Topic 4 Stoichiometry PDF

CHM012 Chemistry for Engineers Topic 4: Stoichiometry Bryan M. Montalban, M.Sc., R.Ch. Sept 21, 2020 Department of Chemistry College of Science and Mathematics MSU-Iligan Institute of Technology bry...

CHM012 Chemistry for Engineers Topic 4: Stoichiometry Bryan M. Montalban, M.Sc., R.Ch. Sept 21, 2020 Department of Chemistry College of Science and Mathematics MSU-Iligan Institute of Technology [email protected] Tel. | Page  Questions to Consider ⮚ Can atoms be counted? ⮚ Does the number of atoms in a reaction affect the result? ⮚ Can the mass of a sample be used to determine the number of moles it comprises? Topic 4: Stoichiometry Montalban, B. | Page 2  Average Mass ✔ It is used to determine the number of atoms in a set quantity of a substance. ✔ It is determined using a sample of the substance ✔ Atoms do not need to be identical in order to be counted by weighing Topic 4: Stoichiometry Montalban, B. | Page 3  Atomic Mass ✔ The modern system of determining atomic masses was first used in 1961, based on 12C - 12 atomic mass units (u). ✔ The most accurate method currently used to compare the masses of atoms involves the use of the mass spectrometer Mass Spectrometer ⮚ Atoms are passed into a stream of high-speed electrons that convert them into positive ions. ⮚ Ions are passed through a magnetic field. ⮚ Accelerating ions create their own magnetic field, resulting in a change in the path travelled. ⮚ The degree of deviation depends on the mass of the ion. Ions with higher masses deviate the least. ⮚ Deviated ions hit the detector plate. Comparing the deflected position of each ion gives their mass. Topic 4: Stoichiometry Montalban, B. | Page 4  Determining Atomic Masses ❖ Consider the analysis of 12C and 13C in a mass spectrometer. ❖ Based on the definition of the atomic mass unit. Exact number by definition ❖ The average mass for an element is also referred to as the average atomic mass or atomic mass. Topic 4: Stoichiometry Montalban, B. | Page 5  Example When a sample of natural copper is vaporized and injected into a mass spectrometer, the results shown in Fig. 5-3 are obtained. Use these data to compute the average mass of natural copper. (The mass values for 63Cu and 65Cu are 62.93 u and 64.93 u, respectively.) Solution Topic 4: Stoichiometry Montalban, B. | Page 6  Mole ✔ The mole is defined as the number equal to the number of carbon atoms in exactly 12 grams of pure 12C. 12 g of 12C = 6.022 × 1023 atoms of 12C (Also called Avogadro’s number) ✔ One mole of a substance contains 6.022 × 1023 units of that substance. ✔ A sample of a natural element with a mass equal to the element’s atomic mass expressed in grams contains 1 mole of atoms. Element Number of Atoms Present Mass of Sample (g) Aluminum 6.022 × 1023 26.98 Copper 6.022 × 1023 63.55 Iron 6.022 × 1023 55.85 Sulfur 6.022 × 1023 32.07 Iodine 6.022 × 1023 126.9 Mercury 6.022 × 1023 200.6 Topic 4: Stoichiometry Montalban, B. | Page 7  Example Americium is an element that does not occur naturally. It can be made in very small amounts in a device known as a particle accelerator. Compute the mass in grams of a sample of americium containing six atoms. Solution Given: 1 mole Americium = 6.022x1023 atoms Americium = 243 g Americium Topic 4: Stoichiometry Montalban, B. | Page 8  Molar Mass ✔ It is the mass of one mole of the compound measured in grams. Traditionally called molecular weight. ✔ Formula unit - Used for compounds that do not contain molecules. Example: NaCl - Sodium chloride CaCO3 - Calcium carbonate Topic 4: Stoichiometry Montalban, B. | Page 9  Example Juglone, a dye known for centuries, is produced from the husks of black walnuts. It is also a natural herbicide (weed killer) that kills off competitive plants around the black walnut tree but does not affect grass and other noncompetitive plants. The formula for juglone is C10H6O3. a) Calculate the molar mass of juglone. b) A sample of 1.56 × 10−2 g of pure juglone was extracted from black walnut husks. How many moles of juglone does this sample represent? Solution Topic 4: Stoichiometry Montalban, B. | Page 10  Percent Composition of a Compound ⮚ In terms of the numbers of its constituent atoms ⮚ In terms of the percentages (mass) of its elements ⮚ Consider ethanol (C2H5OH) Topic 4: Stoichiometry Montalban, B. | Page 11  Mass Percent ⮚ It is calculated by comparing the mass percent of the element in one mole of the compound with the total mass of one mole