CHM101 General Chemistry I Lecture Notes PDF

DEPARTMENT OF APPLIED CHEMISTRY FEDERAL UNIVERSITY DUTSIN-MA, KATSINA. CHM101 (GENERAL CHEMISTRY I) LECTURE NOTE ONE Learning Objectives: At the end of this lecture note, the students should be able to effectively understand the concept of:...

DEPARTMENT OF APPLIED CHEMISTRY FEDERAL UNIVERSITY DUTSIN-MA, KATSINA. CHM101 (GENERAL CHEMISTRY I) LECTURE NOTE ONE Learning Objectives: At the end of this lecture note, the students should be able to effectively understand the concept of:  Definition and career prospect of Chemistry  Sub-divisions of Chemistry  Elements, Atoms, molecules and compounds,  Chemical reactions,  Chemical equations  Stoichiometry; 1. Definition and career prospects of Chemistry Chemistry is the branch of science which studies matter, its composition, properties, and changes. Scientists attempt to discover and describe matter, then determine why these kinds of matter have particular characteristics and why changes in this matter occur. The discoveries of chemistry have greatly helped expand the life span of mankind, increase crop yields, and produce thousands of products that facilitate a higher standard of living. Most of today’s medicines, plastics, synthetic fibers, alloys, pesticides, fertilizers, and many more products that enhance our lives are from the many discoveries of chemistry. Chemistry provides the foundation for the study and understanding of the biological sciences because without chemical reactions, life would not exist. Neither food nor clothing would be available. The energy needed for the human body to move and operate as well as all other biological processes would not exist without chemical reactions. CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 1 2. Sub-divisions of Chemistry Chemistry may be subdivided into several branches. These branches are not separate but overlap considerably. For example 1. Analytical chemistry deals with the separation, identification, and composition of all kinds of matter. Within analytical chemistry, qualitative analysis deals with the separation and identification of the individual components of materials while 2. Quantitative analysis determines how much of each component is present. 3. Biochemistry includes the study of materials and processes in living organisms. 4. Inorganic chemistry covers the chemistry of elements and their compounds except those containing organic carbon. 5. Organic chemistry is the study of carbon-compounds present in living matter. 6. Physical chemistry investigates the laws and theories of all branches of chemistry, especially the structure and transformation of matter and interrelationships of energy and matter. 7. Nuclear chemistry deals with radioactivity, nuclear processes and transformations in the nuclei of atoms. Generally, we can say that, Chemistry is so fundamental to our world, it plays a role in everyone’s live and torches almost every aspect of our existent. Chemistry is essential for meeting our basic needs of food, clothing, shelter, health, energy and clean air, water and soil. 3. Elements, Atoms, Molecules and Compound Element is defined as any substance that cannot be splitted into simpler substances by ordinary chemical processes. Elements are the fundamental materials of which all matter is composed. At present there are 118 known chemical elements.  Eleven (11) elements are gases at room temperature. These includes Hydrogen, Nitrogen, Oxygen, Fluorine, Chlorine, and the six Noble Gases. CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 2   Two (2) elements (Bromine and Mercury) are liquids, Cesium and Gallium, melt at slightly above room temperature, and the rest are solids. Atom: Is a smallest particle of an element that take part in a chemical reaction. Atoms are very tiny particles from which elements are formed. The number of atoms present in each molecule of an element or compound is given by the Avogadro’s number as 6.02×1023atoms. Atoms or groups of atoms that carry positive (+) or negative (-) electric charges are said to be Ions. Example of ions includes Sodium ion ( Na  ), Chlorine ion ( Cl  ), Trioxonitrate ion ( NO3 ), Tetraoxosulphate ion ( SO42 ) etc. Molecules: are groups of two or more atoms held together by the forces of chemical bonds. The ratio of the numbers of atoms that can be bonded together to form molecules is fixed; Example of molecules