Chemistry Textbook (Republic of Iraq) PDF

Republic of Iraq Ministry of Education General Directorate of curricula CHEMISTRY 4 AUTHORS A. Prof. Ammar Hani Al- Dujaili A. Prof. Muhannad Jamil Mahmoud Dr. Samir Hakeem Kareem Salim Mohammad Said N...

Republic of Iraq Ministry of Education General Directorate of curricula CHEMISTRY 4 AUTHORS A. Prof. Ammar Hani Al- Dujaili A. Prof. Muhannad Jamil Mahmoud Dr. Samir Hakeem Kareem Salim Mohammad Said Nasrawi Dr. Huda Salah Kareem Khuloud Mahdi Salim Majid Hussain Khalaf Al- Jassani Basil Ibrahim Al- Shuk Osama Mortadha Al- Khalisi Kareem Abdl- Hussain Al- kanani Third Translated Edition 2021 / 1442  Scientific Supervisor: Khulood Mahdi Salim Design Supervisor: Tayseer Abdul-Ilah Ibraheem Translated by Fezalar Educational publications Institution Translated from 10 th Arabic Edition ‫يوزع مجان ًا ومينع بيعه وتداوله في األسواق‬ ّ ‫استناداً إلى القانون‬  CONTENTS CHAPTER 1 - BASIC CONCEPTS IN CHEMISTRY 5 CHAPTER 2 - GASES 26 CHAPTER 3 - CHEMICAL EQUATIONS & CALCULATIONS 56 CHAPTER 4 - ORGANIC CHEMISTRY 78 CHAPTER 5 - NUCLEAR CHEMISTRY 107  PREFACE Chemistry is an interesting and fundamental branch of science because it gives us the chance to explain the secrets of nature. What is water? What do we use in our cars as fuel? What is aspirin? What are perfumes made of? These kinds of questions and their answers are all part of the world of chemistry. Chemists work everyday to produce new compounds to make our lives easier with the help of this basic knowledge. All industries depend upon chemical substances, including the petroleum, pharmaceuticals, garment, aircraft, steel, and electronics industries, etc. This book helps everyone to understand nature. However, one does not need to be a chemist or scientist to understand the simplicity within the complexity around us. The aim was to write a modern, up-to-date book where students and teachers can get concise information about chemical reactions and inorganic compounds. Sometimes reactions are given in detailed form, but, in general, excessive detail has been omitted. Throughout the book, different figures, colourful tables, important reactions are used to help explain ideas. We hope that after studying this book, you will find chemistry in every part of your life. The Author BASIC CONCEPTS IN CHAPTER-1 CHEMISTRY ACHIEVEMENTS After studing this chapter, the student will be able to: * Understand the atomic theory of Dalton and its hypotheses and use the laws of the Chemical Union to identify the formation of chemical compounds and the ratios of fixed elements in them. *Understand the purpose of the Gay-Lussac's law for combining gas volume, and its relationship to the Avogadro's hypothesis. *Define the basic terms: valance, atomic mass, equivalent mass and the relationship between them and define the con- cept of the mol, the molar mass, Avogadro’s number and the relationship between them. *Find out the chemical formula, Empirical and molecular formula for the compounds to find out them and how by percentages of elements in compounds.  Chapter - 1 1-1-Dalton’s Atomic Theory Scientific and chemical researchs and discoveries, which took place at the end of the eighteenth century and the beginning of the 19th century, led to the knowledge that the substance consisted of atoms, and that the difference in type and number of the atom determines the properties and type of molecules that formate it, i.e. the specifications of the substance. This helped the English scientist John Dalton to declare the atomic theory of the matter in 1803 A.D. and named the Dalton's atomic theory, which included the following hypotheses: 1-The substance consists of small, indivisible particles called "atoms" (which the scientists later managed to fragment 2-Atoms cannot be synthesized, within the human abilities. 3- Atoms of the same element are similar in all their physical and chemical properties and differ from the atoms of other elements. 4- Compound atoms (as called by Dalton) are formed by the union of atoms of the elements atoms with simple ratios. Eight years later, some modifications were made, replacing the expression John Dalton «1766 A.D.-1844 A.D.» "compound atoms" with "molecules" by the Italian scientist Avogadro and our Was born of Poor Weaver father. He understanding of these hypotheses must be in the period Time that Dalton came has worked at an early age. However, in, in the first half of the 19th century, where scientific progress. he had spent much of his spare time 1-2-LaWs of chemical union studying Latin mathematics and the natural sciences. He took care of the As a result of the development of scientific knowledge and the new existing meteorology study. From 1787 A.D. discoveries which based on practical experiments and scientific conculasions until his death, more than 200,000 led to the interpretation of the composition of the material and the drafting of notes were recorded, and his interest the laws of the Chemical Union in the second half of the eighteenth century and the beginning of the 19th century. led him to investigate the properties of the gases, and he discovered the law The first of these laws was the law of mass conservation, which answered the of partial pressure, and also concluded question of what happens to the substance during its chemical reaction? Can that the dissolution of gas in a com- it be destroyed or created? is the mass of reactant substances differ from the bination of gases proportional to the mass of the material resulting from the reaction, or is it equal? To answer all partial pressure and since 1803 A.D., those questions, the French scientist Lavoisier did the oxidation of tin in a closed receptacle, and he found that the mass of the closed receptacle remained he worked to develop his atomic the- unchanged, because a chemical reaction was made between tin and oxygen and ory, he added the double-descent new molecules were formed of tin oxide(II). law as well as the concept of relative atomic mass. Since the mass of the atom does not depend on the nature of the other atoms with which it unites, it is axiomatic that all the oxygen and tin atoms involved in the chemical reaction maintain their mass unchanged, and the Lavoisier ex- periments indicated that: "The mass of material cannot be destroyed or created during the chemical re- action" i.e., :- The mass of reactants = the mass of products from the reaction. The Arab scientist Abu Qasem Almajriti «950 A.D.-1007 A.D.» is considered first to prove the validity of this law, he noticed when heating a weighted amount of elemental mercury in a closed glass container with air presence will turn mercury into a fine red powder without a change inthe total mass of react- ants within the vessel. He repeated Lavoisier and reached the same conclusion and then laid down a law of conservation of mass. 6  Basic Concepts in Chemistry Example 1-1 73g of HCl gas was passed into a solution containing 158 g of sodium thiosulfate, as a result 117 g of table salt, 64 g of SO2 gas, 32 g of sulfer and 18 g of water has been produced. Prove that these results support the law of conservation of mass. Solution: The sumation of the reactants' mass= 73+158=231 g The summation of the products' mass=117+64+32+18=231 g The summation of the reactants' mass=The summation of the products' mass which is support the law of conservation of mass Lavoisier 1743 A.D. -1794 A.D. The other chemical union laws can be summed up with a single idea, the union The law of Conservation of Mass of elements with constant weight ratios for compound formation. The scientist Prust was the first to develop the law of constant composition, which stipulated that "all specimens of a particular compound have the same proportions of its constituent elements" Let's take the following example: If the water is dissociated we will find that 16 g of oxygen in the sample is present versus 2 g of hydrogen, or the ratio of oxygen mass to hydrogen: 16 g (O) The ratio= =8 2 g (H) This ratio will be found in each sample of pure water regardless of the source from which it was taken or in what way it was prepared (Fig. 1-1) The com- pounds have a fixed structure. The water in any way has a constant percentage of hydrogen to oxygen regardless of which source it came from. Ebu El Qasim Mesleme bin Ahmed El Majriti (950-1007). He was born in Andalusia and died there. He was accepted as the pioneers of chemistry at his time. One of his important is Ghayet El Hakim. He observed that when he put 1/4 pound of Mercury in closed container with air inside and heated it, a soft and red powder was formed. There occurred no change in mass. This substance is called mercury (II) oxide today. 