Chemistry For Engineers 1st Semester 2021-2022 PDF

University of Science and Technology of Southern Philippines Cagayan de Oro City 1st Semester SY 2021 - 2022 TOPIC 1 BASICS OF CHEMISTRY 2 Outline I. Basics of Chemistry The Study of C...

University of Science and Technology of Southern Philippines Cagayan de Oro City 1st Semester SY 2021 - 2022 TOPIC 1 BASICS OF CHEMISTRY 2 Outline I. Basics of Chemistry The Study of Chemistry The Science of Chemistry: Observations and Models Numbers and Measurements in Chemistry Problem Solving in Chemistry and Engineering Periodic Table of the Elements Chemical Reactions: Equations Mass and Moles of Substance Determining Chemical Formulas Stoichiometry 3 Learning Outcomes Explain the usefulness of the macroscopic, microscopic, and symbolic perspectives in understanding chemical systems. Draw pictures to illustrate simple chemical phenomena (like the differences among solids, liquids, and gases) on the molecular scale. Explain the difference between inductive and deductive reasoning. Use appropriate ratios to convert measurements from one unit to another. Express the results of calculations using the correct number of significant figures. Recognize the portions of the periodic table that contain the metals, nonmetals, and metalloids (semimetals). Identify the reactants and products in a chemical equation. 4 Learning Outcomes Write chemical equations using appropriate phase labels, symbols of reaction conditions, and the presence of a catalyst. Determine if a chemical reaction is balanced. Master the techniques for balancing chemical equations. Establish a critical relationship between the mass of a chemical substance and the quantity of that substance (in moles). Explore how the percentage composition and mass percentage of the elements in a chemical substance can be used to determine the chemical formula. Develop a molar interpretation of chemical equations, which then allows for calculation of the quantities of reactants and products. 5 The Study of Chemistry Chemistry is the study of matter and the changes that matter undergoes. The study of chemistry involves three levels of understanding, or three perspectives. Macroscopic Microscopic Symbolic 6 The Macroscopic Perspective Matter is anything that has mass and can be observed. Matter is observed through two types of changes. Physical changes Chemical changes 7 The Macroscopic Perspective Physical properties are variables of matter that we can measure without changing the identity of the substance being observed. Aluminum metal is a highly malleable metal; it can withstand large amounts of stress before it breaks or crumbles. The density of an object is a ratio of its mass to its volume. Other physical properties include: mass, color, viscosity, hardness, and temperature. 8 The Macroscopic Perspective Chemical properties are determined only by observing how a substance changes its identity in chemical reactions. Pure aluminum metal reacts with acid, such as in soft drinks, to form an aluminum salt and hydrogen gas 2Al(s) + 6HCl(aq) -----> 2AlCl3(aq) + 3H2(g) The ability of a compound to burn in oxygen, or combustion, is another chemical property. Fuel + O2 → CO2 + H2O The degradation of metals in the presence of air and moisture, or corrosion, is another common chemical property. 4Fe+ 3O2 → 2Fe2O3 9 The Macroscopic Perspective Three phases of matter: Solids are hard and do not change their shapes easily at ordinary temperatures. Liquids assume the shape of the portion of the container they fill. Gases expand to occupy the entire volume of their containers. 10 The Microscopic, or Particulate Perspective Matter is composed of unimaginably small particles called atoms that retain the chemical identity of the element they represent. An element is composed of atoms with identical physical and chemical properties. Molecules are groups of atoms held together by attractive forces whose properties are distinguishable from those of the individual elements. 