Chapter 6 Guide for Introductory Chemistry PDF

Introductory Chemistry Sixth Edition Nivaldo J. Tro Chapter 6 Chemical Composition...

Introductory Chemistry Sixth Edition Nivaldo J. Tro Chapter 6 Chemical Composition Prepared by John N. Beauregard Based on a presentation by Dr. Sylvia Esjornson Southwestern Oklahoma State University Weatherford, OK © 2015 Pearson Education, Inc. Section 6.1 How Much Sodium? Summary In this chapter, you'll learn how chemists count atoms and molecules by measuring mass. In addition, you will learn how to use a chemical formula, together with information from the periodic table to calculate the amount of any constituent element in a chemical compound. Dietary Sodium Sodium is an essential dietary mineral involved in the regulation of body fluids. However, eating too much sodium can lead to high blood pressure. The primary source of sodium in our diet is sodium chloride (table salt). The FDA recommends a person consume less than 2.4 g (2400 mg) of sodium per day. However, the mass of sodium we consume less than the mass of sodium chloride that we eat. © 2015 Pearson Education, Inc. Example: The FDA recommends a person consume a maximum of 2.4 g of sodium per day. Use the chemical formula of sodium chloride (NaCl), together with atomic mass data from the periodic table, to determine the maximum mass of NaCl that should be consumed in a healthy diet. Ø The chemical formula of sodium chloride is NaCl, since it contains one Na+ ion per each Cl– ion. So, Na atoms make up ½ of the atoms in NaCl. Ø However, since Na atoms (22.99 amu) are less massive than Cl atoms (35.45 amu), Na makes up less than half the mass of any given sample of NaCl. Ø Specifically, every 58.44 g of NaCl (22.99 + 35.45) contains only 22.99 g of Na. (Note: the mass ratio is the same in grams as it is in amu.) Ø Therefore, the amount of NaCl needed to obtain 2.4 g of Na is given by © 2015 Pearson Education, Inc. Section 6.2 Counting Nails by the Pound Section 6.3 Counting Atoms by the Gram Summary These sections review the ways by which chemists express amounts of chemical elements. In the laboratory, this is typically done by measuring mass. However, it's often important to express the amount of an element in terms of either a number of individual atoms or a number of moles of atoms. Important Vocabulary Terms and Concepts mole: SI unit of amount. Defined as the number of atoms in exactly 12 grams of carbon–12; it represents 6.022 x 1023 (Avogadro’s number) of individual things. Avogadro’s number (NA) is the conversion factor between the number of moles of something and the number of individual things. atomic mass: average mass of a single atom of a given element; usually expressed in atomic mass units (amu). molar mass (of an element): mass (in grams) of one mole of atoms of a given element. It's the conversion factor between the number of moles of atoms in a sample of an element and the mass of the sample. © 2015 Pearson Education, Inc.  Some hardware stores sell nails by Scientists take a similar approach by the pound, which is easier than calculating how many atoms are in a selling them by the nail. given mass of an element. Example: A customer buys 2.60 lb of medium-sized nails, and a dozen of these nails weigh 0.150 lb. How many nails did the customer buy? © 2015 Pearson Education, Inc. Ø In the previous problem, the conversion factor for the first step is the weight per dozen nails: 0.150 lb nails = 1 doz nails Note: this conversion factor will be different for each type of nail. For instance, a dozen large nails will have a bigger mass than a dozen small nails Ø The conversion factor for the second step is the number of nails in one dozen. 1 doz nails = 12 nails Note: this conversion factor will the same for any type of nail, as 1 dozen of anything is always equal to 12 of that thing. © 2015 Pearson Education, Inc.  