7  Chapter - 1 Example 1-2 Two carbon dioxide samples were obtained from two different sources. They were dissociated into components. The first sample contained 4.8 g of oxygen and 1.8 g of carbon, while the other sample contained 17.1 g of oxygen 6.4 g of carbon. Show that these results correspond to the law of constant compositions. Solution: The ratio of oxygen mass to carbon in the first sample 4.8g(O) Ratio = = 2.7 1.8g (C) The ratio of oxygen mass to carbon in the second sample 17.1 g (O) Ratio = = 2.7 6.4 g (C) The ratio of H and O is constant in Since the ratio is the same for two, it means that these results are consistent water with regard to source of it with the law of constant compositions. 1-3-Gay-Lussac's LaW of Combining Gas Volumes The French scientist Gay-Lussac (Joseph Gay-Lussac) has worked very much in the reaction of gases and saw that there is a correlation between the volumes Exercise 1-1 of gases that involved in the chemical reaction and resulting from it under the same conditions of pressure and temperature, and the measurement of the Two carbon monoxide samples volumes of reactive gases resulting from the reaction formulated the results of were analysed and obtained from his investigation in the year 1808 AC in the following announcement: two different sources. The first sample contained 4.3 g of oxygen "The volumes of gases involved in or resulting from the chemical re- and 3.2 g of carbon. The second action correspond to each other by a simple numerical proportional- sample contained 7.5 g oxygen and ity if measured under the same conditions of pressure and temperature." 5.6 g of carbon. Do these results achieve the law of constant com- For example; positions. 1- One volume of hydrogen is combined with one volume of chlorine, and two volumes of hydrogen chloride are formed, the ratio between the two united gaseous volumes and the volume of gas produced is 1:1: 2, as in the following chemical equation: H2 + Cl2 → 2HCl 2- When the water is electrically analysed, the volume of the freed hydrogen is twice the volume of the oxygen, and it combines two volumes of hydrogen with one volume of oxygen and produces two volumes of water vapour. 2H2 + O2 → 2H2O The ratio between the two volumes united gaseous sizes and the volume of water vapour produced is 2:1:2, which is a simple numerical ratio. The above can be represented in the following forms: 8  Basic Concepts in Chemistry Figure 1-4 Reaction of gases with a constant ratio 2 Volumes H2 1 Volume O2 2 Volumes H2O 1 Volume H2 2 Volume Cl2 2 Volumes HCl 1-4-AVOGADRO’S HYPOTHESIS In 1811, Italian scientist Amedeo Avogadro found out that gas molecules can have more than one atom. In other words, gases can have diatomic molecules. Avogadro defined molecule as the smallest independent sample of matter. He also defined atom as the smallest part of an element which can be found in molecules of different compounds. The basic hypothesis of Avogadro is as fol- lows: (Equal volumes of gases contain equal number of molecules under the same pressure and temperature conditions.) Avogadro’s hypothesis explained the simple ratios between volumes of react- ant and product gases and also provided significant results with the atomic numbers of gases with complex molecules. According to Avogadro, the num- ber of atoms forming a molecule of an element is constant. This is also valid for compounds. Only the types of atoms are different which form compound molecules. For example, if 1 volume of hydrogen gas reacts with 1 volume of chlorine gas, 2 volumes of HCl gas is formed. H2 + Cl2 → 2HCl From the reaction of 2 volumes of hydrogen gas and 1 volume of oxygen gas, 2 volumes of H2O are obtained. 2H2 + O2 → 2H2O This isn’t contradictory with Dalton’s atom model. According to this, a hydro- gen molecule consists of 2 atoms. Similarly, oxygen and chlorine molecules consist two atoms. HCl molecule consists of 1 chlorine and 1 hydrogen atom. Water molecule consists 2 hydrogen and 1 oxygen atom. 9  Chapter - 1 Figure 1-1 Balls represent atom in molecules formed from combination 1-5-VALANCE Chemical formulas of compounds weren’t assigned arbitrarily but they were assigned according to bonding of atoms in molecules of compounds. An atom of element needs to have a capability in order to be bonded to atoms of other elements. The capability of bonding of an atom or number of hydrogen atoms that an atom of an element can combine is called as valence. The bonding capabilities of elements are different. Valence concept was introduced to chemistry in 19th century. We can define valence of an element as follows: “It is the number of electrons which an element loses, gains or shares during a chemical reaction.” For example, the valence of hydrogen is 1. Because it has an electron in its Atomic mass unit (AMU or amu) outer shell which can be shared. Valence of oxygen in water is 2 because it has An atomic mass unit (symbolized 6 electrons in its outer shell. Therefore it needs 2 electrons to satisfy its outer shell. Valence of Na is 1. It has an electron to lose in its outer shell. Valence of AMU or amu) is defined as pre- Mg is 2 because it can lose 2 electrons from its outer shell. Valence of Cl is 1 cisely 1/12 the mass of an atom of because it needs one electron to satisfy its outer shell. carbon-12. The carbon-12 (C-12) 1-6-ATOMIC MASS atom has six protons and six neut- rons in its nucleus. Atoms on the degree of precision and smallness so it was difficult to estimate their atomic masses, however it has been possible to assign their mass with great accuracy, for example, it has found that a hydrogen atom mass reach 1.64 × 10-24 g. it has also been possible to obtain relative masses of atoms by setting the mass of elements with another element, provided that the relative number of atoms in the compounds is known, so atomic mass used to express an ele- ment mass for another element atom mass and it is agreed to use in determining the relative masses of all the elements of the periodic table. In 1961 in Geneva held a Conference of the International Union of Pure and Applied Chemistry (IUPAC) which was agreed on the definition of the standard unit of atomic mass and Atomic Mass Unit named after (AMU) as equivalent to one of twelve part of carbon isotope atom mass 12 and that it's atomic mass considered equal to 12 units and therefore:- Atomic mass unit (amu) = (mass of an atom of carbon isotope 12)/12 i.e.; 1 (amu)= (1/12) mass of an atom of carbon isotope 12 10  Basic Concepts in Chemistry as the mass of carbon 12 isotope atom = 12 12 Avogadro's number = (6.023×1023) So 1 12 1 (amu) = × 12 6.023×1023 1 (amu) = 1.66×10-24 g Thus the atomic mass we use today and found in the periodic table are not ac- tual masses, but relative masses show the relationship in terms of atomic mass between different atoms. For example, the atomic mass of hydrogen isotope 1 is (1/12) the atomic mass of the isotope carbon-12 which is about 1 amu, either nucleus oxygen 16 have mass equal to(16/12) or (4/3) from the mass of carbon 12 isotope, so when atomic mass estimates in grams it called the