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Making Chemistry learning easier M for Moles Concept, calculations and applications. Do I have anything to do with Chemistry? Sample copy (some pages are missing) 1 M for Mole, Copyright 2009 Etacude.com Etacude.com M for Moles (Electronic PDF version 1.1) Sample copy You can assess some of the content on-line free at http://sciencepark.etacude.com/chemistry/introduction.php and http://sciencepark.etacude.com/lzone You can purchase your own full version PDF registered copy. Details can be found at http://shop.etacude.com/m_for_moles.php Benefits of getting a registered PDF copy: (1) Provide detailed work-out for answers to the Checkpoint questions (2) One free upgrade Disclaimer Great effort has been made to ensure the accuracy of the PDF electronic version of the M for Moles from Etacude.com. We are not responsible for any loss or damage that results from assessing or downloading any of the information and any of its derivatives. We disclaim any responsibility for any failure, inaccuracy, harm and damage arise to your project, school, examination, theoretical or experimental works as a result of using the information supplied from this e-book. 2 M for Mole, Copyright 2009 Etacude.com Table of content Some Common Equations 1. Basic Concept 2. Chemical Equation 3. Balancing Equation 4. Moles 5. Answers List of Relative Atomic Mass How to use this e-book Finding calculations in chemistry difficult to learn? Don’t know how/when to or why use moles in chemistry? Confuse about molecular weights and their relationships to moles and chemical equations? This e-book teaches you how to do mole calculations that are often encountered in chemistry. Readers are assumed to have only a basic science concept and with minimum knowledge in chemistry. This e-book will show you how to solve those tricky moles calculations with plenty of examples, showing in clear and concise ways by following through stepwise and simple procedures. The first three chapters are foundations essential to mole calculations. This include topics such as chemical formula, balancing equations and molecular weight calculations. However, this e-book is emphasised in solving mathematical questions in chemistry and its concept is kept to a minimum. For instance, this book will show you how to determine the chemical formula for a chemical substance but will not show you why the formula is as such. It is particularly suitable for pupils from Year 10 onwards who are taking GCSE or A/S levels in UK or Ninth Grade onwards in the US. Words in bold - learn it, pay particular attention and make sure you understand them. Equations and statements come with Questions come with 3 learn it! practice it! M for Mole, Copyright 2009 Etacude.com Some common equations (best to memorise it) This section is just a quick reminder of some of the important equations. Please refer to the appropriate chapters to see how they are used. Chapter 2 (1) Percentage by mass of an element in a molecule Total relative mass of atom Percentage by mass = X 100 % Relative molecular mass (2) Step-by-step procedures to calculate empirical formula (a) Write down element symbols. (b) Write down the mass or percentage of each element. (c) Divide the mass by the relative atomic mass. (d) Divide the mole numbers to get the smallest whole ratio. (e) You get the empirical formula. Chapter 4 (1) Avogadro’s constant (NA) = 6.022 x 1023 mol-1 (2) Mole equation (mol) Mass of substance Mole = (3) Relative molecular mass Density = mass volume (4) Concentration (molarity, M) Moles of solute concentration = Volume of solvent in dm3 (5) 1 dm3 = 1000 cm3 = 1000 ml = 1 l (liter) 4 M for Mole, Copyright 2009 Etacude.com Sample copy 5 M for Mole, Copyright 2009 Etacude.com 1. Basic Concept This Chapter gives definitions of some basic keywords that that is relevant to chemistry. You need to familiarise yourself before going into other parts of this book. 1. Atom - It is the smallest unit of a substance. The identity of a substance will be destroyed if its atom is further divided. Different substances will have different types of atoms. All atoms are made up of a number of protons, neutrons and electrons. Proton (+1) Electron (-1) Neutron (0) Schematic representation of an atom (in this case is a helium atom) A proton has a positive charge (+1) and an electron has a negative charge (-1). Neutron does not have a charge. It is neutral (0). All atoms have equal number of protons and electrons. In other words, they are always neutral. Charged atoms, or ions, always contain a different number of protons and electrons. For example a +2 (positive 2 charged) ion contains two electrons less than the number of protons. A positively charged ion is called a cation, while a negatively charged ion is called an anion. Checkpoint 1: (a) An ion with charge -1 contain _____ number of protons then electrons. By how many? ______ (b) A +2 charge ion contain _____ number of protons then electrons. By how many? ______ 6 M for Mole, Copyright 2009 Etacude.com 2. Element - It is a substance that cannot be further resolved into simpler substance components by chemical means. It consists of a single type of atom of same number of protons. For example, gold and copper are elements which consists of only gold atoms and only copper atoms, respectively. 3. Compound - It is a substance made of more than one type of elements. They are usually formed by a chemical process and atoms of different types are bound together by chemical bonds. 