Chemical Names and Formulas PDF

Section 1 Chemical Names and Chapter 7 Formulas Objectives Explain the significance of a chemical formula. Determine the formula of an ionic compound formed between two given ions. Name an ionic compound given its formula. Using prefixes, name a binary molecular compound from i...

Section 1 Chemical Names and Chapter 7 Formulas Objectives Explain the significance of a chemical formula. Determine the formula of an ionic compound formed between two given ions. Name an ionic compound given its formula. Using prefixes, name a binary molecular compound from its formula. Write the formula of a binary molecular compound given its name.  Section 1 Chemical Names and Chapter 7 Formulas Significance of a Chemical Formula A chemical formula indicates the relative number of atoms of each kind in a chemical compound. For a molecular compound, the chemical formula reveals the number of atoms of each element contained in a single molecule of the compound. example: octane — C8H18 The subscript after the C The subscript after the H indicates that there are 8 indicates that there are 18 carbon atoms in the hydrogen atoms in molecule. the molecule.  Section 1 Chemical Names and Chapter 7 Formulas Significance of a Chemical Formula, continued The chemical formula for an ionic compound represents one formula unit—the simplest ratio of the compound’s positive ions (cations) and its negative ions (anions). example: aluminum sulfate — Al2(SO4)3 Parentheses surround the polyatomic ion to identify it as a unit. The subscript 3 refers to the unit. Note also that there is no subscript for sulfur: when there is no subscript next to an atom, the subscript is understood to be 1.  Section 1 Chemical Names and Chapter 7 Formulas Monatomic Ions Many main-group elements can lose or gain electrons to form ions. Ions formed form a single atom are known as monatomic ions. example: To gain a noble-gas electron configuration, nitrogen gains three electrons to form N3– ions. Some main-group elements tend to form covalent bonds instead of forming ions. examples: carbon and silicon  Section 1 Chemical Names and Chapter 7 Formulas Monatomic Ions, continued Naming Monatomic Ions Monatomic cations are identified simply by the element’s name. examples: K+ is called the potassium cation Mg2+ is called the magnesium cation For monatomic anions, the ending of the element’s name is dropped, and the ending -ide is added to the root name. examples: F– is called the fluoride anion N3– is called the nitride anion  Section 1 Chemical Names and Chapter 7 Formulas Common Monatomic Ions  Section 1 Chemical Names and Chapter 7 Formulas Common Monatomic Ions  Section 1 Chemical Names and Chapter 7 Formulas Binary Ionic Compounds Compounds composed of two elements are known as binary compounds. In a binary ionic compound, the total numbers of positive charges and negative charges must be equal. The formula for a binary ionic compound can be written given the identities of the compound’s ions. example: magnesium bromide Ions combined: Mg2+, Br–, Br– Chemical formula: MgBr2  Section 1 Chemical Names and Chapter 7 Formulas Binary Ionic Compounds, continued A general rule to use when determining the formula for a binary ionic compound is “crossing over” to balance charges between ions. example: aluminum oxide 1) Write the symbols for the ions. Al3+ O2– 2) Cross over the charges by using the absolute value of each ion’s charge as the subscript for the other ion.  Section 1 Chemical Names and Chapter 7 Formulas Binary Ionic Compounds, continued example: aluminum oxide, continued 3) Check the combined positive and negative charges to see if they are equal. (2 × 3+) + (3 × 2–) = 0 The correct formula is Al2O3  Section 1 Chemical Names and Chapter 7 Formulas Writing the Formula of an Ionic Compound  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds The nomenclature, or naming system, or binary ionic compounds involves combining the names of the compound’s positive and negative ions. The name of the cation is given first, followed by the name of the anion: example: Al2O3 — aluminum oxide For most simple ionic compounds, the ratio of the ions is not given in the compound’s name, because it is understood based on the relative charges of the compound’s ions.  Section 1 Chemical Names and Chapter 7 Formulas Naming Ionic Compounds Click below to watch the Visual Concept. Visual Concept  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued Sample Problem A Write the formulas for the binary ionic compounds formed between the following elements: a. zinc and iodine b. zinc and sulfur  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued Sample Problem A Solution Write the symbols for the ions side by side. Write the cation first. a. Zn2+ I− b. Zn2+ S2− Cross over the charges to give subscripts. a. b.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued Sample Problem A Solution, continued Check the subscripts and divide them by their largest common factor to give the smallest possible whole-number ratio of ions. a. The subscripts give equal total charges of 1 × 2+ = 2+ and 2 × 1− = 2−. The largest common factor of the subscripts is 1. The smallest possible whole-number ratio of ions in the compound is 1:2. The formula is ZnI2.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued Sample Problem A Solution, continued b. The subscripts give equal total charges of 2 × 2+ = 4+ and 2 × 2− = 4−. The largest common factor of the subscripts is 2. The smallest whole-number ratio of ions in the compound is 1:1. The formula is ZnS.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued The Stock System of Nomenclature Some elements such as iron, form two or more cations with different charges. To distinguish the ions formed by such elements, scientists use the Stock system of nomenclature. The system uses a Roman numeral to indicate an ion’s charge. examples: Fe2+ iron(II) Fe3+ iron(III)  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued The Stock System of Nomenclature, continued Sample Problem B Write the formula and give the name for the compound formed by the ions Cr3+ and F–.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued The Stock System of Nomenclature, continued Sample Problem B Solution Write the symbols for the ions side by side. Write the cation first. Cr3+ F− Cross over the charges to give subscripts.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued The Stock System of Nomenclature, continued Sample Problem B Solution, continued The subscripts give charges of 1 × 3+ = 3+ and 3 × 1− = 3−. The largest common factor of the subscripts is 1, so the smallest whole number ratio of the ions is 1:3. The formula is CrF3.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued The Stock System of Nomenclature, continued Sample Problem B Solution, continued Chromium forms more than one ion, so the name of the 3+ chromium ion must be followed by a Roman numeral indicating its charge. The compound’s name is chromium(III) fluoride.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued Compounds Containing Polyatomic Ions Many common polyatomic ions are oxyanions— polyatomic ions that contain oxygen. Some elements can combine with oxygen to form more than one type of oxyanion. example: nitrogen can form or. The name of the ion with the greater number of oxygen atoms ends in -ate. The name of the ion with the smaller number of oxygen atoms ends in -ite. nitrate nitrite  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued Compounds Containing Polyatomic Ions, continued Some elements can form more than two types of oxyanions. example: chlorine can form , , or. In this case, an anion that has one fewer oxygen atom than the -ite anion has is given the prefix hypo-. An anion that has one more oxygen atom than the -ate anion has is given the prefix per-. hypochlorite chlorite chlorate perchlorate  Section 1 Chemical Names and Chapter 7 Formulas Polyatomic Ions  Section 1 Chemical Names and Chapter 7 Formulas Naming Compounds with Polyatomic Ions  Section 1 Chemical Names and Chapter 7 Formulas Understanding Formulas for Polyatomic Ionic Compounds  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued Compounds Containing Polyatomic Ions, continued Sample Problem C Write the formula for tin(IV) sulfate.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued Compounds Containing Polyatomic Ions, continued Cross over the charges to give subscripts. Add parentheses around the polyatomic ion if necessary. Cross over the charges to give subscripts. Add parentheses around the polyatomic ion if necessary.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Ionic Compounds, continued Compounds Containing Polyatomic Ions, continued Sample Problem C Solution, continued The total positive charge is 2 × 4+ = 8+. The total negative charge is 4 × 2− = 8−. The largest common factor of the subscripts is 2, so the smallest whole-number ratio of ions in the compound is 1:2. The correct formula is therefore Sn(SO4) 2.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Molecular Compounds Unlike ionic compounds, molecular compounds are composed of individual covalently bonded units, or molecules. As with ionic compounds, there is also a Stock system for naming molecular compounds. The old system of naming molecular compounds is based on the use of prefixes. examples: CCl4 — carbon tetrachloride (tetra- = 4) CO — carbon monoxide (mon- = 1) CO2 — carbon dioxide (di- = 2)  Section 1 Chemical Names and Chapter 7 Formulas Prefixes for Naming Covalent Compounds  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Molecular Compounds, continued Sample Problem D a. Give the name for As2O5. b. Write the formula for oxygen difluoride.