11 Periodic Table of the Elements 12 The Microscopic Perspective Solid: particles maintain a regular ordered structure; maintains size and shape. Liquid: particles remain close but no longer ordered; takes shape of container. Gas: particles are widely separated and move independently of one another; fills available volume of container. 13 The Microscopic Perspective During a physical change, chemical composition does not change. Heating liquid water to make gaseous water (steam). 14 The Microscopic Perspective During a chemical change, a chemical reaction occurs that changes the chemical composition of the matter involved. Using electricity to convert water into oxygen and hydrogen molecules. 15 Example Problem 1: PROBLEM: A candle suspended above boiling water could be used to test a hypothesis about the chemical composition of the bubbles that rise from boiling water. What would be observed if the bubbles were composed of: a) Water b) Hydrogen c) Oxygen 16 Example Problem 1: SOLUTION: a) Water If the bubbles coming out of the liquid contain water, we would expect the flame to diminish in size or be extinguished. Water does not sustain the chemical reaction of combustion (as oxygen does), so if the bubbles are water, the flame should not burn as brightly. b) Hydrogen You should have been able to find (on the web, for example) that hydrogen tends to burn explosively. If the bubbles coming out of the water were hydrogen gas, we would expect to see the flame ignite the gas with some sort of an explosion. (Hopefully, a small one.) c) Oxygen If the bubbles were oxygen, the fl ame should burn more brightly. The amount of fuel would remain the same, but the bubbles would increase the amount of oxygen present and make the reaction more intense. 17 The Symbolic Representation Element abbreviations are used to represent: pure aluminum, Al aluminum oxide, Al2O3 18 The Symbolic Representation Particulate level Particulate level representation for representation for aluminum oxide, pure aluminum, Al. Al2O3, in bauxite. 19 Macroscopic, Microscopic, and Symbolic Levels 20 The Science of Chemistry: Observations and Models Chemistry is an empirical science and is studied by: Measuring physical properties and observing chemical reactions. Models are created to explain observations and organize collected data. 21 Observations in Science Observations are recorded via measurements. Accuracy - how close the observed value is to the “true” value. Precision - the spread in values obtained from measurements; the reproducibility of values. 22 Observations in Science Observations are recorded via measurements. Accuracy - how close the observed value is to the “true” value. Precision - the spread in values obtained from measurements; the reproducibility of values. 23 Example Problem 2: PROBLEM: Suppose a quality control chemist at a pharmaceutical company is tasked with checking the accuracy and precision of three different machines that are meant to dispense 10 ounces (296 mL) of cough syrup into storage bottles. She proceeds to use each machine to fill five bottles and then carefully determines the actual volume dispensed, obtaining the results tabulated in Table below. Identify whether the results are accurate and/or precise. 24 Example Problem 2: SOLUTION: Dispenser #1 is precise (values all close to one another, within a few tenths of a milliliter) but not accurate (none of the values are close to the target value of 296 mL, each being more than 10 mL too low). Dispenser #2 represent improved accuracy (each volume is less than 3 mL away from 296 mL) but worse precision (volumes vary by more than 4 mL). Dispenser #3 is working well, dispensing cough syrup both accurately (all volumes within 0.1 mL of the target volume) and precisely (volumes differing from each other by no more than 0.2 mL). 