gram-atomic mass. The gram-atomic mass for oxygen = 16 g and for silver= 107.9 g and each mass of these masses has Avogadro's number of atoms which is equal to 6.023×1023 atoms, for example: 1 g of hydrogen contains 6.023×1023 atoms of hydrogen 39 g of potassium contain 6.023×1023 atoms of potassium 207 g of lead contain 6.023×1023 atoms of lead The absolute atomic mass is the mass of one atom of the element. i.e.; The absolute atomic mass of an element=gram-atomic mass of the element/ Avogadro's number Example 1-3: Calculate the absolute atomic mass of the oxygen given that it's atomic mass is equal to 16. Solution: The absolute atomic mass of an element=gram-atomic mass of the element/ Avogadro's number 16 = (6.023×1023) =2.656×10-23g 1-7-Equivalent Mass The study of the law of mass ratio in which the various elements combined led to find out the equivalent masses, where Dalton was the first one who calculate these masses, he assumed that the element mass that combined with one atom of hydrogen's mass is the mass equivalent to the element, because the hydrogen component failure in the formation of compounds with the most other ele- ments, or to the fact that most elements do not combine directly with hydrogen, and oxygen combine with it directly. Oxygen has been adopted as a basis for the calculation of equivalent masses, and it's combined mass considered equal to «eight 8». This is the elements are not limited to the combination with each other by equivalent quantities only, but replace each other in their compounds with equivalent masses. Thus the definition of equivalent mass for an element is:- "The mass of that element which combine with eight mass parts of oxygen or pull these quantities from their compounds" 11  Chapter - 1 The equivalence mass concept has enabled drafting of the following law, called the law of equivalent masses:- " The elements combine together in quantities correspond with their equivalent masses." When the equivalent mass estimated in grams it is called then the (Gram Equi- valent), for example, the gram equivalent for oxygen = 8 g , chlorine = 35.5 g, hydrogen = 1 g and silver = 107.9 g, and so on. The equivalent masses can be estimated from the data on the analysis of different compounds, or replace an element, and it is not necessary to specify the equivalent masses to start from compounds containing oxygen or with other equivalent mass element informa- tion depending on the following: (Mass of 1st element/it's equivalent mass)=(mass of 2nd element/it's equivalent mass) Example 1-4 3.5 g of iron combined with sulphur to form 5.5 g of iron sulphide (II). Calcu- Exercise 1-2 late the equivalent mass of iron given that the equivalent mass of sulphur=16 g. 1.31 g of Cu was obtained from re- Solution: duction of 1.64 g of copper oxide with Mass of S = 5.5 - 3.5 = 2 g hydrogen. Calculate equivalent mass of Cu. (equivalent mass of O = 8g) mass of 1st element _________________________ mass of 2nd element _________________________ = equivalent mass of 1st element equivalent mass of 2nd element Exercise 1-3 2 ___ 3.5 = ___________________ 16 equivalent mass of Fe What is the equivalent mass of an el- ement if its atomic mass is 55.85 and Equivalent mass of Fe = 28 g its valence is 3? 1-8-The Relationship betWeen Atomic Mass, Equivalent Mass and Valence While measuring atomic masses, O atom was accepted as 16 units. While measuring equivalent masses, O atom was accepted as 8 units. According to this, there is a mathematical relationship between the two. In order to calculate equivalent mass of an element, mass of the element is divided to the number of atoms which the element can be bonded to or replace. Here the common denominator is the valence of the element. Equivalent mass (the result) is obtained by dividing mass of element to its valence. atomic mass of element equivalent mass of element = valence of element Example 1-5 What is the valence of Al if its atomic mass is 27 and equivalent mass is 9? Solution: atomic mass of Al 27 valence of element = = =3 equivalent mass Al 9 12  Basic Concepts in Chemistry 1-9-Density of Gases The density can be defined by the following relationship: Density (kg/m3) = [the mass (kg)/the volume (m3)] ρ (kg/m3) = m (kg) / V (m3) The density unit can be(g/cm3), or (g/mL) for solids and liquids as for gases, the mass of 1 ml be too small to deal in practice, so it had taken a litter «L» as unit volume for gas density measurement. The volumes of gases are greatly influenced by pressure and temperature, so the circumstances must be specified under which measure the density of gases, and the conditions that gas is meas- ured in the temperature of zero degrees Celsius (0 °C) and pressure (1 atm)are called standard conditions or (Standard Temperature and Pressure) (STP). Example 1-6 If you know that a density of a gas is equal to 0.7 g/L and it occupies a volume of 490 cm3 at STP. What is the mass of this gas? Exercise 1-4 Solution: The mass of a gas is 0.4 g and its Firstly, we need to convert cm3 to L. volume is ¼ L. What is its density under standard conditions? We use the following relationship to calculate mass of gas. 1-10-Mole Concept Chemical reactions occur between large number of particles, and these particles may be in the form of atoms, molecules or ions and each of these particles has their own relative mass, but there is no fit in between the amounts of each mat- ter and its mass. For example, if a student was asked to compare 1 g of H2 gas and 1 g of N2 gas and 1 g of O2 gas in terms of the number of particles it cannot solved for the simple reason that the molecular masses of these elements vary from each other, the molecular mass of hydrogen gas 2, nitrogen 28 and for oxygen 32. If the mass of 1 g is divide for each item on the molecular mass. Then values obtained can be used for comparison. So, years ago, the need arose for independent units to express the amount of substance and this unit had found general acceptance, this unit is mole (mole) and symbolizes (n) and it is one of the basic units in the international system of units, the mole defined as the quantity of material containing the same number of particles (molecules, atoms or ions) containing in 12 g of carbon isotope 12 (12C) (where this isotope is used also as a measure to calculate atomic mass as previously mentioned) and this number of particles called the Avogadro's Number) which equals to 13  Chapter - 1 6.023 × 1023 and symbolizes NA. It must be emphasized that Mole is the actual amount of the matter and is not the mass. The Mole is one of the most important basic concepts in general chemistry and which adopted by scientists to consolidate their theory to many important is- sues in chemistry. And we can apply the concept of mole on atoms, molecules, ions or electrons so it is always necessary to specify the type of particle that we deal with, for example, Mass of one mole of atoms of the isotope carbon-12 is 12 g Mass of one mole of atoms of silver is 107.868 g Mass of one mole of H2 molecules are 2 g Mass of one mole of ions SO42- is 96 g And number of moles (n) is calculated using the following relationship: mass (m) (g) n (mol) = ____________________ molar mass (M) (g/mol) 1-10-1-Molar Mass Since the particles are a group of atoms chemically joined together the final mass for these molecules known from the masses of their atoms i.e., we use relative mass to compare various molecules in terms of mass. Meaning that:- Molar mass of matter =Total relativistic masses of atoms which make up matter Molar mass is the mass of 1 mole of a compound in gram unit. (Molar mass used to be called as molecular mass formerly.) For example, if we want to calculate the molar mass of methane gas (CH4) or if we ask what is the mass of 1 mole of methane gas, as each methane molecule has 1 C and 4 H atoms, 1 mole of methane molecule contains 1 mole of C atom and 4 moles of H atoms. According to this, we can calculate the mass of 1 mole of methane gas as follows: Mass of 1 mole of C = 1 × 12 = 12 g/mol Mass of 4 moles of H = 4 × 1= 4 g/mol Mass of 1 mole of CH4 = 16 g/mol In the same manner, we can calculate molar mass of H2SO4 acid. Mass of 2 moles of H = 2 × 1= 2 g/mol Mass of 1 mole of S = 1 × 32 = 32 g/mol Mass of 4 moles of O = 4 × 16 = 64 g/mol Mass of 1 mole of H2SO4= 98 g/mol If 16g of methane gas makes up 1 mole of CH4 and 98 g of sulfuric acid makes up 1 mole of H2SO4, logically we can call those masses as molar mass. As we understand from previous calculations we have made, in chemistry, all atoms and molecules are bound with the following : a) Atomic mass and molar mass b) Mole c) Avogadro’s number 14  Basic Concepts in Chemistry Avogadro’s number or Avogadro’s constant is the number of atoms in 1 mole of an element or number of molecules in 1 mole of a molecule. It is shown with NA. Previously, we have mentioned that atoms, molecules or ions as many as Avogadro’s number are called as a mole. Example 1-7 Find the molar mass for the following compounds:- a) Aqueous sodium sulfate Na2SO4. 