4. Molecules - This is the smallest unit of a compound. For example, a water molecule is made of two hydrogen atoms and one oxygen atom which bind together by chemical bonds called the covalent bonds. Oxygen atom A chemical bond Hydrogen atom Schematic representation of a water molecule Another example: butane gas, which is used as a camping stove. Butane is a compound which consists of carbon and hydrogen elements. Butane molecule is made of four carbon atoms and ten hydrogen atoms. Carbon atom A chemical bond Hydrogen atom Schematic representation of a butane molecule 7 M for Mole, Copyright 2009 Etacude.com Sample copy 8 M for Mole, Copyright 2009 Etacude.com 9. Periodic Table of the Elements - All known elements are arranged systematically in a table called the Periodic Table of the Elements. They are arranged according to the number of protons and grouped according to their chemical properties. A modern layout of the Periodic Table of the Elements. The one or two alphabets are element symbols with the number of protons shown above for each element. To date, there are around 120 known types of atoms or elements. Of these, about 90 elements can be found in nature. All matters are made of these elements. The rest, usually those of heavier ones (from uranium with atomic number 92 onwards) no longer exist or are found only in traces. However, these heavier elements can be produced in a nuclear reactor. Checkpoint 2: It is useful to remember the symbols of the first twenty elements beginning with hydrogen. Note also the element symbols for other common but heavier elements such as: Iron (Fe), Copper (Cu), Zinc (Zn), Lead (Pb) and Bromine (Br) 9 M for Mole, Copyright 2009 Etacude.com 2. Chemical Equation 2.1 Atomic symbol A complete description of an isotope element can be simplified by means of using the following notation: A ZE For example: 238 92 U where E is the atom symbol shown in the Periodic Table, Z is the number of protons and A is the atomic mass number. Since the mass number is the sum of protons and neutrons, the number of neutrons, N, can be worked out as: Number of neutron = Atomic mass number Or N= A Z number of protons The example given above is the atomic symbol for the most common isotope of uranium with atomic number 92 and mass number 238. There are 238 - 92 = 146 neutrons in the atomic nucleus. Since Z is unique for each atom, it is common to write as 238U or 'uranium-238' in the text. In other words, the atomic number (Z) is left out (since the symbol U is sufficient to identify the element as uranium). This effectively refer to a particular isotope of a uranium atom. The complete notation shown above are more commonly used in radiochemistry, the study of radiation, nuclear fission and stuff like that. Otherwise, it is more commonly written as simply 'U', unless there is a need to refer to a particular isotope. Checkpoint 3: Find the number of proton neutron and electron for the following isotopes (Clue: If the proton number is not given, look at the periodic table of the Elements): 212Bi 83 208Pb 82 127I 234U 4He 12C 13C Notice that certain elements has more than one mass numbers. They are called isotopes. 10 M for Mole, Copyright 2009 Etacude.com Sample copy 11 M for Mole, Copyright 2009 Etacude.com Example 1 A methane molecule contains one carbon and four hydrogen atoms. This can be illustrated as follows: Or in molecular formula: CH4 The ‘ball-and-stick’ model diagram on the right is called structural formula while CH4 is the corresponding chemical (molecular) formula. The lines represent chemical bonds and notice that the carbon atom is connected to four hydrogen atoms, whereas each hydrogen atom can only have one bond, which is connected to the carbon atom. Example 2. The graphical representation of a water molecule can be shown as Or in molecular formula as: H2O Once again, the oxygen atom can only form two bonds, each to a hydrogen atoms, which in turn formed one bond with the oxygen atom. As in the case of methane molecule, the little number 2 indicates there are two hydrogen atoms. Atomic symbols without any number on the right means there is only one of the atom. Example 3. If one were to describe two water molecules, how is this represented in molecular formula? The formula symbol for one water molecule is H2O. For two water molecules, this is 2H2O. 2H2O 12 M for Mole, Copyright 2009 Etacude.com Example 4. Hydrogen peroxide has a formula of H2O2. Is this correct? Hydrogen can only form one bond and oxygen can form two bonds. This is illustrated below: Example 5. Carbon dioxide has a chemical formula of CO2. How is this possible? This is illustrated as follows: In this case, the double-line of one side indicates two bonds are required to connect between the carbon atom and an oxygen atom. Notice that the total number of allowable bonds for both carbon and oxygen is still maintained - four for carbon and two for oxygen. Example 6. Ethyne has a chemical formula of C2H2. What is the corresponding structural formula? Notice that there are three bonds connected between the carbon atom and the total number of connecting bonds for each atom is still the same: Four bonds for carbon and one bond for hydrogen. A single line is usually referred to as a single bond. A two-line symbol as a double bond and a three-line symbol as a triple bond. Checkpoint 4: Table (a) can be used to deduce the chemical formula of some of the simple covalent