Janice G. Smith-General, Organic, And Biological Chemistry .
5CHAPTER OUTLINE5.1Introduction to Chemical Reactions5.2Balancing Chemical Equations5.3The Mole and Avogadro’s Number5.4Mass to Mole Conversions5.5Mole Calculations in ChemicalEquations5.6Mass Calculations in ChemicalEquations5.7Percent Yield5.8Oxidation and Reduction5.9FOCUS ON HEALTH & MEDICINE:PacemakersCHAPTER GOALSIn this chapter you will learn how to:➊ Write and balance chemical equations➋ Define a mole and use Avogadro’snumber in calculations➌ Calculate formula weight and molarmass➍ Relate the mass of a substance to itsnumber of moles➎ Carry out mole and mass calculationsin chemical equations➏ Calculate percent yield➐ Define oxidation and reduction andrecognize the components of a redoxreaction➑ Give examples of common or usefulredox reactionsThread for suturing wounds is made from nylon, one of the countless products synthesized by thechemical industry using chemical reactions.CHEMICAL REACTIONSHAVING learned about atoms, ionic compounds, and covalent molecules in Chapters 2–4, we now turn our attention to chemical reactions. Reactions are at the heartof chemistry. An understanding of chemical processes has made possible the conversion of natural substances into new compounds with different and sometimes superiorproperties. Aspirin, ibuprofen, and nylon are all products of chemical reactions utilizing substances derived from petroleum. Chemical reactions are not limited to industrialprocesses. The metabolism of food involves a series of reactions that both forms newcompounds and also provides energy for the body’s maintenance and growth. Burninggasoline, baking a cake, and photosynthesis involve chemical reactions. In Chapter 5 welearn the basic principles about chemical reactions.121
CHEMICAL REACTIONS1225.1 INTRODUCTION TO CHEMICAL REACTIONSNow that we have learned about compounds and the atoms that compose them, we can betterunderstand the difference between the physical and chemical changes that were first discussedin Section 1.2. A physical change alters the physical state of a substance without changing itscomposition.Changes in state—such as melting and boiling—are familiar examples of physical changes.When ice (solid water) melts to form liquid water, the highly organized water molecules inthe solid phase become more disorganized in the liquid phase, but the chemical bonds do notchange. Each water molecule (H2O) is composed of two O H bonds in both the solid and liquidphases.solid H2Oliquid H2OphysicalchangemeltingH2O molecules areunchanged beforeand after melting.2 O–H bonds2 O–H bonds A chemical change—chemical reaction—converts one substance into another.Chemical reactions involve breaking bonds in the starting materials, called reactants, andforming new bonds in the products. The combustion of methane (CH4), the main constituent ofnatural gas, in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O) is anexample of a chemical reaction. The carbon–hydrogen bonds in methane and the oxygen–oxygenbond in elemental oxygen are broken, and new carbon–oxygen and hydrogen–oxygen bonds areformed in the products.CO2CH4chemicalreactionnew bonds formedO2H2Oreactantsproductsnew bonds formedA chemical reaction may be accompanied by a visible change: two colorless reactants can forma colored product; a gas may be given off; two liquid reactants may yield a solid product. Sometimes heat is produced so that a reaction flask feels hot. A reaction having a characteristic visiblechange occurs when hydrogen peroxide (H2O2) is used to clean a bloody wound. An enzymein the blood called catalase converts the H2O2 to water (H2O) and oxygen (O2), and bubbles ofoxygen appear as a foam, as shown in Figure 5.1. A chemical equation is an expression that uses chemical formulas and other symbols toillustrate what reactants constitute the starting materials in a reaction and what productsare formed.
