The Mole and Chemical Table of Contents Chemical Formulas Bellringer List as many common counting units as you can. Determine how many groups of each unit in your list are present in each of the following amounts: 500 goldfish 150 unicycles 50 jet planes Objectives Identify the mole as the unit used to count particles, whether atoms, ions, or molecules. Use Avogadro s number to convert between amount in moles and number of particles. Solve problems converting between mass, amount in moles, and number of particles using Avogadro s number and molar mass. Avogadro s Number and the Mole The SI unit for amount is called the mole (mol). A mole is the number of atoms in exactly 12 grams of carbon-12. Scientists use the mole to make counting large numbers of particles easier. The number of particles in a mole is called Avogadro s Number. Avogadro s number is 6.02214199 10 23 units/mole. Visual Concepts Avogadro s Number and the Mole, continued The Mole Is a Counting Unit The mole is used to count out a given number of particles, whether they are atoms, molecules, formula units, ions, or electrons. The Mole The mole is just one kind of counting unit: 1 dozen = 12 objects 1 hour = 3600 seconds 1 mole = 6.022 10 23 particles
Avogadro s Number and the Mole, continued Amount in Moles Can Be Converted to Number of Particles Counting units are used to make conversion factors. The definition of one mole is The conversion factor is 6.022 10 23 particles = 1 mol 23 6.022 10 particles 1mol = 1 Avogadro s Number and the Mole, continued Choose the Conversion Factor That Cancels the Given Units All conversion factors are equal to 1, so you can use them to convert among different units. You can tell which conversion factor to use, because the needed conversion factor should cancel the units of the given quantity to give you the units of the answer or the unknown quantity. Converting Between Amount in Moles and Number of Particles Converting Amount in Moles to Number of Particles Sample Problem A Find the number of molecules in 2.5 mol of sulfur dioxide. Converting Amount in Moles to Number of Particles Sample Problem A Solution 2.5 mol SO 2? =? molecules SO 2 You are converting from the unit mol to the unit molecules. The conversion factor must have the units of molecules/mol. You use 6.022 10 23 molecules/1 mol. Converting Amount in Moles to Number of Particles Sample Problem A Solution, continued 2.5 mol SO 2? =? molecules SO 2 23 6.022 10 molecules SO2 2.5 mol SO =? molecules SO 1 mol SO 2 2 2 2.5 mol SO 2 = 1.5 10 24 molecules SO 2
Avogadro s Number and the Mole, continued Number of Particles Can Be Converted to Amount in Moles The reverse calculation is similar to that in Sample Problem A but the conversion factor is inverted to get the correct units in the answer. example: How many moles are 2.54 10 22 iron(iii) ions? Converting Number of Particles to Amount in Moles Sample Problem B A sample contains 3.01 10 23 molecules of sulfur dioxide, SO 2. Determine the amount in moles. 2.54 10 ions Fe 3+ 1molFe 23 6.022 10 ions Fe 0.0422 mol Fe + = 3 22 3 3 + + Converting Number of Particles to Amount in Moles Sample Problem B Solution Converting Number of Particles to Amount in Moles Sample Problem B Solution, continued 3.01 10 23 molecules SO 2? =? mol SO 2 You are converting from the unit molecules to the unit mol. The conversion factor must have the units of mol/molecules. You use 1 mol/6.022 10 23 molecules. 3.01 10 23 molecules SO 2? =? mol SO 2 3.01 10 molecules SO 1molSO2 23 6.022 10 molecules SO =? mol SO 23 2 2 3.01 10 23 molecules SO 2 = 0.500 mol SO 2 2 Visual Concepts Molar Mass Relates Moles to Grams Amount in Moles Can Be Converted to Mass The molar mass is the mass in grams of one mole of an element or compound. Molar Mass Molar mass is numerically equal to the atomic mass of monatomic elements and the formula mass of compounds and diatomic elements. The units for molar mass are g/mol. Molar mass can be used as a conversion factor in problems converting between mass and amount.