The Mole: A Convenient Unit for Counting Atoms Some commonly used units of amount: 1 pair = 2 (a convenient unit for counting shoes and gloves) 1 dozen (doz) = 12 (convenient unit for counting eggs and donuts) 1 baker’s dozen = 13 1 gross = 144 1 ream = 500 (commonly used to count pieces of paper) 1 mole (mol) = 6.022 x 1023 (a convenient unit for counting atoms, since even something as small as a grain of sand contains an incredibly large number of atoms.) Twenty-two “old” copper pennies (pre- Two large helium balloons contain 1982) contain about 1 mol of copper approximately 1 mol of helium (He) (Cu) atoms. atoms. © 2015 Pearson Education, Inc.  Avogadro’s Number The value 6.022 x 1023 is also called Avogadro’s number, named after Amadeo Avogadro (1776–1856). NA = 6.022 X 1023 (GIVEN ON TESTS) One mole of anything is 6.022 × 1023 units of that thing. – Example: 1 mole of marbles corresponds to 6.022 × 1023 marbles, just like 1 dozen marbles corresponds to 12 marbles. – Example: 1 mole of neon atoms (Ne) corresponds to 6.022 × 1023 Ne atoms, just like 1 a pair of Ne atoms corresponds to 2 Ne atoms. © 2015 Pearson Education, Inc.  Converting Moles to Number of Atoms Example: How many individual He atoms are in 3.5 moles of helium? GIVEN: 3.5 mol He FIND: He atoms SOLUTION MAP: SOLUTION: © 2015 Pearson Education, Inc.  Converting Number of Atoms to Moles Example: Calculate the number of moles of silver (Ag) in 1.1 1022 Ag atoms. GIVEN: 1.1 1022 Ag atoms FIND: mol Ag SOLUTION MAP: SOLUTION: © 2015 Pearson Education, Inc.  Molar Mass: Defined Relative to Carbon–12 Recall that 1 atomic mass unit (amu) is defined as 1/12 the mass of an atom of the isotope carbon-12 (a C atom with 6 protons, 6 neutrons, and 6 electrons) Similarly, the the numerical value of 1 mole is defined as the number of atoms in exactly 12 g of pure carbon-12, which turns out to be 6.022 x 1023. This means that there are 6.022 x 1023 atomic mass units in 1 gram: 1 g = 6.022 x 1023 amu This connection between the definitions of the atomic mass unit and mole is very convenient, as it allows us to use the same numerical value for both the molar mass of an element and the atomic mass of the same element. Example: The atomic mass of neon is 20.18 amu. (i.e. the average mass of an individual Ne atom is 20.18 amu) The molar mass of neon is 20.18 g. (i.e. the total mass of 1 mole of Ne atoms is 20.18 g) © 2015 Pearson Education, Inc. The Mass of 1 Mole of Atoms of an Element is the Element’s Molar Mass The lighter the atom, the less mass there is in 1 mole of that atom. For example, the picture below shows 1 mole of S atoms (atomic mass = 32.07 amu) on the left and 1 mole of C atoms (atomic mass 12.01 amu) on the right. So the molar mass of sulfur is 32.07 g (MS = 32.07 g/mol), and the molar mass of carbon is 12.01 g (MC = 12.01 g/mol). © 2015 Pearson Education, Inc.  The molar mass (i.e. mass of 1 mole of atoms) changes for different elements. (The molar mass value for each element is found in the periodic table.) However, the number of atoms in 1 mole of an element is always the same. (It’s Avogadro’s number, NA = 6.022 × 1023 atoms) Some examples are as follows: 32.07 g sulfur = 1 mol sulfur = 6.022 × 1023 S atoms MS = 32.07 g/mol 12.01 g carbon = 1 mol carbon = 6.022 × 1023 C atoms MC = 12.01 g/mol 6.94 g lithium = 1 mol lithium = 6.022 × 1023 Li atoms MLi = 6.94 g/mol © 2015 Pearson Education, Inc.  Mass–Mole Conversion Example: Calculate the number of moles of carbon in 0.58 grams of diamond. SOLUTION MAP: RELATIONSHIP USED: 12.01 g C = 1 mol C (from periodic table) SOLUTION: © 2015 Pearson Education, Inc.  Mass-Mole-Number of Atoms Conversions Example: How many aluminum atoms are in an aluminum can with a mass of 16.2 g? SOLUTION MAP: RELATIONSHIPS USED: 26.98 g Al = 1 mol Al (from periodic table) 6.022 1023 = 1 mol (Avogadro’s number, NA) SOLUTION: © 2015 Pearson Education, Inc. What you need to be able to do (Sections 6.2 & 6.3 Material): Convert back and forth between the number of individual atoms of an element, the number of moles of atoms of the element, and mass of an element. Highly Recommended for Practice (Sections 6.2 & 6.3 Material) EXAMPLES: 6.1 & 6.14(p. 192), 6.2, 6.15(p. 193), 6.3 & 6.16(p. 193) SKILLBUILDERS: 6.1, 6.2, & 6.3 (Answers on p. 205) CONCEPTUAL CHECKPOINTS: 6.1, 6.2, & 6.3 (Answers on p. 205) On Line Resource—The Mole Concept in Chemistry: Click on the image on the left to view a TED-Ed video on YouTube that reviews the mole concept—the amount, not the furry little animal. © 2015 Pearson Education, Inc. On Line Resource—Converting Between Mass of an Element and Moles of that Element: Click on the image below to view a YouTube video review of the concept of the molar mass of an element. Also provided are sample calculations, converting back and forth between the mass of an element and the number of moles of that element. This corresponds to going back and forth on the left side of Mass–Mole–Atoms conversion diagram shown above: On Line Resource—Converting From Mass to Moles to the Number of Atoms of the Element: Click on the image below to view a YouTube video on mass-mole-number of atom conversions. © 2015 Pearson Education, Inc. Section 6.4 Counting Molecules by the Gram Summary In this section, the ways used in Section 6.3 for expressing the amount of an atomic element are extended to molecular elements, molecular compounds, and ionic compounds. Important Vocabulary Terms and Concepts formula mass (molecular substance or ionic compound): average mass of one molecule (molecular element or molecular compound), or the mass of one formula unit (ionic compound). Usually expressed in atomic mass units (amu). molar mass (of a molecular substance or ionic compound): mass of one mole of molecules (molecular substance) or one mole of formula units (ionic compound)—usually expressed in grams. Ø The molar mass (M) of a molecular substance is used to convert between the number of moles of molecules in a sample of the substance and the mass of the sample. Ø Likewise, the molar mass (M) of an ionic compound is used to convert between the number of moles of formula units in a sample of the compound and the mass of the sample. © 2015 Pearson Education, Inc. Some Examples: Molecular bromine has the formula Br2. – formula mass of Br2 = 2 x 79.90 amu = 159.8 amu. (i.e. the average mass of a single Br2 molecule is 159.8 amu.) – So, the molar mass of Br2 is 159.8 g. (i.e. the mass of 1 mole of Br2 molecules is 159.8 g) MBr2 = 159.8 g/mol The molecular compound methane has the formula CH4. – formula mass of CH4 = (1 x 12.01 amu) + (4 x 1.01 amu) = 16.05 amu. (i.e. the average mass of a single CH4 molecule is 16.05 amu.) – So, the molar mass of CH4 is 16.05 g. (i.e. the mass of 1 mole of CH4 molecules is 16.05 g) MCH4 = 16.05 g/mol The ionic compound magnesium chloride has the formula MgCl2. – formula mass of MgCl2 = (1 x 24.30 amu) + (2 x 35.45 amu) = 95.20 amu. (i.e. the average mass of a formula unit of MgCl2 is 95.20 amu.) – So, the molar mass of MgCl2 is 95.20 g. (i.e. the mass of 1 mole of MgCl2 formula units is 95.20 g) © 2015 Pearson Education, Inc. MMgCl2 = 95.20 g/mol  Converting Between Grams and Moles of a Compound Example: Calculate the mass in grams of 1.75 moles of water. GIVEN: 1.75 mol H2O FIND: g H2O RELATIONSHIP USED: MH2O = 2 (1.01) + 1 (16.00) = 18.02 g/mol SOLUTION MAP: SOLUTION: © 2015 Pearson Education, Inc. Converting Between Number of Molecules and Mass of a Compound Example: What is the mass of 4.78 1024 NO2 molecules? GIVEN: 4.78 1024 NO2 molecules FIND: g NO2 RELATIONSHIPS USED: 6.022 1023 molecules = 1 mol (Avogadro’s number) MNO2 = 1 (14.01) + 2(16.00) = 46.01 g/mol SOLUTION MAP: SOLUTION: © 2015 Pearson Education, Inc. What you need to be able to do (Section 6.4 Material): Calculate the molar mass (or formula mass) of a substance based on its chemical formula. Convert back an forth between the moles of a molecular substance and the number of individual molecules. Also convert between the number of moles of molecules and the mass of a molecular substance. Convert back an forth between the moles of an ionic compound and the number of individual formula units. Also convert between the number of moles of formula units and the mass of an ionic compound. Highly Recommended for Practice (Section 6.4 Material) EXAMPLES: 6.4 & 6.5 SKILLBUILDERS: 6.4 & 6.5 (Answers on p. 205) CONCEPTUAL CHECKPOINT: 6.4 (Answer on