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Molecular Compounds, continued Sample Problem D Solution a. A molecule of the compound contains two arsenic atoms, so the first word in the name is diarsenic. The five oxygen atoms are indicated by adding the prefix pent- to the word oxide. The complete name is diarsenic pentoxide.  Section 1 Chemical Names and Chapter 7 Formulas Naming Binary Molecular Compounds, continued Sample Problem D Solution, continued b. Oxygen is first in the name because it is less electronegative than fluorine. Because there is no prefix, there must be only one oxygen atom. The prefix di- in difluoride shows that there are two fluorine atoms in the molecule. The formula is OF2.  Section 1 Chemical Names and Chapter 7 Formulas Covalent-Network Compounds Some covalent compounds do not consist of individual molecules. Instead, each atom is joined to all its neighbors in a covalently bonded, three-dimensional network. Subscripts in a formula for covalent-network compound indicate smallest whole-number ratios of the atoms in the compound. examples: SiC, silicon carbide SiO2, silicon dioxide Si3N4, trisilicon tetranitride.  Section 1 Chemical Names and Chapter 7 Formulas Acids and Salts An acid is a certain type of molecular compound. Most acids used in the laboratory are either binary acids or oxyacids. Binary acids are acids that consist of two elements, usually hydrogen and a halogen. Oxyacids are acids that contain hydrogen, oxygen, and a third element (usually a nonmetal).  Section 1 Chemical Names and Chapter 7 Formulas Acids and Salts, continued In the laboratory, the term acid usually refers to a solution in water of an acid compound rather than the acid itself. example: hydrochloric acid refers to a water solution of the molecular compound hydrogen chloride, HCl Many polyatomic ions are produced by the loss of hydrogen ions from oxyacids. examples: sulfuric acid H2SO4 sulfate nitric acid HNO3 nitrate phosphoric acid H3PO4 phosphate  Section 1 Chemical Names and Chapter 7 Formulas Acids and Salts, continued An ionic compound composed of a cation and the anion from an acid is often referred to as a salt. examples: Table salt, NaCl, contains the anion from hydrochloric acid, HCl. Calcium sulfate, CaSO4, is a salt containing the anion from sulfuric acid, H2SO4. The bicarbonate ion, , comes from carbonic acid, H2CO3.  Section 1 Chemical Names and Chapter 7 Formulas Salt Click below to watch the Visual Concept. Visual Concept  Section 1 Chemical Names and Chapter 7 Formulas Prefixes and Suffixes for Oxyanions and Related Acids Click below to watch the Visual Concept. Visual Concept  Section 2 Oxidation Numbers Chapter 7 Preview Lesson Starter Objectives Oxidation Numbers Assigning Oxidation Numbers Using Oxidation Numbers for Formulas and Names  Section 2 Oxidation Numbers Chapter 7 Lesson Starter It is possible to determine the charge of an ion in an ionic compound given the charges of the other ions present in the compound. Determine the charge on the bromide ion in the compound NaBr given that Na+ has a 1+ charge. Answer: The total charge is 0, so Br – must have a charge of 1– in order to balance the 1+ charge of Na+.  Section 2 Oxidation Numbers Chapter 7 Lesson Starter, continued Numbers called oxidation numbers can be assigned to atoms in order to keep track of electron distributions in molecular as well as ionic compounds.  Section 2 Oxidation Numbers Chapter 7 Objectives List the rules for assigning oxidation numbers. Give the oxidation number for each element in the formula of a chemical compound. Name binary molecular compounds using oxidation numbers and the Stock system.  Section 2 Oxidation Numbers Chapter 7 Oxidation Numbers The charges on the ions in an ionic compound reflect the electron distribution of the compound. In order to indicate the general distribution of electrons among the bonded atoms in a molecular compound or a polyatomic ion, oxidation numbers are assigned to the atoms composing the compound or ion. Unlike ionic charges, oxidation numbers do not have an exact physical meaning: rather, they serve as useful “bookkeeping” devices to help keep track of electrons.  Section 2 Oxidation Numbers Chapter 7 Assigning Oxidation Numbers In general when assigning oxidation numbers, shared electrons are assumed to “belong” to the more electronegative atom in each bond. More-specific rules are provided by the following guidelines. 1. The atoms in a pure element have an oxidation number of zero. examples: all atoms in sodium, Na, oxygen, O2, phosphorus, P4, and sulfur, S8, have oxidation numbers of zero.  Section 2 Oxidation Numbers Chapter 7 Assigning Oxidation Numbers, continued 2. The more-electronegative element in a binary compound is assigned a negative number equal to the charge it would have as an anion. Likewise for the less-electronegative element. 3. Fluorine has an oxidation number of –1 in all of its compounds because it is the most electronegative element.  Section 2 Oxidation Numbers Chapter 7 Assigning Oxidation Numbers, continued 4. Oxygen usually has an oxidation number of –2. Exceptions: In peroxides, such as H2O2, oxygen’s oxidation number is –1. In compounds with fluorine, such as OF2, oxygen’s oxidation number is +2. 5. Hydrogen has an oxidation number of +1 in all compounds containing elements that are more electronegative than it; it has an oxidation number of –1 with metals.  Section 2 Oxidation Numbers Chapter 7 Assigning Oxidation Numbers, continued 6. The algebraic sum of the oxidation numbers of all atoms in an neutral compound is equal to zero. 7. The algebraic sum of the oxidation numbers of all atoms in a polyatomic ion is equal to the charge of the ion. 8. Although rules 1 through 7 apply to covalently bonded atoms, oxidation numbers can also be applied to atoms in ionic compounds similarly.  Section 2 Oxidation Numbers Chapter 7 Rules for Assigning Oxidation Numbers Click below to watch the Visual Concept. Visual Concept  Section 2 Oxidation Numbers Chapter 7 Assigning Oxidation Numbers, continued Sample Problem E Assign oxidation numbers to each atom in the following compounds or ions: a. UF6 b. H2SO4 c.  Section 2 Oxidation Numbers Chapter 7 Assigning Oxidation Numbers, continued Sample Problem E Solution a.Place known oxidation numbers above the appropriate elements. Multiply known oxidation numbers by the appropriate number of atoms and place the totals underneath the corresponding elements.  Section 2 Oxidation Numbers Chapter 7 Assigning Oxidation Numbers, continued The compound UF6 is molecular. The sum of the oxidation numbers must equal zero; therefore, the total of positive oxidation numbers is +6. Divide the total calculated oxidation number by the appropriate number of atoms. There is only one uranium atom in the molecule, so it must have an oxidation number of +6.  Section 2 Oxidation Numbers Chapter 7 Assigning Oxidation Numbers, continued Sample Problem E Solution, continued b. Hydrogen has an oxidation number of +1. Oxygen has an oxidation number of −2. The sum of the oxidation numbers must equal zero, and there is only one sulfur atom in each molecule of H2SO4. Because (+2) + (−8) = −6, the oxidation number of each sulfur atom must be +6.  Section 2 Oxidation Numbers Chapter 7 Assigning Oxidation Numbers, continued Sample Problem E Solution, continued c. The total of the oxidation numbers should equal the overall charge of the anion, 1−. The oxidation number of a single oxygen atom in the ion is −2. The total oxidation number due to the three oxygen atoms is −6. For the chlorate ion to have a 1− charge, chlorine must be assigned an oxidation number of +5.  Section 2 Oxidation Numbers Chapter 7 Using Oxidation Numbers for Formulas and Names As shown in the table in the next slide, many nonmetals can have more than one oxidation number. These numbers can sometimes be used in the same manner as ionic charges to determine formulas. example: What is the formula of a binary compound formed between sulfur and oxygen? From the common +4 and +6 oxidation states of sulfur, you could predict that sulfur might form SO2 or SO3. Both are known compounds.  Section 2 Oxidation Numbers Chapter 7 Common Oxidation States of Nonmetals  Section 2 Oxidation Numbers Chapter 7 Using Oxidation Numbers for Formulas and Names, continued Using oxidation numbers, the Stock system, introduced in the previous section for naming ionic compounds, can be used as an alternative to the prefix system for naming binary molecular compounds. Prefix system Stock system PCl3 phosphorus trichloride phosphorus(III) chloride PCl5 phosphorus pentachloride phosphorus(V) chloride N 2O dinitrogen monoxide nitrogen(I) oxide NO nitrogen monoxide nitrogen(II) oxide Mo2O3 dimolybdenum trioxide molybdenum(III) oxide  Section 3 Using Chemical Formulas Chapter 7 Preview Lesson Starter Objectives Formula Masses Molar Masses Molar Mass as a Conversion Factor Percentage Composition  Section 3 Using Chemical Formulas Chapter 7 Lesson Starter The chemical formula for water is H2O. How many atoms of hydrogen and oxygen are there in one water molecule? How might you calculate the mass of a water molecule, given the atomic masses of hydrogen and oxygen? In this section, you will learn how to carry out these and other calculations for any compound.  Section 3 Using Chemical Formulas Chapter 7 Objectives Calculate the formula mass or molar mass of any given compound. Use molar mass to convert between mass in grams and amount in moles of a chemical compound. Calculate the number of molecules, formula units, or ions in a given molar amount of a chemical compound. Calculate the percentage composition of a given chemical compound.  