Problem Solving Mole Concept Workbook

Name Class Date Skills Worksheet Problem Solving Mole Concept Suppose you want to carry out a reaction that requires combining one atom of iron with one atom of sulfur. How much iron should you use? How much sulfur? When you look around the lab, there is no device that can count numbers of atoms. Besides, the merest speck (0.001 g) of iron contains over a billion billion atoms. The same is true of sulfur. Fortunately, you do have a way to relate mass and numbers of atoms. One iron 23 atom has a mass of 55.847 amu, and 55.847 g of iron contains 6.022 137 10 atoms of iron. Likewise, 32.066 g of sulfur contains 6.022 137 10 23 atoms of sulfur. Knowing this, you can measure out 55.847 g of iron and 32.066 g of sulfur and be pretty certain that you have the same number of atoms of each. The number 6.022 137 10 23 is called Avogadro’s number. For most purposes it is rounded off to 6.022 10 23. Because this is an awkward number to write over and over again, chemists refer to it as a mole (abbreviated mol). 6.022 10 23 objects is called a mole, just as you call 12 objects a dozen. Look again at how these quantities are related. 55.847 g of iron 32.066 g of sulfur 6.022 6.022 10 10 23 23 iron atoms sulfur atoms 1 mol of iron 1 mol of sulfur General Plan for Converting Mass, Amount, and Numbers of Particles 1 Mass of substance Convert using the molar mass of the substance. 2 48 Amount of substance in moles 3 Use Avogadro's number for conversion. Number of atoms, molecules, or formula units of substance Mole Concept Name Class Date Problem Solving continued PROBLEMS INVOLVING ATOMS AND ELEMENTS Sample Problem 1 A chemist has a jar containing 388.2 g of iron filings. How many moles of iron does the jar contain? Solution ANALYZE What is given in the problem? mass of iron in grams What are you asked to find? amount of iron in moles Items Data Mass of iron 388.2 g Molar mass of iron* 55.85 g/mol Amount of iron ? mol * determined from the periodic table PLAN What step is needed to convert from grams of Fe to number of moles of Fe? The molar mass of iron can be used to convert mass of iron to amount of iron in moles. 1 Mass of Fe in g multiply by the inverse molar mass of Fe 2 Amount of Fe in mol given g Fe 1 molar mass Fe 1 mol Fe 55.85 g Fe mol Fe COMPUTE  388.2 g Fe 1 mol Fe 55.85 g Fe 6.951 mol Fe EVALUATE Are the units correct? Yes; the answer has the correct units of moles of Fe. 49 Mole Concept Name Class Date Problem Solving continued Is the number of significant figures correct? Yes; the number of significant figures is correct because there are four significant figures in the given value of 388.2 g Fe. Is the answer reasonable? Yes; 388.2 g Fe is about seven times the molar mass. Therefore, the sample contains about 7 mol. Practice 1. Calculate the number of moles in each of the following masses: a. 64.1 g of aluminum ans: 2.38 mol Al b. 28.1 g of silicon ans: 1.00 mol Si c. 0.255 g of sulfur ans: 7.95 ⴛ 10 ⴚ3 mol S d. 850.5 g of zinc ans: 13.01 mol Zn 50 Mole Concept Name Class Date Problem Solving continued Sample Problem 2 A student needs 0.366 mol of zinc for a reaction. What mass of zinc in grams should the student obtain? Solution ANALYZE What is given in the problem? amount of zinc needed in moles What are you asked to find? mass of zinc in grams Items Data Amount of zinc 0.366 mol Molar mass of zinc 65.39 g/mol Mass of zinc ?g PLAN What step is needed to convert from moles of Zn to grams of Zn? The molar mass of zinc can be used to convert amount of zinc to mass of zinc. 2 1 Amount of