Chemistry Atoms First Julia Burdge Jason Overby Chapter
Chemistry: Atoms First Julia Burdge & Jason Overby Chapter 5 Ionic and Covalent Compounds Kent L. Mc. Corkle Cosumnes River College Sacramento, CA Copyright (c) The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.
5 Ionic and Covalent Compounds 5. 1 Compounds 5. 2 Lewis Dot Symbols 5. 3 Ionic Compounds and Bonding 5. 4 Naming Ions and Ionic Compounds Formulas of Ionic Compounds Naming Ionic Compounds 5. 5 Covalent Bonding and Molecules Molecular Formulas Empirical Formulas 5. 6 Naming Molecular Compounds Specifying Numbers of Atoms Compounds Containing Hydrogen Organic Compounds
5 Ionic and Covalent Compounds 5. 7 Covalent Bonding in Ionic Species Polyatomic Ions Oxoacids Hydrates Familiar Inorganic Compounds 5. 8 Molecular and Formula Masses 5. 9 Percent Composition of Compounds 5. 10 Molar Mass Interconverting Mass, Moles, and Number of Particles Determination of Empirical Formula and Molecular Formula from Percent Composition
5. 1 Compounds A compound is a substance composed of two or more elements combined in a specific ratio and held together by chemical bonds. Familiar examples of compounds are water and salt (sodium chloride).
5. 2 Lewis Dot Symbols When atoms form compounds, it is their valence electrons that actually interact. A Lewis dot symbol consists of the element’s symbol surround by dots. Each dot represents a valence electron. • • B • Boron 1 s 22 p 1 3 valence electrons • B • • Lewis dot symbol for boron • B • other reasonable Lewis dot symbols for boron
Lewis Dot Symbols Atoms combine in order to achieve a more stable electron configuration. Maximum stability results when a chemical species is isoelectronic with a noble gas. Na: 1 s 22 p 63 s 1 Na+: 1 s 22 p 6 10 electrons total, isoelectronic with Ne Cl: 1 s 22 p 63 s 23 p 5 Cl‒: 1 s 22 p 63 s 23 p 6 18 electrons total, isoelectronic with Ar
Lewis Dot Symbols Lewis dot symbols of the main group elements.
Lewis Dot Symbols Dots are not paired until absolutely necessary. • • B • 1 s 22 p 1 • • C • • 1 s 22 p 2 • • • N • • 1 s 22 p 3 5 valence electrons; first pair formed in the Lewis dot symbol Na • For main group metals such as Na, the number of dots is the number of electrons that are lost. • • • O • • • For nonmetals in the second period, the number of unpaired dots is the number of bonds the atom can form.
Lewis Dot Symbols Ions may also be represented by Lewis dot symbols. Remember the charge Na • Na 1 s 22 p 63 s 1 Valence electron lost in the formation of the Na+ ion. • • • O • • • O 1 s 22 p 4 Na+ 1 s 22 p 6 Core electrons not represented in the Lewis dot symbol • • O • • 2‒ O 2‒ 1 s 22 p 6
Worked Example 5. 1 Write Lewis dot symbols for (a) fluoride ion (F-), (b) potassium ion (K+), and (c) sulfide ion (S 2 -). Strategy Starting with the Lewis dot symbols for each element, add dots (for anions) or remove dots (for cations) as needed to achieve the correct charge on each ion. Don’t forget to include the appropriate charge on the Lewis dot symbol. Solution (a) (b) K+ (c) Think About It For ions that are isoelectronic with noble gases, cations should have no dots remaining around the element symbol, whereas anions should have eight dots around the element symbol. Note, too, that for anions, we put square brackets around the Lewis dot symbol and place the negative charge outside the brackets. Because the symbol for a common cation such as the potassium ion has no remaining dots, square brackets are not necessary.
5. 3 Ionic Compounds and Bonding Ionic bonding refers to the electrostatic attraction that holds oppositely charged ions together in an ionic compound. Na • • • Cl • + e− • • Cl • + Na • • • Na+ + e− − • • Cl • • Na+ + • • • • • Cl • • • − The attraction between the cation and anion draws them together to form Na. Cl
Ionic Compounds and Bonding The resulting electrically neutral compound, sodium chloride, is represented with the chemical formula Na. Cl. The chemical formula, or simply formula, of an ionic compound denotes the constituent elements and the ratio in which they combine.
Ionic Compounds and Bonding A three-dimensional array of oppositely-charged ions is called a lattice. Lattice energy is the amount of energy required to convert a mole of ionic solid to its constituent ions in the gas phase. − + + − Na. Cl(s) Na+(g) + Cl−(g) − − + + − − + Hlattice = +788 k. J/mol +
Ionic Compounds and Bonding The magnitude of lattice energy is a measure of an ionic compound’s stability. Lattice energy depends on the magnitudes of the charge and on the distance between them. Q 1 d Q = amount of charge d = distance of separation Q 2
Ionic Compounds and Bonding
Ionic Compounds and Bonding The magnitude of lattice energy is a measure of an ionic compound’s stability. Lattice energy depends on the magnitudes of the charge and on the distance between them.
Ionic Compounds and Bonding
Worked Example 5. 2 Arrange Mg. O, Ca. O, and Sr. O in order of increasing lattice energy. Strategy Consider the charges on the ions and the distances between them. Apply Coulomb’s law to determine the relative lattice energies. All three compounds contain O 2 - and all three cations are +2. Recalling that lattice energy increases as the distance between ions decreases, we need only consider the radii of the cations as all three contain the same anion. From Figure 4. 13, the ionic radii are 0. 72 Å (Mg 2+), 1. 00 Å (Ca 2+), and 1. 18 Å (Sr 2+). Solution Mg. O has the smallest distance between ions, whereas Sr. O has the largest distance between ions. Therefore, in order of increase lattice energy: Sr. O < Ca. O < Mg. O. Think About It Mg, Ca, and Sr are all Group 2 A metals, so we could have predicted this result without knowing their radii. Recall that ionic radii increase as we move down a column in the periodic table, and charges that are farther apart are more easily separated (meaning the lattice energy will be smaller. ) The lattice energies of Sr. O, Ca. O, and Mg. O are 3217, 3414, and 3890 k. J/mol, respectively.