INTRODUCTION TO CHEMICAL REACTIONS123 FIGURE 5.1Treating Wounds with Hydrogen Peroxide—A VisibleChemical ReactionO2H2OH2OH 2 O2H2O2(aq)The enzyme catalase in red blood converts hydrogen peroxide (H2O2) to water and oxygengas, which appears as a visible white foam on the bloody surface. Hydrogen peroxide does notfoam when it comes in contact with skin because skin cells do not contain the catalase neededfor the reaction to occur.Chemical equations are written with the reactants on the left and the products on the right,separated by a horizontal arrow—a reaction arrow—that points from the reactants to the products. In the combustion of methane, methane (CH4) and oxygen (O2) are the reactants on the leftside of the arrow, and carbon dioxide (CO2) and water (H2O) are the products on the right side.coefficientcoefficientChemicalequationCH4 2 O2reactantsCO2 2 H2OproductsThe numbers written in front of any formula are called coefficients. Coefficients show thenumber of molecules of a given element or compound that react or are formed. When nonumber precedes a formula, the coefficient is assumed to be “1.” In the combustion of methane,the coefficients tell us that one molecule of CH4 reacts with two molecules of O2 to form onemolecule of CO2 and two molecules of H2O.When a formula contains a subscript, multiply its coefficient by the subscript to give the totalnumber of atoms of a given type in that formula.2 O2 4 O atoms2 H2O 4 H atoms 2 O atomsCoefficients are used because all chemical reactions follow a fundamental principle of nature, thelaw of conservation of mass, which states: Atoms cannot be created or destroyed in a chemical reaction.
CHEMICAL REACTIONS124TABLE 5.1 Symbols Usedin Chemical EquationsSymbol (s)(l)(g)(aq)MeaningAlthough bonds are broken and formed in reactions, the number of atoms of each element in thereactants must be the same as the number of atoms of each type in the products. Coefficients areused to balance an equation, making the number of atoms of each element the same on bothsides of the equation.Reaction arrowHeatSolidLiquidGasAqueous solutionCH42 O2 CO2Atoms in the reactants: 1 C atom 4 H atoms 4 O atoms2 H2O Atoms in the products: 1 C atom 4 H atoms 4 O atomsTwo other important features are worthy of note. If heat is needed for a reaction to occur, the Greekletter delta ( ) may be written over the arrow. The physical states of the reactants and products aresometimes indicated next to each formula—solid (s), liquid (l), or gas (g). If an aqueous solutionis used—that is, if a reactant is dissolved in water—the symbol (aq) is used next to the reactant.When these features are added, the equation for the combustion of methane becomes:Combustionof methaneCH4(g) 2 O2(g) CO2(g) 2 H2O(g)The symbols used for chemical equations are summarized in Table 5.1.SAMPLE PROBLEM 5.1Label the reactants and products, and indicate how many atoms of each type of element arepresent on each side of the equation.C2H6O(l) 3 O2(g)2 CO2(g) 3 H2O(g)ANALYSISReactants are on the left side of the arrow and products are on the right side in a chemicalequation. When a formula contains a subscript, multiply its coefficient by the subscript to givethe total number of atoms of a given type in the formula.SOLUTIONIn this equation, the reactants are C2H6O and O2, while the products are CO2 and H2O. If nocoefficient is written, it is assumed to be “1.” To determine the number of each type of atomwhen a formula has both a coefficient and a subscript, multiply the coefficient by the subscript.1 C2H6O 2 C’s 6 H’s 1 O3 O2 6 O’s2 CO2 2 C’s 4 O’s3 H2O 6 H’s 3 O’sMultiply the coefficient 3 by the subscript 2.Multiply the coefficient 2 by each subscript;2 1 C 2 C’s; 2 2 O’s 4 O’s.Multiply the coefficient 3 by each subscript;3 2 H’s 6 H’s; 3 1 O 3 O’s.Add up the atoms on each side to determine the total number for each type of element.C2H6O(l) 3 O2(g)2 CO2(g) 3 H2O(g)OHCAtoms in the reactants: 2 C’s 6 H’s 7 O’sAtoms in the products: 2 C’s 6 H’s 7 O’s