Zn in mol given mol Zn multiply by the molar mass of Zn Mass of Zn in mol molar mass Zn 65.39 g Zn 1 mol Zn g Zn COMPUTE 0.366 mol Zn 65.39 g Zn 1 mol Zn 23.9 g Zn EVALUATE Are the units correct? Yes; the answer has the correct units of grams of Zn. Is the number of significant figures correct? Yes; the number of significant figures is correct because there are three significant figures in the given value of 0.366 mol Zn. Is the answer reasonable? Yes; 0.366 mol is about 1/3 mol. 23.9 g is about 1/3 the molar mass of Zn. 51 Mole Concept Name Class Date Problem Solving continued Practice 1. Calculate the mass of each of the following amounts: a. 1.22 mol sodium ans: 28.0 g Na b. 14.5 mol copper ans: 921 g Cu c. 0.275 mol mercury ans: 55.2 g Hg d. 9.37 10 3 mol magnesium ans: 0.228 Mg 52 Mole Concept Name Class Date Problem Solving continued Sample Problem 3 24 How many moles of lithium are there in 1.204 ⴛ 10 lithium atoms? Solution ANALYZE What is given in the problem? number of lithium atoms What are you asked to find? amount of lithium in moles Items Data Number of lithium atoms 1.204 10 24 atoms Avogadro’s number—the number of atoms per mole 6.022 10 23 atoms/mol Amount of lithium ? mol PLAN What step is needed to convert from number of atoms of Li to moles of Li? Avogadro’s number is the number of atoms per mole of lithium and can be used to calculate the number of moles from the number of atoms. 3 2 Number of Li atoms multiply by the inverse of Avogadro's number 1 Avogadro's number given atoms Li Amount of Li in mol 6.022 1 mol Li 1023 atoms Li mol Li COMPUTE 1.204 10 24 atoms Li 6.022 1 mol Li 23 10 atoms Li 1.999 mol Li EVALUATE Are the units correct? Yes; the answer has the correct units of moles of Li. Is the number of significant figures correct? Yes; four significant figures is correct. Is the answer reasonable? 24 Yes; 1.204 ⴛ 10 is approximately twice Avogadro’s number. Therefore, it is reasonable that this number of atoms would equal about 2 mol. 53 Mole Concept Name Class Date Problem Solving continued Practice 1. Calculate the amount in moles in each of the following quantities: a. 3.01 10 23 atoms of rubidium ans: 0.500 mol Rb b. 8.08 10 22 atoms of krypton ans: 0.134 mol Kr c. 5 700 000 000 atoms of lead ans: 9.5 ⴛ 10 d. 2.997 ⴚ15 mol Pb 10 25 atoms of vanadium ans: 49.77 mol V 54 Mole Concept Name Class Date Problem Solving continued CONVERTING THE AMOUNT OF AN ELEMENT IN MOLES TO THE NUMBER OF ATOMS In Sample Problem 3, you were asked to determine the number of moles in 1.204 10 24 atoms of lithium. Had you been given the amount in moles and asked to calculate the number of atoms, you would have simply multiplied by Avogadro’s number. Steps 2 and 3 of the plan for solving Sample Problem 3 would have been reversed. Practice 1. Calculate the number of atoms in each of the following amounts: 23 a. 1.004 mol bismuth ans: 6.046 ⴛ 10 24 b. 2.5 mol manganese ans: 1.5 ⴛ 10 atoms Mg 17 c. 0.000 000 2 mol helium ans: 1 ⴛ 10 d. 32.6 mol strontium ans: 1.96 ⴛ 10 atoms Bi 25 55 atoms He atoms Sr Mole Concept Name Class Date Problem Solving continued Sample Problem 4 How many boron atoms are there in 2.00 g of boron? Solution ANALYZE What is given in the problem? mass of boron in grams What are you asked to find? number of boron atoms Items Data Mass of boron 2.00 g Molar mass of boron 10.81 g/mol Avogadro’s number—the number of boron atoms per mole of boron 6.022 Number of boron atoms ? atoms 10 23 atoms/mol PLAN What steps are needed to convert from grams of B to number of atoms of B? First, you must convert the mass of boron to moles of boron by using the molar mass of boron. Then you can use Avogadro’s number to convert amount in moles to number of atoms of boron. 