p. 205) © 2015 Pearson Education, Inc. On Line Resource—Molar Masses of Elements and Compounds: Click on the image to the right to view YouTube video tutorial on distinguishing between the molar mass of an element (i.e. type of atom) and the molar mass of a compound (e,g, type of molecule).The video also gives a sample calculation for the molar mass of a compound based on its chemical formula and the molar masses of its atoms (obtained from the periodic table.) On Line Resource—Mass–Mole–“Number of Individuals” Conversions: Click on the image below to view a YouTube video reviewing the mole concept in chemistry. It also give sample calculations involving conversions between mass, moles, and the number of individual particles: first for an element and then for a compound. © 2015 Pearson Education, Inc.  Section 6.5 Chemical Formulas as Conversion Factors Summary The chemical formula of a molecular compound gives the number of each type of atom found in one molecule of the compound, while the chemical formula of an ionic compound provides the number of each type of atom found in one formula unit of the 3 leaves : 1 clover compound. In this section, chemical formulas are used as conversion factors to determine the amount (individual atoms, moles, or mass) of a element in a specified amount (individual molecules or formula units, moles, or mass) of a compound. Important Vocabulary Terms and Concepts chemical formula (molecular substance): gives the type(s) of atom(s) and number of each type of atom present in one molecule of the substance. chemical formula (ionic compound): gives the types of atoms and the number of each type of atom present in one formula unit of the compound. © 2015 Pearson Education, Inc. 3-Leaf Clover Analogy: How Many Leaves on 14 Clovers? 3 leaves : 1 clover 1 stem : 1 clover A Molecular Example: How Many O Atoms are in 11 CO2 Molecules? 2 O atoms : 1 CO2 molecule 1 C atom : 1 CO2 molecule © 2015 Pearson Education, Inc.  The Mole Relationships from Chemical Formulas The formula for carbon dioxide (CO2) means there are two O atoms and 1 C atom per one CO2 molecule. We can write this as follows: 2 O atoms : 1 CO2 molecule 1 C atom : 1 CO2 molecule These ratios are maintained no matter what unit of amount is used 2 pairs of O atoms : 1 pair of CO2 molecules 2 dozen O atoms : 1 dozen CO2 molecules 2 gross of O atoms : 1 gross of CO2 molecules 2 moles O : 1 mole CO2 © 2015 Pearson Education, Inc.  Using a Chemical Formula as a Conversion Factor The relationships contained in a chemical formula allow us to convert between moles of the compound and moles of a constituent element (and vice versa). For instance, the number of Cl atoms in any sample of CCl4 will always be 4 times as large as the number of CCl4 molecules. Example: 8 moles of CCl4 contains 32 moles of Cl. © 2015 Pearson Education, Inc.  Converting Between Moles of a Compound and Moles of a Constituent Element Example: Find the number of moles of O in 1.7 mol CaCO3. GIVEN: 1.7 mol CaCO3 FIND: mol O SOLUTION MAP: RELATIONSHIP USED: 3 mol O : 1 mol CaCO3 (from chemical formula) SOLUTION: © 2015 Pearson Education, Inc.  Converting Between Grams of a Compound and Grams of a Constituent Element Example: Find the mass of sodium in 15 g of NaCl. GIVEN: 15 g NaCl FIND: g Na SOLUTION MAP: SOLUTION: © 2015 Pearson Education, Inc. What you need to be able to do (Section 6.5 Material): Use the chemical formula of a molecular substance or ionic compound to set up relationships between (i) the number of atoms of a particular element per molecule/formula unit or (ii) the number of moles of atoms of a particular element per mole of molecules/formula units. Use these relationships to convert between an amount of the compound (individual molecules/formula units, moles of molecules/formula units, or mass) of a compound and the amount (individual atoms, moles of atoms, or mass) of a specific element. Highly Recommended for Practice (Section 6.5 Material) EXAMPLES: 6.6, 6.17(p. 194), 6.7 & 6.18(p. 194) SKILLBUILDERS: 6.6 & 6.7 (Answers on p. 203) CONCEPTUAL CHECKPOINTS: 6.5 & 6.6 (Answers on