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Chemistry Module 1- Lesson 4 Chemistry lesson 4 Study guide - Questions In this lesson we enter the world of equations and moles. This allows us to calculate number of atoms and mass of reactants and products. The concept of the mole allows us to actually calculate quantities of reactant and products from our balanced chemical equations. This branch of chemistry is called Stochiometry. Your weekly program. Develop your own method by all means but this will get you going. Read the notes attached here. All questions and the final assessment are based on what is in the notes. Read the chapter sections in the reading section ( the text has much greater detail than what you are expected to know) especially look at diagrams and figures At this stage not much will make sense but that is OK With your text as reference and this study sheet go through the Power point presentation. Make notes where needed. MANDATORY watch all of the embedded video links (you will need internet access to do this) Optional - Listen to the audio file of a live lesson; be aware that there will be long pauses with not much going on at certain points. Refine your notes / mind maps on the key concepts outlined in these weekly study sheets. Check your understanding Now go through your homework questions and answer those Your study resources Textbook Audio files of the actual lessons Weekly study sheet Power Point slides YouTube links and YouTube as a general resource College forum site Molecule kit Chemistry Module 1- Lesson 4 Textbook reference Bettleheim Edition 9 Please read these sections and note take Chapter 4 sections 4.1 to 4.4 Key Concepts to understand Balancing chemical equations Formula and molecular weight The mole concept Basic Stochiometry o Moles to grams and vice versa o Molar ratios in equations o Limiting reagents Use the links in the power point presentation. Good sources for chemistry videos are Bozeman science http://www.bozemanscience.com/ Crash course chemistry – YouTube Tyler de Witt socratic.org/ Chemistry Module 1- Lesson 4 Study Notes In order to tackle Stochiometry problems we need to master some foundational concepts first. These are balancing equations, the molar concept and finding molecular weights. Chemical equations Chemical equations do not come already balanced. This must be done before the equation can be used in any chemically meaningful way. A balanced equation has equal numbers of each type of atom on each side of the equation. The Law of Conservation of Mass (lesson 2) is the rationale for balancing a chemical equation. The essence of this law is - "Matter is neither created nor destroyed." Therefore, we must finish our chemical reaction with as many atoms of each element as when we started. Let’s step through a balancing exercise. Problem #1: Balance the following equation: H2 + O2 ---> H2O It is an unbalanced equation. This means that there are UNEQUAL numbers of at least one atom on each side of the arrow. In the example equation, there are two atoms of hydrogen on each side, BUT there are two atoms of oxygen on the left side and only one on the right side. Remember this: A balanced equation MUST have EQUAL numbers of EACH type of atom on BOTH sides of the arrow. An equation is balanced by changing coefficients in a somewhat trial-and-error fashion. It is important to note that only the coefficients can be changed, NEVER a subscript. The coefficient times the subscript gives the total number of atoms. Sub examples of finding the number of atoms. (a) 2 H2 - there are 2 x 2 atoms of hydrogen (a total of 4). Chemistry Module 1- Lesson 4 (b) 2 H2O - there are 2 x 2 atoms of hydrogen (a total of 4) and 2 x 1 atoms of oxygen (a total of 2). (c) 2 (NH4)2S - there are 2 x 1 x 2 atoms of nitrogen (a total of 4), there are 2 x 4 x 2 atoms of hydrogen (a total of 16), and 2 x 1 atoms of sulphur (a total of 2). Two things you CANNOT do when balancing an equation. 1) You cannot change a subscript. For example -You cannot change the oxygen's subscript in water from one to two, as in: H2 + O2 ---> H2O2 True, this is a balanced equation, but you have changed the substances in it. H2O2 is a completely different substance from H2O. So, it's not the answer to the question that was asked. 2) You cannot place a coefficient in the middle of a formula. The coefficient goes at the beginning of a formula, not in the middle, as in: H2 + O2 ---> H22O Water only comes as H2O and you can only use whole formula units of it. Two more points: 1) Make sure that your final set of coefficients are all whole numbers with the lowest common factor. For example, this equation is balanced: 4 H2 + 2 O2 ---> 4 H2O However, all the coefficients have the common factor of two. Divide through to eliminate common factors like this. This means getting the coefficient to the lowest common number. 