of the compound and multiplying the result by 100%. Example: Calculating the mass of carbon in ethanol Topic 4: Stoichiometry Montalban, B. | Page 12  Example Carvone is a substance that occurs in two forms having different arrangements of atoms but the same molecular formula (C10H14O) and mass. One type of carvone gives caraway seeds their characteristic smell, and the other type is responsible for the smell of spearmint oil. Compute the mass percent of each element in carvone. Solution Topic 4: Stoichiometry Montalban, B. | Page 13  Determining the Formula of a Compounds ⮚ Determined by using a weighed sample and one of the following techniques ✔ Decomposing it into its component elements ✔ Introducing oxygen to produce substances such as CO2, H2O, etc., which are collected and weighed Analyzing for Carbon and Hydrogen Topic 4: Stoichiometry Montalban, B. | Page 14  Empirical and Molecular Formula ⮚ Empirical formula (EF) - the simplest whole-number ratio of the various types of atoms in a compound. Can be obtained from the mass percent of elements in a compound ⮚ The molecular formula (MF) varies for molecules and ions. ✔ For molecular substances, it is the formula of the constituent molecules. Always an integer multiple of the empirical formula. ✔ For ionic substances, it is the same as the empirical formula Topic 4: Stoichiometry Montalban, B. | Page 15  Steps in Empirical Formula Determination 1. Since mass percentage gives the number of grams of a particular element per 100 grams of compound, base the calculation on 100 grams of compound. Each percent will then represent the mass in grams of that element. 2. Determine the number of moles of each element present in 100 grams of compound using the atomic masses of the elements present 3. Divide each value of the number of moles by the smallest of the values. If each resulting number is a whole number (after appropriate rounding), these numbers represent the subscripts of the elements in the empirical formula 4. If the numbers obtained in the previous step are not whole numbers, multiply each number by an integer so that the results are all whole numbers Topic 4: Stoichiometry Montalban, B. | Page 16  Example Determine the empirical and molecular formulas for a compound that gives the following percentages on analysis (in mass percents): 71.65% Cl, 24.27% C, 4.07% H. The molar mass is known to be 98.96 g/mol Solution Empirical Formula CH2Cl (48.47 g/mol) Molecular Formula C2H4Cl2 (48.47 g/mol) Topic 4: Stoichiometry Montalban, B. | Page 17  Chemical Reactions ⮚ Involve the reorganization of atoms in one or more substances. ⮚ In a chemical equation, the reactants are situated on the left side of an arrow, and the products are located on the right Bonds are broken and new bonds are formed Both sides possess the same number of atoms Topic 4: Stoichiometry Montalban, B. | Page 18  Balancing Chemical Equation ✔ Atoms are neither created nor destroyed in a chemical reaction. The number of atoms on each side of the equation must be the same. Consider the reaction between methane and oxygen Topic 4: Stoichiometry Montalban, B. | Page 19  Meaning of a Chemical Equation ❖ It contains information on: State Symbol ✔ The nature of the reactants and products Solid (s) ✔ The relative numbers of each Liquid (l) ❖ Equations also provide the physical state of the Gas (g) reactants and products Dissolved in water (in aqueous solution) (aq) Reactants Products CH4 (g) + 2 O2(g) CO2 (g) + 2 H2O(g) 1 molecule + 2 molecules 1 molecule + 2 molecules 1 mol + 2 mol 1 mol + 2 mol 6.022 × 1023 molecules + 2 (6.022 × 1023 molecules) 6.022 × 1023 molecules + 2 (6.022 × 1023 molecules) 16 g + 2 (32 g) 44 g + 2 (18 g) 80 g reactants 80 g products Topic 4: Stoichiometry Montalban, B. | Page 20  Steps in Balancing a Chemical Equation Step 1. Start with the most complicated molecules. Consider the following unbalanced equation Step 2. The most complicated molecule is C2H5OH. Balancing carbon Balancing hydrogen Balancing oxygen Topic 4: Stoichiometry Montalban, B. | Page 21  Steps in Balancing a Chemical Equation Step 3. Verifying the results Topic 4: Stoichiometry Montalban, B. | Page 22  Example Chromium compounds exhibit a variety of bright colors. When solid ammonium dichromate, (NH4)2Cr2O7, a vivid orange compound, is ignited, a spectacular reaction occurs. Although the reaction is actually somewhat more complex, let’s assume here that the products are solid chromium (III) oxide, nitrogen gas (consisting of N2 molecules), and water vapor. Balance the equation for this reaction. Solution The unbalanced equation is: Note that nitrogen and chromium are balanced (two nitrogen atoms and two chromium atoms on each side), but hydrogen and oxygen are not. A coefficient of 4 for H2O balances the hydrogen atoms. Topic 4: Stoichiometry Montalban, B. | Page 23  Stochiometric Calculations Consider the following reaction between propane and oxygen. ⮚ Calculating the number of moles of propane present in 96.1 grams (the molar mass of propane is 44.1). ⮚ Constructing a mole ratio ⮚ Calculating the moles of O2 needed ⮚ Calculating the number of grams oxygen needed to burn 96.1 grams of propane. (the molar mass of oxygen gas is 32.0). Topic 4: Stoichiometry Montalban, B. | Page 24  Stochiometric Calculations In order to determine the mass of carbon dioxide produced, conversion between moles of propane and moles of carbon dioxide is required ⮚ Forming the mole ratio ⮚ The conversion is ⮚ Calculating the mass of CO2 produced using the molar mass of CO2 (44.0 g/mol) Topic 4: Stoichiometry Montalban, B. | Page 25  Problem-Solving Strategy - Calculating Masses of Reactants and Products Topic 4: Stoichiometry Montalban, B. | Page 26  Example Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide can be absorbed by 1.00 kg of lithium hydroxide? Solution The balanced chemical equation: Topic 4: Stoichiometry Montalban, B. | Page 27  Limiting Reactant ✔ Limiting reactant – is the reactant that is used the most, limiting the quantity of the product formed. In this example, the number of sandwiches prepared has been limited by the number of cheese slices, and the meats as the limiting ingredient Topic 4: Stoichiometry Montalban, B. | Page 28  Stoichiometric Mixture Consider the reaction between nitrogen and hydrogen, forming ammonia ⮚ Two molecules of ammonia possesses: 1 N2 molecule 3 H2 molecules ⮚ There are just enough molecules to ensure that all are paired ⮚ A stoichiometric mixture is one that possesses equivalent amounts of reactants that match the numbers in the balanced equation ⮚ The presence and quantity of the limiting reactant determines the amount of the product formed Topic 4: Stoichiometry Montalban, B. | Page 29  Determination of Limiting Reactant Consider a reaction in which 25.0 kg of nitrogen is mixed with 5.0 kg of hydrogen to form ammonia. Identify the limiting reactant. Step 1. Determination of the moles of the reactants. Step 2. Calculating the total amount of moles of H2 that react with 8.93×102 moles of N2 Also, 2.48 × 103 moles of H2 requires 8.27 × 102 moles of N2. Therefore, Nitrogen is in excess (8.93 × 102 moles). Hydrogen is the limiting reactant. Topic 4: Stoichiometry Montalban, B. | Page 30  Example Nitrogen gas can be prepared by passing gaseous ammonia over solid copper (II) oxide at high temperatures. The other products of the reaction are solid copper and water vapor. If a sample containing 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N2 will be formed? Solution The balanced chemical equation: Therefore, CuO is the limiting reactant Topic 4: Stoichiometry Montalban, B. | Page 31  Theoretical Yield and Percent Yield ✔ Theoretical yield is the highest quantity of product that can be obtained from a given amount of limiting reactant ✔ Percent yield is the actual amount obtained. Less than theoretical yield Problem-Solving Strategy 1. Write and balance the equation for the reaction 2. Convert the known masses of substances to moles 3. Determine which reactant is limiting 4. Using the amount of the limiting reactant and the appropriate mole ratios, compute the number of moles of the desired product 5. Convert from moles to grams, using the molar mass Topic 4: Stoichiometry Montalban, B. | Page 32  Problem-Solving Strategy - Problems Involving Masses of Reactants and Products Topic 4: Stoichiometry Montalban, B. | Page 33  Example Methanol (CH3OH), also called methyl alcohol, is the simplest alcohol. It is used as a fuel in race cars and is a potential replacement for gasoline. Methanol can be manufactured by combining gaseous carbon monoxide and hydrogen. Suppose 68.5 kg CO(g) is reacted with 8.60 kg H2(g). Calculate the theoretical yield of methanol. If 3.57 × 104 g CH3OH is actually produced, what is the percent yield of methanol? Solution The balanced chemical equation: Therefore, H2 is the limiting reactant Topic 4: Stoichiometry Montalban, B. | Page 34