p. 205) © 2015 Pearson Education, Inc. Section 6.6 Mass Percent Composition of Compounds Section 6.7 Mass Percent Composition from a Chemical Formula Summary Like a chemical formula, the mass-percent compositions of the elements in a chemical formula provide a recipe for making the chemical compound. However, the %–mass of an element gives the amount of that element in terms of its relative mass in the compound rather than by the number of atoms of that element. Important Vocabulary Terms and Concepts mass percent composition of an element in a compound: the percentage of the total mass of a compound that is made up of a given element. © 2015 Pearson Education, Inc. Finding Mass Percent Composition Based on Mass Data Example: A 0.358-g sample of chromium metal reacts with oxygen to form 0.523 g of the metal oxide. Determine the mass-percent of chromium in this compound. © 2015 Pearson Education, Inc. Using Mass Percent Composition as a Conversion Factor Ø The mass percent composition of sodium in sodium chloride is 39%. Ø This can be written as follows: 39 g sodium : 100 g sodium chloride Ø This same relationship can also be written in fractional form: These fractions are conversion factors between g Na and g NaCl. Example 1: Determine the mass of Na in 55 grams of NaCl. Example 2: Find the mass of NaCl needed to obtain 5.0 g of Na. © 2015 Pearson Education, Inc.  Mass Percent Composition from a Chemical Formula Example 1: Based on its chemical formula, determine the mass-percent of element Cl in the compound CCl2F2. GIVEN: CCl2F2 & molar masses FIND: Mass % Cl in CCl2F2 from the periodic table Plan of Action: Solution: MCCl2F2 = 1 (12.01) + 2 (35.45) + 2 (19.00) = 120.91 g/mol © 2015 Pearson Education, Inc. What you need to be able to do (Sections 6.6 & 6.7 Material): Given mass data for a chemical compound, calculate the mass–% composition of each element in the compound, and vice versa. Calculate the %–mass composition for any given element in a compound given the chemical formula of the compound. Highly Recommended for Practice (Section 6.6 & 6.7 Material) EXAMPLES: 6.8, 6.19(p. 195), 6.9 & 6.20(p. 195) SKILLBUILDERS: 6.8 & 6.9 (Answers on p. 205) CONCEPTUAL CHECKPOINT: 6.7 (Answers on p. 205) On Line Resource—Mass–Percent Composition of Chemical Compounds: Click on the image below to view a YouTube video tutorial on mass–% composition. The tutorial also shows a sample calculation. © 2015 Pearson Education, Inc. Section 6.8 Calculating Empirical Formulas for Compounds Section 6.9 Calculating Molecular Formulas for Compounds Summary These sections describe how experimentally obtainable data for relative masses of the elements in a compound are used to obtain the chemical formula for the compound. Important Vocabulary Terms and Concepts empirical formula: gives the elements contained in a compound in their lowest whole– number ratio by atom. empirical molar mass: mass of one mole of empirical formula units. When the empirical molar mass is divided into the actual molar mass of the compound, a whole–number positive integer (n) is obtained: The empirical formula of the compound is multiplied by this integer to obtain the molecular formula of the compound. molecular formula: contains the actual number of each type of atom found in one molecule of a compound: Molecular Formula = Empirical Formula x n © 2015 Pearson Education, Inc.  Empirical Formula Determined based on the percent-mass composition data for a compound. The subscripts in an empirical formula gives the lowest whole-number ratio of atoms for the elements in the compound Here are some examples: Molecular Formula Empirical Formula H2O H2O H2O2 HO C2H6 CH3 C12H22O11 C12H22O11 C6H12O6 CH2O © 2015 Pearson Education, Inc.  Empirical Formulas from Elemental Mass Data Example: When a 27-gram sample of water is decomposed a in the laboratory, it produces 3.0 g of hydrogen and 24 g of oxygen. Determine an empirical formula from these data? STEP 1: Convert the mass of each element to moles of that element STEP 2: Use the result from STEP1 to create a “preliminary formula” for the compound H3.0O1.5 STEP 3: To get whole-number subscripts in the formula, divide all the subscripts by the smallest subscript (in this case 1.5) Note: in the case of water, the empirical formula also happens to be the molecular formula. © 2015 Pearson Education, Inc.  