25 Observations in Science Measurements contain one of two types of errors: Random Error - may make a measurement randomly too high or too low. (e.g., variation associated with equipment limitations) Systematic Error - may make a measurement consistently too high or too low. (e.g., the presence of an impurity) 26 Interpreting Observations Inductive and deductive reasoning are used to interpret collected data and observations. Inductive reasoning - begins with a series of specific observations and attempts to generalize to a larger, more universal conclusion. Deductive reasoning - takes two or more statements or assertions and combines them so that a clear and irrefutable conclusion can be drawn. 27 Models in Science Models refer to a largely empirical description. Gas pressure is proportional to temperature. Theories are explanations grounded in some more fundamental principle or assumption about the behavior of a system. Relationship between gas pressure and temperature explained using kinetic energy. Laws are sufficiently refined, well tested, and widely accepted theories. 28 Numbers and Measurements in Chemistry Units - designate the type of quantity measured. Prefixes - provide scale to a base unit. Significant Figures - indicate the amount of information that is reliable when discussing a measurement. 29 Units The base unit designates the type of quantity being measured. SI units (from French Système International) are the base units of science. Some units comprise combinations of these base units and are termed derived units. 1 J = 1 kg m2s-2 30 Units Prefixes are used with base units to report and understand quantities of any size. 31 SI Prefixes Prefixes are based on multiples of 10. 32 Temperature Temperature is measured using the Fahrenheit, Celsius, and Kelvin (absolute) temperature scales. 33 Temperature Scale Conversions oF = (1.8 x oC) + 32 oC = ( oF -32)/1.8 K = oC + 273.15 oC = K - 273.15 34 Numbers and Significant Figures Scientific notation is used to easily write very small and very large numbers. Factor out powers of ten 54,000 = 5.4 x 104 0.000042 = 4.2 x 10-5 35 Numbers and Significant Figures All digits reported are considered significant except for certain types of zeros. When a zero establishes the decimal place, it is not significant. 51,300 m (3 significant figures) 0.043 g (2 significant figures) A zero is significant when it follows a decimal point or when it occurs between other significant figures. 4.30 mL (3 significant figures) 304.2 kg (4 significant figures) All numbers are significant when written in correct scientific notation. 36 Example Problem 3: PROBLEM: An alloy contains 2.05% of some impurity. How many significant figures are reported in this value? SOLUTION: All digits reported are significant, so there are three significant figures in the number. 37 Numbers and Significant Figures For calculated values, the number of significant figures should be consistent with the data used in the calculation. For multiplication and division, the number of significant figures in a result must be the same as the number of significant figures in the factor with the fewest significant figures. e.g. 0.24 kg x 4621 m = 1100 kg m or 1.1x 103 kg m For addition and subtraction, the number of significant figures are determined from the position of the first uncertain digit. e.g. 4.882 m + 0.3 m__ 5.2 m 38 Example Problem 4: PROBLEM: Report the result for the indicated arithmetic operations using the correct number of significant figures. Assume all values are measurements and not exact numbers. (a) 4.30 × 0.31 (b) 4.033 + 88.1 (c) 5.6/1.732 x 104 39 Example Problem 4: SOLUTION: (a) 4.30 × 0.31 = 1.3 (b) 4.033 + 88.1 = 92.1 (c) 5.6/1.732 x 104 = 3.2 X 10-4 40 Numbers and Significant Figures When counting discrete objects, the result has no ambiguity. Such measurements use exact numbers. They have infinite significant figures. e.g. two pennies would be 2.000000… Exactly defined terms, such as metric prefixes, are also considered exact numbers. 