1 3 Mass of B in g multiply by the inverse of the molar mass of boron given gB Number of B atoms Amount of B in mol 1 molar mass B 1 mol B 10.81 g B COMPUTE 2.00 g B multiply by Avogadro's number 2 1 mol B 10.81 g B 6.022 Avogadro's number 6.022 1023 atoms B 1 mol B 10 23 atoms B 1 mol B atoms B 1.11 10 23 atoms B EVALUATE Are the units correct? Yes; the answer has the correct units of atoms of boron. Is the number of significant figures correct? Yes; the mass of boron was given to three significant figures. 56 Mole Concept Name Class Date Problem Solving continued Is the answer reasonable? Yes; 2 g of boron is about 1/5 of the molar mass of boron. Therefore, 2.00 g boron will contain about 1/5 of an Avogadro’s constant of atoms. Practice 1. Calculate the number of atoms in each of the following masses: 24 a. 54.0 g of aluminum ans: 1.21 ⴛ 10 23 b. 69.45 g of lanthanum ans: 3.011 ⴛ 10 c. 0.697 g of gallium ans: 6.02 ⴛ 10 21 atoms La atoms Ga d. 0.000 000 020 g beryllium ans: 1.3 ⴛ 10 57 atoms Al 15 atoms Be Mole Concept Name Class Date Problem Solving continued CONVERTING NUMBER OF ATOMS OF AN ELEMENT TO MASS Sample Problem 4 uses the progression of steps 1 → 2 → 3 to convert from the mass of an element to the number of atoms. In order to calculate the mass from a given number of atoms, these steps will be reversed. The number of moles in the sample will be calculated. Then this value will be converted to the mass in grams. Practice 1. Calculate the mass of the following numbers of atoms: a. 6.022 b. 3.01 c. 1.506 d. 1.20 10 24 atoms of tantalum ans: 1810. g Ta 10 21 atoms of cobalt ans: 0.295 g Co 10 24 atoms of argon ans: 99.91 g Ar 10 25 atoms of helium ans: 79.7 g He 58 Mole Concept Name Class Date Problem Solving continued PROBLEMS INVOLVING MOLECULES, FORMULA UNITS, AND IONS How many water molecules are there in 200.0 g of water? What is the mass of 15.7 mol of nitrogen gas? Both of these substances consist of molecules, not single atoms. Look back at the diagram of the General Plan for Converting Mass, Amount, and Numbers of Particles. You can see that the same conversion methods can be used with molecular compounds and elements, such as CO 2 , H 2O, H2 SO4 , and O 2 . For example, 1 mol of water contains 6.022 10 23 H2O molecules. The mass of a molecule of water is the sum of the masses of two hydrogen atoms and one oxygen atom, and is equal to 18.02 amu. Therefore, 1 mol of water has a mass of 18.02 g. In the same way, you can relate amount, mass, and number of formula units for ionic compounds, such as NaCl, CaBr 2 , and Al 2(SO 4) 3. Sample Problem 5 How many moles of carbon dioxide are in 66.0 g of dry ice, which is solid CO 2 ? Solution ANALYZE What is given in the problem? mass of carbon dioxide What are you asked to find? amount of carbon dioxide Items Data Mass of CO2 66.0 g Molar mass of CO 2 Amount of CO 2 PLAN What step is needed to convert from grams of CO The molar mass of CO 2 44.0 g/mol ? mol 2 to moles of CO 2? can be used to convert mass of CO 1 2 to moles of CO 2. 2 Mass of CO2 in g multiply by the inverse of the molar mass of CO2 given g CO2 Amount of CO2 in mol 1 molar mass CO2 1 mol CO2 44.01 g CO2 mol CO2 1 mol CO 2 44.01 g CO2 1.50 mol CO 2 COMPUTE 66.0 g CO 2 59 Mole Concept Name Class Date Problem Solving continued EVALUATE Are the units correct? Yes; the answer has the correct units of moles CO 2. Is the number of significant figures correct? Yes; the number of significant figures is correct because the mass of CO given to three significant figures. 