7H2O b) Juglone C10H6O3 c) Sulfur dioxide SO2 Solution: a) M (Na2SO4. 7H2O) = (2×23) + (1×32) + (4×16) + 7(2×1+1×16) = 268 g/mol b) M (C10H6O3) = (10×12) + (6×1) + (3×16) = 174 g/mol c) M (SO2) = (1×32) + (2×16) = 64 g/mol *You can refer to Table 3 at the end of the book to learn atomic masses. Carbon Example 1-8 Oxygen Find out the mole numbers of the following compounds. Hydrogen a) 9.6 g of SO2 b) 85 g of NH3 Juglone is an organic compound. It is used as a pesticide. It is also a natural dye. Solution: a) Molar mass of SO2: b) Molar mass of NH3: 15  Chapter - 1 Exercise 1-5 Perform the following conversions. a) 3.2 g Cu = ……….mol Cu (Cu = 64 g/mol) b) 0.2 mol Ag=……..g Ag Example 1-9 (Ag = 108g/mol) Calculate the mass of 0.7 mol of manganese dioxide (MnO2). c) 1 silver atom = …… g Solution: d) 3.01×1022 Fe atoms =………..mol Fe =…………g Fe (Fe = 56 g/mol) e) 6.4 g S =…….mol S =…….S atoms (S = 32 g/mol) 1-14-2-Applications of Mole Concept As we have told before, the mass of 1 mole of carbon atom is 12 grams. Ac- cording to this, it is possible to calculate the mass of 1 carbon atom. Exercise 1-6 a) What is the mass of 0.04 mole We can write the following equation: of N2? b) What is the number of moles in Number of particles (atoms, ions or molecules) 5.6 g of PCl5? Number of moles = c) Calculate the molar mass of Avogadro's number the gas which has 22.54 g in 0.23 mole. Example 1-10 a) Calculate the number of moles of 3.01×1025 water molecules. b) Calculate the number of molecules in 0.02 mole of CO2. Solution: number of molecules a) Number of moles = Avogadro's number (N A ) 3.01 × 1025 = ___________ = 50 moles H2O 6.023 × 1023 b) Number of molecules = number of moles x Avogadro’s number = 0.02 × 6.023 × 1023 = 1.2 × 1022 CO2 molecules 16  Basic Concepts in Chemistry Example 1-11 Exercise 1-7 Calculate the number of molecules in 170 g of H2S gas. Calculate the number of molecules of SiO2 found in 1mg of dust. (As- Solution: sume that dust particles are made up of 100% SiO2.) M (H2S) = (2×1) + (1×32) = 34 g/mol m (g) __________ 170 (g) __________ n (mol) = = = 5 mol M (g/mol) 34 (g/mol) Number of molecules = number of moles × Avogadro’s number = 5 × 6.023 × 1023 = 3.01 × 1024 H2S molecules 1-11-Mass Percentages of Elements in a Compound There are two ways to describe molecular structures for compounds, first; knowing how many atoms of each element involved in the composition of the compound and second; know percentages in terms of mass of elements in this structure, i.e., element in g 100 grams of the compound, the percentage can be Exercise 1-8 found for each element in the composition of compound as follows: Calculate the number of moles A- Find the molar mass of a compound of molecular form. of each of the following. a) 3.01 × 1022 N2 molecules B- Set and find the mass of each element in a molecule, i.e., product of atomic mass for each element x number of its atoms b) 4.82 × 1024 iron atoms C- Find percentage for the item in the compound by the following relationship: Mass percentage of Molar mass of element × number of atoms ______________________________ an element in a compound = × 100 Molar mass of the compound Example 1-12 Calculate the percentage for each carbon, hydrogen and oxygen in the com- pound of isopentyl acetate (C7H14O2) (substance released by the bee insect). Solution: Molar mass of C7H14O2: M(C7H14O2) = (7×12) + (14×1) + (2×16) = 130 g/mol According to the equation above, we can find out mass percentages of ele- ments. Isopentyl is a substance made by bees. Sum of mass percentages of elements is equal to 100%. 17  Chapter - 1 Exercise 1-9 Example 1-13 Find out mass percentages of elements a) Find out mass percentages of elements in C2H2O4 acid. in CH3COOH acid. b) What is the mass percentage of crystal water in C2H2O4.2H2O compound? Solution: a) Molar mass of C2H2O4 M (H2C2O4)= (2×1) + (2×12) + (4×16)= 90 g/mol 2×12 24 C% = = ×100% = 26.67% Exercise: 1-10 90 90 50% of by mass of a 2.2 moles mixture of He and Ar gasses is He. Calculate the mole number of He 2×1 2 H% = = × 100% = 2.22% in the mixture? He = 4, Ar = 40 90 90 4×16 64 O% = = ×100% = 71.11% 90 90 b) For C2H2O4. 2H2O: M ( C2H2O4. 2H2O) = 2 ×12 +2 ×1 +4 ×16+ 2 (2 ×1 +1× 16) = 126 g/ mol For crystal water: 2 ×18 H2O% = × 100% = 28.57% 126 In order to find out the mass of any element in a compound with a known mass, molar mass of compound and atomic mass of elements are used. This calcula- tion can be expressed with the following formula. Atomic mass of element number of element in the compound × in the compound × mass of compound _____________________________________________________ Mass of element = Molar mass of compound Example 1-14 Find out the mass of Ca in 20 g of Ca3(PO4)2. Solution: Molar mass of Ca3(PO4)2: M (Ca3(PO4)2) = (3×40) + 2(31 + 4×16) = 310 g/mol If we insert the values in the formula above; Mass of Ca = 18  Basic Concepts in Chemistry Example 1-15 Exercise 1-11 In 10 g of CuSO4.5H2O, Calculate the mass of Na in 25 g of Na2CO3.10H2O. a) Find out mass of Cu. b) Find out mass percentage of crystal water. Solution: a) Molar mass of CuSO4.5H2O: b) Using the same equation; 1-12-Chemical Formula The chemical composition of compounds represented with formulas, which are its constituent elements symbols set with the number of atoms of those elements in the molecule and the composition of known chemical matter can Benzene molecule be expressed in different formats including: 1-12-1 Empirical Formula It is the simplest formula that gives the minimum limit of absolute information about the compound, it sets the rational number of the atoms included in the compound. For example, one molecule of benzene consist of 6 carbon atoms and 6 hydrogen atoms, so the greatest common divisor for these numbers is 6, and by dividing the number of atoms over 6 we will get the empirical formula for benzene which is CH. The same for the water molecule which is consist of two atoms of hydrogen and one atom of oxygen, so the empirical formula will be H2O, and so for one molecule of glucose which consist of 6 carbon atoms, 12 hydrogen atoms and 6 oxygen atoms, so the greatest common division for these numbers is 6, and by dividing he atom numbers by 6 the empirical formula will be CH2O. How to find the empirical formula for compounds? We should follow the steps below to find an empirical formula of a compound: a) The elements forming a compound are analyzed. b) The ratio of number of atoms is found by dividing mass of element or its Glucose percent ratio with atomic mass. Mass of element or percent ratio Ratio of no. atoms of element = Atomic mass 19  Chapter - 1 c) In order to get the simplest ratio of number of atoms, we need to divide num- ber of atoms of element to the number of atoms with lowest