Visual Concepts Molar Mass as a Conversion Factor Molar Mass Relates Moles to Grams, continued The Mole Plays a Central Part in Chemical Conversions To convert from number of particles to mass, you must use a two-part process: First, convert number of particles to amount in moles. Second, convert amount in moles to mass in grams. One step common to many problems in chemistry is converting to amount in moles. Converting Between Mass, Amount, and Number of Particles Converting Number of Particles to Mass Sample Problem C Find the mass in grams of 2.44 10 24 atoms of carbon, whose molar mass is 12.01 g/mol. Converting Number of Particles to Mass Sample Problem C Solution First part: 2.44 10 24 atoms? =? mol Select the conversion factor that will take you from number of atoms to amount in moles. You use 1 mol/6.022 10 23 atoms. Converting Number of Particles to Mass Sample Problem C Solution, continued Second part:? mol? =? g Select the conversion factor that will take you from amount in moles to mass in grams. You use the molar mass of carbon, 12.01 g C/1 mol.
Converting Number of Particles to Mass Sample Problem C Solution, continued 24 1 mol 12.01 g C 2.44 10 atoms 23 6.022 10 atoms 1 mol = 48.7 g C Molar Mass Relates Moles to Grams, continued Mass Can Be Converted to Amount in Moles Converting from mass to number of particles is the reverse of the operation in the previous problem. To convert from mass to number of particles, you must use a two-part process: First, convert mass in grams to amount in moles. Second, convert amount in moles to number of particles. Converting Mass to Number of Particles Sample Problem D Find the number of molecules present in 47.5 g of glycerol, C 3 H 8 O 3. The molar mass of glycerol is 92.11 g/mol. Converting Mass to Number of Particles Sample Problem D Solution First part: 47.5 g? =? mol Select the conversion factor that will take you from mass in grams to amount in moles. You use the inverse of the molar mass of glycerol: 1mol 92.11 g C H O 3 8 3 Converting Mass to Number of Particles Sample Problem D Solution, continued Second part:? mol? =? molecules Select the conversion factor that will take you from amount in moles to number of particles. 23 6.022 10 molecules You use 1mol. Converting Mass to Number of Particles Sample Problem D Solution, continued 23 1 mol 6.022 10 molecules 47.5 g C3H8O3 92.11 g C 3 H 8 O 3 1 mol = 3.11 10 23 molecules
Molar Mass Relates Moles to Grams, continued Remember to Round Consistently Remember that an answer must never be given to more significant figures than is appropriate. Round molar masses from the periodic table to two significant figures to the right of the decimal point. Bellringer Compare the masses of a roll of pennies and a roll of dimes. Both contain 50 coins. Why are the masses of the rolls different when both rolls contain the same number of coins? If given a roll of mixed coins, what information would you need to determine the mass of the roll? Objectives Use a periodic table or isotopic composition data to determine the average atomic masses of elements. Infer information about a compound from its chemical formula. Determine the molar mass of a compound from its formula. Average Atomic Mass and the Periodic Table Most Elements are a Mixture of Isotopes Isotopes are atoms that have different numbers of neutrons than other atoms of the same element do. Average atomic mass is a weighted average of the atomic mass of an element s isotopes. If you know the abundance of each isotope, you can calculate the average atomic mass of an element. Visual Concepts Average Atomic Mass Calculating Average Atomic Mass Sample Problem E The mass of a Cu-63 atom is 62.94 amu, and that of a Cu-65 atom is 64.93 amu. Using the data below, find the average atomic mass of copper. abundance of Cu-63 = 69.17% abundance of Cu-65 = 30.83%
Calculating Average Atomic Mass Sample Problem E Solution The contribution of each isotope is equal to its atomic mass multiplied by the fraction of that isotope. contribution of Cu-63: 62.94 amu 0.6917 contribution of Cu-65: 64.93 amu 0.3083 Average atomic mass is the sum of the individual contributions: (62.94 amu 0.6917) + (64.93 amu 0.3083) = 63.55 amu and Moles Formulas Express A compound s chemical formula tells you which elements, as well as how much of each, are present in a compound. Formulas for covalent compounds show the elements and the number of atoms of each element in a molecule. Formulas for ionic compounds show the simplest ratio of cations and anions in any pure sample. and Moles, continued Formulas Express, continued Any sample of compound has many atoms and ions, and the formula gives a ratio of those atoms or ions. and Moles, continued Formulas Give Ratios of Polyatomic Ions Formulas for polyatomic ions show the simplest ratio of cations and anions. They also show the elements and the number of atoms of each element in each ion. For example, the formula KNO 3 indicates a ratio of one K + cation to one NO 3 anion. Understanding Formulas for Polyatomic Ionic Compounds and Moles, continued Formulas Are Used to Calculate Molar Masses The molar mass of a molecular compound is the sum of the masses of all the atoms in the formula expressed in g/mol. The molar mass of an ionic compound is the sum of the masses of all the atoms in the formula expressed in g/mol.