2 was Is the answer reasonable? Yes; 66 g is about 3/2 the value of the molar mass of CO the sample contains 3/2 (1.5) mol. 2. It is reasonable that Practice 1. Calculate the number of moles in each of the following masses: a. 3.00 g of boron tribromide, BBr 3 ans: 0.0120 mol BBr 3 b. 0.472 g of sodium fluoride, NaF ans: 0.0112 mol NaF c. 7.50 2 10 g of methanol, CH 3OH ans: 23.4 mol CH d. 50.0 g of calcium chlorate, Ca(ClO 3 )2 60 3OH ans: 0.242 mol Ca(ClO 3) 2 Mole Concept Name Class Date Problem Solving continued CONVERTING MOLES OF A COMPOUND TO MASS Perhaps you have noticed that Sample Problems 1 and 5 are very much alike. In each case, you multiplied the mass by the inverse of the molar mass to calculate the number of moles. The only difference in the two problems is that iron is an element and CO 2 is a compound containing a carbon atom and two oxygen atoms. In Sample Problem 2, you determined the mass of 1.366 mol of zinc. Suppose that you are now asked to determine the mass of 1.366 mol of the molecular compound ammonia, NH 3 . You can follow the same plan as you did in Sample Problem 2, but this time use the molar mass of ammonia. Practice 1. Determine the mass of each of the following amounts: a. 1.366 mol of NH 3 ans: 23.28 g NH 3 b. 0.120 mol of glucose, C 6H 12O6 ans: 21.6 g C c. 6.94 mol barium chloride, BaCl d. 0.005 mol of propane, C 3 H8 2 6 H 12 O 6 ans: 1.45 ⴛ 10 ans: 0.2 g C 61 3 g or 1.45 kg BaCl 2 3 H8 Mole Concept Name Class Date Problem Solving continued Sample Problem 6 Determine the number of molecules in 0.0500 mol of hexane, C 6 H 14 . Solution ANALYZE What is given in the problem? amount of hexane in moles What are you asked to find? number of molecules of hexane Items Data Amount of hexane 0.0500 mol Avogadro’s number—the number of molecules per mole of hexane Molecules of hexane 6.022 10 23 molecules/mol ? molecules PLAN What step is needed to convert from moles of C C6H 14? 6H14 to number of molecules of Avogadro’s number is the number of molecules per mole of hexane and can be used to calculate the number of molecules from number of moles. 2 3 Amount of C6H14 in mol given mol C6H14 6.022 COMPUTE 0.0500 mol C 6H 14 multiply by Avogadro's number 6.022 Number of C6H14 molecules Avogadro's number 1023 molecules C6H14 1 mol C6H14 molecules C6H14 23 10 molecules C 6 H14 1 mol C 6H14 3.01 10 22 molecules C 6H 14 EVALUATE Are the units correct? Yes; the answer has the correct units of molecules of C 6H 14 . Is the number of significant figures correct? Yes; three significant figures is correct. Is the answer reasonable? Yes; multiplying Avogadro’s number by 0.05 would yield a product that is a factor 22 of 10 less with a value of 3 ⴛ 10 . 62 Mole Concept Name Class Date Problem Solving continued Practice 1. Calculate the number of molecules in each of the following amounts: a. 4.99 mol of methane, CH 4 24 ans: 3.00 ⴛ 10 b. 0.005 20 mol of nitrogen gas, N 2 molecules CH ans: 3.13 ⴛ 10 c. 1.05 mol of phosphorus trichloride, PCl 3 molecules N ans: 6.32 ⴛ 10 d. 3.5 10 5 mol of vitamin C, ascorbic acid, C cules C 6H 8 O 6 63 21 4 6 H8 O6 23 2 molecules PCl ans: 2.1 ⴛ 10 19 3 mole- Mole Concept Name Class Date Problem Solving continued USING FORMULA UNITS OF IONIC COMPOUNDS Ionic compounds do not exist as molecules. A crystal of sodium chloride, for example, consists of Na ions and Cl ions in a 1:1 ratio. Chemists refer to a combination of one Na ion and one Cl ion as one formula unit of NaCl. A