ratio and round it up to nearest integer. Number of atoms of each element Simplest ratio of number of atoms = The lowest value of number of atoms We get the empirical formula this way. Example 1-16 It is found that one of the gases consist of 20% of hydrogen and 80% of carbon. Find the empirical formula for this gas. Solution: 1) Divide each percent of element on its atomic mass, i.e., Percent ratio of element Number of atoms of element = Atomic mass 20 Number of atoms of H = = 20 1 80 Number of atoms of C = = 6.60 12 2) Dividing the above ratios over the smallest one and round to the nearest integer Number of atoms of each element Simplest ratio of number of atoms = The lowest value of number of atoms 20 _____ The simplest number of atoms of H = =3 6.60 6.60 _____ The simplest number of atoms of C = =1 6.60 The empirical formula of gas : CH3 Example 1-17 Cholesterol is an organic compound and it is found in almost every tissue of human body. Therefore it causes occlusion in veins. It contains 83.87% C, 11.99% H and 4.14% O. Find out the empirical formula of cholesterol. Solution: 1) Divide each percent of element over its atomic mass, i.e. Percent ratio of element Number of atoms of element = , Atomic mass 11.99 Number of atoms of H = 1 = 11.99 20  Basic Concepts in Chemistry 83.87 Exercise 1-12 Number of atoms of C = = 6.989 12 After burning a medicine, it was found to contain 74.27% carbon, 7.47% hy- 4.14 drogen, 12.99% nitrogen and 4.95% Number of atoms of O = = 0.258 oxygen. Write down the empirical 16 formula of this medicine. 2) We divide the results by the smallest value and get it close to the nearest ….. number. Number of atoms of each element The simplest number of atoms = The lowest value of number of atoms Exercise 1-13 What is the empirical formula of a compound consisting of 7.8 g of 11.99 K, 7.1 g of Cl and 9.6 g of O? K The simplest number of atoms of H= = 46 0.258 = 39 Cl = 35.5 O = 16 6.989 The simplest number of atoms of C= = 27 0.258 0.258 The simplest number of atoms of O = =1 0.258 The empirical formula of cholesterol: C 27 H 46 O 1-12-2-Molecular Formula It is the chemical formula that gives the real number of the atoms of elements that involve in the composition of one molecule of the matter. For example the molecule of ethane consist of 2 carbon atoms and 6 hydrogen atoms so the molecular formula will be (C2H6), according to that, the molecular formula is twice the empirical formula (CH3). The same for the molecular formula of the water which is (H2O) that means the water composes by the bonding of two hydrogen atoms with one oxygen atom, and it is the same of its empirical for- mula of water (H2O) so: Molecular formula = empirical formula x empirical formula unit To find the molecular formula of a substance, we follow these steps: a) Find out the empirical formula as mentioned before. b) Calculate the molar mass for the empirical formula by summation of atomic masses of their elements. c) Find the molar mass of the substance (molecular formula). d) Divide the molar mass of the molecular formula over the molar mass of the empirical formula to obtain the units of the empirical formula. Molar mass of molecular formula Empirical formula unit = Molar mass of empirical formula e) The result of division is multiplied with empirical formula and molecular formula is determined. 21  Chapter - 1 Example 1-18 The molar mass of an organic acid = 60 g/mol and contains 0% carbon, 6.7% hydrogen and the rest is oxygen. Find the molecular formula of the organic acid. Solution: The percentage of oxygen is 100-(40+6.7) = 53.3% 1) For each element, percent ratio is divided to atomic mass: Percent ratio of element Number of atoms of element = Atomic mass Number of atoms of H= 6.7 = 6.7 1 Number of atoms of C= 40 =3.3 12 53.3 Number of atoms of O = =3.3 16 2) These values are divided to the smallest among them and rounded up to the nearest integer. Number of atoms of each element Simplest ratio of number of atoms = The lowest value of number of atoms 6.7 The simplest no. of atoms of H = =2 3.3 3.3 =1 The simplest no. of atoms of C = 3.3 3.3 =1 The simplest no. of atoms of O = 3.3 For the empirical formula CH2O Thus, molar mass of CH2O: M ( CH2O) = (1 × 12) + (2 × 1) + (1 × 16) = 30 g/mol Molar mass of molecular formula Empirical formula unit = Molar mass of empirical formula 60 = =2 30 (CH 2 O)x 2 Molecular Formula Molecular formula = C2H4O2 22  Basic Concepts in Chemistry Example 1-19 Exercise 1-14 The empirical formula of an organic compound is known as C2H4O.As molar Caffeine is found in tea, coffee and mass of this compound is 88g/mol, find out its molecular formula. chocolate. It contains 49.48% C, 5.15% H, 16.49% O and 28.88% N. Solution: As its molar mass is 194 g/mol, find C2H4O out its molecular formula. M (C2H4O) = (2 × 12) + (4 × 1) + (1 × 16) = 44 g/mol Molar mass of molecular formula 88 Empirical formula unit = = =2 Molar mass of empirical formula 44 Molecular formula: C4H8O2 Basic concepts Law of Conservation of Mass: Matter is neither created nor destroyed Law of constant composition: The composition of elements are constant in all samples of a compound. Law of Combining Gas Volumes: There is a simple ratio between the volumes of reactant and product gases in a chemical reaction under the same temperature and pressure. Caffeine Avogadro’s Hypothesis: Equal volumes of gases contain equal number of molecules under the same pressure and temperature conditions. Valence: It is the number of electrons which an element loses, gains or shares during a chemical reaction. Atomic Mass Unit:The mass of C-12 isotope is accepted as 12.0000 and 1/12 of this is called as 1 atomic mass unit (amu). Temperature:Degree of temperature is used to tell to express that some ob- jects have high or low temperatues; temperature tells which way heat flows. Heat flow occure from a cold object to a hot object spontaneously. Temperat- ure has got three units. Centigrade (C˚), Kelvin (K) and Fahrenheit (F˚) Empirical Formula: It is the simplest formula of chemical compounds. It shows relative numbers of elements which form a compound. Molecular Formula: It is the formula which gives detaied and clear informa- tion about elements which form a chemical compound. Avogadro’s Number: It is either number of atoms in 1 mole of an element or number of molecules in 1 mole of a compound. Its value is 6.023×1023. Mlar mass: It is the unit of mass of 1 mole of atom or 1 mole of molecule in gram unit. Its unit is g/ mol. Mole (n): The amount of substance which contains, molecules or ions as many as Avogadr’s number is called as 1 mole of substance. 23  Chapter - 1 Questions of Chapter -1 1.1) What is Dalton’ atomic model? What is the relationship of it to the law of mass conservation? 1.2) It was observed that when H2 and Cl2 reacted, there was a simple ratio between the resulting gas (HCl) and the elements forming it. How do you explain the results according to the law of constant composition? 1.3)When two samples of CS2 samples were decomposed to its components, the first sample was found to have 8.68g S and 1.51g C whereas the second sample was found to have 31.3g S and 3.85g C. Verify if these results are consistent with the law of constant composition. 1.4) When two samples of sodium chloride were decomposed to their constituents , the first sample was found to have 4.65g Na and 7.16g Cl whereas the second sample was found to have 7.45g Na and 11.5g Cl. Verify if these results are consistent with the law of constant composition. 1.5) Na/F ratio in sodium fluoride is 1.21. When a sample of NaF was decomposed, it was found to have 34.5g of Na. Calculate the amount of fluoride that the sample contained in grams. 