GUC Pharmaceutical Chemistry Department Chemistry for Engineering (Chem102) Tutorial 1 PDF

Pharmaceutical Chemistry Department Chemistry for Engineering (Chem102) Tutorial 1 Atomic theory WS 2024 CHEM 102 1 Course evaluation In class assessment: - Punctuality + In-class participa...

Pharmaceutical Chemistry Department Chemistry for Engineering (Chem102) Tutorial 1 Atomic theory WS 2024 CHEM 102 1 Course evaluation In class assessment: - Punctuality + In-class participation : (4% + 6%) More than 15 min late considered absent. More than 3 absences, course will be dropped. - 2 Activities : (10%) Best quiz out of 2 : (10%) Midterm exam : (30%) Final term exam: (40%) WS 2024 CHEM 102 2  Course plan Lecture 1 Lecture 2 Lecture 3 Lecture 4 During the semester: Lecture 5 Lecture 6 2 Quizzes (Best 1 out of 2) Midterm 30% 10% Lecture 7 Activity 10% Lecture 8 Lecture 9 Punctuality 4% Lecture 10 Lecture 11 In class Participation 6% Lecture 12 Final 40% 3 WS 2024 CHEM 102 LEARNING OUTCOMES a. Knowledge & Understanding a1. Define basic principles of analytical chemistry and physicochemical properties of compounds and types of reactions used in preparations of different pharmaceutical products. a2. Outline the characteristics of atoms, isotopes, ions and molecules. a3. Give examples of metals, nonmetals and metalloids. b. Professional & Practical skills b1. Use the proper abbreviations & symbols related to inorganic Chemistry. c. Intellectual skills c1. Apply basic analytical Chemistry knowledge in Chemistry practice. c2. Illustrate qualitative analytical methods used for identifying elements and compounds, and balance different types of chemical equations. d. General and transferable skills d1. Communicate clearly by verbal & written means through the open discussions of the tutorial, and the oral questions asked by the tutors. d2. Retrieve information from the assigned text book, the study guide, and the assigned website. d3. Work effectively in the tutorials in groups to answer the assignments. d5. Practice independent learning needed for continuous professional development through searching the internet sources for the answers of the advanced pop questions of the lectures. WS 2024 CHEM 102 4  Atomic theory laws 1. Law of Conservation of mass 2. Law of Definite Proportion 3. Law of Multiple Proportions WS 2024 CHEM 102 5  1.Law of Conservation of mass “In an ordinary chemical reaction, mass is neither created nor destroyed” WS 2024 CHEM 102 6  1.Law of Conservation of mass Example In a combustion reaction, 46.0 g of ethanol react with 96.0 g of oxygen to produce water and carbon dioxide. If 54.0 g of water was produced, how much carbon dioxide is produced? C 2 H 5OH  3O 2 2CO 2  3H 2 O Answer Total mass reacted = total mass produced Mass of ethanol + mass of oxygen = mass of water + mass of carbon dioxide 46.0 g + 96.0 g = 54.0 g + g carbon dioxide Amount of carbon dioxide produced =142.0 g – 54.0g = 88.0 g carbon WS 2024 dioxide produced CHEM 102 7  2.Law of Definite Proportion Regardless of preparation or amount of sample, a pure“ compound will ALWAYS contain the same elements in the ”same proportion by mass Example Samples A and B were prepared in different ways and both contain sodium and oxygen, do the samples (A &B) consist of pure compounds or not? and do they obey the law of Definite Proportion? Sample Total mass Mass of sodium Mass of oxygen A 1.020g 0.757g 0.263g B 1.548g 1.149g 0.399g WS 2024 CHEM 102 8  2.Law of Definite Proportion Answer 1. Do the samples (A &B) consist of Pure compounds or not? Sample Total mass Mass of Mass of