Section 3 Using Chemical Formulas Chapter 7 A chemical formula indicates: the elements present in a compound the relative number of atoms or ions of each element present in a compound Chemical formulas also allow chemists to calculate a number of other characteristic values for a compound: formula mass molar mass percentage composition  Section 3 Using Chemical Formulas Chapter 7 Formula Masses The formula mass of any molecule, formula unit, or ion is the sum of the average atomic masses of all atoms represented in its formula. example: formula mass of water, H2O average atomic mass of H: 1.01 amu average atomic mass of O: 16.00 amu average mass of H2O molecule: 18.02 amu  Section 3 Using Chemical Formulas Chapter 7 Formula Masses The mass of a water molecule can be referred to as a molecular mass. The mass of one formula unit of an ionic compound, such as NaCl, is not a molecular mass. The mass of any unit represented by a chemical formula (H2O, NaCl) can be referred to as the formula mass.  Section 3 Using Chemical Formulas Chapter 7 Formula Mass Click below to watch the Visual Concept. Visual Concept  Section 3 Using Chemical Formulas Chapter 7 Formula Masses, continued Sample Problem F Find the formula mass of potassium chlorate, KClO3.  Section 3 Using Chemical Formulas Chapter 7 Formula Masses, continued Sample Problem F Solution The mass of a formula unit of KClO3 is found by adding the masses of one K atom, one Cl atom, and three O atoms. Atomic masses can be found in the periodic table in the back of your book. In your calculations, round each atomic mass to two decimal places.  Section 3 Using Chemical Formulas Chapter 7 Formula Masses, continued Sample Problem F Solution, continued formula mass of KClO3 = 122.55 amu  Section 3 Using Chemical Formulas Chapter 7 The Mole Click below to watch the Visual Concept. Visual Concept  Section 3 Using Chemical Formulas Chapter 7 Molar Masses The molar mass of a substance is equal to the mass in grams of one mole, or approximately 6.022 × 1023 particles, of the substance. example: the molar mass of pure calcium, Ca, is 40.08 g/mol because one mole of calcium atoms has a mass of 40.08 g. The molar mass of a compound is calculated by adding the masses of the elements present in a mole of the molecules or formula units that make up the compound.  Section 3 Using Chemical Formulas Chapter 7 Molar Masses, continued One mole of water molecules contains exactly two moles of H atoms and one mole of O atoms. The molar mass of water is calculated as follows. molar mass of H2O molecule: 18.02 g/mol A compound’s molar mass is numerically equal to its formula mass.  Section 3 Using Chemical Formulas Chapter 7 Molar Mass Click below to watch the Visual Concept. Visual Concept  Section 3 Using Chemical Formulas Chapter 7 Calculating Molar Masses for Ionic Compounds  Section 3 Using Chemical Formulas Chapter 7 Molar Masses, continued Sample Problem G What is the molar mass of barium nitrate, Ba(NO3)2?  Section 3 Using Chemical Formulas Chapter 7 Molar Masses, continued Sample Problem G Solution One mole of barium nitrate, contains one mole of Ba, two moles of N (1 × 2), and six moles of O (3 × 2). molar mass of Ba(NO3)2 = 261.35 g/mol  Section 3 Using Chemical Formulas Chapter 7 Molar Mass as a Conversion Factor The molar mass of a compound can be used as a conversion factor to relate an amount in moles to a mass in grams for a given substance. To convert moles to grams, multiply the amount in moles by the molar mass: Amount in moles × molar mass (g/mol) = mass in grams  Section 3 Using Chemical Formulas Chapter 7 Mole-Mass Calculations  Section 3 Using Chemical Formulas Chapter 7 Molar Mass as a Conversion Factor Click below to watch the Visual Concept. Visual Concept  Section 3 Using Chemical Formulas Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem H What is the mass in grams of 2.50 mol of oxygen gas?  Section 3 Using Chemical Formulas Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem H Solution Given: 2.50 mol O2 Unknown: mass of O2 in grams Solution: moles O2 grams O2 amount of O2 (mol) × molar mass of O2 (g/mol) = mass of O2 (g)  Section 3 Using Chemical Formulas Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem H Solution, continued Calculate the molar mass of O2. Use the molar mass of O2 to convert moles to mass.  Section 3 Using Chemical Formulas Chapter 7 Converting Between Amount in Moles and Number of Particles  Section 3 Using Chemical Formulas Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem I Ibuprofen, C13H18O2, is the active ingredient in many nonprescription pain relievers. Its molar mass is 206.31 g/mol. a. If the tablets in a bottle contain a total of 33 g of ibuprofen, how many moles of ibuprofen are in the bottle? b. How many molecules of ibuprofen are in the bottle? c. What is the total mass in grams of carbon in 33 g of ibuprofen?  Section 3 Using Chemical Formulas Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem I Solution Given: 33 g of C13H18O2 molar mass 206.31 g/mol Unknown: a. moles C13H18O2 b. molecules C13H18O2 c. total mass of C Solution: a. grams moles  Section 3 Using Chemical Formulas Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem I Solution, continued b. moles molecules c. moles C13H18O2 moles C grams C  Section 3 Using Chemical Formulas Chapter 7 Molar Mass as a Conversion Factor, continued Sample Problem I Solution, continued a. b. c.  Section 3 Using Chemical Formulas Chapter 7 Percentage Composition It is often useful to know the percentage by mass of a particular element in a chemical compound. To find the mass percentage of an element in a compound, the following equation can be used. The mass percentage of an element in a compound is the same regardless of the sample’s size.  Section 3 Using Chemical Formulas Chapter 7 Percentage Composition, continued The percentage of an element in a compound can be calculated by determining how many grams of the element are present in one mole of the compound. The percentage by mass of each element in a compound is known as the percentage composition of the compound.  Section 3 Using Chemical Formulas Chapter 7 Percentage Composition of Iron Oxides  Section 3 Using Chemical Formulas Chapter 7 Percentage Composition Click below to watch the Visual Concept. Visual Concept  Section 3 Using Chemical Formulas Chapter 7 Percentage Composition Calculations  Section 3 Using Chemical Formulas Chapter 7 Percentage Composition, continued Sample Problem J Find the percentage composition of copper(I) sulfide, Cu2S.  Section 3 Using Chemical Formulas Chapter 7 Percentage Composition, continued Sample Problem J Solution Given: formula, Cu2S Unknown: percentage composition of Cu2S Solution: formula molar mass mass percentage of each element  Section 3 Using Chemical Formulas Chapter 7 Percentage Composition, continued Sample Problem J Solution, continued Molar mass of Cu2S = 159.2 g  Section 3 Using Chemical Formulas Chapter 7 Percentage Composition, continued Sample Problem J Solution, continued  Section 4 Determining Chemical Chapter 7 Formulas Preview Lesson Starter Objectives Calculation of Empirical Formulas Calculation of Molecular Formulas  Section 4 Determining Chemical Chapter 7 Formulas Lesson Starter Compare and contrast models of the molecules NO2 and N2O4. The numbers of atoms in the molecules differ, but the ratio of N atoms to O atoms for each molecule is the same.  Section 4 Determining Chemical Chapter 7 Formulas Objectives Define empirical formula, and explain how the term applies to ionic and molecular compounds. Determine an empirical formula from either a percentage or a mass composition. Explain the relationship between the empirical formula and the molecular formula of a given compound. Determine a molecular formula from an empirical formula.  Section 4 Determining Chemical Chapter 7 Formulas Empirical and Actual Formulas  Section 4 Determining Chemical Chapter 7 Formulas An empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number mole ratio of the different atoms in the compound. For an ionic compound, the formula unit is usually the compound’s empirical formula. For a molecular compound, however, the empirical formula does not necessarily indicate the actual numbers of atoms present in each molecule. example: the empirical formula of the gas diborane is BH3, but the molecular formula is B2H6.  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Empirical Formulas To determine a compound’s empirical formula from its percentage composition, begin by converting percentage composition to a mass composition. Assume that you have a 100.0 g sample of the compound. Then calculate the amount of each element in the sample. example: diborane The percentage composition is 78.1% B and 21.9% H. Therefore, 100.0 g of diborane contains 78.1 g of B and 21.9 g of H.  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Empirical Formulas, continued Next, the mass composition of each element is converted to a composition in moles by dividing by the appropriate molar mass. These values give a mole ratio of 7.22 mol B to 21.7 mol H.  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Empirical Formulas, continued To find the smallest whole number ratio, divide each number of moles by the smallest number in the existing ratio. Because of rounding or experimental error, a compound’s mole ratio sometimes consists of numbers close to whole numbers instead of exact whole numbers. In this case, the differences from whole numbers may be ignored and the nearest whole number taken.  