Visual Concepts Formula Mass Calculating Molar Mass for Ionic Compounds Calculating Molar Mass of Compounds Sample Problem F Find the molar mass of barium nitrate, Ba(NO 3 ) 2. Calculating Molar Mass of Compounds Sample Problem F Solution Find the # of moles of each element in 1 mol Ba(NO 3 ) 2 : Each mole has 1 mol Ba, 2 mol N, and 6 mol O. Use the periodic table to find the molar mass of each element in the formula: molar mass of Ba: 137.33 g/mol; molar mass of N: 14.01 g/mol and molar mass of O: 16.00 g/mol Calculating Molar Mass of Compounds Sample Problem F Solution, continued Multiply the molar mass of each element by the number of moles of each element. Add these masses to get the total molar mass of Ba(NO 3 ) 2. mass of 1 mol Ba = 1 137.33 g/mol = 137.33 g/mol mass of 2 mol N = 2 14.01 g/mol = 28.02 g/mol + mass of 6 mol O = 6 16.00 g/mol = 96.00 g/mol Bellringer Brainstorm a list of what you know about percentages. molar mass of Ba(NO 3 ) 2 = 261.35 g/mol
Objectives Determine a compound s empirical formula from its percentage composition. Determine the molecular formula or formula unit of a compound from its empirical formula and its formula mass. Calculate percentage composition of a compound from its molecular formula or formula unit. Using Analytical Data The percentage composition is the percentage by mass of each element in a compound. Percentage composition helps verify a substance s identity. Percentage composition also can be used to compare the ratio of masses contributed by the elements in two different substances. Percentage of Iron Oxides Using Analytical Data, continued Determining Empirical Formulas An empirical formula is a chemical formula that shows the simplest ratio for the relative numbers and kinds of atoms in a compound. An actual formula shows the actual ratio of elements or ions in a single unit of a compound. For example, the empirical formula for ammonium nitrate is NH 2 O, while the actual formula is NH 4 NO 2. Empirical and Actual Formulas Using Analytical Data, continued Determining Empirical Formulas, continued You can use the percentage composition for a compound to determine its empirical formula. 1. Convert the percentage of each element to g. 2. Convert from g to mol using the molar mass of each element as a conversion factor. 3. Compare these amounts in mol to find the simplest whole-number ratio among the elements.
Visual Concepts Percentage Determining an Empirical Formula from Percentage Sample Problem G Chemical analysis of a liquid shows that it is 60.0% C, 13.4% H, and 26.6% O by mass. Calculate the empirical formula of this substance. Determining an Empirical Formula from Percentage Sample Problem G Solution Assume that you have a 100.0 g sample, and convert the percentages to grams. for C: for H: for O: 60.0% 100.0 g = 60.0 g C 13.4% 100.0 g = 13.4 g H 26.6% 100.0 g = 26.6 g O Determining an Empirical Formula from Percentage Sample Problem G Solution, continued Convert the mass of each element into the amount in moles, using the reciprocal of the molar mass. 1molC 60.0gC 5.00molC 12.01 g C = 1molH 13.4 g H 13.3 mol H 1.01 g H = 1molO 26.6 g O 1.66 mol O 16.00 g O = Determining an Empirical Formula from Percentage Sample Problem G Solution, continued The formula can be written as C 5 H 13.3 O 1.66, but you divide by the smallest subscript to get whole numbers. 5.00 mol C 1.66 13.3 mol H 1.66 1.66 mol O 1.66 = 3.01 mol C = 8.01 mol H = 1.00 mol O The empirical formula is C 3 H 8 O. Using Analytical Data, continued Molecular Formulas Are Multiples of Empirical Formulas The formula for an ionic compound shows the simplest whole-number ratio of the large numbers of ions in a crystal of the compound. A molecular formula is a whole-number multiple of the empirical formula. The molar mass of any compound is equal to the molar mass of the empirical formula times a whole number, n.