sodium oxygen A 1.020g 0.757g 0.263g B 1.548g 1.149g 0.399g Observation: Sum of masses of both sodium and oxygen in each sample add up to the total mass of the sample Conclusion: Both sample A and B consist only of sodium and oxygen WS 2024 CHEM 102 9  2.Law of Definite Proportion Answer 2.Do they obey the law of Definite Proportion? Sample Total mass Mass of Mass of sodium oxygen A 1.020g 0.757g 0.263g B 1.548g 1.149g 0.399g Proportion by mass expressed as Mass Percent % : Mass % of ANY element =( Mass of element / Total Mass of sample) × 100 Proportion by mass for sample A: Mass % sodium = ( 0.757g / 1.020g )×100 = 74.2 % Mass % oxygen = ( 0.263g / 1.020g )×100 = 25.8 % OR 100.0% - 74.2% = 25.8% Proportion by mass for sample B: Mass % sodium = ( 1.149g / 1.548g ) ×100 = 74.2 % Mass % oxygen = ( 0.399g / 1.548g ) ×100 = 25.8 % OR 100.0% - 74.2% = 25.8% Conclusion: Samples A and B are the SAME COMPOUND WS 2024 CHEM 102 10  3.Law of Multiple Proportions “When two elements (B,N) combine to form more than one compound (BN, BN2, BN3…) with the masses of one element remaining fixed (B), then the ratios of the masses of the other element (N) in the other compounds exist in SMALL WHOLE NUMBERS (not parts or fractions)“. - It refers to cases where two elements are combined to each other with more than one way to form two or more different compounds. - This law allows us to discover the ratio by which those elements are combined by using information from the masses of samples. WS 2024 CHEM 102 11  3.Law of Multiple Proportions Examp le Is the law of multiple proportion obeyed in this example?? Oxygen Carbon Carbon monoxide 57.1% 42.9% (CO) Carbon dioxide 72.7% 27.3% (CO2) Answer Assume samples have a total mass of 100g, then percentages become grams Carbon monoxide(CO) Carbon dioxide(CO2) 42.9g C  57.1g oxygen 27.3 g C  72.7 g oxygen 1.00g C  ? g O 1.00g C  ? g O O= 1.33 g O= 2.66 g WS 2024 CHEM 102 12  3.Law of Multiple Proportions Answer Oxygen Carbon Carbon monoxide 57.1% 42.9% (CO) Carbon dioxide 72.7% 27.3% (CO2) monoxide(CO) Carbon Carbon dioxide(CO2) 42.9g C  57.1g oxygen 27.3 g C  72.7 g oxygen 1.00g C  ? g O 1.00g C  ? g O O= 1.33 g O= 2.66 g Mass to Mass ratio of Oxygen in the 2 compounds: Oxygen in CO2 / Oxygen in CO =2.66g oxygen / 1.33g oxygen, This is a ″ratio of small whole numbers″ that illustrates the Ratio law = 2/1 of multiple proportions. Conclusion : CO and CO2 obey the rule of law of Multiple Proportions. WS 2024 CHEM 102 13  Problems Exercise 1 A sample of chloroform is found to contain 12.0 gmof carbon, 106.4 gm of chlorine, and 1.01 gm of hydrogen. If another sample of chloroform is found to contain 30.0 gm of carbon, how many grams of chlorine and grams of hydrogen does it contain? The law of definite Composition ″A compound always contains the same elements in the same proportions by mass ″. Strategy: The ratio of the masses of the elements in chloroform are constant. Therefore, if the amount of carbon is increased by a factor of 2.5 (30.0 g / 12.0 g) then the same must hold true for chlorine and hydrogen. Answer Carbon is increased by a factor of 2.5 (30.0 gm/12.0 gm). Amount in grams of Chlorine = 106.4 gm x 2.5 = 266.0 gm Amount in grams of Hydrogen = 1.01 gm x 2.5 = 2.53 gm WS 2024 CHEM 102 14  Exercise 2 The chemical composition of a large sample of CaCO3 was determined to be 40.04 gm calcium, 12.01 gm carbon, and 47.96 gm oxygen. Thermal decomposition of a smaller CaCO3 sample produced 3.203 gm calcium, 0.9608 gm carbon, and an unknown amount of oxygen. What mass of oxygen Calcium Carbon Oxygen was present in the