are: hydrogen molecule (H2), Oxygen molecule (O2) water molecule (H2O), Carbon dioxide (CO2) etc. Compound is a substance formed by chemical bonding of two or more atoms of different elements. Molecules are called compounds when they contains two or more atoms of different elements in their molecular structure. Thus water (H2O) and carbon dioxide (CO2) above are some examples of compound. The type of bond keeping the elements in a compound together may vary as covalent bonds or ionic bonds. Exercise 1.1 In a tabular form, classify the following in to atoms, molecules, compounds and ions  H2O2,  Cl2,  F-,  AgBr,  O3,  Ur,  Mg2+,  CO  NO42-  Pb, and CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 3   C6H12O6 4. Chemical Reactions Chemical reaction: This is a process in which substances called reactants are converted to one or more different substances, called products. The substances may be either chemical elements or compounds. Chemical reactions are an integral part of technology, of culture, and indeed of life itself. For example, Burning fuels in automobiles, smelting of iron, making glass and pottery, brewing beer, and making wine and cheese are among many examples of activities incorporating chemical reactions that have been known and used for thousands of years. Chemical reactions must be distinguished from physical changes. Physical changes is a change of state, just like ice melting to water and water evaporating to vapour. If a physical change occurs, the physical properties of a substance will change, but its chemical identity will remain the same. Common example of physical changes is given below   H 2O(l )   H 2O( g ) oC oC O 100 H 2O(ice)   Classification of chemical reactions Many chemical reactions can be broadly classified into five basic types. Having a thorough understanding of these types of reactions will be useful for predicting the products of an unknown reaction. The five basic types of chemical reactions are: a. Combination Reactions: A combination reaction, also known as a synthesis reaction, is a reaction in which two or more substances combine to form a single new substance. The general form of a combination reaction is: A + B → AB A good example of combination reaction is; solid sodium metal reacts with chlorine gas to produce a solid sodium chloride. 2Na(s) + Cl2(g) → 2NaCl(s) CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 4 Carbon dioxide combine with water to produce glucose in a process called photosynthesis 6CO2(g) + 6H2O(l) → C6H12O6(s) +O2(g) b. Decomposition Reactions A decomposition reaction is a reaction in which a substance breaks down into two or more simpler substances. The general form of a decomposition reaction is: AB → A + B Most decomposition reactions require an input of energy in the form of heat, light, or electricity. The simplest example of decomposition reaction is when a binary compound decomposes into its elements. For example; Mercury (II) oxide, a red solid, decomposes when heated to produce mercury and oxygen gas. 2HgO( s )  2Hg(l )  O2( g ) Limestone decomposes when heated into calcium oxide and carbon dioxide. CaCO3( s )  CaO( s )  CO2( g ) c. Displacement Reactions A displacement reaction is a reaction in which one element replaces a similar element in a compound. The general form of a displacement reaction is: A + BC → AC + B For example: Magnesium is a more reactive metal than copper. When a strip of magnesium metal is placed in an aqueous solution of copper (II) nitrate, it displaces the copper. The products of the reaction are aqueous magnesium nitrate and solid copper metal. Mg(s) + Cu(NO3)2(aq) → Mg(NO3)2(aq) + Cu(s) 2. Metals react with acids to produce hydrogen gas by displacement reaction. For example; Zinc reacts with hydrochloric acid to produce aqueous zinc chloride and hydrogen. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 5 d. Double-Displacement Reactions: A double-replacement reaction is a reaction in which the positive and negative ions of two ionic compounds exchange places to form two new compounds. The general form of a double-replacement (also called double-displacement) reaction is: AB + CD → AD+CB This type of reaction is also known as precipitation reaction as it usually produce precipitate when ions interchanged. A common example is the reaction of potassium iodide and lead nitrate which produce a yellow precipitate of lead iodide 2KI(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbI2(s) e. Combustion Reactions A combustion reaction is a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve O2 as one reactant. For example; the combustion of hydrogen gas produces water vapor. 