41 Problem Solving in Chemistry and Engineering There are several categories of problems: Calculations involving ratios. Conceptual understanding of particulate level. Visualization of phenomena on different levels. 42 Using Ratios Ratios represent the relationship between two quantities and can be expressed two ways: 43 Example Problem 5: PROBLEM: Suppose that your supermarket is offering 20-count shrimp for $5.99 per pound. How much should you expect to pay for one dozen shrimp? SOLUTION: 44 Ratios in Chemistry Calculations Mass Density - ratio of an object’s mass to its volume. Temperature- and compound-specific. Allows conversion between mass and volume. e.g. Units of measurement can be used to determine how to write the appropriate ratio by “canceling” out; called dimensional analysis or the factor-label method. 45 Example Problem 6: PROBLEM: What is the wavelength, in meters, of orange light of wavelength 615 nm? SOLUTION: 1 m = 1 × 109 nm = 46 Example Problem 7: PROBLEM: The density of water at 25ºC is 0.997 g per mL. A child’s swimming pool holds 346 L of water at this temperature. What mass of water is in the pool? SOLUTION: 47 Conceptual Chemistry Problems Conceptual problems focus on the particulate perspective of chemistry. Depictions of atoms and molecules are used to visualize molecular phenomena. 48 Example Problem 8: PROBLEM: Draw a picture that shows what carbon dioxide molecules might look like as a solid and as a gas. 49 Example Problem 8: SOLUTION: 50 Example Problem 9: PROBLEM: Identify the group and period to which each of the following elements belongs. Then decide whether the element is a metal, nonmetal, or metalloid. a) Se b) Cs c) Fe d) Cu e) Br 51 Example Problem 9: SOLUTION: a) Se: Group VIA, Period 4; nonmetal b) Cs: Group IA, Period 6; metal c) Fe: Group VIIIB, Period 4; metal d) Cu: Group IB, Period 4; metal e) Br: Group VIIA, Period 4; nonmetal 52 Chemical Reactions: Equations Chemical formulas provide a concise way to represent chemical compounds. e.g. Nitroglycerin becomes C3H5N3O9. A chemical equation builds upon chemical formulas to concisely represent a chemical reaction. 53 Writing Chemical Equations Chemical equations represent the transformation of one or more chemical species into new substances. Reactants are the original materials and are written on the left hand side of the equation. Products are the newly formed compounds and are written on the right hand side of the equation. REACTANTS → PRODUCTS 54 Writing Chemical Equations Chemical formulas represent reactants and products. Phase labels follow each formula. solid = (s) liquid = (Ɩ) gas = (g) aqueous (substance dissolved in water) = (aq) Some reactions require an additional symbol placed over the reaction arrow to specify reaction conditions. Thermal reactions: heat (Δ) Photochemical reactions: light (hν) 55 Writing Chemical Equations 2NaNO3 2NaNO2 + O2 2NaNO3(s) 2NaNO2(s) + O2(g) 2NaNO3(s) Δ 2NaNO2(s) + O2(g) 56 Writing Chemical Equations Different representations for the reaction between hydrogen and oxygen to produce water. 57 Balancing Chemical Equations The law of conservation of matter: matter is neither created nor destroyed. Chemical reactions must obey the law of conservation of matter. The same number of atoms for each element must occur on both sides of the chemical equation. A chemical reaction simply rearranges the atoms into new compounds. 58 Balancing Chemical Equations Balanced chemical equation for the combustion of methane. 59 Balancing Chemical Equations Chemical equations may be balanced via inspection, which really means by trial and error. Numbers used to balance chemical equations are called stoichiometric coefficients. The stoichiometric coefficient multiplies the number of atoms of each element in the formula unit of the compound that it precedes. Stoichiometry refers to the various quantitative relationships between reactants and products. 