Visual Concepts Comparing Molecular and Empirical Formulas Determining a Molecular Formula from an Empirical Formula Sample Problem H The empirical formula for a compound is P 2 O 5. Its experimental molar mass is 284 g/mol. Determine the molecular formula of the compound. Determining a Molecular Formula from an Empirical Formula Sample Problem H Solution Determining a Molecular Formula from an Empirical Formula Sample Problem H Solution, continued Find the molar mass of the empirical formula P 2 O 5. 2 molar mass of P = 61.94 g/mol + 5 molar mass of O = 80.00 g/mol molar mass of P 2 O 5 = 141.94 g/mol n = experimental molar mass of compound molar mass of empirical formula 284 g/mol n = = 2.00 141.94 g/mol n (empirical formula) = 2 (P 2 O 5 ) = P 4 O 10 Using Analytical Data, continued Can Give Percentage If you know the chemical formula of any compound, then you can calculate the percentage composition. Using Analytical Data, continued Can Give Percentage CO and CO 2 are both made up of C and O, but they have different percentage compositions. From the subscripts, determine the mass contributed by each element and add these to get molar mass. Divide the mass of each element by the molar mass. Multiply by 100 to find the percentage composition of that element.
Using a Chemical Formula to Determine Percentage Sample Problem I Calculate the percentage composition of copper(i) sulfide, Cu 2 S, a copper ore called chalcocite. Using a Chemical Formula to Determine Percentage Sample Problem I Solution Find the molar mass of Cu 2 S. 2 mol 63.55 g Cu/mol = 127.10 g Cu + 1 mol 32.07 g S/mol = 32.07 g S molar mass of Cu 2 S = 159.17 g/mol Using a Chemical Formula to Determine Percentage Sample Problem I Solution, continued Calculate the fraction that each element contributes to the total mass by substituting the masses into the equations below and rounding correctly. mass of 2 mol Cu mass % Cu = 100 = molar mass of Cu S 127.10 g Cu 100 = 159.17 Cu S 2 2 79.852% Cu Using a Chemical Formula to Determine Percentage Sample Problem I Solution, continued mass of 1 mol S mass % S = 100 = molar mass of Cu S 32.07 g S 100 = 159.17 g Cu S 2 2 20.15% S 1. Element A has two isotopes. One has an atomic mass of 120 and constitutes 60%; the other has an atomic mass of 122 and constitutes 40%. Which range below includes the average atomic mass of Element A? A. less than 120 B. between 120 and 121 C. between 121 and 122 D. greater than 122 1. Element A has two isotopes. One has an atomic mass of 120 and constitutes 60%; the other has an atomic mass of 122 and constitutes 40%. Which range below includes the average atomic mass of Element A? A. less than 120 B. between 120 and 121 C. between 121 and 122 D. greater than 122
2. Which of the following can be determined from the empirical formula of a compound alone? F. the true formula of the compound G. the molecular mass of the compound H. the percentage composition of the compound I. the arrangement of atoms within a molecule of the compound 2. Which of the following can be determined from the empirical formula of a compound alone? F. the true formula of the compound G. the molecular mass of the compound H. the percentage composition of the compound I. the arrangement of atoms within a molecule of the compound 3. How many ions are in 0.5 moles of NaCl? A. 1.204 10 23 B. 3.011 10 23 C. 6.022 10 23 D. 9.033 10 23 3. How many ions are in 0.5 moles of NaCl? A. 1.204 10 23 B. 3.011 10 23 C. 6.022 10 23 D. 9.033 10 23 4. How many moles of calcium (mass = 40.1) are in a serving of milk containing 290 mg of calcium? 4. How many moles of calcium (mass = 40.1) are in a serving of milk containing 290 mg of calcium? Answer: 7.23 10 3 mol