1.6) When a sample of magnesium fluoride was decomposed, it was found to have 1.65kg of magnesium and 2.57kg of fluoride. Another sample contained 1.32kg of magnesium. Calculate the amount of fluoride inside it. 1.7) When two samples of carbon tetrachloride (CCl4) were decomposed, the first sample was found to have 32.4g of C and 373g of Cl whereas the second sample contained 12.3 g of C and 112 g of Cl. Verify if these results are consistent with the law of constant composition. 1.8) Define the following terms Valence, atomic mass unit (amu), equivalent mass, atomic mass, Avogadro’s theory 1.9) 2 L of gas was obtained from the reaction of 1L of Cl2 and 3L of F2 gases. As the volumes of these gases were measured at constant pressure and temperature, what is the formula of the product gas? 1.10) When 1.55 g of silver was heated with chlorine gas, 2.05 g of silver chloride (AgCl) was formed. As the equival- ent mass of chlorine is 35.5 g, calculate the equivalent mass of silver. 1.11) When 0.72g of zinc was added to lead acetate, lead precipitated. After this precipitate was washed and dried, its mass was measured as 2.29g. What is the equivalent mass of lead? (Equivalent mass of zinc = 32.5g) 1.12) If the valence of an element is 2 and its equivalent mass is 32.7g, find out the atomic mass of this element. 1.13) If the atomic mass of an element is 55.85g and its valence is 3, find out its equivalent mass. 1.14) Calculate the number of moles of the following. a) 7 g of NaHCO3 b) 10 mg of Fe c) 16g of CO2 24  Basic Concepts in Chemistry 1.15) a) Calculate the number of moles of 5 g of silver and its number of atoms. b) A piece of diamond contains 5.0×1021 C atoms. Calculate the no. of moles of C and its mass in grams. 1.16) Calculate the amount of the following. a) The mass of 3.8×1020 NO2 molecules. b) The number of atoms of Cl in 0.0425g of C2H4Cl2. 1.17) Calculate the molar mass percentages of the elements forming the following compounds. a- NaClO3 b- CuSO4.5H2O c- (NH4)3PO4 d- Al2(SO4)3 e- Ca(C2H3O2)2 1.18) Calculate the molar mass of magnesium and crystal water in MgSO4.7H2O compound. 1.19) A sample of urea contains 1.121 g of N, 0.161 g of H, 0.4808 g of C and 0.640 g of O. Find out the empirical formula of this compound. 1.20) A compound contains C, H and N elements. When 35mg of this compound was burnt, 33.5mg of CO2 and 41.1mg of H2O were obtained. Find out the empirical formula of this compound. 1.21) The mass of some white dust was known to be 31.9% K, 39.2% O and 28.9% Cl. Find out the empirical formula and molecular formula of this compound. (Molar mass of the compound is 122.5g.) 1.22) A compound is made up of 24.27% C, 4.07% H and 71.65% Cl and its molar mass is known to be 99g/mol. What is the molecular formula of this compound? 1.23) 52.2% of a compound is C and 13.1% H and the remaining is O. Find out the molecular formula of this com- pound. (Molar mass of the compound is 46 g/mol) 1.24) Make the following calculations. a) The number of moles of O in 7.2 moles of H2SO4 , b) The number of atoms of Zn with a mass of 48.3 g, c) The mass of 6.73 moles of Al in grams, d) The mass of Fe in 79.2 g of Fe2O3. 25  GASES CHAPTER-2 ACHIEVEMENTS After studying this chapter, students will be able to. *Learn gas state of matter and its properties. *Learn the factors affecting gas state of matter. *Learn gas laws. *Can explain diffusion of gases. *Can differentiate ideal gases and real gases. *Can explain the effect of pressure on liquid vapor and boiling point.  Gases 2-1-PREFACE We live in the layer of the atmosphere in which gases are found most densely is called as troposphere. It consists of 78% N2, 21% O2 and 1% other gases. Most of this 1% is CO2.In nature,some substances are found in gas form at 25 °C temperature and 1 atm pressure conditions. In the following table 2-1, those gases and their symbols are given. Table 2-1 Some Elements and Compound in gas form at room temperature Chemical Symbol Element Chemical Formula Compound H2 Hydrogen HF Hydrogen flouride N2 Nitrogen HCl Hydrogen Chloride O2 Oxygen HBr Hydrogen Bromide F2 Fluorine HI Hydrogen Iodide Cl2 Clorine CO Caron monoxide Ne Neon CO2 Carbon dioxide Ar Argon NH3 Ammonia Kr Krypton NO Nitrogen Monoxide Xe Xenon NO2 Nitrogen Dioxide Rn radon N2O Dinitrogen monoxide SO2 Sulfur dioxide H2S Hydrogen Sulfide Molecules of matter can be studied most easily when they are in gas form. Gas molecules occupy 0.1% of the medium in which they are found; the remaining is space. Therefore, gas molecules move independently. As gases have com- pressibility, their volumes can be reduced, they can be liquefied by pressure or cooling. 2-2-VOLUme Volume is the amount of space matter takes up. The volume of gases is as much as the volume of the container they are inside. Volume is shown with V and measured with l, ml or cm3. To convert units of volume; 1 L = 1000 cm3 1 L = 1000 mL 1 cm3 = 1 mL conversion are used 27  Chapter - 2 Exercise 2-1 Example 2-1 What is the volume of a 0.125 L What is the volume of an 800 cm3 sample of NO2 gas in liters? sample of O2 gas in millilitres(ml)? Solution: 2-3-TemperatUre PV = nRT In Chapter 1, when we have mentioned temperature, we have told that there Do you know that? are different temperature scales, the Celsius (°C) and Kelvin (K) following formula to convert Celsius (°C) scale to Kelvin (K). The highest temperature was recorded in Mexico as +85°C T(K) = t(°C ) + 273 whereas the lowest temperature was recorded in Antarctica as Example 2-2 -88°C. The temperature of water in a container is 80°C whereas the temperature of water in another container is -13°C. What are their temperatures in Kelvin scale? Exercise 2-2 Solution: a) -100°C b)1°C c)127°C Write down the temperatures First container: above in Kelvin scale. Do you know that? According to football rules, Second container: mass of a football must be between 410 and 450 grams and pressure inside must be 2-4-PressUre between 0.6 and 1.1 atm. If pressure is higher than those, Pressure is the force (F) applied to per unit area (A). symbolizes it(P). While there is a risk of blast. pressure in open air is measured with a barometer, pressure of a closed con- tainer is measured with a manometer. F (Force) = = Pa (Pascal) A (area) Exercise 2-3 P= 1N(Newton) m2 What is the equivalent of 1.5 atm pressure in Torr? Units of pressure are Pascal (Pa), atmosphere (atm) and Torr. The relationship between them is shown below. 1 atm = 101325 Pa 1 atm = 760 mmHg 1 atm = 760 Torr 1 Torr = 1 mmHg 1 atm = 76 cmHg = 760 mmHg = 760 Torr 28  Gases Do you know that? Example 2-3 What is the equivalent of 688 Torr pressure in atm? Solution: 2-5-THE GAS LAWS Air bags are actually inflated by the equivalent of a solid rocket booster. 2-5-1-Volume-Pressure Relationship (Boyle’s Law) Sodium azide (NaN3) and potas- sium nitrate (KNO3) react very British scientist Robert Boyle made the first experiment on pressure and vol- ume relationship. for gases He used a U-shaped glass tube with one arm shorter. quickly to produce a large pulse of The short arm is closed and filled with some gases and when mercury is added hot nitrogen gas. This gas inflates from the open and long arm, it starts to apply pressure on the gas. Until the gas the bag, which literally bursts out is compressed to some volume, the amount of mercury increases and as shown of the steering wheel or dashboard in Fig. 2-1, the volume of gas has decreased. as it expands. Figure-2-1 A) Pay attention to the air when some mercury is added. B) We can observe that some more mercury is added ,vol- ume of air reduced to half Do you know that? Boyle observed that as pressure increased, volume of air decreased at con- stant temperature and the gas amount. Thus, Boyle’s Law is stated “as long as The pressure on us is the pres- temperature and amount of gas are kept constant, volume of gas is inversely sure of air surrounding the Earth. proportional to pressure applied to it.” This relationship is shown mathemati- The thickness of the atmosphere cally as follows: is 500 miles. In other words, we live under the ocean of