2) NO fractions allowed in the final answer, only whole numbers. Chemistry Module 1- Lesson 4 The final balanced equation is as follows Some more balanced equation examples Example 1 shows the numerical logic in the process Chemistry Module 1- Lesson 4 Example 2 the combustion of methane Example 3 respiration of glucose for energy Example 4 Balance the following equation: Na + H2O ---> NaOH + H2 You will notice that Na and the O are already balanced. So, we look only at the H. As H must come in twos on the left-hand side, we must have an even number of hydrogen on the right-hand side. We do this: Na + H2O ---> 2NaOH + H2 We then balance the H on the left like this: Na + 2H2O ---> 2NaOH + H2 Notice that the first placing of a 2 messed up the balance of the Na and the O. In addition, notice that the second placing of a 2 corrected the imbalance in the oxygen. The last step is to balance the Na: 2Na + 2H2O ---> 2NaOH + H2 Chemistry Module 1- Lesson 4 The mole In chemistry the mole is a fundamental unit in the SI system, and it is used to measure the amount of substance. In Latin mole means a "massive heap" of material. It is convenient to think of a chemical mole as such. A mole is the same number concept as a dozen eggs A dozen always means 12 of anything and a mole always means many atoms or molecules. This means six multiplied by 10 twenty three times consecutively. Notice the exponential notation. It is commonly known as Avogadro’s number. Amedeo Avogadro (1776–1856) was an Italian chemist who in working with gases postulated this mole concept. Avogadro's number is equal to 602,214,199,000,000,000,000,000 or more simply, 6.02214199 × 10 23. This is a large number! This scale of number allows us to manipulate minute atoms or molecules by number or mass accurately. Formula or Molecular weight This is the weight in grams of an ionic (formula) or covalent compound (molecular). To be able to calculate this we need to understand molar mass. First, it is a mass, so it has units of mass, commonly the gram. Second, it concerns the mole (Avogadro's number). Whether you're dealing with elements or compounds, the molar mass of a species is the mass in grams of one mole (6.022 x 1023) of that species: one mole of atoms, one mole of molecules, or one mole of formula units. To find the molar mass, add the atomic masses of all of the atoms in the molecule. To find the atomic mass for each element, use the mass given in the Periodic Table. Multiply the subscript (number of atoms) times the atomic mass of that element and add the masses of all of the elements. Chemistry Module 1- Lesson 4 Example 1: Sodium chloride (NaCl) has a formula weight of 58.443 (atomic weight of Na + atomic weight of Cl) and a molar mass of 58.443 g/mol. Formaldehyde (CH 2 O) has a molecular weight of 30.03 (atomic weight of C + 2 [atomic weight of H]) + atomic weight of O] and a molar mass of 30.03 g/mol. Example 2: Calcium phosphate has the formula Ca3(PO4)2. Calculate the mass of one mole of calcium phosphate. 3 mole of Ca 2 mole of P 8 mole of O. 1 mole of calcium is 40.08g, so 3 mole is 120.24 g 1 mole of phosphorus is 30.9738g, so 2 mole is 61.9476g 1 mole of oxygen is 15.9994g, so 8 mole is 127.9952g 1 mole of Ca3(PO4)2 is 310.18 g So why does this work? (Optional section) Carbon 12 is set as the standard reference for molar mass- one mole of this element was defined as having Avogadro’s number of atoms. By international agreement, the atomic mass (atomic weight) is the mass of an atom expressed in atomic mass unit (amu). One atomic mass unit (amu) is defined as a mass exactly equal to one-twelfth (1/12th) of one carbon-12 atom. This means that the mass of one carbon-12 is assigned as exactly 12 amu and 12 grams. The mass of other atoms are determined based on their relative weight. For example, the hydrogen atom weighs only 8.40 percent of carbon-12 atom. Then the mass of hydrogen atom is calculated as follows. Atomic mass of hydrogen = mass of one atom carbon-12 x 8.40% = 1.008 amu Chemistry Module 1- Lesson 4 Basic Stochiometry Now let’s use the knowledge we have covered to work out mass and moles from an equation. Once we have a balanced equation then the elements and compounds in the equation have a numerical relationship. This is called a mole ratio. Please note it is a ratio of moles not weight. Definition: A mole ratio is ratio between the amounts in moles of any two compounds involved in a chemical reaction. Mole ratios are used as