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1 Chapter Three Stoichiometry: Chemical Calculations Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 2 Molecular Masses and Formula Masses • Molecular mass: sum of the masses of the atoms represented in a molecular formula. • Simply put: the mass of a molecule. • Molecular mass is specifically for molecules. • Ionic compounds don’t exist as molecules; for them we use … • Formula mass: sum of the masses of the atoms or ions present in a formula unit. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 3 Molecular Mass Molecular mass is the sum of the masses of the atoms represented in a molecular formula. 1 Oxygen atom Example: water - H2O 2(1.0079 u) + 15.9994 u = 18.0152 u 2 Hydrogen atoms EOS Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 4 Formula Mass Formula mass is the sum of the masses of the atoms or ions present in a formula unit. ClNa+ Cl- Na+ Cl- One Na+ and one Cl– make a formula unit for sodium chloride Na+ Cl- Na+ The mass of one formula unit is: Crystal of = 22.9898 u + 35.4527 u sodium chloride Prentice Hall © 2005 = 58.4425 u General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry EOS Chapter Three 5 Example 3.1 Calculate the molecular mass of glycerol (C3H8O3). Example 3.2 Calculate the formula mass of ammonium sulfate, a fertilizer commonly used by home gardeners. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 6 Determining the Formula Mass of Ammonium Sulfate Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 7 The Mole & Avogadro’s Number • Mole (mol): amount of substance that contains as many elementary entities as there are atoms in exactly 12 g of the carbon-12 isotope. • Atoms are small, so this is a BIG number … • Avogadro’s number (NA) = 6.022 × 1023 mol–1 • 1 mol = 6.022 × 1023 “things” (atoms, molecules, ions, formula units, oranges, etc.) – A mole of oranges would weigh about as much as the earth! • Mole is NOT abbreviated as either M or m. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 8 Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 9 One Mole of Four Elements One mole each of helium, sulfur, copper, and mercury. How many atoms of helium are present? Of sulfur? Of copper? Of mercury? Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 10 The Mole and Molar Mass • Molar mass is the mass of one mole of a substance. • Molar mass is numerically equal to atomic mass, molecular mass, or formula mass. However … • … the units of molar mass are grams (g/mol). • Examples: 1 atom Na = 22.99 u 1 mol Na = 22.99 g 1 molecule CO2 = 44.01 u 1 mol CO2 = 44.01 g 1 formula unit KCl = 74.56 u 1 mol KCl = 74.56 g Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 11 Conversions involving Mass, Moles, and Number of Atoms/Molecules 1 mol Na = 6.022 × 1023 Na atoms = 22.99 g Na We can use these equalities to construct conversion factors, such as: 1 mol Na ––––––––– 22.99 g Na 22.99 g Na ––––––––– 1 mol Na 1 mol Na –––––––––––––––––– 6.022 × 1023 Na atoms Note: preliminary and follow-up calculations may be needed. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 12 We can read formulas in terms of moles of atoms or ions. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 13 Hints for Doing Mass/Mole/Atom Problems • HINT #1: If you see, ANYWHERE AT ALL IN THE PROBLEM, “grams”, you will NEED to make a SAMT chart! • HINT #2: If you see, ANYWHERE AT ALL IN THE PROBLEM, “atoms/molecules/etc”, you will NEED to use Avogadro’s #. • HINT #3: If you see, MOLE in the given or find, you will only do a 2-step problem. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 14 Examples • Given: 3.50 mol Cu; Find: # g of Cu. • Given: 11.5 g of H2O; Find: # mol of H2O. • Given: 0.55 mol of Al; Find: # atoms of Al. • Given: 3.73 x 10 23 molecules of HCl; Find: # mol of HCl. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 15 Examples – Part 2 • These are more difficult, and are 3-steps: • Given = 11.5 g of Ca; Find = # atoms of Ca. • Given = 1.22 x 10 23 molecules of NO; Find = # g of NO. • Given = 13.5 g of HCl; Find = # particles of HCl. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 16 Examples • Calculate the mass in milligrams of 1.34 x 10-4 mol Ag. • Calculate the number of oxygen atoms in 20.5 mol O2. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 17 Examples • Calculate both the (a) # of moles and (b) the # of atoms of Al in a cube of aluminum metal 5.5 cm on an edge. The density of aluminum is known to be 2.70 g/cm3. • Calculate the volume occupied by 4.06 x 1024 Br atoms present as Br2 molecules in liquid bromine. (d = 3.12 g/mL) Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 18 Example 3.4 Determine (a) the number of NH4+ ions in a 145-g sample of (NH4)2SO4 and (b) the volume of 1,2,3propanetriol (glycerol, d = 1.261 g/mL) that contains 1.00 mol O atoms. Example 3.5 An Estimation Example Which of the following is a reasonable value for the number of atoms in 1.00 g of helium? (a) 4.1 × 10–23 (c) 1.5 × 1023 (b) 4.0 (d) 1.5 × 1024 Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 19 Mass Percent Composition from Chemical Formulas The mass percent composition of a compound refers to the proportion of the constituent elements, expressed as the number of grams of each element per 100 grams of the compound. In other words … X g element X % element = –––––––––––––– OR … 100 g compound g element % element = ––––––––––– × 100 g compound Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 20 Percentage Composition of Butane Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 21 Example 3.6 Calculate, to four significant figures, the mass percent of each element in ammonium nitrate. Example 3.7 How many grams of nitrogen are present in 46.34 g ammonium nitrate? Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 22 Chemical Formulas from Mass Percent Composition • We can “reverse” the process of finding percentage composition. • First we use the percentage or mass of each element to find moles of each element. • Then we can obtain the empirical formula by finding the smallest whole-number ratio of moles. – Find the whole-number ratio by dividing each number of moles by the smallest number of moles. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 23 Examples • Analysis shows that a compound contains 32.31% sodium, 22.67% sulfur, and 45.02% oxygen. Find the empirical formula for this compound. • HINT = Whenever you are given %s, we assume we ALWAYS have a 100-gram sample, so immediately convert you %s to grams. • ANS = Na2SO4 Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 24 Examples (cont.) • Find the empirical formula of a compound found to contain 26.56% K, 35.41% Cr, and the remainder O. • ANS = K2Cr2O7 • Analysis of a 10.150-g sample of a compound is known to contain only phosphorus and oxygen. There is a phosphorus content of 4.433-g. What is the empirical formula of this compound? • ANS = P2O5 Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 25 Example 3.9 Phenol, a general disinfectant, has the composition 76.57% C, 6.43% H, and 17.00% O by mass. Determine its empirical formula. Example 3.10 Diethylene glycol, used in antifreeze, as a softening agent for textile fibers and some leathers, and as a moistening agent for glues and paper, has the composition 45.27% C, 9.50% H, and 45.23% O by mass. Determine its empirical formula. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 26 Molecular Formulas • Molecular formulas are the actual formulas for the compound. • Example: In our last example, we determined that the empirical formula was P2O5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula? • Hint = The molar mass given in the problem statement is the numerator!!!! • ANS = P4O10 Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 27 Relating Molecular Formulas to Empirical Formulas • A molecular formula is a simple integer multiple of the empirical formula. • That is, an empirical formula of CH2 means that the molecular formula is CH2, or C2H4, or C3H6, or C4H8, etc. • So: we find the molecular formula by: molecular formula mass = integer (nearly) empirical formula mass We then multiply each subscript in the empirical formula by the integer. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 28 Examples • Determine the molecular formula of the compound with an empirical formula of CH and a formula mass of 78.110 amu (g/mol). • ANSWER = C6H6 • A sample of a compound with a formula mass of 34.00 amu (g/mol) is found to consist of 0.44 gH and 6.92 gO. Determine (a) empirical and (b) molecular formulas for this compound. • ANSWER = (a) HO; (b) H2O2 Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 29 Example 3.11 The empirical formula of hydroquinone, a chemical used in photography, is C3H3O, and its molecular mass is 110 u. What is its molecular formula? Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 30 Elemental Analysis … • … is one method of determining empirical formulas in the laboratory. • This method is used primarily for simple organic compounds (that contain carbon, hydrogen, oxygen). – The organic compound is burned in oxygen. – The products of combustion (usually CO2 and H2O) are weighed. – The amount of each element is determined from the mass of products. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 31 Elemental Analysis (cont’d) … H2O, which is absorbed by MgClO4, and … The sample is burned in a stream of oxygen gas, producing … Prentice Hall © 2005 … CO2, which is absorbed by NaOH. General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 32 Elemental Analysis (cont’d) If our sample were CH3OH, every two molecules of CH3OH … Prentice Hall © 2005 … would give two molecules of CO2 … General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry … and four molecules of H2O. Chapter Three 33 Example 3.12 Burning a 0.1000-g sample of a carbon– hydrogen–oxygen compound in oxygen yields 0.1953 g CO2 and 0.1000 g H2O. A separate experiment shows that the molecular mass of the compound is 90 u. Determine (a) the mass percent composition, (b) the empirical formula, and (c) the molecular formula of the compound. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 34 Writing Chemical Equations • A chemical equation is a shorthand description of a chemical reaction, using symbols and formulas to represent the elements and compounds involved. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 35 Writing Chemical Equations • Sometimes additional information about the reaction is conveyed in the equation. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 36 Balancing Equations Illustrated How can we tell that the equation is not balanced? … not by changing the equation … … and not by changing the formulas. The equation is balanced by changing the coefficients … Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 37 Chemical Reactions • What is a chemical reaction? • A chemical reaction is any process by which 1 or more substances are changed into one or more different substances. • A + B C + D • A and B are referred to as REACTANTS; C and D are referred to as PRODUCTS. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 38 Chemical Equation • Chemical reactions are described by chemical equations. • Chemical equation = an equation that represents, with symbols and formulas, the identities and relative molecular amounts of the reactants and products in a chemical reaction. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 39 Indications of a Chemical Reaction • The following methods are ways we tell a chemical reaction occurs: • #1. Evolution of energy as heat and/or light. • #2. Production of a gas. • #3. Formation of a precipitate. • Precipitate = a solid that is produced as a result of a chemical reaction of 2 or more solutions. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 40 Characteristics of a Chemical Equation • 1. The equation must represent known facts. All reactants and products MUST be identified. • 2. The equation must contain the correct formulas for the reactants and products. • 3. The law of conservation of mass must be satisfied. • NOTE = we are ONLY allowed to change coefficients, NOT subscripts in a chemical equation! Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 41 Coefficients vs. Subscripts Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 42 Word and Formula Equations • __H2O __H2 + __O2 • What are the reactant(s)? • What are the product(s)? • What are the coefficients for each material? Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 43 Word Equations • A word equation is an equation in which the reactants products in a chemical reaction are represented by words. • Remember – what are our diatomic elements? • Ex#1: Solid aluminum reacts with gaseous oxygen to produce solid aluminum oxide. • Ex#2: Methane, in the presence of oxygen, combusts into water and carbon dioxide. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 44 A Few More Examples… • Example #3: • Solid sodium oxide is added to water at room temperature and forms sodium hydroxide. • Example #4: • Hydrazine, N2H4, is used as rocket fuel. Hydrazine reacts violently with oxygen to produce gaseous nitrogen and water. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 45 A Few Hints… • When balancing chemical equations, here are a few hints: • #1. Balance most elements OTHER THAN hydrogen and oxygen 1st, then go with hydrogen, and save oxygen for LAST. • #2. Sometimes, it is easier to call water HOH. • #3. Look for polyatomic ions on both sides of the equation. If they are there, you can use a polyatomic as a “pseudo” element. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 46 What Do I Mean by Hint #3? • Example: • Magnesium phosphate reacts with lithium chloride to form magnesium chloride and lithium phosphate. • Calcium hydroxide and sodium fluoride react to form calcium fluoride and sodium hydroxide. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 47 Pop Quiz…. • Try the following….think about what you would get: • #1. __Li + __O2 ??? • #2. __NaCl ??? • #3. __H2 + __F2 • #4. Prentice Hall © 2005 __MgCO3 ??? ??? General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 48 5 Basic Types of Chemical Reactions • There are several ways to classify chemical reactions. • • • • • • We will look at the 5 basic types of reactions: 1. synthesis (composition) 2. decomposition 3. single displacement (replacement) 4. double displacement (replacement) 5. combustion Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 49 Synthesis Reactions • Synthesis reactions are AKA composition reactions. • Synthesis reactions = a reaction in which 2 or more substances combine to form a new compound. • A + Prentice Hall © 2005 X AX General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 50 Types of Synthesis Reactions • Type #1: Reactions of elements with oxygen and sulfur. • Remember, oxygen is diatomic (O2). Sulfur is odd – it can be S8. • __Rb + __S8 __Rb2S • __Fe + __O2 Prentice Hall © 2005 __Fe2O3 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 51 Types of Synthesis Rxns (cont.) • Type #2: Reactions of metals with halogens. • __Na + __Br2 __NaBr • __F2 + __Mg __MgF2 • Remember – ALL halogens are diatomic! Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 52 Types of Synthesis Rxns. (cont.) • Type #3: Synthesis reactions with oxides. • __CaO + __H2O • __SO2 + __H2O __Ca(OH)2 __H2SO4 • Remember, when balancing, call water HOH. • Usually, you get a metal hydroxide. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 53 Decomposition Reactions • In a decomposition reaction, a single compound undergoes a reaction that produces 2 or more simpler substances. • Decomposition reactions are usually the opposite of synthesis reactions. • AX A + X • Decomposition reactions usually require heat or electricity in order to occur. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 54 Types of Decomposition Reactions • Type #1: Decomposition of a Binary Compound • 2H2O 2 H2 + O2 (electrolysis) • 2HgO 2Hg + O2 (under heat) Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 55 Types (cont.) • Type #2: Decomposition of Metal Carbonates • What IS a metal carbonate? • CaCO3 CaO + CO2 • Breaks down into a metal oxide and carbon dioxide gas. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 56 Types (cont.) • Type #3: Decomposition of a Metal Hydroxide • What IS a metal hydroxide? • Ca(OH)2 CaO + HOH • Breaks down into a metal oxide and water. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 57 Types (cont.) • Type #4: Decomposition of a Metal Chlorate • What IS a metal chlorate? • 2KClO3 2KCl + 3O2 • Breaks down into a metal chloride and oxygen. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 58 Types (cont.) • Type #5: Decomposition of Acids • Think acids that are NOT binary for these! • H2CO3 CO2 + H2 O • H2SO4 SO3 + H2 O • Breaks down into water and “what’s left”. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 59 Single Displacement Reactions • Single displacement reactions are also known as single replacement reactions. • We will look at this type of reaction more in the next section… Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 60 Double Displacement Reactions • In double displacement (replacement) reactions, the ions of 2 compounds exchange places in an aqueous solution to form 2 new compounds. • AX + BY • __KI +__Pb(NO3)2__PbI2 +__KNO3 Prentice Hall © 2005 AY + BX General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 61 Combustion Reactions • In a combustion reaction, a hydrocarbon reacts with oxygen , releasing a LARGE amount of energy, as well as water and carbon dioxide. • __CH4 + __O2 __CO2 + __H2O • __C3H8 + __O2 __CO2 + __H2O Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 62 Example 3.13 Balance the equation Fe + O2 Fe2O3 (not balanced) Example 3.14 Balance the equation C2H6 + O2 CO2 + H2O Example 3.15 Balance the equation H3PO4 + NaCN HCN + Na3PO4 Example 3.16 A Conceptual Example Write a plausible chemical equation for the reaction between water and a liquid molecular chloride of phosphorus to form an aqueous solution of hydrochloric acid and phosphorus acid. The phosphorus-chlorine compound is 77.45% Cl by mass. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 63 Stoichiometric Equivalence and Reaction Stoichiometry • A stoichiometric factor or mole ratio is a conversion factor obtained from the stoichiometric coefficients in a chemical equation. • In the equation: CO(g) + 2 H2(g) CH3OH(l) – 1 mol CO is chemically equivalent to 2 mol H2 – 1 mol CO is chemically equivalent to 1 mol CH3OH – 2 mol H2 is chemically equivalent to 1 mol CH3OH 1 mol CO ––––––––– 2 mol H2 Prentice Hall © 2005 1 mol CO ––––––––––––– 1 mol CH3OH 2 mol H2 ––––––––––––– 1 mol CH3OH General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 64 Concept of Stoichiometric Equivalence One car may be equivalent to either 25 feet or 10 feet, depending on the method of parking. One mole of CO may be equivalent to one mole of CH3OH, or to one mole of CO2, or to two moles of CH3OH, depending on the reaction(s). Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 65 Outline of Simple Reaction Stoichiometry Note: preliminary and/or follow-up calculations may be needed. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 66 Example 3.17 When 0.105 mol propane is burned in an excess of oxygen, how many moles of oxygen are consumed? Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 67 Outline of Stoichiometry Involving Mass … we’ve added a conversion from mass at the beginning … To our simple stoichiometry scheme … Substances A and B may be two reactants, two products, or reactant and product. … and a conversion to mass at the end. Think: If we are given moles of substance A initially, do we need to convert A to grams? Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 68 Examples • Magnesium reacts with oxygen to produce magnesium oxide. • Given = 2.00 mol Mg • Find = # grams of MgO Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 69 Examples • Tin reacts with hydrofluoric acid to produce tin (II) fluoride and hydrogen gas. • Given = 30.00 grams of HF • Find = # grams of SnF2 Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 70 Example 3.18 The final step in the production of nitric acid involves the reaction of nitrogen dioxide with water; nitrogen monoxide is also produced. How many grams of nitric acid are produced for every 100.0 g of nitrogen dioxide that reacts? Example 3.19 Ammonium sulfate, a common fertilizer used by gardeners, is produced commercially by passing gaseous ammonia into an aqueous solution that is 65% H2SO4 by mass and has a density of 1.55 g/mL. How many milliliters of this sulfuric acid solution are required to convert 1.00 kg NH3 to (NH4)2SO4? Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 71 Limiting Reactants • Many reactions are carried out with a limited amount of one reactant and a plentiful amount of the other(s). • The reactant that is completely consumed in the reaction limits the amounts of products and is called the limiting reactant, or limiting reagent. • The limiting reactant is not necessarily the one present in smallest amount. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 72 Limiting Reactant Analogy If we have 10 sandwiches, 18 cookies, and 12 oranges … Prentice Hall © 2005 … how many packaged meals can we make? General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 73 Limiting vs. Excess • When looking at any chemical reaction, the reaction is rarely carried out as per the balanced equation. • We will be looking at reactions with 2 reactants. • (A) limiting reactant = the reactant in the chemical reaction that is completely “used up” – none will remain when the reaction is complete. • (B) excess reactant = the reactant that is NOT completely used up in a chemical reaction. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 74 What to Do • Silicon dioxide, aka quartz, is usually quite unreactive but reacts readily with hydrogen fluoride to produce silicon tetrafluoride and water. • If 6.0g of HF is added to 4.5g of SiO2, which is the limiting reactant? • Step #1: BALANCE THE EQUATION! • __SiO2 + __HF Prentice Hall © 2005 __SiF4 + General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry __H2O Chapter Three 75 What to Do (cont.) • SiO2 + 4HF SiF4 + 2H2O • Step #2: Pick a product, any product (if the reactant is limiting for 1 product, it will be limiting for the other) = I pick SiF4. • Step #3: Convert BOTH reactants to grams of product. • Step #4: The smaller value gives us the reactant that is LIMITING! Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 76 Example • Fe2O3 is a substance that can be made in the chem lab via the following, UNBALANCED, equation: • __Fe +__H2O __Fe2O3 +__H2 • (a) Balance the equation • (b) If 36.0 g water is mixed with 67.0 g Fe, determine the limiting reactant. • (c) What mass of iron oxide is produced, based on the limiting reactant? • (d) What mass of excess reactant remains when the reaction is completed? Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 77 Molecular View of the Limiting Reactant Concept 1. Why is ethylene left over, when we started with more bromine than ethylene? (Hint: count the molecules.) 2. What mass of ethylene is left over after reaction is complete? (Hint: it’s an easy calculation; why?) When 28 g (1.0 mol) ethylene reacts with … Prentice Hall © 2005 … 128 g (0.80 mol) bromine, we get … … 150 g of 1,2dibromoethane, and leftover ethylene! General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 78 Recognizing and Solving Limiting Reactant Problems • • • We recognize limiting reactant problems by the fact that amounts of two (or more) reactants are given. One way to solve them is to perform a normal stoichiometric calculation of the amount of product obtained, starting with each reactant. The reactant that produces the smallest amount of product is the limiting reactant. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 79 Example 3.20 Magnesium nitride can be formed by the reaction of magnesium metal with nitrogen gas. (a) How many grams of magnesium nitride can be made in the reaction of 35.00 g of magnesium and 15.00 g of nitrogen? (b) How many grams of the excess reactant remain after the reaction? Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 80 Yields of Chemical Reactions • The theoretical yield of a chemical reaction is the calculated quantity of product in the reaction. • The actual yield is the amount you actually get when you carry out the reaction. • Actual yield will be less than the theoretical yield, for many reasons … can you name some? actual yield Percent yield = ––––––––––––– × 100 theoretical yield Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 81 Actual Yield of ZnS Is Less than the Theoretical Yield Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 82 Example 3.21 Ethyl acetate is a solvent used as fingernail polish remover. What mass of acetic acid is needed to prepare 252 g ethyl acetate if the expected percent yield is 85.0%? Assume that the other reactant, ethanol, is present in excess. The equation for the reaction, carried out in the presence of H2SO4, is CH3COOH + HOCH2CH3 CH3COOCH2CH3 + H2O Acetic acid Ethanol Ethyl acetate Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 83 Solutions and Solution Stoichiometry • Solute: the substance being dissolved. • Solvent: the substance doing the dissolving. • Concentration of a solution: the quantity of a solute in a given quantity of solution (or solvent). – A concentrated solution contains a relatively large amount of solute vs. the solvent (or solution). – A dilute solution contains a relatively small concentration of solute vs. the solvent (or solution). – “Concentrated” and “dilute” aren’t very quantitative … Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 84 Molar Concentration Molarity (M), or molar concentration, is the amount of solute, in moles, per liter of solution: moles of solute Molarity = –––––––––––––– liters of solution • A solution that is 0.35 M sucrose contains 0.35 moles of sucrose in each liter of solution. • Keep in mind that molarity signifies moles of solute per liter of solution, not liters of solvent. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 85 Preparing 0.01000 M KMnO4 Weigh 0.01000 mol (1.580 g) KMnO4. Prentice Hall © 2005 Dissolve in water. How much water? Doesn’t matter, as long as we don’t go over a liter. General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Add more water to reach the 1.000 liter mark. Chapter Three 86 Example 3.23 What is the molarity of a solution in which 333 g potassium hydrogen carbonate is dissolved in enough water to make 10.0 L of solution? Example 3.24 We want to prepare a 6.68 molar solution of NaOH (6.68 M NaOH). (a) How many moles of NaOH are required to prepare 0.500 L of 6.68 M NaOH? (b) How many liters of 6.68 M NaOH can we prepare with 2.35 kg NaOH? Example 3.25 The label of a stock bottle of aqueous ammonia indicates that the solution is 28.0% NH3 by mass and has a density of 0.898 g/mL. Calculate the molarity of the solution. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 87 Dilution of Solutions • Dilution is the process of preparing a more dilute solution by adding solvent to a more concentrated one. • Addition of solvent does not change the amount of solute in a solution but does change the solution concentration. • It is very common to prepare a concentrated stock solution of a solute, then dilute it to other concentrations as needed. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three Visualizing the Dilution of a Solution We start and end with the same amount of solute. Prentice Hall © 2005 88 Addition of solvent has decreased the concentration. General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 89 Dilution Calculations … • • • • … couldn’t be easier. Moles of solute does not change on dilution. Moles of solute = M × V Therefore … Mconc × Vconc = Mdil × Vdil Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 90 Example 3.26 How many milliliters of a 2.00 M CuSO4 stock solution are needed to prepare 0.250 L of 0.400 M CuSO4? Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 91 Solutions in Chemical Reactions • Molarity provides an additional tool in stoichiometric calculations based on chemical equations. • Molarity provides factors for converting between moles of solute (either reactant or product) and liters of solution. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three Adding to the previous stoichiometry scheme … 92 If substance A is a solution of known concentration … If substance B is in solution, then … … we can start with molarity of A times volume (liters) of the solution of A to get here. Prentice Hall © 2005 … we can go from moles of substance B to either volume of B or molarity of B. How? General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 93 Example 3.27 A chemical reaction familiar to geologists is that used to identify limestone. The reaction of hydrochloric acid with limestone, which is largely calcium carbonate, is seen through an effervescence—a bubbling due to the liberation of gaseous carbon dioxide: CaCO3(s) + 2 HCl(aq) CaCl2(aq) + H2O(l) + CO2(g) How many grams of CaCO3(s) are consumed in a reaction with 225 mL of 3.25 M HCl? Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three 94 Cumulative Example The combustion in oxygen of 1.5250 g of an alkane-derived compound composed of carbon, hydrogen, and oxygen yields 3.047 g CO2 and 1.247 g H2O. The molecular mass of this compound is 88.1 u. Draw a plausible structural formula for this compound. Is there more than one possibility? Explain. Prentice Hall © 2005 General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry Chapter Three