2H 2( g )  O2( g )   H 2O( g ) Hydrocarbon fuels undergo combustion reaction to produce carbon dioxide, water and large amount of heat energy. A good example is the combustion of propane in a gas cooker which is been used as a domestic food cooking C3 H8( g )  O2( g )   CO2( g )  H 2O( g )  Heat energy Other types of chemical reactions include; Neutralization reaction, Polymerization reaction, Oxidation-reduction reaction etc. Exercise 1.2: Classify the following chemical reactions according to their types 1. 2NaOH(s) → Na2O(s) + H2O(g) 2. Mg(s) + Cu(NO3)2(aq) → Mg(NO3)2(aq) + Cu(s) 3. 2Na(s) + Cl2(g) → 2NaCl(s) CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 6 4. C5H12 + 8O2(g) → 5CO2(g) + 6H2O(g) 5. C6H12O6 → CO2(g) + 6H2O(l) 6. KNO3( aq )  AgCl( aq )  AgNO3( s )  KCl( aq ) 5. Chemical equation: A chemical equation is a shorthand representation of a chemical reaction using the symbols and formulae of substance involved in the chemical reaction. In Chemistry, all chemical reactions are represented by chemical equations. A chemical reaction occurs when starting substances react to produce new substances. a. Starting substances are called reactants. b. New substances formed are called products. c. The reactants are written at the left-hand side whereas the products are at the right- hand side. For example 2 H 2( g )  O2( g )  2 H 2O( l ) Re ac tan t Re ac tan t Pr oduct Constructing chemical equations Based on the law of conservation of mass, matter can neither be created nor destroyed. This means that the numbers of atoms before and after a chemical reaction are the same. Therefore, a chemical equation must be balanced. Example 1 Magnesium burns in oxygen to give magnesium oxide. Write the balanced equation for the reaction. Solution Here, the reactants are magnesium and oxygen. The product is magnesium oxide. Step 1: The word equation is Magnesium + Oxygen ⟶ Magnesium oxide Step 2: Replacing the names with symbols and formulae, we get the chemical equation as CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 7  Mg + O2 ⟶ MgO Step 3: Balancing the chemical equation: To balance oxygen on both sides, multiply RHS by 2, i.e Mg + O2 ⟶ 2MgO Now, the number of oxygen atoms is balanced but the number of magnesium atoms is not. Therefore, multiply magnesium on the LHS by 2. Thus, the equation becomes 2Mg + O2 ⟶ 2MgO Example 2 When copper (II) carbonate is heated, it decomposes into copper (II) oxide and carbon dioxide. The presence of carbon dioxide is detected by the limewater. Write the equation for the reaction Solution: The balanced equation for the heating of copper (II) carbonate is CuCO3( s )   CuO( s )  CO2( g ) Copper ( ii ) carbonate Copper (ii ) oxide Carbon (iv ) oxide Since the number atoms in the LHS and RHS are the same, this equation is already balanced, so we don’t need to multiply with any number. Exercises 1.3 1. The concentrated hydrochloric acid and concentrated ammonia solution are left for a few minutes to produce hydrogen chloride gas and ammonia gas respectively. When the hydrogen chloride gas and ammonia gas are brought together, they react to form fine white solids of ammonium chloride. These are seen as thick white fumes. Write and balance the chemical equation for the reaction. 2. When a colourless lead (II) nitrate solution is added to a colourless potassium iodide solution, yellow precipitate of lead (II) iodide is produced. At the same time, CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 8  colourless potassium nitrate solution is also produced. Construct a chemical equation for the above reaction. 