compounds. However, it is useful to remember the chemical formula of some of the covalent substances: Water: H2O Hydrogen chloride: HCl Ammonia: NH3 Methane: CH4 Hydrogen gas : H2 13 Carbon dioxide: CO2 Carbon monoxide: CO Sulfur dioxide: SO2 Ethane: C2H4 Chlorine gas: Cl2 M for Mole, Copyright 2009 Etacude.com Sample copy 14 M for Mole, Copyright 2009 Etacude.com Example 1. Sodium hydroxide is made of one sodium cation and one hydroxide anion: Na+ and OHThere is one sodium atom, one oxygen atom and one hydrogen atom for the sodium hydroxide molecule. It is not necessary to show the chemical formula as Na+OH- . This is because the charge is balanced and the chemical formula is simply written as NaOH Example 2. Two sodium hydroxide molecules. In this case, this is written as: 2NaOH Note: Do not confuse this as ‘two sodium atoms, one oxygen atom and one hydrogen atom’. The correct total number of atoms are two sodium atoms, two oxygen atoms and two hydrogen atoms. Example 3. What is the chemical formula for calcium hydroxide? From Table (b) a calcium ion has positive two charge, whereas the hydroxide ion has negative one charge. To get a net zero charge, you need two hydroxide anion to cancel out the charge on the calcium cation. So, the chemical formula is Ca(OH)2 The bracket enclosed with a little two means there are two sets of OH-. The bracket is not needed if there is only one set of the group, as shown in the Example 1 above. 15 M for Mole, Copyright 2009 Etacude.com Sample copy 16 M for Mole, Copyright 2009 Etacude.com Checkpoint 5: Determine the chemical formula of the following compounds. It is best to remember the symbols and charge values of these common ions, as shown in Table (b). Sodium oxide, Na+ and O2- Iron(II) sulfate, Fe2+ and SO42- Magnesium oxide, Mg2+ and O2- Iron(III) sulfate, Fe3+ and SO42- Potassium bromide, K+ and BrAmmonium sulfate, NH4+ and SO422.3 Relative molecular mass It is also known as the molecular weight or formula weight. The relative molecular mass is essential in mole calculations. For instance, to determine the amount of products formed in a chemical reaction. It is determined by summing up the relative atomic mass of all the atomic constituents in a chemical formula. The atomic mass values can be found in any standard Periodic Table of the Elements. Example 1. What is the molecular mass for water? The chemical formula of water is H2O, which consists of two hydrogen and one oxygen atoms, with relative atomic mass of 1.0 and 16.0 respectively. Therefore: Molecular mass of H2O = 2 x (H) + 1 x (O) = 2 x 1 + 1 x 16 = 18.0 Example 2. What is the molecular mass for sodium chloride? The chemical formula of sodium chloride is NaCl, which consists of one sodium and one chloride respectively. 17 M for Mole, Copyright 2009 Etacude.com Sample copy 18 M for Mole, Copyright 2009 Etacude.com 2.4 Percentages of elements A molecule consists of a number of atoms in different proportions. The percentage content of an element in a compound can be determined from the total relative atomic mass of the atom in the molecular formula and the relative molecular mass of the compound. Using the following relationship: Total relative mass of atom Percentage by mass = X 100 % Relative molecular mass Note: The total percentage of all the elements contained in a compound has to be 100 %. Example 1. Work out the percentage by mass of sodium and chlorine in sodium chloride. From the Periodic Table, relative atomic masses of sodium (Na) and Chlorine (Cl) are 23 and 35.45 respectively. Molecular mass of NaCl = 1 x (Na) + 1 x (Cl) = 1 x 23 + 1 x 35.45 = 58.45 Percentage of sodium in NaCl = 23 58.45 X 100 % = 39.35 % Percentage of chlorine in NaCl = 35.45 58.45 X 100 % = 60.65 % Example 2. Work out the percentage by mass of of all constituent atoms in dry sodium carbonate. The molecular symbol for dry (anhydrous) sodium carbonate is Na2CO3. The relative atomic mass for Na = 23, C = 12 and O = 16. 19 M for Mole, Copyright 2009 Etacude.com Therefore: Molecular mass of Na2CO3 = 2 x (Na) + 1 x (C) + 3 x (O) = 2 X 23.0 + 1 x 12.0 + 3 x 16.0 = 106.0 Since there are two Na atoms, the total relative atomic mass of sodium atoms in Na2CO3 is 2 x 23 = 46. Therefore: Percentage of sodium in Na2CO3 = 46 106 X 100 % = 43.40 % Percentage of carbon in Na2CO3 = 12 106 X 100 % = 11.32 % The total percentage mass of all the atoms in the molecule is 100%. Therefore Percentage of oxygen in Na2CO3 = 100 % - 43.40 % - 11.32 % = 45.28% Checkpoint 7: Determine percentage by masses of the following compounds. Potassium bromide, KBr Glucose, C6H12O6 Potassium dichromate, K2Cr2O7 Ethanoic (acetic) acid, CH3COOH Ammonium sulfate, (NH4)2SO4 Also, work out what is the percentage mass of ammonium anion in ammonium sulfate? 20 M for Mole, Copyright 2009 Etacude.com 2.5 Empirical Formulae The empirical formula of a substance is chemical formula in its simplest form. It shows the atomic constituent of a chemical substance in the simplest whole number ratio. This is in contrast to its normal chemical formula which gives the actual number of atoms of each type of element that made up a molecule. Example 1. The chemical formula for ethane is C2H6. What is its empirical formula? The ratio of carbon to hydrogen is 2:6. To find out the empirical formula, the whole number ratio must be reduced to its simplest and smallest whole number form. Carbon 2 Hydrogen 6 Divide this by two 2 2 6 2 Simplest ratio 1 3 Note: always divide with the same Number for all atoms. The empirical formula for ethane is therefore CH3. Example 2. What is the empirical formula for butene (C4H8) ? Write down the ratio: Divide this by four Simplest ratio Carbon 4 Hydrogen 8 4 4 1 8 4 2 The empirical formula for butene is CH2. Note: If you divide the ratio by 2, you get a ratio of carbon 2 and hydrogen 4, which is not the simplest ratio as it can still be further divided. 