conversion factors between products and reactants in many chemistry problems. Example 1: 2 H2(g) + O2(g) → 2 H2O(g) The mole ratio between O2 and H2O is 1:2. For every 1 mole of O2 used, 2 moles of H2O are formed. The mole ratio between H2 and H2O is 1:1. For every two moles of H2 used, 2 moles of H2O is formed. Using our skill we can also work out which reactants will be used up first in a particular process – this is called finding the limiting reagent. The key skills in Stochiometry problems are o Balance the equation o Be able to determine the mole ratios in the equation o Be able to convert moles to grams and vice versa Let’s get to work on real problem The reaction of powdered aluminium and iron(II)oxide Here is the balanced equation for the reaction 2Al(s) + Fe2O3(s) Al2O3(s) + 2Fe(l) Chemistry Module 1- Lesson 4 This reaction produces so much heat the iron that forms is molten. Because of this, railroads use the reaction to provide molten steel to weld steel rails together when laying track. Note the brackets tell us the state of the chemical species (s) means solid. In the products section the (l) means the Iron is liquid hence the heat. In the title of the question (ii) against the Iron means it has an overall charge of 2- for the ion. Here is the question Suppose that in one batch of reactants 4.20mol Al was mixed with 1.75mol Fe2O3. Which reactant, if any, was the limiting reactant? Calculate the mass of iron (in grams) that can be formed from this mixture of reactants. How do we approach this question – firstly determine what has been asked of you. Note that there are two parts to the question! Part 1 - Which reactant is the limiting reactant – this means which reactant runs out first in the process. Part 2 – Calculate the mass in grams of iron that can be formed with this mixture of reactants Notice that we are given information in moles and asked to give the answer in grams. This means moles need to be converted to grams. Also we will need to find out the molar ration between Fe2O3 and Al. With all chemistry problems and most scientific problems you arrive at the answer by using logical steps rather than guessing. It’s important to set out your work neatly, step wise so that you can check your calculations at the end and also record your reasoning process. In an assignment the reasoning and calculations hold as much weight as the actual correct answer. In your text book dimensional analysis is used for these problems but I will outline an easier method for you to use. First step – balance the equation, this has already been done. Done Chemistry Module 1- Lesson 4 Second step - work out the molar ration between the two compounds mentioned. In this question it is Fe2O3 and Al The molar ration is 2:1 Al to Fe2O3 Third step - use the molar ratio to work out which reactant is not fully used up. The other species is then the limiting reagent. This will give us the answer to the first part of the question. Taking the 1.75 moles of Fe2O3 as one part and multiplying by two gives 3.5 moles of Al needed to react with it. We are given 4.2 moles of Al so there is some left over. The limiting reactant is the Fe2O3 as it is completely used up. Fourth step – Find out the molar ratio between the limiting reagent and the product 2Fe(l) Looking at the balanced equation we see the ratio is 1:2 Fe2O3 to 2Fe Fifth step - Find the amount of moles of 2Fe formed by the reaction of Fe2O3. 1.75 moles of Fe2O3 multiplied by 2 gives 3.5 moles of 2Fe using the molar ratio. Sixth step - Now convert the moles into grams for the final part of the answer. 3.5 moles of Fe with formula weight of (55.8) 111.6 grams = 3.5 x 55.8 = 195.3 grams. The given mixture of reactants produces 195.3 grams of Iron Fe. Note: the answer is presented as a concluding statement with the number of the answer and the scientific units in this case Chemistry Module 1- Lesson 4 Home work Question 1: Write out the molar ratios for these questions Example A: N2 + 3H2 ---> 2NH3 Write the molar ratios for N2 to H2 and NH3 to H2. Example B: 2SO2 + O2 ---> 2SO3 Write the molar ratios for O2 to SO3 and O2 to SO2. Example C: PCl3 + Cl2 ---> PCl5 Write the molar ratios for PCl3 to Cl2 and PCl3 to PCl5. Question 2: Determine the mass in grams of the following: 1.35 moles Fe Question 3: Calculate the number of moles of this compound: 21.5 g CaCO3 Question 4: Using the following equation answer three questions about the reaction. N2 + H2 ---> NH3 Question A – Balance the equation Question B – What is the molar ratio between N2 and NH3 Question C – If you have 2.00 mol of N2 reacting with sufficient H2, how many moles of NH3 will be produced? Chemistry Module 1- Lesson 4