BALANCING CHEMICAL EQUATIONSPROBLEM 5.1125Label the reactants and products, and indicate how many atoms of each type of element arepresent on each side of the following equations.a. 2 H2O2(aq)2 H2O(l) O2(g)16 CO2 18 H2Ob. 2 C8H18 25 O2c. 2 Na3PO4(aq) 3 MgCl2(aq)Mg3(PO4)2(s) 6 NaCl(aq)PROBLEM 5.2One term in a balanced chemical equation contained the coefficient 3 in front of the formulaAl2(SO4)3. How many atoms of each type of element does this represent?PROBLEM 5.3Write a chemical equation from each of the following descriptions of reactions.a. One molecule of gaseous methane (CH4) is heated with four molecules of gaseous chlorine(Cl2), forming one molecule of liquid carbon tetrachloride (CCl4) and four molecules ofgaseous hydrogen chloride (HCl).b. One molecule of liquid methyl acetate (C3H6O2) reacts with two molecules of hydrogengas (H2) to form one molecule each of liquid ethanol (C2H6O) and methanol (CH4O).ENVIRONMENTAL NOTEBALANCING CHEMICAL EQUATIONS5.2In order to carry out a reaction in the laboratory, we must know how much of each reactantwe must combine to give the desired product. For example, if we wanted to synthesize aspirin(C9H8O4) from a given amount of salicylic acid (C7H6O3), say 100 g, we would have to determine how much acetic acid would be needed to carry out the reaction. A calculation of this sortbegins with a balanced chemical equation.HHCCCHThe reaction of propane with oxygenforms carbon dioxide, water, anda great deal of energy that can beused for cooking, heating homesand water, drying clothes, andpowering generators and vehicles.The combustion of propane andother fossil fuels adds a tremendousamount of CO2 to the atmosphereeach year, with clear environmentalconsequences (Section 12.8).HOW TOEXAMPLEStep [1]HCOCHH OCCHOHHsalicylic acidC7H6O3CHHOCCOHacetic acidC2H4O2CCHOCOCCOCCHOCH3 H2OHaspirinC9H8O4In this example, the equation is balanced as written and the coefficient in front of each formula is“1.” Thus, one molecule of salicylic acid reacts with one molecule of acetic acid to form one molecule of aspirin and one molecule of water. More often, however, an equation must be balancedby adding coefficients in front of some formulas so that the number of atoms of each elementis equal on both sides of the equation.Balance a Chemical EquationWrite a balanced chemical equation for the reaction of propane (C3H8) with oxygen (O2) to form carbon dioxide(CO2) and water (H2O).Write the equation with the correct formulas. Write the reactants on the left side and the products on the right side of the reaction arrow, and check if theequation is balanced without adding any coefficients.C3H8 O2CO2 H2OContinued
CHEMICAL REACTIONS126How To, continued. . . This equation is not balanced as written since none of the elements—carbon, hydrogen, and oxygen—has thesame number of atoms on both sides of the equation. For example, there are 3 C’s on the left and only 1 C onthe right. The subscripts in a formula can never be changed to balance an equation. Changing a subscript changes theidentity of the compound. For example, changing CO2 to CO would balance oxygen (there would be 2 O’s onboth sides of the equation), but that would change CO2 (carbon dioxide) into CO (carbon monoxide).Step [2]Balance the equation with coefficients one element at a time. Begin with the most complex formula, and start with an element that appears in only one formula on both sidesof the equation. In this example, begin with either the C’s or H’s in C3H8. Since there are 3 C’s on the left, placethe coefficient 3 before CO2 on the right.C3H8 3 CO2O23 C’s on the leftH2OPlace a 3 to balance C’s. To balance the 8 H’s in C3H8, place the coefficient 4 before H2O on the right.C3H8 3 CO2O2 4 H2O8 H’s on the leftPlace a 4 to balance H’s.(4 2 H’s in H2O 8 H’s) The only element not balanced is oxygen, and at this point there are a total of 10 O’s on the right—six fromthree CO2 molecules and four from four H2O molecules. To balance the 10 O’s on the right, place the coefficient5 before O2 on the left.C3H8 5 O23 CO2Place a 5 to balance O’s.Step [3] 4 H2O10 O’s on the rightCheck to make sure that the smallest set of whole numbers is used.C3H8 5 O23 CO2 4 H2OOHCAtoms in the reactants: 3 C’s 8 H’s 10 O’s (5 2 O’s)Atoms in the products: 3 C’s 8 H’s 10 O’s [(3 2 O’s) (4 1 O)] This equation is balanced because the same number of C’s, O’s, and H’s is present on both sides of the equation. Sometimes an equation is balanced but the lowest set of whole numbers is not used as coefficients. Say, forexample, that balancing yielded the following equation:2 C3H8 10 O26 CO2 8 H2O This equation has the same number of C’s, O’s, and H’s on both sides, but each coefficient must be divided bytwo to give the lowest set of whole numbers for the balanced equation, as drawn in the first equation in step [3].Sample Problems 5.2–5.4 illustrate additional examples of balancing chemical equations. SampleProblem 5.3 gives an example that uses fractional coefficients in balancing. Sample Problem 5.4illustrates how to balance an equation that contains several polyatomic anions.