mole of an ionic compound consists of 6.022 10 23 formula units. The mass of one formula unit is called the formula mass. This mass is used in the same way atomic mass or molecular mass is used in calculations. Practice 1. Calculate the number of formula units in the following amounts: a. 1.25 mol of potassium bromide, KBr ans: 7.53 ⴛ 10 2 ans: 3.01 ⴛ 10 2CO 3 ans: 1.5 ⴛ 10 b. 5.00 mol of magnesium chloride, MgCl MgCl 2 c. 0.025 mol of sodium carbonate, Na Na 2CO 3 d. 6.82 10 6 mol of lead(II) nitrate, Pb(NO units Pb(NO 3 ) 2 64 23 3) 2 formula units KBr 24 22 formula units formula units ans: 4.11 ⴛ 10 18 formula Mole Concept Name Class Date Problem Solving continued CONVERTING NUMBER OF MOLECULES OR FORMULA UNITS TO AMOUNT IN MOLES In Sample Problem 3, you determined the amount in moles of the element lithium. Suppose that you are asked to determine the amount in moles of copper(II) hydroxide in 3.34 10 34 formula units of Cu(OH) 2. You can follow the same plan as you did in Sample Problem 3. Practice 1. Calculate the amount in moles of the following numbers of molecules or formula units: a. 3.34 10 34 formula units of Cu(OH) b. 1.17 10 16 molecules of H 2S ans: 1.94 ⴛ 10 2 ans: 5.55 ⴛ 10 ⴚ8 65 mol Cu(OH) 2 mol H 2 S c. 5.47 10 21 formula units of nickel(II) sulfate, NiSO NiSO 4 d. 7.66 10 19 molecules of hydrogen peroxide, H H 2 O2 10 2O 2 4 ans: 9.08 ⴛ 10 ans: 1.27 ⴛ 10 ⴚ4 ⴚ3 mol mol Mole Concept Name Class Date Problem Solving continued Sample Problem 7 22 What is the mass of a sample consisting of 1.00 ⴛ 10 MgSO 4 ? formula units of Solution ANALYZE What is given in the problem? number of magnesium sulfate formula units What are you asked to find? mass of magnesium sulfate in grams Items Data Number of formula units of magnesium sulfate 1.00 10 22 Avogadro’s number—the number of formula units of magnesium sulfate per mole 6.022 10 Molar mass of magnesium sulfate 120.37 g/mol Mass of magnesium sulfate ?g formula units 23 PLAN What steps are needed to convert from formula units of MgSO MgSO4? formula units/mol 4 to grams of First, you must convert the number of formula units of MgSO 4 to amount of MgSO 4 by using Avogadro’s number. Then you can use the molar mass of MgSO to convert amount in moles to mass of MgSO 4. 3 1 Number of MgSO4 formula units Mass of MgSO4 in g multiply by the inverse of Avogadro's number multiply by the molar mass of MgSO4 2 Amount of MgSO4 in mol 1 Avogadro's number given formula units MgSO4 6.022 1 mol MgSO4 1023 formula units MgSO4 molar mass MgSO4 120.37 g MgSO4 1 mol MgSO4 66 g MgSO4 Mole Concept 4 Name Class Date Problem Solving continued COMPUTE 1.00 10 22 formula units MgSO 4 6.022 1 mol MgSO4 23 10 formula units MgSO 120.37 g MgSO4 1 mol MgSO 4 4 2.00 g MgSO4 EVALUATE Are the units correct? Yes; the answer has the correct units of grams of MgSO 4. Is the number of significant figures correct? Yes; the number of significant figures is correct because data were given to three significant figures. Is the answer reasonable? Yes; 2 g of MgSO 4 is about 1/60 of the molar mass of MgSO 4 . Therefore, 2.00 g MgSO 4 will contain about 1/60 of an Avogadro’s number of formula units. Practice 1. Calculate the mass of each of the following quantities: a. 2.41 10 24 molecules of hydrogen, H 2 ans: 8.08 g H 2 b. 5.00 10 21 formula units of aluminum hydroxide, Al(OH) Al(OH) 3 c. 8.25 10 22 molecules of bromine pentafluoride, BrF d. 1.20 10 23 formula units of sodium oxalate, Na 67 3 ans: 0.648 g 5 ans: 24.0 g BrF 2 C 2 O4 ans: 26.7 g Na 5 2 C 2 O4 Mole Concept Name Class Date Problem Solving continued