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Empirical Formulas, continued Sample Problem L Quantitative analysis shows that a compound contains 32.38% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula of this compound.  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Empirical Formulas, continued Sample Problem L Solution Given: percentage composition: 32.38% Na, 22.65% S, and 44.99% O Unknown: empirical formula Solution: percentage composition mass composition composition in moles smallest whole-number mole ratio of atoms  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Empirical Formulas, continued Sample Problem L Solution, continued  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Empirical Formulas, continued Sample Problem L Solution, continued Smallest whole-number mole ratio of atoms: The compound contains atoms in the ratio 1.408 mol Na:0.7063 mol S:2.812 mol O. Rounding yields a mole ratio of 2 mol Na:1 mol S:4 mol O. The empirical formula of the compound is Na2SO4.  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Molecular Formulas The empirical formula contains the smallest possible whole numbers that describe the atomic ratio. The molecular formula is the actual formula of a molecular compound. An empirical formula may or may not be a correct molecular formula. The relationship between a compound’s empirical formula and its molecular formula can be written as follows. x(empirical formula) = molecular formula  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Molecular Formulas, continued The formula masses have a similar relationship. x(empirical formula mass) = molecular formula mass To determine the molecular formula of a compound, you must know the compound’s formula mass. Dividing the experimental formula mass by the empirical formula mass gives the value of x. A compound’s molecular formula mass is numerically equal to its molar mass, so a compound’s molecular formula can also be found given the compound’s empirical formula and its molar mass.  Section 4 Determining Chemical Chapter 7 Formulas Comparing Empirical and Molecular Formulas  Section 4 Determining Chemical Chapter 7 Formulas Comparing Molecular and Empirical Formulas Click below to watch the Visual Concept. Visual Concept  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Molecular Formulas, continued Sample Problem N In Sample Problem M, the empirical formula of a compound of phosphorus and oxygen was found to be P2O5. Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula?  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Molecular Formulas, continued Sample Problem N Solution Given: empirical formula Unknown: molecular formula Solution: x(empirical formula) = molecular formula  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Molecular Formulas, continued Sample Problem N Solution, continued Molecular formula mass is numerically equal to molar mass. molecular molar mass = 283.89 g/mol molecular formula mass = 283.89 amu empirical formula mass mass of phosphorus atom = 30.97 amu mass of oxygen atom = 16.00 amu empirical formula mass of P2O5 = 2 × 30.97 amu + 5 × 16.00 amu = 141.94 amu  Section 4 Determining Chemical Chapter 7 Formulas Calculation of Molecular Formulas, continued Sample Problem N Solution, continued Dividing the experimental formula mass by the empirical formula mass gives the value of x. 2 × (P2O5) = P4O10 The compound’s molecular formula is therefore P4O10. End of Chapter 7 Show  Section 1 Introduction to Chapter 9 Stoichiometry Lesson Starter Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) If 2 mol of HCl react, how many moles of H2 are obtained? How many moles of Mg will react with 2 mol of HCl? If 4 mol of HCl react, how many mol of each product are produced? How would you convert from moles of substances to masses?  Section 1 Introduction to Chapter 9 Stoichiometry Objective Define stoichiometry. Describe the importance of the mole ratio in stoichiometric calculations. Write a mole ratio relating two substances in a chemical equation.  Section 1 Introduction to Chapter 9 Stoichiometry Stoichiometry Definition Stoichiometry is the process of using a balanced chemical equation to calculate relationships between reactants and products. -How many moles of one thing react with how many moles of another to make something new. -This is fundamental because it allows us to calculate real-world quantities, like how much of a chemical you need to produce a certain amount of product.  Section 1 Introduction to Chapter 9 Stoichiometry Mole Ratio A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction Example: 2Al2O3(l) → 4Al(s) + 3O2(g) Mole Ratios: 2 mol Al2O3 2 mol Al2O3 4 mol Al 4 mol Al , 3 mol O2 3 mol O2 ,  Section 1 Introduction to Chapter 9 Stoichiometry Stoichiometry Calculations  Section 2 Ideal Stoichiometric Chapter 9 Calculations Objective Calculate the amount in moles of a reactant or a product from the amount in moles of a different reactant or product. Calculate the mass of a reactant or a product from the amount in moles of a different reactant or product.  Section 2 Ideal Stoichiometric Chapter 9 Calculations Objectives, continued Calculate the amount in moles of a reactant or a product from the mass of a different reactant or product. Calculate the mass of a reactant or a product from the mass of a different reactant or product.  Section 2 Ideal Stoichiometric Chapter 9 Calculations Conversions of Quantities in Moles, continued Sample Problem A In a spacecraft, the carbon dioxide exhaled by astronauts can be removed by its reaction with lithium hydroxide, LiOH, according to the following chemical equation. CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O(l) How many moles of lithium hydroxide are required to react with 20 mol CO2, the average amount exhaled by a person each day?  Section 2 Ideal Stoichiometric Chapter 9 Calculations Conversions of Quantities in Moles, continued Sample Problem A Solution CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O(l) Given: amount of CO2 = 20 mol Unknown: amount of LiOH (mol) Solution: mol ratio  Section 2 Ideal Stoichiometric Chapter 9 Calculations Conversions of Amounts in Moles to Mass, continued Sample Problem B In photosynthesis, plants use energy from the sun to produce glucose, C6H12O6, and oxygen from the reaction of carbon dioxide and water. What mass, in grams, of glucose is produced when 3.00 mol of water react with carbon dioxide?  Section 2 Ideal Stoichiometric Chapter 9 Calculations Conversions of Amounts in Moles to Mass, continued Sample Problem B Solution Given: amount of H2O = 3.00 mol Unknown: mass of C6H12O6 produced (g) Solution: Balanced Equation: 6CO2(g) + 6H2O(l) → C6H12O6(s) + 6O2(g) mol ratio molar mass factor 90.1 g C6H12O6  Section 2 Ideal Stoichiometric Chapter 9 Calculations Conversions of Mass to Amounts in Moles, continued Sample Problem D The first step in the industrial manufacture of nitric acid is the catalytic oxidation of ammonia. NH3(g) + O2(g) → NO(g) + H2O(g) (unbalanced) The reaction is run using 824 g NH3 and excess oxygen. a. How many moles of NO are formed? b. How many moles of H2O are formed?  Section 2 Ideal Stoichiometric Chapter 9 Calculations Conversions of Mass to Amounts in Moles, continued Sample Problem D Solution Given: mass of NH3 = 824 g Unknown: a. amount of NO produced (mol) b. amount of H2O produced (mol) Solution: Balanced Equation: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) molar mass factor mol ratio a. b.  Section 2 Ideal Stoichiometric Chapter 9 Calculations Conversions of Mass to Amounts in Moles, continued Sample Problem D Solution, continued molar mass factor mol ratio a. b.  Section 2 Ideal Stoichiometric Chapter 9 Calculations Mass-Mass to Calculations  Section 2 Ideal Stoichiometric Chapter 9 Calculations Mass-Mass Calculations Click below to watch the Visual Concept. Visual Concept  Section 2 Ideal Stoichiometric Chapter 9 Calculations Mass-Mass to Calculations, continued Sample Problem E Tin(II) fluoride, SnF2, is used in some toothpastes. It is made by the reaction of tin with hydrogen fluoride according to the following equation. Sn(s) + 2HF(g) → SnF2(s) + H2(g) How many grams of SnF2 are produced from the reaction of 30.00 g HF with Sn?  Section 2 Ideal Stoichiometric Chapter 9 Calculations Mass-Mass to Calculations, continued Sample Problem E Solution Given: amount of HF = 30.00 g Unknown: mass of SnF2 produced (g) Solution: molar mass factor mol ratio molar mass factor = 117.5 g SnF2  Section 3 Limiting Reactants and Chapter 9 Percentage Yield Objectives Describe a method for determining which of two reactants is a limiting reactant. Calculate the amount in moles or mass in grams of a product, given the amounts in moles or masses in grams of two reactants, one of which is in excess. Distinguish between theoretical yield, actual yield, and percentage yield. Calculate percentage yield, given the actual yield and quantity of a reactant.  Section 3 Limiting Reactants and Chapter 9 Percentage Yield Limiting Reactants The limiting reactant is the reactant that limits the amount of the other reactant that can combine and the amount of product that can form in a chemical reaction. The excess reactant is the substance that is not used up completely in a reaction.  Section 3 Limiting Reactants and Chapter 9 Percentage Yield Limited Reactants, continued Sample Problem F Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equation. SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l) If 6.0 mol HF is added to 4.5 mol SiO2, which is the limiting reactant?  