21 M for Mole, Copyright 2009 Etacude.com Sample copy 22 M for Mole, Copyright 2009 Etacude.com Example 1. A sample compound was determined to contain only 24 g of carbon and 32 g of oxygen. What is its empirical formula? Refer to the step-by-step procedures from the previous page: (a) (b) (c) (d) divide by 2 (e) smallest ratio C 24 g 24/12 = 2 2/2 = 1 1 O 32 g 32/16 = 2 2/2 = 1 1 The empirical formula of the substance is CO. (Do not confuse this as carbon monoxide!) Example 2. A compound is made of 48 g of carbon, 8 g of hydrogen and 64 g of oxygen. What is its empirical formula? Refer to the step-by-step procedures from the previous page: (a) (b) (c) (d) divide by 4 (e) smallest ratio C 48 g 48/12 = 4 4/4 = 1 1 H 8g 8/1 = 8 8/4 = 2 2 O 64 g 64/16 = 4 4/4 = 1 1 The empirical formula of the substance is CH2O. Example 3. A compound is made of 40% carbon, 6.67% hydrogen and 53.33 % oxygen. What is its empirical formula? The percentage composition is proportional to the relative mass of the substance. The same step-by-step procedures can be used. (a) (b) (c) (d) Divide by 3.33 (e) smallest ratio C 40% 40/12=3.33 3.33/3.33=1 1 H 6.67% 6.67/1=6.67 6.67/3.33=2 2 O 53.33% 53.33/16=3.33 3.33/3.33=1 1 The empirical formula is CH2O. Note that to get the smallest whole number, always look for smallest number in step (c). 23 M for Mole, Copyright 2009 Etacude.com Sample copy 24 M for Mole, Copyright 2009 Etacude.com 3. Balancing Equation A chemical equation is an important piece of information for most of the moles calculations. A chemical equation shows a complete summary description of a chemical reaction. However, before the equation can be used, it must be properly balanced. Balancing a chemical equation involves two parts: (1) Balance the individual chemical formula of a compound. (2) Balance the overall compounds involve in a chemical reaction. Remember that balancing chemical equations always involving atoms rearrangement. The total number of atoms at the left is always the same as that on the right. Virtually all simple gas molecules are made of two-atom pairs. They are called diatomics. For example, H2, Cl2, N2 and F2. The exceptions are those belong to Group 0. These are called inert gases and they exist as a single atom entity or monoatomic. For instance He, Ar and Xe. Example 1 The hydrogen can react explosively with oxygen to give water. Write the chemical equation of this reaction. So, first of all, pick up the component compounds that involved in this reaction: Oxygen + hydrogen water O2 + H2 H2O Remember that gases always exist as pairs as mentioned above. You can see that the equation is not balanced: There are one less oxygen atom on the right hand side of the equation. The only way to balance the equation is to change the number of molecules involved the equation. Start by making two water molecules and the extra hydrogen atoms introduced on the right hand side is balanced by making two hydrogen molecules on the left hand side: O2 + 2H2 Graphical illustrations: 25 2H2O + M for Mole, Copyright 2009 Etacude.com oxygen hydrogen Example 2 A displacement reaction occurs when a piece of zinc metal is added into a solution of copper sulfate: Zinc enters into solution to become zinc cation, while copper ions in solution are changed into the copper metal which coats the zinc metal piece. Write the chemical equation for the reaction. The chemical equation for the reaction is shown below: Zn(s) + CuSO4(aq) Cu(s) + ZnSO4(aq) Notice that the total number of atoms and atom types are the same for both sides and no further balancing is required. The brackets describe the state of a compound in a reaction, or the state symbol. There are usually four different states: solid (s), liquid (l), aqueous (aq) and gas (g). Since the sulfate anion, SO42-, does not actually involve in the reaction (remains the same from both left hand and right hand sides), the equation can also be simplified as follows: Zn(s) + Cu2+(aq) Zn2+(aq) + Cu(s) This type of equation is also called the half equation because the charge is not balanced and usually only one type of charge is shown. Example 3 Magnesium metal burns brilliantly with oxygen to give white magnesium oxide. Write the chemical equation for the reaction. First of all write down all the chemical components involve in the reaction: Magnesium + oxygen Magnesium oxide Mg(s) + O2(g) MgO(s) Next, balance the equation. Again, start looking for unbalance atoms. Here, we have one less oxygen atom on the right hand side. Start by writing two magnesium oxide molecules (so as to make use of the two oxygen atoms on the left) and then write two for the magnesium in order to balance the equation: 2Mg(s) + O2(g) 2MgO(s) 26 M for Mole, Copyright 2009 Etacude.com Sample copy 27 M for Mole, Copyright 2009 Etacude.com Checkpoint 10: Balancing an equation requires experience. The more you do it, the more familiar it becomes to you. Obviously, it is also important to know the chemical formula of the compounds involve in a chemical reaction. Otherwise, the equation can never be balanced properly. Balance the following equations. ZnO(s) + HCl(aq) H2(g) + Cl2(g) Cu(s) + AgNO3(aq) CH4(g) + O2(g) HCl(aq) + MnO2(s) ZnCl2(aq) + H2O(l) HCl(g) Cu(NO3)2(aq) + Ag(s) CO2(g) + H2O(l) MnCl2(aq) + H2O(l) + Cl2(g) Copper + oxygen Copper(II) oxide Calcium hydroxide + carbon dioxide Calcium carbonate + water Hydrochloric acid + zinc Zinc chloride + hydrogen Hydrochloric acid + calcium carbonate Calcium chloride + carbon dioxide + water 28 M for Mole, Copyright 2009 Etacude.com 4. Moles This chapter marks the beginning of showing you how moles are used in chemistry calculations. You are assumed to have some basic understanding of chemistry. The first three chapters highlight some of these that are relevant to mole calculations. Please make sure that you are familiar with them before proceed any further. 