atmos- phere. Exercise 2-4 That means the product of pressure and volume is equal to a constant. Accord- If the pressure of 50 L of gas in a ing to this law, at constant temperature, for a certain amount of sample gas balloon was reduced from 1 atm under P1 pressure and V1 volume is taken and if we would like to change P2 to 0.9 atm, what will be the final pressure or V2 volume of this gas, we can use the following equation: volume of the gas? P V = k (k is constant) 29  Chapter - 2 P1 × V1 = P2 × V2 (constant temperature and constant amount of gas) Example 2-4 The pressure of perfume in a 0.5 L bottle is 3atm. If applied pressure were 4 atm, what would be its volume? Solution: Example 2-5 a)1 L of a gas was filled in a cylinder under 1 atm pressure. When a weight was put on it, the volume decreased to 0.5 L. Assuming the tem- perature was constant, calculate the final pressure. b) Divers are subject to 1 atm pressure over the surface of water. At constant temperature, in 20 m depth, how much pressure is applied to divers? (Due to the weight of water, assume that pressure increases by 1 atm at every 10m) Solution: b) As the pressure on a diver increases by 1 atm at every 10 m, the pres- sure in 20 meter depth is 2 atm. Therefore, total pressure applied to the diver is 3 atm to take the pressure over the surface into account. 2-5-2-Temperature and Volume Relationship (Charles’ Law) When temperature of all gases increase, also their volumes increase. To measure volume increase by rise in temperature, a gas with a certain mass is filled into a cylinder with a moving piston as in Figure 2-2. 30  Gases Do you know that? Respiration in humans occurs according to Boyle’s Law. Contraction of diaphragm causes expansion of lungs and decrease in pressure in them. Therefore, they let air in and inhaling occurs. inhalation B A increase in volume of Figure 2-2 lugn A) Volume of heated gas increase and piston moves up. lung B) Volume of cooled gas decreases and piston moves down. When the mass over the piston is constant, the gas is under constant pressure. When the gas is heated, the piston moves upward and the volume of the gas increases. According to Charles’ Law, temperature (in Kelvin) and volume of Contraction of gases change in direct proportion when mass and pressure are constant. Math- diaphragm ematically, Expansion of diaphragm causes decrease in volume of ( k is constant ) lungs and increase of pressure in them. By exit of air from lungs, exhaling occurs. Generally, for a certain mass of gas, at T1 and T2 temperatures, V1 and V2 vol- umes are considered. In this case, at constant temperature, temperature-volume relationship is as follows: inhalation T V V At constant pressure and amount of gas = 1 or 1 = 2 T2 T1 T2 decrease in vo- lume of lung Example 2-6 A balloon was filled with air until its volume reached 4 L at 27°C. What will be its volume when we put it in a fridge at 0°C? (Pressure is the same for both lungs temperatures.) Solution: We convert temperature from °C to K. Expansion diaphragm 31  Chapter - 2 After conversion, we find the volume by Charles’ Law. Do you know that? Bicycle pumps are examples for Charles’ Law. When we use them, we notice that they get warm be- cause air molecules in a pump are forced to fit into a smaller area. Therefore, they collide with the walls of the pump more frequently and they warm the surroundings. 2-5-3-Pressure-Temperature Relationship (Gay-Lussac’s Law) When with constant volume is heated , Do you know that? its pressure increases When an inflated balloon is im- mersed in cold water, rates of crash of air molecules in the balloon de- crease and therefore volume of the When with constant volume is cooled , balloon decreases. its pressure decrease According to Gay-Lussac’s Law, for an ideal gas with constant volume and mass, its pressure is directly proportional to its temperature in Kelvin scale. This expression is shown as follows: The principle of a spray mechanism (k : constant) is that the difference between air pressure and pressure in spray can makes the spray work. For gases at two different temperatures (T1 and T2) and two different pressures (P1 and P2), the following equation are used: 32  Gases P1 P2 P1 T1 Exercise 2-5 = or = 1 L of CO2 is found in a balloon at T1 T2 P2 T2 (Volume and amount of gas con- 27°C. What happens when we put stant) the balloon in a pool at -3°C? Example 2-7 Why shouldn’t deodorant and spray cans be disposed of in fire? We can under- stand the reason better with the following question. The pressure of a spray can is 3 atm at 17°C. What will be its pressure at Exercise 2-6 187°C? The air pressure of tires of a car is 1.8 atm at 20°C. The car’s owner Solution: wants to go to Basra. When he ar- When temperature increases, kinetic energy of molecules in the can increases. rives at Basra, what will be the air Therefore, number of collisions of gas molecules with the walls of the can and pressure of tires at 36°C? also pressure increase. This causes can to burst. The following steps are applied for solution: We convert °C to K. We find out P2 through Gay-Lussac Law. 2-6-The COmBined Gas LaW If we re-write mathematical expressions of gas laws; Boyle’s Law.........................................P.V = k Charles’ Law....................................... Gay-Lussac’s Law................................ A combined law can be written from those 3 laws: PV =K K(Constant) T We can write the change in a gas with constant amount from P1, V1and T1 con- ditions to P2, V2 and T2 conditions. 33  Chapter - 2 Do you know that? Inside a refrigerator, cooling liquid is constantly cycled in the tubes. It rapidly vaporizes and turns into gas while passing through a thin The equation above is also called as equation of state for a constant amount of tube. When it is transformed into gas. gas, it absorbs heat in the refrigera- tor and it cools the surroundings. Example 2-8 The gas goes to condenser and it is transformed into liquid again. A bubble of 2.1 mL in volume is going up from the bottom of a lake at 8°C This condensation is made through temperature and 6.4 atm pressure. At the surface, the temperature is 25°C and the pressure is 1 atm. Calculate the volume of the bubble at the surface. pressure. Solution: We convert °C to K: k: constant Do you know that? We need to check air pressure of tires to avoid unequal corrosion on the surfaces of them. (The volume of the bubble at the surface) 2-7-The ReLatiOnship BetWeen AmOUnt Of Gas and VOLUme (AvOgadrO’s LaW) Italian scientist Avogadro discovered that volume of a gas and its amount is in direct proportion under constant temperature and pressure. The amount of gas is expressed with number of moles (n). According to this: 34  Gases For two different amounts of gases (n1, n2) and two different volumes (V1,V2) of them, we can use the following equation. ( at constant temperature and pressure) Avogadro’s Law is expressed as “equal volumes of all gases, at the same tem- perature and pressure, have the same number of moles.” We can observe this in Fig 2-3. Figure-2-3 gas tube When we withdraw gas from the environment, so, number of moles Withdrawing gas Adding gas decreases. When we add some more gas, number of moles in- Volume decreases Volume increases creases and gas volume increases. open air 2-7-1-MOLar VOLUme The value obtained by dividing volume to no. of moles is called as molar vol- ume. As no. of moles is shown with (n) and volume with (V), Vm expresses molar volume. Molar volume of a gas is equal to 22.4 L (22414cm3) at standard temperature and pressure (STP). (1atm = 760 Torr, 0°C = 273 K). Molar mass (M) is found by dividing mass (m) to no. of moles (n) as we have studied in Chapter 1.0.0. Exercise 2-7 Example 2-9 The volume of a sample of CO2 is If the volume of 1 mole of hydrogen gas is 22.4 L at standard temperature and 4 L at 66°C temperature and 1.2 pressure, what is the volume of 3 moles of hydrogen gas at the same condi- atm pressure. At 42°C, the volume tions? becomes 1.7 L. Calculate the pres- sure at this temperature. (The no. Solution: of moles hasn’t changed.) 