6. Stoichiometry Stoichiometry is the study of relationship between the quantities of substances involved in a chemical reaction and the substances that are produced as a result of that chemical reaction. That is to say, Stoichiometry is the relationship between the reactants and products in a chemical reaction. Stoichiometry is the fundamental mathematical principle used to describe the law of conservation of mass, which states that: matter cannot be created nor destroyed, but only converted from one form to another. Balanced equations and mole ratios A common type of stoichiometric relationship is the mole ratio, which relates the amounts in moles of any two or more substances in a chemical reaction. We can write a mole ratio for a pair of substances by looking at the coefficients in front of each species in the balanced chemical equation. For example, consider the equation for the reaction between iron (III) oxide and aluminum metal: Fe2O3( aq )  2 Al( s )  2Fe( aq )  Al2O3( s ) The coefficients in the equation tells that, 1 mole of Fe2O3 react with 2 mol of Al metal, forming 2 mol of Fe and 1 mole of Al2O3. Thus, the stoichiometry of the reaction is written as 1 mol Fe2O3 : 2 mol Al. Using the stoichiometric ratio, we can calculate how many moles of Al are needed to fully react with a certain amount of Fe2O3 or vice versa. In general, mole ratios can be used to convert between amounts of reactants. Example: Using a mole ratios to calculate mass of a reactant: Consider the following equation: CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 9  NaOH( aq )  H 2 SO4( aq )  H 2O(l )  Na2 SO4( aq ) How many grams of NaOH are required to fully consume 3.10 g of H2SO4? Solution First, we need to balance the equation. In this case, we have 1 Na atom and 3 H atoms on the reactant side and 2 Na atoms and 2 H atoms on the product side. 2NaOH(aq) + H2SO4(aq) →2H2O(l) + Na2SO4(aq) After the equation is balanced, we then, go ahead solving the problem. Remember, we asked to find the mass of NaOH that is needed to completely react 3.10 g of H2SO4. To solve this stoichiometry problem, the following steps are needed: Step 1 Convert the mass of the reactants to mole: In order to relate the amount of H2SO4 to mole ratio, we first need to know the quantity of H2SO4 in moles; Mass of H2SO4 given = 3.10g Molar mass of H2SO4 = 2 + 32 + 16(4) = 98g/mol mass 3.10 № of mole of H2SO4 =   0.032mol molar mass 98.0 Step 2 Use the mole ratio relationship from the balanced equation for the reaction to determine the number of mole of the other reactant. From the equation below 2NaOH(aq) + H2SO4(aq) →2H2O(l) + Na2SO4(aq) 2 moles of NaOH reacted with I mol of H2SO4 x mol of NaOH will react with 0.032 mol of H2SO4 2  0.032mol x  0.063 1 CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 10   no. of mol NaOH  0.063mol Step 3 Convert the mole of the other reactant to mass in gram In the question, we were asked for the mass of NaOH in grams, so the last step is to convert the 0.063 mol to gram. Nowthe number of mol NaOH  0.063mol Mass of NaOH  number of mol  Molar mass 40.0 g  0.063mol   2.52 g mol Therefore, 2.52g of NaOH is required to fully consume 3.10g of H2SO4 in the reaction. Be sure to pay extra attention to the units if you take this approach. Example 2.5 Sulphuric acid (H2SO4) reacts with ammonia (NH3) to produce ammonium sulphate fertilizer (NH4)2SO4 according to the following equation H2SO4(aq) + 2NH3(g) → (NH4)2SO4(aq) A factory worker carries out the above reaction (using 2.0 kg of sulfuric acid and 1.0 kg of ammonia) and gets 2.5 kg of ammonium sulfate. What is the percentage yield of the reaction? Solution In the above fertilizer production, we were told that, 2.0 kg of H2SO4 is combined with 1.0 kg of NH3, therefore: The total reacting mass is 2.0 kg + 1.0 kg = 3.0 kg Mass of the product formed = 2.5 kg Mass of product Percentage yield  100% Mass of reac tan ts CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 11  2.5 kg Percentage yield  100% 3.0 kg  83.3% Exercises 1.4 1. Given the following reaction: Fe2O3(s) + CO(g)→ Fe2O4(s) + CO2(g) If 2.3 kg of Fe2O3 and 1.7 kg of CO is used, what is the maximum mass of Fe2O4 that can be produced? 2. Sodium nitrate decomposes on heating to produce sodium nitrite and oxygen according to the equation below: NaNO3(s) → NaNO2(s) + O2(g) A student carries out the above reaction using 50 g of sodium nitrate. He found to get 36 g of sodium nitrite. What is the percentage yield? 3. Cuprite is a minor ore of copper. Cuprite is mainly composed of copper (I) oxide (Cu2O). Jennifer wants to know how much copper oxide is in a sample of cuprite. She has a sample of cuprite that weight 7.7 g. She performs some experiments and finds that, the mass of iron oxide and crucible (a container that is used to heat compounds) is 7.4g. The mass of the crucible is 0.2 g. What is the percentage purity of the sample of cuprite? CHM101 LECTURE ONE (Mr. M. S. DARMA) Page 12