Section 3 Limiting Reactants and Chapter 9 Percentage Yield Limited Reactants, continued Sample Problem F Solution SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l) Given: amount of HF = 6.0 mol amount of SiO2 = 4.5 mol Unknown: limiting reactant Solution: mole ratio  Section 3 Limiting Reactants and Chapter 9 Percentage Yield Limited Reactants, continued Sample Problem F Solution, continued SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(l) HF is the limiting reactant.  Section 3 Limiting Reactants and Chapter 9 Percentage Yield Percentage Yield The theoretical yield is the maximum amount of product that can be produced from a given amount of reactant. The actual yield of a product is the measured amount of that product obtained from a reaction. The percentage yield is the ratio of the actual yield to the theoretical yield, multiplied by 100.  Section 3 Limiting Reactants and Chapter 9 Percentage Yield Percentage Yield, continued Sample Problem H Chlorobenzene, C6H5Cl, is used in the production of many important chemicals, such as aspirin, dyes, and disinfectants. One industrial method of preparing chlorobenzene is to react benzene, C6H6, with chlorine, as represented by the following equation. C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g) When 36.8 g C6H6 react with an excess of Cl2, the actual yield of C6H5Cl is 38.8 g. What is the percentage yield of C6H5Cl?  Section 3 Limiting Reactants and Chapter 9 Percentage Yield Percentage Yield, continued Sample Problem H Solution C6H6 (l) + Cl2(g) → C6H5Cl(l) + HCl(g) Given: mass of C6H6 = 36.8 g mass of Cl2 = excess actual yield of C6H5Cl = 38.8 g Unknown: percentage yield of C6H5Cl Solution: Theoretical yield molar mass factor mol ratio molar mass  Section 3 Limiting Reactants and Chapter 9 Percentage Yield Percentage Yield, continued Sample Problem H Solution, continued C6H6(l) + Cl2(g) → C6H5Cl(l) + HCl(g) Theoretical yield Percentage yield  AP CHEMISTRY Test Booklet Unit 4 Practice Exam 1. Which of the following has the bonds arranged in order of decreasing polarity? (A) H−F > N−F > F−F (B) H−I > H−Br > H−F (C) O−N > O−S > O−Te (D) Sb−I > Sb−Te > Sb−Cl 2. Which of the following can be inferred from the diagram above that shows the dependence of potential energy on the internuclear distance between two atoms? (A) The atoms form a bond with a bond length of. (B) The atoms form a bond with a bond length of. (C) The net force between the atoms is attractive at. (D) The net force between the atoms is attractive at. 3. Which of the following bonds represented in the diagram above is the most polar? (A) (B) (C) (D) 4. Which of the following scientific claims about the bond in the molecular compound is most likely to be  true? AP Chemistry Page 1 of 6 Test Booklet Unit 4 Practice Exam (A) There is a partial negative charge on the atom. (B) Electrons are shared equally between the and atoms. (C) The bond is extremely weak. (D) The bond is highly polar. 5. A particle-level diagram of a metallic element is shown above. Typically, metals are both malleable and ductile. The best explanation for these properties is that the electrons involved in bonding among metal atoms are (A) unequally shared and form nondirectional bonds (B) unequally shared and form highly directional bonds (C) equally shared and form nondirectional bonds (D) equally shared and form highly directional bonds 6. Which of the following compounds contains both ionic and covalent bonds? (A) C2H5OH (B) MgF2 (C) H2S (D) NH4Cl 7. Two pure elements react to form a compound. One element is an alkali metal, , and the other element is a halogen,. Which of the following is the most valid scientific claim that can be made about the compound? (A) It has the formula. (B) It does not dissolve in water. (C) It contains ionic bonds. (D) It contains covalent bonds. 8. The elements and have the same electronegativity value, 2.55. Which of the following claims about the compound that forms from and is most likely to be true? Page 2 of 6 AP Chemistry Test Booklet Unit 4 Practice Exam (A) The carbon-to-selenium bond is unstable. (B) The carbon-to-selenium bond is nonpolar covalent. (C) The compound has the empirical formula. (D) A molecule of the compound will have a partial negative charge on the carbon atom. 9. Based on the diagrams shown above, which of the following would have the greatest dipole moment, and why? (A) , because the double bonds create an area of higher electron density around the. (B) , because the draw electron density away from the and the bond dipoles do not cancel one another. (C) , because it has the greatest number of polar bonds. (D) , because the large electronegativity difference between and results in very polar bonds. 10. The geometry of the SO3 molecule is best described as (A) trigonal planar (B) trigonal pyramidal (C) bent (D) tetrahedral  11. Which of the following Lewis electron-dot diagrams represents the molecule that contains the smallest bond angle? AP Chemistry Page 3 of 6 Test Booklet Unit 4 Practice Exam (A)  (B) (C) (D) 12. Molecule Bond Dissociation Energy The table above shows the bond dissociation energies for and at. Which of the following correctly shows the potential energy as a function of internuclear distance for two bromine atoms and two iodine atoms? Page 4 of 6 AP Chemistry Test Booklet Unit 4 Practice Exam  (A) (B) (C) (D) AP Chemistry Page 5 of 6 Test Booklet Unit 4 Practice Exam 13. The diagram above shows two resonance structures for a molecule of. The phenomenon shown in the diagram best supports which of the following claims about the bonding in ? (A) In the molecule, all the bonds between the carbon atoms have the same length. (B) Because of variable bonding between its carbon atoms, is a good conductor of electricity. (C) The bonds between carbon atoms in are unstable, and the compound decomposes quickly. (D) The molecule contains three single bonds between carbon atoms and three double bonds between carbon atoms. 14. Which of the following species is NOT planar? (A) CO32- (B) NO3- (C) ClF3 (D) PCl3 Page 6 of 6 AP Chemistry Name: ________________________ Period ___________________ Honors Chemistry 10/22/2024 ID: A Hrs Unit 3 Practice Exam 3 ____ 1. Which has these types of electromagnetic radiation arranged in order of decreasing frequency? A) visible, ultraviolet, x-ray B) radiowaves, visible, ultraviolet C) ultraviolet, visible, infrared D) x-ray, visible, ultraviolet E) gamma, microwave, visible ____ 2. Which has these types of electromagnetic radiation arranged in order of increasing wavelength? A) visible, ultraviolet, x-ray B) radiowaves, visible, ultraviolet C) ultraviolet, visible, infrared D) x-ray, visible, ultraviolet E) none of these ____ 3. Which has these types of electromagnetic radiation arranged in order of decreasing energy? A) visible, ultraviolet, x-ray B) radiowaves, visible, ultraviolet C) ultraviolet, visible, infrared D) x-ray, visible, ultraviolet E) none of these ____ 4. In the quantum theory, the magnetic orbital momentum quantum number is most directly associated with which property of orbitals? A) energy C) orientation B) size D) shape ____ 5. In the quantum theory, the angular orbital momentum quantum number is most directly associated with which property of orbitals? A) energy C) orientation B) size D) shape ____ 6. The ground state electron configuration for Te is: A) [Kr]5s24d105p4 B) [Kr]5s25p64d8 C) [Kr]5s25d105p4 D) [Kr]5s24f14 E) Kr]5s25d105p6 1 Name: ________________________ ID: A ____ 7. Select the correct electron configuration for Cu. A) [Ar]4s23d9 B) [Ar]4s13d10 C) [Ar]4s24p63d3 D) [Ar]4s24d9 E) [Ar]3d10 ____ 8. Which of the following lists of atoms are arranged in order of INCREASING first ionization energy? A) Li < O < N < F B) Li < N < O < F C) F < O < N < Li D) Na < Sr < O < F E) Ca > Cs > S > Se ____ 9. Which of the following lists of atoms are arranged in order of DECREASING atomic radius? A) Li > O > N > F B) Li > N > O > F C) F > O > N > Li D) Na > Sr > O > F E) Ca < Cs < S < Se 10. How many electrons in an atom can have the following quantum numbers? a) n = 3 b) n = 2, l = 0 c) n = 2, l = 2, ml = 0 d) n = 2, l = 0, ml = 0, ms = 1/2 Write the electron configuration for the following: 11. Ag 12. I 13. Identify the following elements as metals, nonmetals, or metalloids: (a) aluminum; (b) carbon; (c) germanium; (d) arsenic; (e) selenium; (f) tellurium.  2 Name: ________________________ ID: A ____ 14. Which form of electromagnetic radiation has the longest wavelengths? A) gamma rays B) microwaves C) radio waves D) infrared radiation E) x-rays ____ 15. A line in the spectrum of atomic mercury has a wavelength of 254 nm. When mercury emits a photon of light at this wavelength, the frequency of this light is A) 8.47  10–16 s–1 B) 7.82  10–19 s–1 C) 1.18  1015 s–1 D) 76.1 s–1 E) none of these ____ 16. The four lines observed in the visible emission spectrum of hydrogen tell us that: A) The hydrogen molecules they came from have the formula H4. B) We could observe more lines if we had a stronger prism. C) There are four electrons in an excited hydrogen atom. D) Only certain energies are allowed for the electron in a hydrogen atom. E) The spectrum is continuous. ____ 17. When a hydrogen electron makes a transition from n = 3 to n = 1, which of the following statements is true? I. Energy is emitted. II. Energy is absorbed. III. The electron loses energy. IV. The electron gains energy. V. The electron cannot make this transition. A) I, IV B) I, III C) II, III D) II, IV E) V ____ 18. How many f orbitals have the value n = 3?  