BALANCING CHEMICAL EQUATIONS127SAMPLE PROBLEM 5.2Write a balanced equation for the reaction of glucose (C6H12O6) with oxygen (O2) to formcarbon dioxide (CO2) and water (H2O).ANALYSISBalance an equation with coefficients, one element at a time, beginning with the most complexformula and starting with an element that appears in only one formula on both sides of theequation. Continue placing coefficients until the number of atoms of each element is equal onboth sides of the equation.SOLUTION[1]Write the equation with correct formulas.C6H12O6 O2CO2 H2Oglucose None of the elements is balanced in this equation. As an example, there are 6 C’s on the leftside, but only 1 C on the right side.[2]Balance the equation with coefficients one element at a time. Begin with glucose, since its formula is most complex. Balance the 6 C’s of glucoseby placing the coefficient 6 before CO2. Balance the 12 H’s of glucose by placing thecoefficient 6 before H2O.Place a 6 to balance C’s.C6H12O6Bagels, pasta, bread, andrice are high in starch, whichis hydrolyzed to the simplecarbohydrate glucose afteringestion. The metabolism ofglucose forms CO2 and H2Oand provides energy for bodilyfunctions. O26 CO2 6 H2OPlace a 6 to balance H’s. The right side of the equation now has 18 O’s. Since glucose already has 6 O’s on the leftside, 12 additional O’s are needed on the left side. The equation will be balanced if thecoefficient 6 is placed before O2.C6H12O6 6 O26 CO2 6 H2OPlace a 6 to balance O’s.[3]Check. The equation is balanced since the number of atoms of each element is the same on both sides.Answer:PROBLEM 5.4 6 O26 CO2 6 H2 OAtoms in the reactants:Atoms in the products: 6 C’s 12 H’s 18 O’s (1 6 O’s) (6 2 O’s) 6 C’s (6 1 C) 12 H’s (6 2 H’s) 18 O’s (6 2 O’s) (6 1 O)Write a balanced equation for each reaction.a. H2 O2H2ONO2b. NO O2PROBLEM 5.5C6H12O6c. Fe O2d. CH4 Cl2Fe2O3CH2Cl2 HClWrite a balanced equation for the following reaction, shown with molecular art.CO
CHEMICAL REACTIONS128 FIGURE 5.2Chemistry of an Automobile Airbagb. An airbag deployed in a head-on collisiona. The chemical reaction that inflates an airbaginflated airbaginflatorcrash sensorNaN2NaN3sodium azideA severe car crash triggers an airbag to deploy when an electric sensor causes sodium azide (NaN3) to ignite, converting it to sodium(Na) and nitrogen gas (N2). The nitrogen gas causes the bag to inflate fully in 40 milliseconds, helping to protect passengers from seriousinjury. The sodium atoms formed in this first reaction are hazardous and subsequently converted to a safe sodium salt. It took 30 years todevelop a reliable airbag system for automobiles.SAMPLE PROBLEM 5.3The airbag in an automobile inflates when ionic sodium azide (NaN3), which is composed ofNa cations and the polyatomic anion, N3– (azide), rapidly decomposes to sodium (Na) andgaseous N2 (Figure 5.2). Write a balanced equation for this reaction.ANALYSISBalance an equation with coefficients, one element at a time, beginning with the most complexformula and starting with an element that appears in only one formula on both sides of theequation. Continue placing coefficients until the number of atoms of each element is equal onboth sides of the equation.SOLUTION[1]Write the equation with correct formulas.NaN3Na N2sodium azide The N atoms are not balanced since there are 3 N’s on the left side and only 2 N’s on the right.[2]Balance the equation with coefficients. To balance the N atoms, we can use a fractional coefficient on the right side. A coefficientof 3/2 before N2 is the equivalent of 3 N atoms.NaN3Na 3 N2 23 N atoms[3]32 2 N atoms1 N2 molecule 3 N atomsCheck and simplify. The equation is balanced since the number of atoms of each element is the same on both sides.NaN3Na 3N2 2Atoms in the reactants:Atoms in the products: 1 Na 3 N’s 1 Na 3 N’s [(3/2) 2 N’s]