CONVERTING MOLECULES OR FORMULA UNITS OF A COMPOUND TO MASS In Sample Problem 4, you converted a given mass of boron to the number of boron atoms present in the sample. You can now apply the same method to convert mass of an ionic or molecular compound to numbers of molecules or formula units. Practice 1. Calculate the number of molecules or formula units in each of the following masses: a. 22.9 g of sodium sulfide, Na 2S ans: 1.77 ⴛ 10 23 ans: 8.96 ⴛ 10 b. 0.272 g of nickel(II) nitrate, Ni(NO Ni(NO 3 ) 2 3) 2 c. 260 mg of acrylonitrile, CH ans: 3.0 ⴛ 10 2CHCN 68 formula units Na 21 20 2S formula units molecules CH 2CHCN Mole Concept Name Class Date Problem Solving continued Additional Problems 1. Calculate the number of moles in each of the following masses: a. 0.039 g of palladium b. 8200 g of iron c. 0.0073 kg of tantalum d. 0.006 55 g of antimony e. 5.64 kg of barium f. 3.37 10 6 g of molybdenum 2. Calculate the mass in grams of each of the following amounts: a. 1.002 mol of chromium b. 550 mol of aluminum c. 4.08 10 8 mol of neon d. 7 mol of titanium e. 0.0086 mol of xenon 10 4 mol of lithium f. 3.29 3. Calculate the number of atoms in each of the following amounts: a. 17.0 mol of germanium b. 0.6144 mol of copper c. 3.02 mol of tin 10 6 mol of carbon d. 2.0 e. 0.0019 mol of zirconium f. 3.227 10 10 mol of potassium 4. Calculate the number of moles in each of the following quantities: a. 6.022 10 24 atoms of cobalt b. 1.06 10 23 atoms of tungsten c. 3.008 10 19 atoms of silver d. 950 000 000 atoms of plutonium e. 4.61 10 17 atoms of radon f. 8 trillion atoms of cerium 5. Calculate the number of atoms in each of the following masses: a. 0.0082 g of gold b. 812 g of molybdenum c. 2.00 10 2 mg of americium d. 10.09 kg of neon e. 0.705 mg of bismuth f. 37 69 g of uranium Mole Concept Name Class Date Problem Solving continued 6. Calculate the mass of each of the following: a. 8.22 10 23 atoms of rubidium b. 4.05 Avogadro’s numbers of manganese atoms c. 9.96 10 26 atoms of tellurium d. 0.000 025 Avogadro’s numbers of rhodium atoms e. 88 300 000 000 000 atoms of radium f. 2.94 10 17 atoms of hafnium 7. Calculate the number of moles in each of the following masses: a. 45.0 g of acetic acid, CH 3COOH b. 7.04 g of lead(II) nitrate, Pb(NO c. 5000 kg of iron(III) oxide, Fe d. 12.0 mg of ethylamine, C 3 )2 2 O3 2H5 NH2 e. 0.003 22 g of stearic acid, C 17 H35COOH f. 50.0 kg of ammonium sulfate, (NH 4 ) 2 SO4 8. Calculate the mass of each of the following amounts: a. 3.00 mol of selenium oxybromide, SeOBr b. 488 mol of calcium carbonate, CaCO c. 0.0091 mol of retinoic acid, C d. 6.00 8 10 3 20H 28O 2 mol of nicotine, C 10H14N 2 e. 2.50 mol of strontium nitrate, Sr(NO f. 3.50 6 10 2 3 )2 mol of uranium hexafluoride, UF 6 9.C alculate the number of molecules or formula units in each of the following amounts: a.4 .27 mol of tungsten(VI) oxide, WO 3 b.0 .003 00 mol of strontium nitrate, Sr(NO 3 )2 c.7 2.5 mol of toluene, C 6H5 CH3 d.5 .11 7 10 mol of -tocopherol (vitamin E), C 29 H50 O2 e.1 500 mol of hydrazine, N 2H4 f.0 .989 mol of nitrobenzene C 6H 5NO 2 10. Calculate the number of molecules or formula units in each of the following masses: a.2 85 g of iron(III) phosphate, FePO 4 b. 0.0084 g of C 5H5 N c. 85 mg of 2-methyl-1-propanol, (CH d. 4.6 10 4 e. 0.0067 g of lithium carbonate, Li 70 3 ) 2 CHCH2OH g of mercury(II) acetate, Hg(C 2 H3 O2 )2 2CO 3 Mole Concept   Name Class Date Problem Solving continued 11. Calculate the mass of each of the following quantities: a. 8.39 b. 6.82 10 23 molecules of fluorine, F 10 24 formula units of beryllium sulfate, BeSO c. 7.004 10 26 2 molecules of chloroform, CHCl 4 3 d. 31 billion formula units of chromium(III) formate, Cr(CHO e. 6.3 f. 8.37 10 18 10 molecules of nitric acid, HNO 25 molecules of freon 114, C 2) 3 3 2 Cl2F 4 12. Precious metals are commonly measured in troy ounces. A troy ounce is equivalent to 31.1 g. How many moles are in a troy ounce of gold? How many moles are in a troy ounce of platinum? of silver? 