5. How is Avogadro's number related to moles? 5. How is Avogadro's number related to moles? Answer: Avogadro's number is the number of particles in a mole. 6. Antimony has two isotopes. One, amounting to 57.3% of the atoms, has a mass of 120.9 amu. The other, 42.7% of the atoms, has a mass of 122.9 amu. What is the average atomic mass of antimony? 6. Antimony has two isotopes. One, amounting to 57.3% of the atoms, has a mass of 120.9 amu. The other, 42.7% of the atoms, has a mass of 122.9 amu. What is the average atomic mass of antimony? Answer: 121.8 amu Reading Skills Read the passage below. Then answer the questions. In 1800 two English chemists, Nicholson and Carlisle, discovered that when an electric current is passed through water, hydrogen and oxygen were produced in a 2:1 volume ratio and a 1:8 mass ratio. This evidence helped to support John Dalton's theory that matter consisted of atoms, demonstrating that water consists of the two elements in a constant proportion. If the same number of moles of each gas occupy the same volume, then each molecule of water must consist of twice as much hydrogen as oxygen, even though the mass of hydrogen is only one-eighth that of oxygen. Reading Skills 7. Based on this experiment, what is the empirical formula of water? F. HO G. H 2 O H. H 2 O 8 I. O 8 H
Reading Skills 7. Based on this experiment, what is the empirical formula of water? F. HO G. H 2 O H. H 2 O 8 I. O 8 H Reading Skills 8. How would the experimental result have been different if hydrogen gas existed as individual atoms while oxygen formed molecules with two atoms bound by a covalent bond? A. The result would be the same. B. The ratio of hydrogen to oxygen would be 1:1. C. The ratio of hydrogen to oxygen would be 1:4. D. The ratio of hydrogen to oxygen would be 4:1. Reading Skills 8. How would the experimental result have been different if hydrogen gas existed as individual atoms while oxygen formed molecules with two atoms bound by a covalent bond? Reading Skills 9. How does the empirical formula for water compare to its molecular formula? A. The result would be the same. B. The ratio of hydrogen to oxygen would be 1:1. C. The ratio of hydrogen to oxygen would be 1:4. D. The ratio of hydrogen to oxygen would be 4:1. Reading Skills 9. How does the empirical formula for water compare to its molecular formula? Use the diagram below to answer questions 10 13. Answer: Both formulas are the same, H 2 O.
10.How many moles of oxygen atoms are there in 100.0 moles of carbon dioxide? F. 66.7 G. 72.7 H. 100.0 I. 200.0 10.How many moles of oxygen atoms are there in 100.0 moles of carbon dioxide? F. 66.7 G. 72.7 H. 100.0 I. 200.0 11.Explain why the percentage of oxygen in carbon dioxide is not twice the percentage of oxygen in carbon monoxide, if there are twice as many oxygen atoms. 11.Explain why the percentage of oxygen in carbon dioxide is not twice the percentage of oxygen in carbon monoxide, if there are twice as many oxygen atoms. Answer: Although there are twice as many oxygen atoms in carbon dioxide, the percentage composition is based on the mass, not the number of atoms. 12.If you did not know the true formulas for carbon monoxide and carbon dioxide, what information would you need beyond what is provided in the illustration in order to calculate them? A. the percentage compositions B. the atomic masses of carbon and oxygen C. the melting and boiling points of each compound D. the number of atoms of each element in the compound 12.If you did not know the true formulas for carbon monoxide and carbon dioxide, what information would you need beyond what is provided in the illustration in order to calculate them? A. the percentage compositions B. the atomic masses of carbon and oxygen C. the melting and boiling points of each compound D. the number of atoms of each element in the compound
13.How many grams of carbon are contained in 200.0 grams of carbon dioxide? F. 27.29 G. 42.88 H. 54.58 I. 85.76 13.How many grams of carbon are contained in 200.0 grams of carbon dioxide? F. 27.29 G. 42.88 H. 54.58 I. 85.76