35  Chapter - 2 Do you know that? 2-8-IdeaL Gas LaW We can obtain a single equation by combining 4 gas laws: Gay-Lussac discovered the rela- tionship between chemical reac- Boyle`s Law tions and volume. But other chem- ists didn’t accept his theory then and they didn’t know composition Charles`s Law of substances. Therefore, chemi- cal equations couldn’t be written at Gay-Lussac`s Law that time. The studies of Gay-Lus- sac and Avogadro enabled writing of chemical equations. Avogadro Law We can get a single equation from the equations above: When we convert the ratio to an equation: Exercise 2-8 Calculate the molar volume of 3 By inserting gas constant (R), we get the final equation will be: moles of a gas with a volume of 36 L. PV = nRT ideal gas law We can apply the equation above can be applied to gases which are applied 4 laws. This kind of gases are called as ideal gases. R is the general constant for ideal gases(R) (0.082). This equation must be used along with P(atm) , V (L), n (number of moles) and T(K) units in mathematical calculations. In order to calculate R value, we take 1 mole of an ideal gas with 22.414 L volume at standard temperature and pressure (0°C and 1 atm). Exercise 2-9 If 0.5 mole of a gas has 11.2 L of volume at standard temperature and pressure (STP), what is the no. of mole when it has a volume of 16.8 L? In order to find R value in International System of Units, we take pressure as 101325 Pa (Pascal), volume as 22.4×10-3 m3, no. of moles as 1 mole and tem- perature as 273K. 36  Gases Pa unit = Kg m.s 2 When we use it in the equation above, Do you know that? If we know pressure, tempera- ture and volume of any gas, Unit of energy we can find out its number of moles by making use of ideal gas law. Example 2-10 The volume of NO gas is 5.6 L. Find out the number of moles under standard temperature and pressure. Solution: Standard pressure and temperature conditions are 1 atm and 273K. 2-8-1-Calculation of Gas Density We can use ideal gas law to calculate gas density. According to the following equation; PV = nRT........................................(1) As , we can insert n in (1). 37  Chapter - 2 or; from density formula Exercise 2-10 Calculate the number of moles of 10 L of oxygen under standard shows density and by using it in (3) conditions. We get By arranging Equation (4), we can calculate density of a gas with known molar mass and pressure at a certain temperature. We can find out the mass of gas or molar mass from Equation (3): By re-arranging mass of gas Example 2-11 Hydrazine (N2H4) is used as rocket propellant. Calculate the density of this substance at standard temperature and pressure (STP). Solution: Molar mass of hydrazine At standard conditions, the pressure is 1 atm and the temperature is 273K 38  Gases Example 2-12 Find out the number of moles of a 700 mL gas at 27°C temperature and 0.8 atm pressure. Solution: The following steps are followed for solution: We convert the volume of the gas from mL to L. We convert the temperature from °C to K. From ideal gas law; Example 2-13 The pressure of a gas in a 3 litre container was found as 5.46 atm at 27°C tem- perature. As the molar mass of the gas is 44 g/mol, find out the mass and mole number of the gas. Solution: The following steps are applied for solution: 1. We convert the temperature from °C to K. 2. Using the following equation; 39  Chapter - 2 3. Calculate the mole number; Exercise 2-11 Example 2-14 At 227°C temperature and 748 Torr pressure, the mass of a gas is 0.6 g in a 500 Calculate the density of oxygen mL container. Calculate the molar mass of the gas. gas (O2) at 373K temperature and 5 atm pressure in g/L unit. Solution: The following steps are applied for solution: 1) We convert the volume from mL to L. We convert the temperature from °C to K. Exercise 2-12 Methane is a gas obtained by oil refining. Find out the volume of 0.5 mole of a sample from meth- ane gas at 27°C temperature and 3 2) We convert the pressure from Torr to atm. atm pressure. 3) From Equation Example 2-15 The molar mass of 0.31 g of a gas is 32 g/mol and it has a pressure of 1.17 atm. What is the temperature when this gas has a volume of 0.23 L? 40  Gases Solution: 2-9-DaLtOn’s LaW Of PartiaL PressUres So far, we have dealt with only one type of gas. But what happens when we handle a gas mixture which doesn’t interact such as air? When Dalton studied a sample of air, he found out that the total gas pressure is equal to the sum of par- tial pressures of gases. Partial pressure is the pressure of a gas in a gas mixture. Closed valve Dalton’s law of partial pressures is expressed as “Total pressure of a gas mix- ture in which gases don’t react is equal to sum of partial pressures Oxygen Hydrogen gas with 1L gas with 1L of all gases.” Mathematically, volume and 2 volume and 2 atm pressure atm pressure PT show total gas pressure whereas P1 , P2 and P3 …. show partial pressures of gases in the mixture. 2-9-1-The Relationship between Total Pressure, open valve Total Number of Moles and Mole Fraction Assume that there is a gas mixture of two gases with known volume and pres- sure. We find the partial pressures using the ideal gas law as follows. When we open the valve , gases diffuse in both bottles. According to Dalton’s Law ,pressure of both gases change. From Dalton’s Law PT =P1 +P2...................................3 We insert Equation 1 and 2 in Equation 3. Exercise 2-13 At 127°C temperature and 3.65 atm pressure, find out the molar mass of By dividing Eq. 1 to Eq. 4; a gas with 900 mL volume and 4.41 g mass. 41  Chapter - 2 Exercise 2-14 By eliminating similar terms; Calculate the molar mass of a gas with 0.4 g mass and 280 mL of vol- ume at standard temperature and pressure (STP). We get the equation nT shows total no. of moles in the mixture. It is found by adding no. of moles of gases in the mixture. nT= n1+ n2 Thus, Equation 6 becomes as follows: P1 n1 =.............................(7) PT n T Similarly, for the second gas in the mixture; Do you know that? P2 n 2 What prevents molecules in the at- =.............................(8) equation is obtained. mosphere from escaping to space? PT n T Gravity is applied on molecules as it is applied on us. Molecules need Mole Fraction: The ratio of no. of moles (for 1st or 2nd) of each component in a to have a velocity of 1.1x103 m/s mixture to total no. of moles of components gives mole fraction. Mole fraction to escape from this gravitational of 1st component is shown as (x1). force. As molecules of helium gas n1 n are much faster than this limit, they x1 = = 1 can escape to space. Therefore, n1 +n 2 n T there isn’t much helium gas in the atmosphere. Mole fraction of 2nd component is shown as (x2). n2 x2 = nT When we insert mole fraction in Equations 7 and 8; we get these equations: We can re-write Eq. 10 as follows: xi shows mole fraction of a component and Pi shows its partial pressure. Be- sides, sum of mole fractions of all gases in a mixture is equal to 1. We can apply this to the previous two-gas mixture: 42  Gases Generally, we can write the following equation: Example 2-16 In a mixture of noble gases, there are 4.46 moles of Ne, 0.74 mole of Ar and 2.15 moles of Xe. Calculate the partial pressures of each gas. (Temperature is constant and total pressure is 2 atm.) Solution: The following steps are applied for solution: 1) Total number of moles is found. 2) Mole fractions of noble gases are found. Exercise 2-15 Gases obtained by oil refining were placed into a container. There are 6 moles of methane, 4 moles of ethane and 2 moles of propane in it. As the total pressure is 6 atm, cal- culate partial pressures of each gas. 3) Using the following formula, partial pressure of eac