smaller sample? Sample 1 40.04 12.01 47.96 gm gm gm Sample 2 3.203 0.9608 ??? gm gm Answer Mass of Oxygen: 3.836 g WS 2024 CHEM 102 15  Exercise 3 Nitric oxide, NO, contains 14 g of nitrogen and 16.0 g of oxygen. Nitrogen dioxide, NO2, contains 14 g of nitrogen and 32.0 grams of oxygen. Show that the above data confirm the law of multiple proportions Answer Since the masses of Nitrogen in both compounds are equal, we can directly get the ratio of masses of oxygen in the two compounds by: Dividing the mass of oxygen in NO2 by its mass in NO : 32 g/ 16 g This = is a ″ratio of small whole numbers″ that illustrates the law 2/1 of multiple proportions. WS 2024 CHEM 102 16  Exercise 4 Sulfur and oxygen can react to form both sulfur dioxide and sulfur trioxide. In sulfur dioxide, 90.00 g of sulfur are combined with 90.00 g of oxygen. While in sulfur trioxide, 32.00 g of sulfur are combined with 48.00 g of oxygen. a. What is the ratio of the masses of oxygen that combine with 1 g of sulfur in both sulfur dioxide and sulfur trioxide? b. How do these data illustrate the law of multiple proportions? Answer Sulfur dioxide (SO2Sulfur ) trioxide (SO3) Mass of S : mass ofMass O of S : mass of O Oxygen of SO2 : Oxygen of 90 : 90 32 : 48 SO3 1 : 1.5 =2:3 90/90 : 90/90 32/32 : 48/32 The ratio is of two whole small numbers 1:1 1 : 1.5 which illustrates the WS 2024 CHEM 102 law of multiple 17 proportions  Atomic number and Mass number  Only the number of protons determines identity of the element; specific for each element. For example, Potassium has 19 protons. If an element has 20 protons, it is calcium. contains 15 protons, 15 electrons and 16 neutrons (31 – 15) 31 15 P In an atom, which is electrically neutral, the number of protons equal the number of electrons 30 Everything would be the same as above except for the number P of neutrons, which would be 15 neutrons 31 P (30 – 15) 15 15 This phosphorous atom is therefore an isotope of WS 2024 CHEM 102 18  Isotopic Abundance Exercise 5 Europium is known to have two stable isotopes, Eu-151 which has a mass of 150.9 amu with relative abundance 47 % and Eu- 153 that has a mass of 152.9 amu with relative abundance 53 %. Calculate the average mass of Europium. Answer Atomic mass (Y) = atomic mass (Y1) X (Y1) % + atomic mass (Y2) X (Y2) % 100 100 Atomic mass (Y) = 150.9 x 47 + 152.9 x 53 = 151.96 amu 100 100 WS 2024 CHEM 102 19  Alkali Earth Metal Halogen Noble Gas Alkali Metal The Modern Periodic Table Group Period Transition metals WS 2024 CHEM 102 20  Atomic number and Mass number Exercise 1 (Z = 29) WS 2024 CHEM 102 21  Exercise 2 Fill in the missing information in the following table Electron Neutron Protons Element Symbol s s 14 6 C C 6 8 6 235 92 U U 92 143 92  U 92 92   238 92 146 U 55  Mn 25  Mn 25 30 25 WS 2024 CHEM 102 22  Molecules and ions ►The molecule SF6 contains 7 atoms (6 fluorine and 1 sulfur) ► Certain compound tends to ionize. For example, CaF2 contains 3 ions: a Ca2+ and 2 F-. We call CaF2 an ionic solid or a salt. ► The positive and negative charges in salts exactly balance. Exercise 3 In CaF2, there are 2+ and 2- charges, yielding a neutral Which of the following is an atom, an ion, or a salts. C6H12O6 molecule? molecule N2 molecule CO32- ion (polyatomic anion ) Ag atom Fe2+ ion (cation ) ion (polyatomic cation ) WS 2024 NH4 + CHEM 102 23  Exercise 4 Fill in the missing information in the following table charge Electron Neutron Protons Symbol s s 32 16 S2  16 16 18  2 137  56  Ba 2 56 81 54 2 37  17  Cl  17 20 18 1 52 24 Cr3 24 28 21 3 WS 2024 CHEM 102 24  Thank you WS 2024 CHEM 102 25