BALANCING CHEMICAL EQUATIONS129 Since a properly balanced equation uses the lowest set of whole numbers, multiply eachcoefficient by 2 to convert the fraction 3/2 to the whole number 3.Answer:2 NaN3 2 Na(2 1)3 N2(2 1)) 2 32 )PROBLEM 5.6Write a balanced equation for the reaction of ethane (C2H6) with O2 to form CO2 and H2O.PROBLEM 5.7The Haber process is an important industrial reaction that converts N2 and H2 to ammonia(NH3), an agricultural fertilizer and starting material for the synthesis of nitrate fertilizers.Write a balanced equation for the Haber process.SAMPLE PROBLEM 5.4Balance the following equation.Ca3(PO4)2 calcium phosphateANALYSISH2SO4CaSO4sulfuric acid H3PO4calcium sulfate phosphoric acidBalance an equation with coefficients, one element at a time, beginning with the most complexformula and starting with an element that appears in only one formula on both sides of theequation. Continue placing coefficients until the number of atoms of each element is equal onboth sides of the equation.SOLUTION[1]Write the equation with correct formulas. The correct formula for each compound is given in the problem statement. When thereactants and products contain polyatomic ions, PO43– and SO42– in this case, balance eachion as a unit, rather than balancing the individual atoms. Thus, phosphate is not balancedin the equation as written, because the left side has two PO43– anions while the right sidehas only one.[2]Balance the equation with coefficients. Begin with Ca3(PO4)2. Balance the 3 Ca’s by placing the coefficient 3 before CaSO4.Balance the 2 PO43– anions by placing the coefficient 2 before H3PO4.Place a 3 to balance Ca’s.Ca3(PO4)2 H2SO43 CaSO4Place a 2 to balance PO4 2 H3PO43 . Two components are still not balanced—H atoms and sulfate anions (SO42–). Both can bebalanced by placing the coefficient 3 before H2SO4 on the left.6 H’sCa3(PO4)26 H’s3 H2SO4 3 CaSO4 2 H3PO43 SO42 in bothPlace a 3 to balance H and SO42 .[3]Ammonium hydrogen phosphate [(NH4)2HPO4], the majorphosphorus fertilizer, is formedfrom phosphoric acid, H3PO4,which is synthesized industrially by the chemical reaction inSample Problem 5.4.Check. The equation is balanced since the number of atoms and polyatomic anions is the same onboth sides.Answer:Ca3(PO4)2 3 H2SO4Atoms or ions in the reactants: 3 Ca’s 6 H’s 2 PO43 3 SO42 3 CaSO4 2 H3PO4Atoms or ions in the products: 3 Ca’s 2 PO43 6 H’s 3 SO42
CHEMICAL REACTIONS130PROBLEM 5.8Balance each chemical equation.a. Al H2SO4b. Na2SO3 H3PO4Al2(SO4)3 H2H2SO3 Na3PO45.3 THE MOLE AND AVOGADRO’S NUMBERAlthough the chemical equations in Section 5.2 were discussed in terms of individual atoms andmolecules, atoms are exceedingly small. It is more convenient to talk about larger quantities ofatoms, and for this reason, scientists use the mole. A mole defines a quantity, much like a dozenitems means 12, and a case of soda means 24 cans. The only difference is that a mole is muchlarger. A mole is a quantity that contains 6.02 1023 items—usually atoms, molecules, or ions.The definition of a mole is based on the number of atoms contained in exactly 12 g of the carbon-12isotope. This number is called Avogadro’s number, after the Italian scientist Amadeo Avogadro,who first proposed the concept of a mole in the nineteenth century. One mole, abbreviated as mol,always contains an Avogadro’s number of particles.1 mole of C atoms 1 mole of CO2 molecules 1 mole of H2O molecules 1 mole of vitamin C molecules PROBLEM 5.96.026.026.026.02 