60 Balancing Chemical Equations Pay attention to the following when balancing chemical equations: Do not change species Do not use fractions (cannot have half a molecule) Make sure you have the same number of atoms of each element on both sides 61 Example Problem 9: PROBLEM: Write a balanced chemical equation describing the reaction between propane, C3H8, and oxygen, O2, to form carbon dioxide and water. 62 Example Problem 9: SOLUTION: 1) 2) 63 Example Problem 9: SOLUTION: 3) 4) 5) 64 Mass and Moles of Substance Molecular mass (MM) of a substance is the sum of the atomic masses of all the atoms in a molecule of the substance. Average mass of a molecule of that substance, expressed in atomic mass units. Formula mass (FM) of a substance is the sum of the atomic masses of all atoms in a formula unit of the compound, whether molecular or not. 65 Example Problem 10: PROBLEM: Calculate the formula mass of each of the following to three significant figures, using a table of atomic masses (AM): a. chloroform, CHCl3; b. iron(III) sulfate, Fe2(SO4)3. 66 Example Problem 10: SOLUTION: a. chloroform, CHCl3 b. b. iron(III) sulfate, Fe2(SO4)3. 67 Example Problem 11: PROBLEM: For the following two compounds, write the molecular formula and calculate the formula mass to four significant figures: 68 Example Problem 11: SOLUTION: a) This molecular model is of a molecule that is composed of two O and two H atoms. For inorganic compounds, the elements in a chemical formula are written in order such that the most metallic element is listed first. (Even though H is not metallic, it is positioned in the periodic table in such a way that it is considered to be a metal when writing formulas.) Hence, the chemical formula is H2O2 and formula mass yields 34.02 amu. b) This molecular model represents a molecule made up of one N atom, three O atoms, and one H atom. The chemical formula is then HNO3. The formula mass is 63.01 amu. 69 Mole and Molar Mass A mole (symbol mol) is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of 12C (carbon-12). e.g. One mole of ethanol contains the same number of ethanol molecules as there are carbon atoms in 12 g of carbon-12. 1 mole contains Avogadro’s number (6.022 x 1023 particles/mole) of particles. The mass of 6.022 x 1023 atoms of any element is the molar mass of that element. 70 Mole and Molar Mass A mole (symbol mol) is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of 12C (carbon-12). e.g. One mole of ethanol contains the same number of ethanol molecules as there are carbon atoms in 12 g of carbon-12. 1 mole contains Avogadro’s number (6.022 x 1023 particles/mole) of particles. The mass of 6.022 x 1023 atoms of any element is the molar mass of that element. e.g. A mole of oxygen molecules (formula O2) contains 6.02 x 1023 O2 molecules—that is, 2 x 6.02 x 1023 O atoms. 71 Mole and Molar Mass A mole (symbol mol) is defined as the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 g of 12C (carbon-12). e.g. One mole of ethanol contains the same number of ethanol molecules as there are carbon atoms in 12 g of carbon-12. 1 mole contains Avogadro’s number (6.022 x 1023 particles/mole) of particles. The mass of 6.022 x 1023 atoms of any element is the molar mass of that element. e.g. A mole of oxygen molecules (formula O2) contains 6.02 x 1023 O2 molecules—that is, 2 x 6.02 x 1023 O atoms. 72 Mole and Molar Mass One mole samples of various elements. All have the same number of particles. 73 Mole and Molar Mass The molar mass of a compound is the sum of the molar masses of all the atoms in a compound. æ 1.0 g H ö æ 16.0 g ö ç 2 mol H ´ ÷ + ç 1 mol O ´ ÷ è 1 mol H ø è 1 mol O ø = 18.0 g/mol H 2 O 74 Determining Molar Mass Balanced chemical reactions also provide mole ratios between reactants and products. 2H 2 (g) + O2 (g) ¾¾ ® 2H 2O(g) 2 moles H2 : 1 mole O2 : 2 moles H2O 75 Example Problem 12: PROBLEM: Determine the molar mass of each of the following compounds, all of which have been used as explosives: a) lead azide, PbN6 b) nitroglycerin, C3H5N3O9 c) Mercury fulminate, Hg(ONC)2 76 Example Problem 12: SOLUTION: a) lead azide, PbN6 b) nitroglycerin, C3H5N3O9 77 Example Problem 12: SOLUTION: c) Mercury fulminate, Hg(ONC)2 78 Calculations Using Moles and Molar Mass Molar mass allows conversion from mass to number of moles, much like a unit conversion. 