Empirical Formulas from Percent-Mass Composition Data Example: A certain compound is composed of 25.9% nitrogen and 74.1% oxygen by mass. Determine the empirical formula of this compound? STEP 1: Assume a 100-gram sample of the compound and convert the mass of each element to moles of that element STEP 2: Use the result from STEP1 to create a “preliminary formula” for the compound N1.85O4.63 STEP 3: Divide all the subscripts by the smallest subscript (in this case 1.85). However, this time this does not give only whole-number subscripts. So multiply thru by 2 to obtain whole numbers for both elements. © 2015 Pearson Education, Inc. Conversion of Fractional Subscripts to Whole Numbers As in the previous problem, dividing by the smallest number of moles doesn’t always give whole-number subscripts. In cases such as these, multiply all the subscripts by the appropriate small whole number to obtain whole-number subscripts. © 2015 Pearson Education, Inc.  Calculating Molecular Formulas for Compounds from Empirical Formulas and Molar Masses The molecular formula is always a whole-number multiple of the empirical formula. We need to find n in the expression: Molecular Formula = Empirical Formula × n We can find n using the equation © 2015 Pearson Education, Inc.  Calculating Molecular Formulas for Compounds from Empirical Formulas and Molar Masses Example: Mass-percent composition data show that the empirical formula of fructose, a sugar found in fruit, to be CH2O. In addition, mass spectrometry experiments show that the molar molar mass of fructose is180.2 g/mol. Determine the molecular formula of fructose. Empirical Molar Mass = 1 (12.01) + 2 (1.01) + 1 (16.00) = 30.03 g/mol Empirical Formula × n = Molecular Formula: So, each fructose molecule has 6 C atoms, 12 H atoms, and 6 O atoms. © 2015 Pearson Education, Inc. What you need to be able to do (Sections 6.8 & 6.9 Material): Be able to transform the molecular formula of a given compound into its respective empirical formula. Derive the empirical formula of a chemical compound given either specific masses or the percent-masses of the elements in the compound. Be able to calculate the molecular formula for a compound given both the empirical formula and molar mass of the compound. Highly Recommended for Practice (Section 6.8 & 6.9 Material) EXAMPLES: 6.10, 6.11, 6.21(p. 196), 6.12 & 6.13 SKILLBUILDERS: 6.10, 6.11, 6.12 & 6.13 (Answers on p. 205) On Line Resource—Empirical Formula Calculation: Click on the image below to view a YouTube video showing the how to solve for the empirical formula of a compound given the masses of all the elements in a sample of the compound: © 2015 Pearson Education, Inc. On Line Resource—Empirical Formula and Molecular Formula Calculations: Click on the image below to view a YouTube video tutorial that distinguishes between the empirical formula of a compound and its molecular formula. The video also gives sample calculations for (i) determining the empirical formula of a compound based on mass data for its elements and (ii) converting the empirical formula to the molecular formula given the molar mass of the compound. On Line Resource—More on Empirical Formula and Molecular Formula: Click on the image below to view a YouTube video with a second tutorial on empirical formulas and molecular formulas, including additional sample calculations. © 2015 Pearson Education, Inc. Recommended for Addition Practice with the Material from Chapter 6 These Questions from the Self-Assessment Quiz (p. 190 of the text): Q1, Q2, Q3, Q4, Q5, Q6, Q7, Q8, Q9, Q10, Q11, Q12, Q13 & Q14 These Supplemental Exercises from the End of the Chapter (Solutions to these will be provided.) EXERCISES: 3, 7, & 13 PROBLEMS: 19, 25, 29, 37, 39, 47, 53, 55, 61, 65, 69, 71, 75, 77, 79, 85, 87, 93, 97, 101, 105, 111, 119, 121, & 127 © 2015 Pearson Education, Inc.