4.1 Introduction Mole is the amount of substance, measures in terms of elementary units, as there are number of atoms in 0.012 kg (12.0 g) of carbon12. The elementary units can be atoms, electrons, molecules, etc. It is abbreviated as mol and is the standard SI unit for the amount of substance. One mole of carbon-12 contains about 6.02214 x 1023 carbon-12 atoms. 1 mol of 12C contains 6.02214 x 1023 carbon atoms This number is also known as the Avogadro’s number (a constant) and is in unit mol-1 (per mole). Avogadro’s number = 6.02214 x 1023 mol-1 Because of the relationship of the atomic mass unit with the Avogadro’s number, one mole of a chemical compound has a mass equal to its relative molecular mass expressed in grams. This is also true for the chemical elements. This is called molar mass. For example, aluminium (Al) has a relative atomic mass of 26.98. This means that there are 6.02214 x 1023 aluminium atoms in 26.98 g of aluminium. 1 mol of Al weighs 26.98 g and contains 6.02214 x 1023 Al atoms Water (H2O) has a mass unit of 18.0. This means that there are 6.02214 x 1023 water molecules in 18.0 g of water. 1 mol of H2O weighs 18 g and contains 6.02214 x 1023 H2O molecules Therefore: 29 M for Mole, Copyright 2009 Etacude.com Mass of substance Mole = Relative molecular mass In fact, you have already come across calculation of moles in finding empirical formula (step (c) in Page 22). Example 1 What is the mole value for 15 g of a pure magnesium sample? The atomic weight of magnesium is 24.305 and the molar mass is 24.305 g/mol. Using the equation shown above 15.0 g = 0.617 mol Mole = 24.305 g/mol Example 2 What is the mole value for 22.3 g of a pure sodium carbonate sample? The molecular weight of sodium carbonate is 106.0 (see Example 3 of Section 2.3) and therefore the molar mass is 106.0 g/mol. Using the equation shown above Mole = 22.3 g = 0.210 mol 106.0 g/mol Example 3 How many molecules are there in 5.6 g of pure potassium chloride? The chemical formula of potassium chloride is KCl and the molecular weight is 39.098 + 35.453 = 74.551 or 74.551 g/mol. Therefore, the mole value is: 5.6 g = 0.075 mol 74.551 g/mol Since there are 6.02214 x 1023 molecules in 1 mol of KCl, for 0.075 mol, there are 6.02214 x 1023 mol-1 x 0.075 mol = 4.517 x 1022 molecules 30 M for Mole, Copyright 2009 Etacude.com Sample copy 31 M for Mole, Copyright 2009 Etacude.com Sample copy 32 M for Mole, Copyright 2009 Etacude.com 4.2 Moles in Chemical Reactions Mole is an important feature to determine the amount of a product being produced in a chemical reaction. Consider the following chemical reaction (see Example 1 of Chapter 3): 2H2(g) + O2(g) 2H2O(l) Here two hydrogen molecules react with one oxygen molecule to produce two water molecules. One may perhaps think that a mass unit can be added to both sides of a reaction, just like a normal mathematical equation. For instance, the above chemical equation wrongly reads ‘two grams of hydrogen react with one grams of oxygen to produce two grams of water’. Clearly this does not make sense as the total mass of the reactants and the product must be the same from both sides of the equation. The correct interpretation of the above reaction is ‘two moles of hydrogen molecule react with one mole of oxygen molecule to produce two moles of water molecules. The actual mass can then be determined by using the mole equation in Chapter 4. Example 1 How many moles of the oxygen is required when 0.4 mol of magnesium ribbon burns? The chemical reaction proceeds as follows: 2Mg(s) + O2(g) 2MgO(s) So the reaction shows two moles of magnesium reacts with one mole of oxygen (ignore magnesium oxide since it is not relevant to the question). To easily work out the answer for 0.4 mol of magnesium, arrange the number in the following way: 2 Mg 0.4 Mg The double arrows mean ‘relate to’ and 33 O2 y O2 y is the answer. M for Mole, Copyright 2009 Etacude.com Cross multiply as the green dotted line shows. This gives Or 2 x y mol = 1 x 0.4 mol 2y = 0.4 mol y = 0.4/2 = 0.2 mol So 0.4 mol of magnesium will react with 0.2 mol of oxygen. The above ‘cross’ method works for any chemical reaction provided it is balanced. Example 2 How many grams of carbon dioxide is produced when 2.5 mol of propane is burnt in the presence of excess air? The reaction of propane with the oxygen in air is as follows (see Example 4 of Chapter 3) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 1 mol of propane will produce 3 mol of carbon dioxide. Using the cross method 1 C3H8 2.5 C3H8 3 CO2 y CO2 And so y x 1 = 3 x 2.5 y = 7.5 mol (of carbon dioxide) Now the molar mass of CO2 is 44 g/mol. Therefore, mass for 7.5 mol of CO2 is (see Example 4 of Chapter 4.1) 7.5 mol