1 mol C7H5N3O6 = 227.133 g C7H5N3O6 1 mol C 7 H 5 N 3O 6 300.0 g C 7 H 5 N 3O6 ´ 227.133 g C 7 H 5 N 3O 6 = 1.320 mol C 7 H 5 N 3O 6 79 Calculations Using Moles and Molar Mass Avogadro’s number functions much like a unit conversion between moles to number of particles. 1 mol C7H5N3O6 = 6.022 × 1023 C7H5N3O6 molecules How many molecules are in 1.320 moles of nitroglycerin? 6.022 ´ 10 23 molecules C 7 H 5 N 3O6 1.320 mol C 7 H 5 N 3O6 ´ 1 mol C 7 H 5 N 3O 6 = 7.949 ´ 10 23 molecules C 7 H 5 N 3O 6 80 Example Problem 13: PROBLEM: A sample of the explosive TNT (C7H5N3O6) has a mass of 650.5 g. How many moles of TNT are in this sample? How many molecules is this? 81 Example Problem 13: SOLUTION: 82 Example Problem 14: PROBLEM: A demolition engineer is planning to use the explosive ethylenedinitramine (C2H6N4O4, also known as halite) to bring down an abandoned building. Calculations show that 315 moles of the compound will provide the necessary explosive force. How many pounds of C2H6N4O4 should be used? 83 Example Problem 14: SOLUTION: 84 Determining Chemical Formulas Elemental Analysis: Determining Empirical and Molecular Formulas Empirical formulas can be determined from an elemental analysis. An elemental analysis measures the mass percentage of each element in a compound. The formula describes the composition in terms of the number of atoms of each element. The molar masses of the elements provide the connection between the elemental analysis and the formula. 85 Elemental Analysis: Determining Empirical and Molecular Formulas Assume a 100 gram sample size Percentage element × sample size = mass element in compound. (e.g., 16% carbon = 16 g carbon) Convert mass of each element to moles using the molar mass. Divide by smallest number of moles to get mole to mole ratio for empirical formula. When division by smallest number of moles results in small rational fractions, multiply all ratios by an appropriate integer to give whole numbers. 2.5 × 2 = 5, 1.33 × 3 = 4, etc. 86 Example Problem 15: PROBLEM: The explosive known as RDX contains 16.22% carbon, 2.72% hydrogen, 37.84% nitrogen, and 43.22% oxygen by mass. Determine the empirical formula of the compound. 87 Example Problem 15: SOLUTION: 88 Example Problem 15: SOLUTION: Empirical formula: CH2N2O2 89 Elemental Analysis: Determining Empirical and Molecular Formulas A molecular formula is a whole number multiple of the empirical formula. Molar mass for the molecular formula is a whole number multiple of the molar mass for the empirical formula. If the empirical formula of a compound is CH2 and its molar mass is 42 g/mol, what is its molecular formula? 90 Example Problem 16: PROBLEM: An alloy contains 70.8 mol % palladium and 29.2 mol % nickel. Express the composition of this alloy as weight percentage (wt %). 91 Example Problem 16: SOLUTION: 92 Molarity Molarity, or molar concentration, M, is the number of moles of solute per liter of solution. Provides relationship among molarity, moles solute, and liters solution. moles of solute Molarity (M ) = liter of solution If we know any two of these quantities, we can determine the third. 93 Example Problem 17: PROBLEM: A solution is prepared by dissolving 45.0 g of NaClO in enough water to produce exactly 750 mL of solution. What is the molarity of this solution? 