Conclusion Physiological solutions Percentage composition w/v% 5% w/v glucose solution w/w% 0.9% w/v saline solution milligram percent Parts per million Moles Avogadro’s Number Balance chemical equations Molarity = moles per litre Dilutions 43 Thank you F O R M O R E I N F O R M AT I O N P L E A S E C O N TA N T D r. D a r r e n G r i ffi t h E M A I L : d g r i ffi t h @ R C S I. C O M 44