x 44 g/mol = 330.0 g Example 3 How many grams of magnesium is needed to produce 50.0 g of magnesium sulfate from the reaction with excess sulfuric acid? The equation is as follows: H2SO4(aq) + Mg(s) 34 MgSO4(aq) + H2O(l) M for Mole, Copyright 2009 Etacude.com Sample copy 35 M for Mole, Copyright 2009 Etacude.com The equation for the chemical process is as follows: CaCO3(s) CaO(s) + CO2(g) From Chapter 4.1, we see that the standard mole definition of a chemical compound bears the relationship of a mass equal to its relative molecular mass expressed in grams (gram-mole). However, we can also express the mole in other mass units, such as kilogram, pound, ton and so on. Of course, any mass units other than gram will have different numbers than the Avogadro’s constant (6.022 x 1023). So, the molecular weight of calcium carbonate is 40.078 + 12.011 + (3 x 16.00) = 100.089 To express the molar mass in kilogram-mole, there is 100.089 kg of calcium carbonate in one mole or 100.089 kg/mol. For 250 kg 250 kg = 2.498 mol 100.089 kg/mol From the equation 1 mol CaCO3 2.498 mol CaCO3 1 mol CaO y mol CaO y = 2.498 mol Therefore The molecular weight of CaO is 56.078, which is 56.078 kg/mol. Therefore, the mass of CaO produced is 56.078 kg/mol x 2.498 mol = 140.08 kg Alternatively, without calculating the mole (as in Example 3) 1 x 100.089 kg of CaCO3 250 kg of CaCO3 x= 1 x 56.078 kg of CaO x kg of CaO 56.078 kg x 250 kg = 140.07 kg 100.089 kg You can see that only the mass unit is changed, but the calculation method remains the same. 36 M for Mole, Copyright 2009 Etacude.com Sample copy 37 M for Mole, Copyright 2009 Etacude.com Checkpoint 12: (1) How many grams of magnesium oxide is produced when 8.6 g of magnesium metal is burnt in air? (2) Aluminium burns in oxygen to produce aluminium oxide: 4Al(s) + 3O2(g) 2Al2O3(s) How much aluminium oxide can be produced if 25.8 g of aluminium is burnt? (3) How grams of calcium carbonate is needed in order to produce 34.9 grams of calcium chloride from a reaction with an excess of hydrochloric acid? CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) (4) Calcium hydroxide (slaked lime) is used in agriculture to reduce soil acidity. It is manufactured in industry by adding water to calcium oxide (quicklime): CaO(s) + H2O(l) Ca(OH)2(s) Given that there is 350 kg of calcium oxide. How much water is needed to convert it to calcium hydroxide? (5) In the industry, iron is extracted by heating iron ore (haematite, iron(III) oxide) with carbon in a blast furnace. One reaction where iron is formed is as follows: 2Fe2O3 + 3C 4Fe + 3CO2 Given that there is 12400 kg of iron(III) oxide contained in the iron ore, how much iron can be produced? (6) Kroll process is an industrial process to produce pure titanium metal. This process basically involves reaction of magnesium with titanium(IV) chloride in a stainless steel retort at 820°C: TiCl4(g) + 2Mg(l) Ti(s) + 2MgCl2(l) If a metal extraction plant can produce 1000 tonnes of titanium in a day, how much magnesium metal is needed in a day, given that there is plenty of supply of TiCl4? 38 M for Mole, Copyright 2009 Etacude.com Sample copy 39 M for Mole, Copyright 2009 Etacude.com Examples below show how a solution of known concentration can be determined, and the amount of chemical substance needed to give such concentration. Example 1 What is the mass of sodium chloride (NaCl) needed to prepare a solution of 0.1 M concentration? A concentration of 0.1 M of sodium chloride solution means 0.1 mol of sodium chloride dissolves in 1 dm3 of water. Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol To find out mass of 0.1 mol of sodium chloride rearrange mole equation Mass of substance = mole x molecular weight (molar mass) = 0.1 mol x 58.44 g/mol = 5.844 g Hence, 5.844 g of sodium chloride is needed to dissolve in 1 dm3 of water to give a concentration of 0.1 M. Example 2 Given that a 250 cm3 volumetric flask is used, what is the amount of sodium chloride (NaCl) needed to prepare a 0.30 M solution? Molecular weight of NaCl = 58.44 g/mol A concentration of 0.30 M requires 0.30 mol of NaCl dissolves in 1 dm3. Therefore, the amount of sodium chloride needed 0.30 mol x 58.44 g/mol = 17.532 g So this is the amount of NaCl needed to dissolve in 1 dm3 of water. However, to make a 250 cm3 of 0.30 M solution, the amount of NaCl needed is obviously less than that. This can be determined by using the same proportionality relationship as first described in Example 1 of Chapter 4.2. Given that 1 dm3 = 1000 cm3: 17.532 g yg 40 1000 cm3 (to make 0.25 M solution) 250 cm3 (to make 0.25 M solution) M for Mole, Copyright 2009 Etacude.com y = 17.532 x 250 = 4.383 g 1000 y x 1000 = 17.532 x 250 Therefore You need 4.383 g of NaCl to dissolve in 250 cm3 of water to make 0.3 M solution. Alternatively, you can determine the amount needed in terms of mole and then converts the answer to grams: 0.30 mol x mol 1000 cm3 (to make 0.25 M solution) 250 cm3 (to make 0.25 M solution) y = 0.30 x 250 = 0.075 mol Therefore 1000 And the amount of sodium chloride needed 0.075 mol x 58.44 g/mol = 4.383 g Example 3 How much sodium hydroxide is needed in order to prepare a 200 cm3 6 M of solution? First of all, work out the molar mass of sodium hydroxide: 22.99 + 16.00 + 1.00 = 39.99 g/mol Use the relationship: 6 mol x mol 1000 cm3 200 cm3 x = 6.0 x 200 1000 (to make 6 M solution) (to make 6 M solution) = 1.2 mol Therefore, the amount of sodium hydroxide needed 1.2 mol x 39.99 g/mol = 47.99 