94 Example Problem 17: SOLUTION: 95 Dilution Dilution is the process in which solvent is added to a solution to decrease the concentration of the solution. The number of moles of solute is the same before and after dilution. Since the number of moles of solute equals the product of molarity and volume (M × V), we can write the following equation, where the subscripts denote initial and final values. M i ´ Vi = M f ´ Vf 96 Example Problem 18: PROBLEM: A chemist requires 1.5 M hydrochloric acid, HCl, for a series of reactions. The only solution available is 6.0 M HCl. What volume of 6.0 M HCl must be diluted to obtain 5.0 L of 1.5 M HCl? 97 Example Problem 18: SOLUTION: To obtain the desired quantity of diluted HCl, the chemist should begin with 1.3 L of the concentrated solution and add enough water to bring the volume up to 5.0 L. 98 Stoichiometry Stoichiometry is a term used to describe quantitative relationships in chemistry. “How much?” of a product is produced or reactant is consumed. A balanced chemical equation is needed. Conversion between mass or volume to number of moles frequently needed. 99 Ratios from a Balanced Chemical Equation Mole ratios are obtained from the coefficients in the balanced chemical reaction. 1 mol CH4 : 2 mol O2 : 1 mol CO2 : 2 mol H2O These ratios can be used in solving problems: 1 mol CH 4 2 mol H 2 O or 2 mol O 2 1 mol CH 4 100 Example Problem 19: PROBLEM: In the combustion of methane, how many moles of O2 are required if 6.75 mol of CH4 is to be completely consumed? 101 Example Problem 19: SOLUTION: 102 Ratios from a Balanced Chemical Equation This flow diagram illustrates the various steps involved in solving a typical reaction stoichiometry problem. No different than unit conversion Usually more than one conversion is necessary Write all quantities with their complete units 103 Example Problem 20: PROBLEM: How many grams of water can be produced if sufficient hydrogen reacts with 26.0 g of oxygen? SOLUTION: 104 Ratios from a Balanced Chemical Equation Solution to Problem 20 using the stoichiometry problem flow diagram. 105 Example Problem 21: PROBLEM: If we have 153 g of S8 and an excess of phosphorus, what mass of P4S3 can be produced in the reaction shown? 8P4 + 3S8 ® 8P4 S3 SOLUTION: Molar mass of S8 is 256.6 g/mol Molar mass of P4S3 is 220.1 g/mol 106 Limiting Reactants In many chemical reactions, one reactant is often exhausted before the other reactants. This reactant is the limiting reactant. Limiting reactant is determined using stoichiometry. The limiting reactant limits the quantity of product produced. 107 Limiting Reactants In many chemical reactions, one reactant is often exhausted before the other reactants. This reactant is the limiting reactant. Limiting reactant is determined using stoichiometry. The limiting reactant limits the quantity of product produced. 108 Limiting Reactants Reaction between 6 H2 and 2H 2 (g) + O2 (g) ¾¾ ® 2H 2O(g) 6 O2 will produce 6 H2O. 6 H2 can produce 6 H2O. 6 O2 can produce 12 H2O. H2 is limiting reactant. 3 O2 left over. 109 Limiting Reactants In many cases, we manipulate the amounts of reactants to ensure that one specific compound is the limiting reactant. For example, a more expensive or scarce reagent is usually chosen to be the limiting reagent. Other times, it is best to have a stoichiometric mixture (equal ratio of moles) to prevent waste. For example, rocket fuel is designed so that no mass is left over, which would add unnecessary weight to the rocket. 110 Example Problem 22: PROBLEM: A solution of hydrochloric acid contains 5.22 g of HCl. When it is allowed to react with 3.25 g of solid K2CO3, the products are KCl, CO2, and H2O. Which reactant is in excess? SOLUTION: So the given quantity of HCl (5.22 g) requires 9.89 g K2CO3, but we have only 3.25 g of K2CO3 available. The reaction will stop once all of the K2CO3 is consumed. K2CO3 is the limiting reactant, and HCl is in excess. 111 Example Problem 23: PROBLEM: If 28.2 g of P4 is allowed to react with 18.3 g of S8, which is the limiting reactant? 8P4 + 3S8 ® 8P4 S3 SOLUTION: So, 28.2 g of P4 requires 21.9 g of S8 to react completely. We have only 18.3 g of S8, so there is not enough S8 to react with all of the P4. Therefore, S8 is the limiting reactant. 