13. A chemist needs 22.0 g of phenol, C moles of phenol is this? 6H 5OH, for an experiment. How many 14. A student needs 0.015 mol of iodine crystals, I of iodine crystals should the student obtain? 2, for an experiment. What mass 15. The weight of a diamond is given in carats. One carat is equivalent to 200. mg. A pure diamond is made up entirely of carbon atoms. How many carbon atoms make up a 1.00 carat diamond? 16. 8.00 g of calcium chloride, CaCl a. How many moles of CaCl present? 2, is dissolved in 1.000 kg of water. 2 are in solution? How many moles of water are b. Assume that the ionic compound, CaCl 2 , separates completely into Ca and Cl ions when it dissolves in water. How many moles of each ion are present in the solution? 2 17. How many moles are in each of the following masses? a. 453.6 g (1.000 pound) of sucrose (table sugar), C 12 H22O11 b. 1.000 pound of table salt, NaCl 18. When the ionic compound NH 4 Cl dissolves in water, it breaks into one ammonium ion, NH 4 , and one chloride ion, Cl . If you dissolved 10.7 g of NH 4 Cl in water, how many moles of ions would be in solution? 19. What is the total amount in moles of atoms in a jar that contains 2.41 10 23 23 atoms of chromium, 1.51 10 atoms of nickel, and 3.01 10 atoms of copper? 24 20. The density of liquid water is 0.997 g/mL at 25°C. a. Calculate the mass of 250.0 mL (about a cupful) of water. b. How many moles of water are in 250.0 mL of water? Hint: Use the result of (a). c. Calculate the volume that would be occupied by 2.000 mol of water at 25°C. d. What mass of water is 2.000 mol of water? 71 Mole Concept Name Class Date Problem Solving continued 21. An Avogadro’s number (1 mol) of sugar molecules has a mass of 342 g, but an Avogadro’s number (1 mol) of water molecules has a mass of only 18 g. Explain why there is such a difference between the mass of 1 mol of sugar and the mass of 1 mol of water. 22. Calculate the mass of aluminum that would have the same number of atoms as 6.35 g of cadmium. 23. A chemist weighs a steel cylinder of compressed oxygen, O 2 , and finds that it has a mass of 1027.8 g. After some of the oxygen is used in an experiment, the cylinder has a mass of 1023.2 g. How many moles of oxygen gas are used in the experiment? 24. Suppose that you could decompose 0.250 mol of Ag 2S into its elements. a. How many moles of silver would you have? How many moles of sulfur would you have? b. How many moles of Ag 2 S are there in 38.8 g of Ag 2 S? How many moles of silver and sulfur would be produced from this amount of Ag 2 S? c. Calculate the masses of silver and sulfur produced in (b). 72 Mole Concept ␮ 4 3 c. 60.4 kg; 1.88 ⫻ 10 dm 3 ⫺4 3 d. 0.94 g/cm ; 5.3 ⫻ 10 m 3 6 3 e. 2.5 ⫻ 10 kg; 2.7 ⫻ 10 cm 3 7. 2.8 g/cm 8. a. 0.72 m 3 b. 2.5 ⫻ 10 atoms 9. 1300 L/min 6 10. 1.3 ⫻ 10 cal/h 3 11. 5.44 g/ cm 4 3 12. 2.24 ⫻ 10 cm 13. 32 000 uses 14. 2500 L 15. 