g 41 M for Mole, Copyright 2009 Etacude.com Sample copy 42 M for Mole, Copyright 2009 Etacude.com Note that this is the number of moles in 300 cm3 of water. To find out the concentration (as at 1000 cm3), use the relationship: 300 cm3 0.054 mol x mol 1000 cm3 x = 1000 x 0.054 = 0.18 mol 300 Therefore, 13.5 g of copper sulfate dissolved in 300 cm3 water will give a solution concentration of 0.18 M. Hence, to prepare 1000 cm3 of 0.18 M solution will need 0.18 mol of copper sulfate. This is equal to 0.18 mol x 249.62 g/mol = 44.93 g of copper sulfate This example demonstrates how concentration of solution remains the same in different volumes but the actual amount of substance dissolved can be different. Example 6 Given a 250 cm3 hydrochloric acid of 2 M concentration, how much water must be added to the acid in order to reduce the concentration to 0.4 M? Remember that the solution concentration is fixed for any given volume. Hence, you can use the following relationship: 250 cm3 y cm3 2M 0.4 M Where y is the corresponding volume of water if the concentration is 0.4 M. However, since the more water is added, the less the concentration becomes (solution becomes more dilute). This means the relationship in this case is inversely proportional. The cross relationship introduced in Example 1 of Chapter 4.2 only applies to situation that is directly proportional. For instance, when more substance is added the higher the concentration becomes, as shown in the previous Examples. 43 M for Mole, Copyright 2009 Etacude.com Therefore, the equation for this problem is now related as the green lines as shown above: y x 0.4 = 250 x 2 Therefore y = 250 x 2 = 1250 cm3 0.4 And so the concentration is reduced from 2 M to 0.4 M as the water is increased from 250 cm3 to 1250 cm3. However, since the original solution has a volume of 250 cm3, the extra volume of water that needs to be added in order to reduce the concentration from 2 M to 0.4 M is 1250 cm3 - 250 cm3 = 1000 cm3 Checkpoint 13: (1) Calculate the concentration of 14.5 g of sodium carbonate (Na2CO3) dissolved in 1000 cm3 of water. (2) What is the concentration of the following solutions: (a) 1.5 g of potassium bromide (KBr) in 100 cm3 of water. (b) 16.5 g of potassium iodide (KI) in 250 cm3 of water. (c) 45.1 g of potassium manganate(VI) (KMnO4) in 2000 cm3 of water. (d) 20.6 g of sodium hydroxide (NaOH) in 125 cm3 of water. (e) 5.7 g of sodium chloride in 175 cm3 of water. (3) What is the amount (in grams) of silver nitrate (AgNO3) in 75 cm3 of 0.05 M of solution? (4) What is the amount (in grams) of sodium chloride (NaCl) in 25 cm3 of 0.15 M of solution? (5) Calculate the mass of sodium sulfate (Na2SO4) needed to make 250 cm3 of 1.5 M solution. 44 M for Mole, Copyright 2009 Etacude.com Sample copy 45 M for Mole, Copyright 2009 Etacude.com Example 1 A 20 cm3 standard sodium hydroxide solution of concentration 0.2 M needed 18.5 cm3 of hydrochloric acid for neutralization. Determine the concentration of the hydrochloric acid. First of all, determine the amount of NaOH in 20 cm3 solution. 1000 cm3 0.2 mol x mol Therefore 20 cm3 x = 0.2 x 20 mol 1000 = 0.004 mol Now, the neutralization reaction is as follows: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) The reaction shows that 1 mol HCl reacts with 1 mol NaOH. Therefore 0.004 mol HCl will react with 0.004 mol NaOH. There are 0.004 mol HCl in 18.5 cm3 solution. To find out the concentration of HCl sample Therefore 0.004 mol 18.5 cm3 y mol 1000 cm3 y= 0.004 x 1000 18.5 = 0.216 mol mol The concentration of HCl is 0.216 M. Alternatively, the equation from the previous page can be used as a quicker way to determine the concentration. Given that Ma and Va are the concentration and volume of NaOH; and Mb and Vb are the concentration and volume of HCl, from Equation (3), it follows that 0.2 M x 20.0 cm3 = 1 1 Mb x 18.5 cm3 And Mb x 18.5 x 1 = 0.2 x 20.0 x 1 Therefore Mb = 46 0.2 x 20.0 18.5 M = 0.216 M M for Mole, Copyright 2009 Etacude.com Sample copy 47 M for Mole, Copyright 2009 Etacude.com Sample copy 48 M for Mole, Copyright 2009 Etacude.com 4.4 Moles in Gases In previous examples, calculations involving gas compounds were described in quantities such as grams and moles. However, sometimes it is more practical to describe the quantity in terms of volumes. In questions 8 and 9 of Checkpoint 11 you will notice that different gases apparently occupy a same volume at a given temperature. This is in fact true for any other gases which makes volume calculations for gases becomes much easier. However, the volume of gas depends on both temperature and pressure. In chemistry, two most commonly defined conditions are as follows: (a) At the temperature of 0°C (273.15 K) and 1 atmospheric pressure, one mole of a gas occupies a volume of 22.414 dm3. Such conditions are called the standard temperature and pressure (s.t.p). (b) At the temperature of 25°C (298.15 K) and 1 atmospheric pressure, one mole of a gas occupies a volume of 24.466 dm3. Such conditions are called the standard room temperature and pressure (s.r.p). For condition (a) the volume occupied is called the standard molar volume. For condition (b) the volume occupied is called the molar volume. And so 1 mole of a gas contains 6.022 x 1023 mol-1 particles and occupies a volume of 22.4 dm3 (standard molar volume) at s.t.p. 1 mole of a gas contains 6.022 x 1023 mol-1 particles and occupies a volume of 24.0 dm3 (molar volume) at r.t.p. Note that the volumes are approximated to 22.4 dm3 /mol and 24.0 dm3 /mol respectively for our calculations. 