112 Example Problem 24: PROBLEM: If 45.0 kg of methanol is allowed to react with 70.0 kg of isobutene, what is the maximum mass (theoretical yield) of MTBE that can be obtained? CH 3OH + (CH 3 )2 C=CH 2 ® (CH 3 )3 COCH 3 Methanol I sobutene M TBE 113 Example Problem 24: SOLUTION: 114 Example Problem 25: PROBLEM: The solid fuel rockets of the space shuttle are based on the following reaction between ammonium perchlorate and aluminum: 3NH 4 ClO4 (s) + 3Al(s) ® Al2O3 (s) + AlCl3 (g) + 3NO(g) + 6H2O(g) If either reactant is in excess, unnecessary mass will be added to the shuttle, so a stoichiometric mixture is desired. What mass of each reactant should be used for every kilogram of the fuel mixture? 115 Example Problem 25: SOLUTION: 116 Example Problem 25: SOLUTION: So each kilogram of fuel should comprise 186.8 g Al and 813.2 g NH4ClO4. 117 Theoretical Yield The maximum mass of a product that can be obtained in a reaction is determined by the limiting reactant. Determine which reactant is the limiting reactant. Calculate the mass of product that can be made from the limiting reactant. This mass is the theoretical yield. In stoichiometric mixtures, however, both reactants are consumed completely, so either could be considered the limiting reactant. 118 Theoretical and Percent Yields Many factors determine the amount of desired product actually produced in a reaction. Temperature of the reaction The possibility of side reactions Further reaction of the product Time 119 Solution Stoichiometry Reaction efficiency is measured with percentage yield. The mass of product obtained is the actual yield. The ideal mass of product obtained from calculation is the theoretical yield. æ actual yield ö Percentage Yield = ç ÷ ´ 100% è theoretical yield ø 120 Example Problem 26: PROBLEM: In a laboratory experiment, a student heats 42.0 g of NaHCO3 and determines that 22.3 g of Na2CO3 is formed. What is the percentage yield of this reaction? ¾® Na 2CO3 (s) + CO2 (g) + H2O(g) 2NaHCO3 (s) ¾heat 121 Example Problem 26: SOLUTION: ¾® Na 2CO3 (s) + CO2 (g) + H2O(g) 2NaHCO3 (s) ¾heat 122 Solution Stoichiometry For reactions occurring in solution, the concentration and volume of reactants and products are often used instead of mass to solve solution stoichiometry problems. n = number of moles; M = mol/L; V = L 123 Example Problem 27: PROBLEM: If 750.0 mL of 0.806 M NaClO is mixed with excess ammonia, how many moles of hydrazine can be formed? NaClO(aq) + 2NH 3 (aq) ® N 2 H 4 (aq) + NaCl(aq) + H2O( ) If the final volume of the resulting solution is 1.25 L, what will be the molarity of hydrazine? 124 Example Problem 27: SOLUTION: NaClO(aq) + 2NH 3 (aq) ® N 2 H 4 (aq) + NaCl(aq) + H2O( ) 125 Solution Stoichiometry A titration is a common laboratory technique that uses solution stoichiometry. A solution-phase reaction is carried out under controlled conditions so that the amount of one reactant can be determined with high precision. An indicator is a dye added to a titration to indicate when the reaction is complete. 126 Solution Stoichiometry 1. A solution of one of the reactants (A) is added to a burette. 2. The burette is positioned above a flask containing the second reactant (B). 3. The burette is used to add A to the flask in a controlled manner; volume is determined from initial and final burette readings. 4. The reaction is complete when the indicator changes color. 127 Example Problem 28: PROBLEM: If 24.75 mL of 0.503 M NaOH solution is used to titrate a 15.00-mL sample of sulfuric acid, H2SO4, what is the concentration of the acid? 128 Example Problem 28: SOLUTION: 129 REFERENCES PRIMARY REFERENCE L. S. Brown & T. A. Holme. Chemistry for Engineering Students. OTHER REFERENCES D.D. Ebing & S.D. Gammon. General Chemistry. W.L. Masterton & C. N. Hurley. Chemistry :Principles and Reactions. M.S. Silberberg. Chemistry: The Molecular Nature of Matter and Change. T.L. Brown, et. al. Chemistry: T h e C e n t r a l S c i e n c e. M. S. Silberberg. Principles of General Chemistry. R. Chang. Chemistry. OpenStax College. (2015). Chemistry. 130