1023 C atoms1023 CO2 molecules1023 H2O molecules1023 vitamin C moleculesHow many items are contained in one mole of (a) baseballs; (b) bicycles; (c) Cheerios; (d) CH4molecules?We can use Avogadro’s number as a conversion factor to relate the number of moles of a substance to the number of atoms or molecules it contains.1 molTwo possible conversion factors:6.02 1023atomsor6.02 1023 atoms1 molThese conversion factors allow us to determine how many atoms or molecules are contained in agiven number of moles. To carry out calculations that contain numbers written in scientific notation, we must first learn how to multiply and divide numbers written in this form.Each sample contains one mole ofthe substance—water (H2O molecules), salt (NaCl, one mole of Na and one mole of Cl–), and aspirin(C9H8O4 molecules). Pictured is amole of aspirin molecules, not a moleof aspirin tablets, which is a quantitytoo large to easily represent. If a moleof aspirin tablets were arranged nextto one another to cover a footballfield and then stacked on top of eachother, they would occupy a volume100 yards long, 53.3 yards wide, andover 20,000,000,000 miles high! To multiply two numbers in scientific notation, multiply the coefficients together and addthe exponents in the powers of 10.Add exponents.(5 2)(3.0 105) (2.0 102) 6.0 107Multiply coefficients.(3.0 2.0) To divide two numbers in scientific notation, divide the coefficients and subtract theexponents in the powers of 10.Divide coefficients.(6.0/2.0)6.0 1022.0 1020Subtract exponents.(2 20) 3.0 10 18
THE MOLE AND AVOGADRO’S NUMBER131Sample Problems 5.5 and 5.6 illustrate how to interconvert moles and molecules. In both problems we follow the stepwise procedure for problem solving using conversion factors outlined inSection 1.7B.For a number written in scientificnotation as y 10x, y is thecoefficient and x is the exponentin the power of 10 (Section 1.6).SAMPLE PROBLEM 5.5How many molecules are contained in 5.0 moles of carbon dioxide (CO2)?ANALYSIS AND SOLUTION[1][2]Identify the original quantity and the desired quantity.5.0 mol of CO2? number of molecules of CO2original quantitydesired quantityWrite out the conversion factors. Choose the conversion factor that places the unwanted unit, mol, in the denominator so thatthe units cancel.1 mol6.02 1023moleculesor6.02 1023 molecules1 molChoose this conversion factor to cancel mol.[3]Set up and solve the problem. Multiply the original quantity by the conversion factor to obtain the desired quantity.Convert to a number between 1 and 10.5.0 mol 6.02 1023 molecules1 molMoles cancel. 30. 1023 molecules 3.0 1024 molecules of CO2Answer Multiplication first gives an answer that is not written in scientific notation since thecoefficient (30.) is greater than 10. Moving the decimal point one place to the left andincreasing the exponent by one gives the answer written in the proper form.PROBLEM 5.10How many carbon atoms are contained in each of the following number of moles: (a) 2.00 mol;(b) 6.00 mol; (c) 0.500 mol; (d) 25.0 mol?PROBLEM 5.11How many molecules are contained in each of the following number of moles?a. 2.5 mol of penicillin moleculesb. 0.25 mol of NH3 moleculesc. 0.40 mol of sugar moleculesd. 55.3 mol of acetaminophen moleculesHow many moles of aspirin contain 8.62 1025 molecules?SAMPLE PROBLEM 5.6ANALYSIS AND SOLUTION[1][2]Identify the original quantity and the desired quantity.8.62 1025 molecules of aspirin? mole of aspirinoriginal quantitydesired quantityWrite out the conversion factors. Choose the conversion factor that places the unwanted unit, number of molecules, in thedenominator so that the units cancel.