A) 0 B) 3 C) 5 D) 7 E) 1 3 Name: ________________________ ID: A ____ 19. How many f orbitals have n = 6? A) 2 B) 7 C) 10 D) 5 E) 18 ____ 20. Mendeleev is given the most credit for the concept of a periodic table of the elements because: A) He had the longest history of research in elemental properties. B) He emphasized its usefulness in predicting the existence and properties of unknown elements. C) His representation of the table was the most understandable. D) His periodic table was arranged in octaves. E) He grouped elements into triads of similar properties. ____ 21. Which of the following atoms or ions has three unpaired electrons? A) N B) O C) Al D) S2– E) Ti2+ ____ 22. The electron configuration for the carbon atom is: A) 1s22s22p2 B) [He]2s4 C) [Ne]2s22p2 D) 1s22p4 E) none of these ____ 23. The complete electron configuration of tin is A) 1s22s22p63s23p64s23d104p65s24d105d105p2 B) 1s22s22p63s23p64s23d104d104p2 C) 1s22s22p63s23p64s24p65s24d105d105p2 D) 1s22s22p63s23p64s23d104p65s24d105p2 E) none of these ____ 24. The statement that "the lowest energy configuration for an atom is the one having the maximum number of unpaired electrons allowed by the Pauli principle in a particular  set of degenerate orbitals" is known as A) the aufbau principle B) Hund's rule C) Heisenberg uncertainty principle D) the Pauli exclusion principle E) the quantum model 4 Name: ________________________ ID: A ____ 25. For which of the following elements does the electron configuration for the lowest energy state show a partially filled d orbital? A) Ti B) Rb C) Cu D) Ga E) Kr ____ 26. Which of the following is the highest energy orbital for a silicon atom? A) 1s B) 2s C) 3s D) 3p E) 3d ____ 27. Which of the following processes represents the ionization energy of bromine? A) Br(s) Br+(g) + e– B) Br(l) Br+(g) + e– C) Br(g) Br+(g) + e– D) Br(s) Br+(s) + e– E) Br2(g) Br2+(g) + e– ____ 28. Order the elements S, Cl, and F in terms of increasing ionization energy. A) S, Cl, F B) Cl, F, S C) F, S, Cl D) F, Cl, S E) S, F, Cl ____ 29. Order the elements S, Cl, and F in terms of increasing atomic radii. A) S, Cl, F B) Cl, F, S C) F, S, Cl D) F, Cl, S E) S, F, Cl 30. When examining the electromagnetic spectrum, why is it more harmful to be exposed  to x-rays than radio waves over a period of time? In your explanation, include the concepts of frequency, waves, and energy. Also, draw waves to assist in your explanation. 31. How does the Bohr theory explain the emission and absorption spectra of hydrogen? 5 ID: A Hrs Unit 3 Practice Exam 3 Answer Section 1. ANS: C PTS: 1 DIF: Easy REF: 7.1 KEY: Chemistry | general chemistry | atomic theory | light | electromagnetic radiation MSC: Conceptual 2. ANS: C PTS: 1 DIF: Easy REF: 7.1 KEY: Chemistry | general chemistry | atomic theory | light | electromagnetic radiation MSC: Conceptual 3. ANS: C PTS: 1 DIF: Easy REF: 7.1 KEY: Chemistry | general chemistry | atomic theory | light | electromagnetic radiation MSC: Conceptual 4. ANS: C PTS: 1 DIF: Easy REF: 7.6 KEY: Chemistry | general chemistry | atomic theory | quantum mechanics | quantum numbers | magnteic quantum number MSC: Conceptual 5. ANS: D PTS: 1 DIF: Easy REF: 7.6 KEY: Chemistry | general chemistry | atomic theory | quantum mechanics | quantum numbers | angular quantum number MSC: Conceptual 6. ANS: A PTS: 1 DIF: Easy REF: 7.11 KEY: Chemistry | general chemistry | atomic theory | electronic structure of atoms | electron configuration | aufbau principle MSC: Conceptual 7. ANS: B PTS: 1 DIF: Easy REF: 7.11 KEY: Chemistry | general chemistry | atomic theory | electronic structure of atoms | electron configuration and the periodic table | exceptions to the aufbau principle MSC: Conceptual 8. ANS: A PTS: 1 DIF: Moderate REF: 7.12 KEY: Chemistry | general chemistry | atomic theory | periodicity of the elements | periodic properties | ionization energy MSC: Conceptual 9. ANS: B PTS: 1 DIF: Easy REF: 7.12 KEY: Chemistry | general chemistry | atomic theory | periodicity of the elements | periodic properties | ionization energy MSC: Conceptual 10. ANS: a) 18; b) 2; c) 0; d) 1  a) The n = 3 level consists of an s, three p, and five d orbitals, each of which may contain 2 electrons, for a total of 18 electrons. b) n = 2, l = 0 describes the 2s orbital, which may contain 2 electrons. c) This set of quantum numbers is impossible, since when n = 2, l can only be 0 or 1. d) This set of four quantum numbers describes one specific electron in the 2s orbital. See Sec. 7.6 and 7.8 in Zumdahl Chemistry. PTS: 1 DIF: Moderate REF: 7.8 KEY: Chemistry | general chemistry | atomic theory | quantum mechanics | quantum numbers MSC: Conceptual 1 ID: A 11. ANS: 1s22s22p63s23p64s23d104p65s14d10 or [Kr]5s14d10 PTS: 1 DIF: Moderate REF: 7.11 KEY: Chemistry | general chemistry | atomic theory | electronic structure of atoms | electron configuration MSC: Conceptual 12. ANS: 1s22s22p63s23p64s23d104p65s24d105p5 or [Kr]5s24d105p5 PTS: 1 DIF: Easy REF: 7.11 KEY: Chemistry | general chemistry | atomic theory | electronic structure of atoms | electron configuration MSC: Conceptual 13. ANS: a. metal b. nonmetal c. metalloid d. metalloid e. nonmetal f. metalloid PTS: 1 14. ANS: C PTS: 1 DIF: Easy REF: 7.1 KEY: Chemistry | general chemistry | atomic theory | light | electromagnetic radiation MSC: Conceptual 15. ANS: C PTS: 1 DIF: Easy REF: 7.1 KEY: Chemistry | general chemistry | atomic theory | light | electromagnetic radiation MSC: Quantitative 16. ANS: D PTS: 1 DIF: Easy REF: 7.3 KEY: Chemistry | general chemistry | atomic theory | light | Bohr theory | atomic line spectra MSC: Conceptual 17. ANS: B PTS: 1 DIF: Easy REF: 7.4 KEY: Chemistry | general chemistry | atomic theory | light | Bohr theory | atomic line spectra MSC: Conceptual 18. ANS: A PTS: 1 DIF: Easy REF: 7.6 KEY: Chemistry | general chemistry | atomic theory | quantum mechanics | quantum numbers MSC: Conceptual 19. ANS: B PTS: 1 DIF: Easy REF: 7.6 KEY: Chemistry | general chemistry | atomic  theory | quantum mechanics | quantum numbers MSC: Conceptual 20. ANS: B PTS: 1 DIF: Easy REF: 7.1 KEY: Chemistry | general chemistry | atomic theory | periodicity of the elements | Mendeleev's predictions MSC: Conceptual 2 ID: A 21. ANS: A PTS: 1 DIF: Moderate REF: 7.11 KEY: Chemistry | general chemistry | atomic theory | electronic structure of atoms | electron configuration | aufbau principle MSC: Conceptual 22. ANS: A PTS: 1 DIF: Easy REF: 7.11 KEY: Chemistry | general chemistry | atomic theory | electronic structure of atoms | electron configuration | aufbau principle MSC: Conceptual 23. ANS: D PTS: 1 DIF: Easy REF: 7.11 KEY: Chemistry | general chemistry | atomic theory | electronic structure of atoms | electron configuration | aufbau principle MSC: Conceptual 24. ANS: B PTS: 1 DIF: Easy REF: 7.11 KEY: Chemistry | general chemistry | atomic theory | electronic structure of atoms | electron configuration | Hund's rule MSC: Conceptual 25. ANS: A PTS: 1 DIF: Moderate REF: 7.11 KEY: Chemistry | general chemistry | atomic theory | electronic structure of atoms | electron configuration MSC: Conceptual 26. ANS: D PTS: 1 DIF: Easy REF: 7.11 KEY: Chemistry | general chemistry | atomic theory | electronic structure of atoms | electron configuration and the periodic table MSC: Conceptual 27. ANS: C PTS: 1 DIF: Easy REF: 7.12 KEY: Chemistry | general chemistry | atomic theory | periodicity of the elements | periodic properties | ionization energy MSC: Conceptual 28. ANS: A PTS: 1 DIF: Easy REF: 7.12 KEY: Chemistry | general chemistry | atomic theory | periodicity of the elements | periodic properties | ionization energy MSC: Conceptual 29. ANS: D PTS: 1 DIF: Easy REF: 7.12 KEY: Chemistry | general chemistry | atomic theory | periodicity of the elements | periodic properties | atomic radius MSC: Conceptual  3 ID: A 30. ANS: X-rays have a shorter wavelength (in the 10–10 m range) and therefore a higher frequency (since c = ). This means that the number of waves that penetrate your body over a period of time will be greater than the number of radio waves that hit your body in that same period of time (since radio waves have a longer wavelength and lower frequency). Refer to the picture below: X rays shorter , higher Radio Waves longer , lower Since more x-ray waves penetrate your body, you are being exposed to more energy in the form of radiation, which can be harmful. See Sec 7.1 and 7.2 of Zumdahl Chemistry. PTS: 1 DIF: Moderate REF: 7.2 KEY: Chemistry | general chemistry | atomic theory | light | electromagnetic radiation MSC: Conceptual  31. ANS: The Bohr model assumes the electron can occupy circular orbits with specific energies at corresponding specific distances from the nucleus. The electron can move from one orbital (energy level) to another by releasing or absorbing the amount of energy corresponding to the difference in energy between the two orbitals. The observed emission and absorption spectra show only the discrete emissions or absorptions for these energy differences. See Sec. 7.4 of Zumdahl, Chemistry. PTS: 1 DIF: Moderate REF: 7.4 KEY: Chemistry | general chemistry | atomic theory | light | Bohr theory | atomic line spectra MSC: Conceptual 4 Hrs Unit 3 Practice Exam 3 [Answer Strip] ID: A C _____ 7. C _____ 17. B _____ 1. B _____ 15. C _____ 4. C _____ C _____ 8. A _____ 16. 5. D 2. C _____ 18. A _____ 19. _____ 6. D _____ _____ 9. B _____ 14. _____ 3. A  B _____ 20. B _____ 26. D _____ _____ 21. 27. C A _____ 22. A _____ 28. _____ 23. A _____ D 29. D _____ 24. B _____ 25. A Name: CHAPTER 21 REVIEW Class: Date: Nuclear Chemistry SECTION 1 SHORT ANSWER Answer the following questions in the space provided. 1. Based on the information about the three elementary particles in the text, which has the greatest mass? (a) the proton (b) the neutron (c) the electron (d) They all have the same mass. 2. The force that keeps nucleons together is 3. 