9.5 L/min MOLE CONCEPT ⫺4 mol Pd 1. a. 3.7 ⫻ 10 b. 150 mol Fe c. 0.040 mol Ta ⫺5 mol Sb d. 5.38 ⫻ 10 e. 41.1 mol Ba ⫺8 f. 3.51 ⫻ 10 mol Mo 2. a. 52.10 g Cr 4 b. 1.5 ⫻ 10 g or 15 kg Al ⫺7 c. 8.23 ⫻ 10 g Ne 2 d. 3 ⫻ 10 g or 0.3 kg Ti e. 1.1 g Xe 5 f. 2.28 ⫻ 10 g or 228 kg Li 25 3. a. 1.02 ⫻ 10 atoms Ge 23 b. 3.700 ⫻ 10 atoms Cu 24 c. 1.82 ⫻ 10 atoms Sn 30 d. 1.2 ⫻ 10 atoms C 21 e. 1.1 ⫻ 10 atoms Zr 14 f. 1.943 ⫻ 10 atoms K 4. a. 10.00 mol Co b. 0.176 mol W ⫺5 c. 4.995 ⫻ 10 mol Ag ⫺15 mol Pu d. 1.6 ⫻ 10 ⫺7 mol Rn e. 7.66 ⫻ 10 ⫺11 mol Ce f. 1 ⫻ 10 19 5. a. 2.5 ⫻ 10 atoms Au 24 b. 5.10 ⫻ 10 atoms Mo 20 c. 4.96 ⫻ 10 atoms Am 26 d. 3.011 ⫻ 10 atoms Ne 18 e. 2.03 ⫻ 10 atoms Bi 16 f. 9.4 ⫻ 10 atoms U 6. a. 117 g Rb b. 223 g Mn 5 c. 2.11 ⫻ 10 g Te ⫺3 g Rh d. 2.6 ⫻ 10 ⫺8 g Ra e. 3.31 ⫻ 10 ⫺5 f. 8.71 ⫻ 10 g Hf 7. a. 0.749 mol CH 3 COOH b. 0.0213 mol Pb(NO 3 )2 324 4 c. 3 ⫻ 10 mol Fe 2 O3 ⫺4 mol C 2 H 5NH2 d. 2.66 ⫻ 10 ⫺5 mol C 17H 35 COOH e. 1.13 ⫻ 10 f. 378 mol (NH 4) 2SO4 8. a. 764 g SeOBr 2 4 b. 4.88 ⫻ 10 g CaCO3 c. 2.7 g C 20H28 O 2 ⫺6 g C10 H14 N 2 d. 9.74 ⫻ 10 e. 529 g Sr(NO 3 ) 2 ⫺3 f. 1.23 ⫻ 10 g UF 6 24 9. a. 2.57 ⫻ 10 formula units WO 3 21 b. 1.81 ⫻ 10 formula units Sr(NO 3)2 25 c. 4.37 ⫻ 10 molecules C 6H 5CH3 17 d. 3.08 ⫻ 10 molecules C 29H50 O2 26 e. 9.0 ⫻ 10 molecules N 2H4 23 f. 5.96 ⫻ 10 molecules C 6H 5NO 2 24 formula units FePO 4 10. a. 1.14 ⫻ 10 19 b. 6.4 ⫻ 10 molecules C 5H 5N 20 c. 6.9 ⫻ 10 molecules (CH 3) 2CHCH2OH 17 formula units Hg(C2 H3O 2)2 19 e. 5.5 ⫻ 10 formula units Li 2CO3 11. a. 52.9 g F 2 3 b. 1.19 ⫻ 10 g or 1.19 kg BeSO 4 5 c. 1.388 ⫻ 10 g or 138.8 kg CHCl 3 ⫺12 g Cr(CHO 2 )3 d. 9.6 ⫻ 10 ⫺4 g HNO3 e. 6.6 ⫻ 10 4 f. 2.38 ⫻ 10 g or 23.8 kg C 2 Cl2 F 4 12. 0.158 mol Au 0.159 mol Pt 0.288 mol Ag 13.0.234 mol C 6 H5OH 14. 3.8 g I 2 22 atoms C 15. 1.00 ⫻ 10 16. a. 0.0721 mol CaCl 2 55.49 mol H 2 O 2⫹ b. 0.0721 mol Ca 0.144 mol Cl ⫺ 17. a. 1.325 mol C 12H22 O 11 b. 7.762 mol NaCl 18. 0.400 mol ions 19. 4.75 mol atoms 20. a. 249 g H 2O b. 13.8 mol H 2 O c. 36.1 mL H 2O d. 36.0 g H 2O 21. The mass of a sugar molecule is much greater than the mass of a water molecule. Therefore, the mass of 1 mol of sugar molecules is much greater than the mass of 1 mol of water molecules. 22. 1.52 g Al d. 8.7 ⫻ 10 Answer Key 23. 0.14 mol O 2 24. a. 0.500 mol Ag 0.250 mol S b. 0.157 mol Ag 2 S 0.313 mol Ag 0.157 mol S c. 33.8 g Ag 5.03 g S PERCENTAGE COMPOSITION 1. a. HNO 3 1.60% H 22.23% N 76.17% O b. NH 3 82.22% N 17.78% H c. HgSO 4 67.616% Hg 10.81% S 21.57% O d. SbF 5 56.173% Sb 43.83% F 2. a. 7.99% Li 92.01% Br b. 94.33% C 5.67% H c. 35.00% N 5.05% H 59.96% O d. 2.15% H 29.80% N 68.06% O e. 87.059% Ag 12.94% S f. 32.47% Fe 13.96% C 16.29% N 37.28% S g. LiC2 H3 O 2 10.52% Li 36.40% C 4.59% H 48.49% O h. Ni(CHO 2 )2 39.46% Ni 16.15% C 1.36% H 43.03% O 3. a. 46.65% N b. 23.76% S c. 89.491% Tl 325 d. 39.17% O e. 79.95% Br in CaBr 2 f. 78.767% Sn in SnO 2 4. a. 1.47 g O b. 26.5 metric tons Al c. 262 g Ag d. 0.487 g Au e. 312 g Se 4 f. 3.1 ⫻ 10 g Cl 5. a. 40.55% H 2O b. 43.86% H2O c. 20.70% H2O d. 28.90% H2O 6. a. Ni(C 2H 3 O2) 2ⴢ4H2O 23.58% Ni b. Na 2CrO4ⴢ4H2O 22.22% Cr c. Ce(SO 4) 2 ⴢ4H2O 34.65% Ce 7. 43.1 kg Hg 8. malachite: 5.75 ⫻ 10 2 kg Cu 2 chalcopyrite: 3.46 ⫻ 10 kg Cu malachite has a greater Cu content 9. a. 25.59% V b. 39.71% Sn c. 22.22% Cl 10. 319.6 g anhydrous CuSO 4 11. 1.57 g AgNO 3 12. 54.3 g Ag 8.08 g S 13. 23.1 g MgSO 4ⴢ7H2O 2 14. 3.27 ⫻ 10 g S EMPIRICAL FORMULAS 1. a. BaCl 2 b. BiO 3H3 or Bi(OH) 3 c. AlN3O9 or Al(NO 3 )3 d. ZnC 4H6O 4 or Zn(CH 3 COO) 2 e. NiN 2S2H 8 O 8 or Ni(NH 4) 2SO 4 f. C2 HBr3O2 or CBr 3COOH 2. a. CuF 2 b. Ba(CN) 2 c. MnSO4 3. a. NiI 2 b. MgN2O 6 or Mg(NO3) 2 c. MgS 2O 3, magnesium thiosulfate d. K2 SnO3, potassium stannate 4. a. As 2S3 b. Re 2O7 c. N2 H4O 3 or NH4 NO3 d. Fe 2Cr 3O12 or Fe 2 (CrO 4 )3 e. C 5 H9N3 f. C6 H5F 2N or C6 H3 F2NH 2 Answer Key