CHEMICAL REACTIONS1326.02 1023 molecules1 molor1 mol6.02 1023 moleculesChoose this conversion factor to cancel molecules.[3]Set up and solve the problem. Multiply the original quantity by the conversion factor to obtain the desired quantity. To divide numbers using scientific notation, divide the coefficients (8.62/6.02) and subtractthe exponents (25 – 23).8.62 1025 molecules 1 mol6.02 1023 moleculesMolecules cancel. 1.43 102 mol 143 mol of aspirinAnswerPROBLEM 5.12How many moles of water contain each of the following number of molecules?a. 6.02 1025 moleculesb. 3.01 1022 moleculesc. 9.0 1024 molecules5.4 MASS TO MOLE CONVERSIONSIn Section 2.3, we learned that the atomic weight is the average mass of an element, reportedin atomic mass units (amu), and that the atomic weight of each element appears just below itschemical symbol in the periodic table. Thus, carbon has an atomic weight of 12.01 amu. We useatomic weights to calculate the mass of a compound. The formula weight is the sum of the atomic weights of all the atoms in a compound,reported in atomic mass units (amu).The term “formula weight” is used for both ionic and covalent compounds. Often the term“molecular weight” is used in place of formula weight for covalent compounds, since theyare composed of molecules, not ions. The formula weight of ionic sodium chloride (NaCl) is58.44 amu, which is determined by adding up the atomic weights of Na (22.99 amu) and Cl(35.45 amu). The stepwise procedure for calculating the formula weight of compounds whosechemical formulas contain subscripts is shown in the accompanying How To.Formula weight of NaCl:Atomic weight of 1 Na 22.99 amuAtomic weight of 1 Cl 35.45 amuFormula weight of NaCl 58.44 amuHOW TOEXAMPLEStep [1]Calculate the Formula Weight of a CompoundCalculate the formula weight for iron(II) sulfate, FeSO4, an iron supplement used to treat anemia.Write the correct formula and determine the number of atoms of each element from the subscripts. FeSO4 contains 1 Fe atom, 1 S atom, and 4 O atoms.Step [2]Multiply the number of atoms of each element by the atomic weight and add the results.1 Fe atom 55.85 amu 55.85 amu1 S atom 32.07 amu 32.07 amu4 O atoms 16.00 amu 64.00 amuFormula weight of FeSO4 151.92 amu
MASS TO MOLE CONVERSIONSPROBLEM 5.13133Calculate the formula weight of each ionic compound.a. CaCO3, a common calcium supplementb. KI, the essential nutrient added to NaCl to make iodized saltPROBLEM 5.14Calculate the molecular weight of each covalent compound.Ha.HHHCCHHCOHb.HCCCCHethanol(alcohol in iseptic)halothane(general anesthetic)MOLAR MASSWhen reactions are carried out in the laboratory, single atoms and molecules are much too smallto measure out. Instead, substances are weighed on a balance and amounts are typically reportedin grams, not atomic mass units. To determine how many atoms or molecules are contained in agiven mass, we use its molar mass. The molar mass is the mass of one mole of any substance, reported in grams per mole.The value of the molar mass of an element in the periodic table (in grams per mole) is thesame as the value of its atomic weight (in amu). Thus, the molar mass of carbon is 12.01 g/mol,since its atomic weight is 12.01 amu; that is, one mole of carbon atoms weighs 12.01 g.6C12.01When a consumer product containsa great many lightweight smallobjects—for example, Cheerios—itis typically sold by weight, not by thenumber of objects. We buy Cheeriosin an 8.9-oz box, not a box thatcontains 2,554 Cheerios.SAMPLE PROBLEM 5.7ANALYSISSOLUTION Carbon’s atomic weight is 12.01 amu. Carbon’s molar mass is 12.01 g/mol. One mole of carbon atoms weighs 12.01 g. The value of the molar mass of a compound in grams equals the value of its formulaweight in amu.Since the formula weight of NaCl is 58.44 amu, its molar mass is 58.44 g/mol. One mole of NaClweighs 58.44 g. Since we know how to calculate a compound’s formula weight, we also knowhow to calculate its molar mass, as shown in Sample Problem 5.7.What is the molar mass of nicotine (C10H14N2), the toxic and addictive stimulant in tobacco?Determine the number of atoms of each element from the subscripts in the chemical formula,multiply the number of atoms of each element by the atomic weight, and add up the results.10 C atoms 12.01 amu14 H atoms 1.01 amu2 N atoms 14.01 amuFormula weight of nicotine 120.1 amu14.14 amu28.02 amu162.26 amu rounded to 162.3 amuAnswer: Since the formula weight of nicotine is 162.3 amu, the molar mass of nicotine is162.3 g/mol.