4. 5. (a) a strong nuclear force.  (b) a weak nuclear force. (c) an electromagnetic force. (d) a gravitational force. The stability of a nucleus is most affected by the (a) number of neutrons. (b) number of protons. (c) number of electrons. (d) ratio of neutrons to protons. If an atom should form from its constituent particles, (a) matter is lost and energy is taken in. (b) matter is lost and energy is released. (c) matter is gained and energy is taken in. (d) matter is gained and energy is released. For atoms of a given mass number, those with greater mass defects, have (a) smaller binding energies per nucleon. (b) greater binding energies per nucleon. (c) the same binding energies per nucleon as those with smaller mass defects. (d) variable binding energies per nucleon. 6. Based on Figure 1.1 of the text, which isotope of He, helium-3 or helium-4, a. has the smaller binding energy per nucleon? b. is more stable to nuclear changes? 7. The number of neutrons in an atom of magnesium-25 is 8. Nuclides of the same element have the same Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor. Modern Chemistry  169 Nuclearchemistry Name: Class: Date: SECTION 1 continued 9. Atom X has 50 nucleons and a binding energy of 4.2 × 10-11 J. Atom Z has 80 nucleons and a binding energy of 8.4 x 10-11 J. a. The mass defect of Z is twice that of X. True or False? b. Which atom has the greater binding energy per nucleon? c. Which atom is likely to be more stable to nuclear transmutations? 10. Identify the missing term in each of the following nuclear equations. Write the element's symbol, its atomic number, and its mass number. a. 14C →e+ b. 3 Cu + H→ 63 29 + He 11. Write the equation that shows the equivalency of mass and energy. 56 12. Consider the two nuclides 26 Fe 14C and a. Determine the number of protons in each nucleus. b. Determine the number of neutrons in each nucleus. 56 26  c. Determine whether the Fe nuclide is likely to be stable or unstable, based on its position in the band of stability shown in Figure 1.2 of the text. PROBLEM Write the answer on the line to the left. Show all your work in the space provided. 13. Neon-20 is a stable isotope of neon. Its actual mass has been found to be 19.992 44 u. Determine the mass defect in this nuclide. Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor. Modern Chemistry 170 Nuclearchemistry Name: CHAPTER 21 REVIEW Class: Date: Nuclear Chemistry SECTION 2 SHORT ANSWER Answer the following questions in the space provided. 1. The nuclear equation 210 84 Po→ 206 Pb+ + He is an example of an equation that represents (a) alpha emission.  2. 3. (b) beta emission. (c) positron emission. (d) electron capture. When Zundergoes electron capture to form a new element X, which of the following best represents the product formed? (a) a-1 X a+1 X a (b) (c) b+iX (d) b-ix Which of the following best represents the fraction of a radioactive sample that remains after four half-lives have occurred? 4 (a ) ( 2 ) x4 (b)  (c) 4 x4 (d ) 2 4. Match the nuclear symbol on the right to the name of the corresponding particle on the left. beta particle proton positron alpha particle 5. Label each of the following statements as True (a) ip (b) 2 He (c) -i B (d) +iB +1 β or False. Consider the U-238 decay series in Figure 2.7 of the text. For the series of decays from U-234 to Po-218, each nuclide a. shares the same atomic number b. differs in mass number from others by multiples of 4 c. has a unique atomic number d. differs in atomic number from others by multiples of 4  Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor. Modern Chemistry 171 Nuclearchemistry Name: SECTION 2 continued 6. Class: Date: Identify the missing term in the following nuclear equation. Write the element's symbol, its atomic number, and its mass number. ? 231Th + He 90 2 7. Lead-210 undergoes beta emission. Write the nuclear equation showing this transmutation. 8. Einsteinium is a transuranium element. Einsteinium-247 can be prepared by bombarding uranium-238 with nitrogen-14 nuclei, releasing several neutrons, as shown by the following equation: 238U + 14N → 247ES + xon 92 Es 99 xó n What must be the value of x in the above equation? Explain your reasoning. PROBLEMS Write the answer on the line to the left. Show all your work in the space provided. 9. Phosphorus-32 has a half-life of 14.3 days. How many days will it take for a sample of phosphorus-32 to decay to one-eighth its original amount? 10. Iodine-131 has a half-life of 8.0 days. How many grams of an original 160 mg sample will remain after 40 days? 11. Carbon-14 has a half-life of 5715 years. It is used to determine the age of ancient objects. If a sample today contains 0.060 mg of carbon-14, how much carbon-14 must have been present in the sample 11 430 years ago? Original content Copyright © by Holt, Rinehart and Winston. Additions and changes to the original content are the responsibility of the instructor. Modern Chemistry 172 Nuclearchemistry  Section 1 Introduction to Chapter 6 Chemical Bonding Objectives Define chemical bond. Explain why most atoms form chemical bonds. Describe ionic and covalent bonding. Explain why most chemical bonding is neither purely ionic nor purely covalent. Classify bonding type according to electronegativity differences.  Section 1 Introduction to Chapter 6 Chemical Bonding Chemical Bond A mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together. Two types of bonding: 1. Ionic Bonding: Involves a transfer of electrons. One element loses electrons and the other gains electrons. 2. Covalent Bonding: Involves a sharing of electrons. Atoms will share in order to reach a stable electron configuration. Two types of bonding: 1. Ionic Bonding: Involves a transfer of electrons. One element loses electrons and the other gains electrons. 2. Covalent Bonding: Involves a sharing of electrons. Atoms will share in order to reach a stable electron configuration. Electronegativity  Section 1 Introduction to Chapter 6 Chemical Bonding Use Electronegativity Difference to Classify Bonding  Section 1 Introduction to Chapter 6 Chemical Bonding Chemical Bonding, continued Sample Problem A Use electronegativity values listed in Figure 20 from the previous chapter in your book, on page 161, and Figure 2 in your book, on page 176, to classify bonding between sulfur, S, and the following elements: hydrogen, H; cesium, Cs; and chlorine, Cl. In each pair, which atom will be more negative?  Section 1 Introduction to Chapter 6 Chemical Bonding Chemical Bonding, continued Sample Problem A Solution The electronegativity of sulfur is 2.5. The electronegativities of hydrogen, cesium, and chlorine are 2.1, 0.7, and 3.0, respectively. In each pair, the atom with the larger electronegativity will be the more-negative atom. Bonding between Electroneg. More-neg- sulfur and difference Bond type ative atom hydrogen 2.5 – 2.1 = 0.4 polar-covalent sulfur cesium 2.5 – 0.7 = 1.8 ionic sulfur chlorine 3.0 – 2.5 = 0.5 polar-covalent chlorine  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Preview Objectives Molecular Compounds Formation of a Covalent Bond Characteristics of the Covalent Bond The Octet Rule Electron-Dot Notation Lewis Structures Multiple Covalent Bonds  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Objectives Define molecule and molecular formula. Explain the relationships among potential energy, distance between approaching atoms, bond length, and bond energy. State the octet rule.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Objectives, continued List the six basic steps used in writing Lewis structures. Explain how to determine Lewis structures for molecules containing single bonds, multiple bonds, or both. Explain why scientists use resonance structures to represent some molecules.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Molecular Compounds A molecule is a neutral group of atoms that are held together by covalent bonds. A chemical compound whose simplest units are molecules is called a molecular compound.  Visual Concepts Chapter 6 Molecule  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Molecular Compounds The composition of a compound is given by its chemical formula. A chemical formula indicates the relative numbers of atoms of each kind in a chemical compound by using atomic symbols and numerical subscripts. A molecular formula shows the types and numbers of atoms combined in a single molecule of a molecular compound.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Chemical Formula Click below to watch the Visual Concept. Visual Concept  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Structure of a Water Molecule Click below to watch the Visual Concept. Visual Concept  Visual Concepts Chapter 6 Comparing Monatomic, Diatomic, and Polyatomic Molecules  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Formation of a Covalent Bond Most atoms have lower potential energy when they are bonded to other atoms than they have as they are independent particles. The figure below shows potential energy changes during the formation of a hydrogen-hydrogen bond.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Formation of a Covalent Bond The electron of one atom and proton of the other atom attract one another. The two nuclei and two electrons repel each other. These two forces cancel out to form a covalent bond at a length where the potential energy is at a minimum.