49 M for Mole, Copyright 2009 Etacude.com Example 1 Given a 2.5 mol carbon dioxide, what is the corresponding volume occupied by the gas at s.t.p.? Since more gas substance will occupy more volume, the proportionality relationship that first introduced in Example 1 of Chapter 4.2 can be used: 1 mol CO2 22.4 dm3 (at s.t.p.) 2.5 mol CO2 x dm3 (at s.t.p.) x = 22.4 x 2.5 dm3 Therefore = 56 dm3 Example 2 Given that 10.5 g of nitrogen gas is produced in a chemical reaction at room temperature. What is the volume occupied by the gas? The molar mass of N2 = 14.0 x 2 = 28 g/mol. Using Mole Equation: Moles of N2 = 10.5 g = 0.375 mol 28.0 g/mol And 1 mol N2 0.375 mol N2 Therefore 24.0 dm3 (at r.t.p.) x dm3 (at r.t.p.) x = 24.0 x 0.375 dm3 = 9 dm3 Example 3 What is the volume of carbon dioxide produced when 50.0 dm3 (measured at room temperature) of propane is burnt in the presence of excess air? If x is the moles of propane gas that occupies 50 dm3 at room temperature, then 50 1 mol propane 24.0 dm3 (at r.t.p.) x mol propane 50.0 dm3 (at r.t.p.) M for Mole, Copyright 2009 Etacude.com Sample copy 51 M for Mole, Copyright 2009 Etacude.com Example 4 Carbon disulfide is a highly flammable liquid that burns in oxygen to produce sulfur dioxide and carbon dioxide. What is the volume of oxygen needed to burn in excess carbon disulfide in order to produce 125 dm3 (measured at 25 °C) of sulfur dioxide? The reaction is shown as follows: CS2(l) + 3O2(g) CO2(g) + 2SO2(g) The above equation shows ‘3 to 2’ relationship between the oxygen and sulfur dioxide. Therefore, to determine the volume of oxygen involves in the reaction: And 3 unit volumes O2 2 unit volumes SO2 z unit volume O2 125 dm3 SO2 2x z = 3 x 125 z = 3 x 125 2 = 187.5 dm3 Example 5 Potassium chlorate will decompose and release oxygen when it is heated to above 400 °C. What is the volume of oxygen gas released (measured at r.t.p.) when 5.4 g of potassium chlorate is completely decomposed? The reaction is shown as follows: 2KClO3(s) 3O2(g) + 2KCl(s) First of all, work out the molar mass of potassium chlorate: 39.098 + 35.45 + 3 x 16.0 = 122.55 g/mol Then work out 5.4 g of potassium chlorate in moles: 5.4 g = 0.044 mol 122.55 g/mol 52 M for Mole, Copyright 2009 Etacude.com Sample copy 53 M for Mole, Copyright 2009 Etacude.com 5. Answers If you want detailed work out for the answers, then please email [email protected] using the same email you have purchased this e-book, including the name that you have registered in (see Page 2). You will be sent the work out free of charge provided that your email and the registered name is matched. Checkpoint 1 (a) ...less...one (b) ...more...two Checkpoint 2 Just memorize it! Checkpoint 3 Number of neutron Isotopes proton electron 222 Bi 83 83 129 83 208 Pb 82 82 126 82 127 I = 127 I 53 53 74 53 234 U = 234 U 92 92 142 92 2 2 2 4 He = 4 2 He Checkpoint 4 Memorize it! Checkpoint 5 Sodium oxide, Na2O Magnesium oxide, MgO Potassium bromide, KBr 54 Iron(II) sulfate, FeSO4 Iron(III) sulfate, Fe2(SO4)3 Ammonium sulfate, (NH4)2SO4 M for Mole, Copyright 2009 Etacude.com Checkpoint 6 Atomic weight (note: if you use different values, your answer will be slightly different than the model answer) K = 39.10 Cr = 52.00 N = 14.01 H = 1.00 Br = 79.90 Na = 22.99 Cl = 35.45 C =12.00 Cu = 63.55 S = 32.07 O = 16.00 KBr = 119.00 CuSO4.5H2O = 249.62 C6H12O6 = 180.00 K2Cr2O7 = 294.20 CH3COOH = 60.00 (NH4)2SO4 = 132.09 NaClO3 = 106.44 Checkpoint 7 Potassium bromide, KBr K = 32.86 % Br = 67.14 % Glucose, C6H12O6 C = 40 % H = 6.67 % O = 53.55 % Potassium dichromate, K2Cr2O7 K = 26.58 % Cr = 35.35 % O = 38.07 % Ethanoic acid, CH3COOH C = 40 % H = 6.67 % O = 53.55 % Ammonium sulfate, (NH4)2SO4 N = 21.21 % H = 6.06 % S = 24.28 % O = 48.45 % Percentage of ammonium = 48.48 % 55 M for Mole, Copyright 2009 Etacude.com Checkpoint 8 Empirical formulae for: Ethyne, C2H2 = CH Benzene, C6H6 =CH Aluminium chloride, Al2Cl6 = AlCl3 Octane, C8H18 = C4H9 Phosphorus pentoxide, P4O10 = P2O5 Ethanedioic (oxalic) acid, (COOH)2 = CO2H Methane, CH4 (empirical formula is the same as its molecular formula) Glucose, C6H12O6 = CH2O Sucrose, C12H22O11 (same as its molecular formula) Checkpoint 9 (1) (2) (3) (4) CH2 Empirical formula: CH3 Empirical formula: C2H6O XeF2O2 Molecular formula: C2H6 Molecular formula: C4H12O2 Checkpoint 10 ZnO + 2HCl ZnCl2 + H2O H2 + Cl2 2HCl Cu + 2AgNO3 Cu(NO3)2 + 2Ag 2CH4 + 4O2 2CO2 + 4H2O 2Al + 2NaOH + 3H2O 2NaAl(OH)4 + 3H2 4HCl + MnO2 MnCl2 + 2H2O + Cl2 (Words equation) Cu + O2 CuO2 Ca(OH)2 + CO2 CaCO3 + H2O 2HCl + Zn ZnCl2 + H2 2HCl + CaCO3 CaCl2 +CO2 + H2O 56 M for Mole, Copyright 2009 Etacude.com Sample copy 57 M for Mole, Copyright 2009 Etacude.com List of relative atomic mass for some elements (in alphabetical order) Elements Aluminium Bromine Calcium Carbon Chlorine Chromium Copper Fluorine Hydrogen Iodine Iron Magnesium Manganese Nitrogen Oxygen Phosphorus Potassium Sodium Sulfur Xenon Zinc Symbols Al Br Ca C Cl Cr Cu F H I Fe Mg Mn N O P K Na S Xe Zn Relative atomic mass 26.98 79.90 40.08 12.00 35.45 52.00 63.55 19.00 1.00 126.9 55.85 24.31 54.94 14.00 16.00 30.97 39.10 22.99 32.07 131.3 65.39 For a complete list please consult a good Periodic Table of the Elements. List of other useful on-line resources The Periodic Table of the Elements Chemistry glossary in Etacude.com’s Science Dictionary On-line chemistry lesson Fun Science Projects 58 M for Mole, Copyright 2009 Etacude.com