CHEMICAL REACTIONS134PROBLEM 5.15What is the molar mass of CaCO3 and KI, whose formula weights were calculated inProblem 5.13?PROBLEM 5.16What is the molar mass of each compound?a. Li2CO3 (lithium carbonate), a drug used to treat bipolar disorderb. C2H5Cl (ethyl chloride), a local anestheticc. C13H21NO3 (albuterol), a drug used to treat asthma5.4BRELATING GRAMS TO MOLESThe molar mass is a very useful quantity because it relates the number of moles to thenumber of grams of a substance. In this way, the molar mass can be used as a conversion factor.For example, since the molar mass of H2O is 18.0 g/mol, two conversion factors can be written.18.0 g H2O1 mol1 molor18.0 g H2OUsing these conversion factors, we can convert a given number of moles of water to grams, ora specific number of grams of water to moles. To solve problems of this sort, we once againreturn to the stepwise procedure for problem solving using conversion factors that was outlinedin Section 1.7B.SAMPLE PROBLEM 5.8Converting moles to mass: How many grams does 0.25 moles of water weigh?ANALYSIS AND SOLUTION[1][2]Identify the original quantity and the desired quantity.0.25 mol of
5.6 Mass Calculations in Chemical Equations 5.7 Percent Yield . The carbon–hydrogen bonds in methane and the oxygen–oxygen bond in elemental oxygen are broken, and new carbon–oxygen and hydrogen–oxygen bonds are . One molecule of liquid methyl acetate (C 3H6O2) reacts with
Text: General, Organic and Biological Chemistry, Janice G. Smith, 3rd ed, 2016, McGraw-Hill. Lab Text: Laboratory Manual for General, Organic and Biochemistry, Karen C. Timberlake, 3rd ed, 2014, Pearson' Labs: All 7 labs count towards your grade. No make-up labs. Late labs will incur a penalty. You MUST wear eye protection during lab!
meth-eth-prop-but-pent-6 8 7 9 10 hex-hept-oct-non-dec-5 3.2 Naming Organic Compounds Organic H 3C CH 2 CH 2 CH 3 H 2C CH2 H 2C CH CH 2 3.3 Naming Organic Compounds Organic H 3C CH CH CH2 H3C CH CH3 CH3 CH 3 CH 3 CH 3 3.4 Naming Organic Compounds Organic Name all
General Introduction to Organic Compounds Most of the foodstuffs that we consume every day such as sugar, fats, starch, vinegar, etc are basically organic compounds. Even though the . hydrocarbons and their derivatives is known as organic chemistry. Browse more Topics under Organic Chemistry Classification of Organic Compounds Isomerism
Lecture Text: REQUIRED Janice G. Smith, General, Organic and Biological Chemistry, 4th ed., 2018, McGraw-Hill. You can acquire a digital text for 30 for the course from the following link: Here is the link for purchasing the 30 ebook (Smith, 4th Ed Chem 30A-ISBN: 9781307601619- 30)
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Cardinal Health Clippers (JHI) 9/1/2018 12/31/2019 Cory Winn Cardinal Health Convertors (Drapes Gowns (JHI) 9/1/2018 12/31/2019 Janice Varney Cardinal Health Exam Gloves (JHI) 9/1/2018 12/31/2019 Janice Varney Cardinal Health Incontinence Care 12/1/2017 11/30/2020 Janice Varney Cardinal Health
We thank Janice Battaglia and the Social Committee for all their efforts over the past few years. The association surely has big shoes to fill after Janice's retirement from her role as Committee Chair. We appreciate that Janice and her Committee have documented their connections and processes for holding events and we look forward
Phone Janice Harvey (808) 974-2603 Mail Journey through the Universe, Janice Harvey, Gemini Observatory, 670 N. A'ohoku Place, Hilo, Hawaii Press Inquiries Janice Harvey, Gemini Observatory, (808) 974-2603, jharvey@gemini.edu Press Releases Journey Through The Universe in February 2009 - Taking Big Island Students and Teachers to the Stars and .