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Formation of a Covalent Bond  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Characteristics of the Covalent Bond The distance between two bonded atoms at their minimum potential energy (the average distance between two bonded atoms) is the bond length. In forming a covalent bond, the hydrogen atoms release energy. The same amount of energy must be added to separate the bonded atoms. Bond energy is the energy required to break a chemical bond and form neutral isolated atoms.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Bond Energy Click below to watch the Visual Concept. Visual Concept  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Bond Length and Stability  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Bond Energies and Bond Lengths for Single Bonds  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Characteristics of the Covalent Bond When two atoms form a covalent bond, their shared electrons form overlapping orbitals. This achieves a noble-gas configuration. The bonding of two hydrogen atoms allows each atom to have the stable electron configuration of helium, 1s2.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds The Octet Rule Noble gas atoms are unreactive because their electron configurations are especially stable. This stability results from the fact that the noble-gas atoms’ outer s and p orbitals are completely filled by a total of eight electrons. Other atoms can fill their outermost s and p orbitals by sharing electrons through covalent bonding. Such bond formation follows the octet rule: Chemical compounds tend to form so that each atom, by gaining, losing, or sharing electrons, has an octet of electrons in its highest energy level.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds The Octet Rule Click below to watch the Visual Concept. Visual Concept  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds The Octet Rule, continued Exceptions to the Octet Rule Exceptions to the octet rule include those for atoms that cannot fit eight electrons, and for those that can fit more than eight electrons, into their outermost orbital. Hydrogen forms bonds in which it is surrounded by only two electrons. Boron has just three valence electrons, so it tends to form bonds in which it is surrounded by six electrons. Main-group elements in Periods 3 and up can form bonds with expanded valence, involving more than eight electrons.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Electron-Dot Notation To keep track of valence electrons, it is helpful to use electron-dot notation. Electron-dot notation is an electron-configuration notation in which only the valence electrons of an atom of a particular element are shown, indicated by dots placed around the element’s symbol. The inner-shell electrons are not shown.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Electron-Dot Notation Click below to watch the Visual Concept. Visual Concept  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Electron-Dot Notation, continued Sample Problem B a. Write the electron-dot notation for hydrogen. b. Write the electron-dot notation for nitrogen.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Electron-Dot Notation, continued Sample Problem B Solution a. A hydrogen atom has only one occupied energy level, the n = 1 level, which contains a single electron. b. The group notation for nitrogen’s family of elements is ns2np3. Nitrogen has five valence electrons.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Lewis Structures Electron-dot notation can also be used to represent molecules. The pair of dots between the two symbols represents the shared electron pair of the hydrogen-hydrogen covalent bond. For a molecule of fluorine, F2, the electron-dot notations of two fluorine atoms are combined.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Lewis Structures The pair of dots between the two symbols represents the shared pair of a covalent bond. In addition, each fluorine atom is surrounded by three pairs of electrons that are not shared in bonds. An unshared pair, also called a lone pair, is a pair of electrons that is not involved in bonding and that belongs exclusively to one atom.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Lewis Structures Click below to watch the Visual Concept. Visual Concept  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Lewis Structures The pair of dots representing a shared pair of electrons in a covalent bond is often replaced by a long dash. example: A structural formula indicates the kind, number, and arrangement, and bonds but not the unshared pairs of the atoms in a molecule. example: F–F H–Cl  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Lewis Structures The Lewis structures and the structural formulas for many molecules can be drawn if one knows the composition of the molecule and which atoms are bonded to each other. A single covalent bond, or single bond, is a covalent bond in which one pair of electrons is shared between two atoms.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Lewis Structures, continued Sample Problem C Draw the Lewis structure of iodomethane, CH3I.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Lewis Structures, continued Sample Problem C Solution 1. Determine the type and number of atoms in the molecule. The formula shows one carbon atom, one iodine atom, and three hydrogen atoms. 2. Write the electron-dot notation for each type of atom in the molecule. Carbon is from Group 14 and has four valence electrons. Iodine is from Group 17 and has seven valence electrons. Hydrogen has one valence electron.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Lewis Structures, continued Sample Problem C Solution, continued 3. Determine the total number of valence electrons available in the atoms to be combined. C 1 × 4e– = 4e– I 1 × 7e– = 7e– 3H 3 × 1e– = 3e– 14e–  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Lewis Structures, continued Sample Problem C Solution, continued 4. If carbon is present, it is the central atom. Otherwise, the least-electronegative atom is central. Hydrogen, is never central. 5. Add unshared pairs of electrons to each nonmetal atom (except hydrogen) such that each is surrounded by eight electrons.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Multiple Covalent Bonds A double covalent bond, or simply a double bond, is a covalent bond in which two pairs of electrons are shared between two atoms. Double bonds are often found in molecules containing carbon, nitrogen, and oxygen. A double bond is shown either by two side-by-side pairs of dots or by two parallel dashes.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Multiple Covalent Bonds A triple covalent bond, or simply a triple bond, is a covalent bond in which three pairs of electrons are shared between two atoms. example 1—diatomic nitrogen: example 2—ethyne, C2H2:  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Comparing Single, Double, and Triple Bonds Click below to watch the Visual Concept. Visual Concept  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Multiple Covalent Bonds Double and triple bonds are referred to as multiple bonds, or multiple covalent bonds. In general, double bonds have greater bond energies and are shorter than single bonds. Triple bonds are even stronger and shorter than double bonds. When writing Lewis structures for molecules that contain carbon, nitrogen, or oxygen, remember that multiple bonds between pairs of these atoms are possible.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Drawing Lewis Structures with Many Atoms  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Drawing Lewis Structures with Many Atoms  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Multiple Covalent Bonds, continued Sample Problem D Draw the Lewis structure for methanal, CH2O, which is also known as formaldehyde.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Multiple Covalent Bonds, continued Sample Problem D Solution 1. Determine the number of atoms of each element present in the molecule. The formula shows one carbon atom, two hydrogen atoms, and one oxygen atom. 2. Write the electron-dot notation for each type of atom. Carbon is from Group 14 and has four valence electrons. Oxygen, which is in Group 16, has six valence electrons. Hydrogen has only one valence electron.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Multiple Covalent Bonds, continued Sample Problem D Solution, continued 3. Determine the total number of valence electrons available in the atoms to be combined. C 1 × 4e– = 4e– O 1 × 6e– = 6e– 2H 2 × 1e– = 2e– 12e–  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Multiple Covalent Bonds, continued Sample Problem D Solution, continued 4. Arrange the atoms to form a skeleton structure for the molecule. Connect the atoms by electron-pair bonds. 5. Add unshared pairs of electrons to each nonmetal atom (except hydrogen) such that each is surrounded by eight electrons.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Multiple Covalent Bonds, continued Sample Problem D Solution, continued 6a.Count the electrons in the Lewis structure to be sure that the number of valence electrons used equals the number available. The structure has 14 electrons. The structure has two valence electrons too many. 6b.Subtract one or more lone pairs until the total number of valence electrons is correct. Move one or more lone electron pairs to existing bonds until the outer shells of all atoms are completely filled.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Multiple Covalent Bonds, continued Sample Problem D Solution, continued Subtract the lone pair of electrons from the carbon atom. Move one lone pair of electrons from the oxygen to the bond between carbon and oxygen to form a double bond.  Section 2 Covalent Bonding and Chapter 6 Molecular Compounds Atomic Resonance Cli