5. 4 Naming Ions and Ionic Compounds A monatomic ion is named by changing the ending of the element’s name to –ide. Cl– is chloride O 2– is oxide Some metals can form cations of more than one possible charge. Fe 2+ : ferrous ion [Fe(II)] Fe 3+ : ferric ion [Fe(III)] Mn 2+ : manganese(II) ion Mn 3+ : manganese(III) ion Mn 4+ : manganese(IV) ion
Naming Ions and Ionic Compounds
Naming Ions and Ionic Compounds Formulas for ionic compounds are generally empirical formulas. Ionic compounds are electronically neutral.
Formulas of Ionic Compounds In order for ionic compounds to be electronically neutral, the sum of the charges on the cation and anion in each formula must be zero. Aluminum oxide: Al 3+ O 2– Al 2 O 3 Sum of charges: 2(+3) + 3(– 2) = 0
Formulas of Ionic Compounds
Naming Ions and Ionic Compounds To name ionic compounds: 1) Name the cation Ø omit the word ion Ø use a Roman numeral if the cation can have more than one charge 2) Name the anion Ø omit the word ion Examples: Na. CN sodium cyanide Fe. Cl 2 iron(II) chloride Fe. Cl 3 iron(III) chloride
Worked Example 5. 3 Name the following ionic compounds: (a) Ca. O, (b) Mg 3 N 2, and (c) Fe 2 S 3. Strategy Begin by identifying the cation and anion in each compound, and then combine the names for each, eliminating the word ion. Solution (a) Ca. O is calcium oxide. (b) Mg 3 N 2 is magnesium nitride. (c) Fe 2 S 3 is iron(III) sulfide. Think About It Be careful not to confuse the subscript in the formula with the charge in the metal ion. In part (c), for example, the subscript on Fe is 2, but this is an iron(III) compound.
Worked Example 5. 4 Deduce the formulas of the following ionic compounds: (a) mercury(II) chloride, (b) lead(II) bromide, and (c) potassium nitride. Strategy Identify the ions in each compound, and determine their ratios of combination using the charges on the cation and anion in each. Solution (a) Mercury(II) chloride is a combination of Hg 2+ and Cl-. To produce a neutral compound, these two ions must combine in a 1: 2 ratio – Hg. Cl 2. (b) Lead(II) bromide is a combination of Pb 2+ and Cl-. These ions combine in a 1: 2 ratio to give Pb. Br 2. (c) Potassium nitride is a combination of K+ and N 3 -. These ions combine in a 3: 1 ratio to give K 3 N. Think About It Make sure that the charges sum to zero in each compound formula. In part (a), for example, Hg 2+ + 2 Cl- = (+2) + 2(-1) = 0; in part (b), (+2) + 2(-1) = 0; and in part (c), 3(+1) + (-3) = 0.
5. 5 Covalent Bonding and Molecules When compounds form between elements with similar properties, electrons are not transferred from one element to another but instead are shared in order to give each atom a noble gas configuration. This approach is known as the Lewis theory of bonding, named for it’s proponent, Gilbert Lewis theory depicts bond formation in H 2 as H∙ + ∙H → H: H This type of arrangement, where two atoms share a pair of electrons, is known as covalent bonding, and the shared pair of electrons constitutes a covalent bond.
Covalent Bonding and Molecules A molecule is a combination of at least two atoms in a specific arrangement held together by chemical forces (chemical bonds). A molecule may be an element or a compound. Different samples of a given compound always contain the same elements in the same ratio. This is known as the law of definite proportions. Sample Mass of O (g) Mass of C (g) Ratio (g O: g C) 123 g carbon dioxide 89. 4 33. 6 2. 66: 1 50. 5 g carbon dioxide 36. 7 13. 8 2. 66: 1 88. 6 g carbon dioxide 64. 4 24. 2 2. 66: 2
Covalent Bonding and Molecules If two elements can for two or more different compounds, the law of multiple proportions tells us that the ratio of masses of one element that combine with a fixed mass of the other element can be expressed in small whole numbers. In addition to carbon dioxide, carbon also combines with oxygen to form carbon monoxide. Sample Mass of O (g) Mass of C (g) Ratio (g O: g C) 16. 3 g carbon monoxide 9. 31 6. 99 1. 33: 1 25. 9 g carbon monoxide 14. 8 11. 1 1. 33: 1 88. 4 g carbon monoxide 50. 5 37. 9 1. 33: 2
Covalent Bonding and Molecules The mass ratio of oxygen to carbon in carbon dioxide is 2. 66: 1, and the ratio of oxygen to carbon in carbon monoxide is 1. 33: 1. The ratio of two such mass ratios can be expressed as small whole numbers. mass ratio of O to C in carbon dioxide 2. 66 = 2: 1 = mass ratio of O to C in carbon monoxide 1. 33
Covalent Bonding and Molecules Diatomic molecules contain two atoms and may be either heteronuclear or homonuclear. Polyatomic molecules contain more than two atoms.
Covalent Bonding and Molecules
Covalent Bonding and Molecules A chemical formula denotes the composition of the substance. A molecular formula shows the exact number of atoms of each element in a molecule. Some elements have two or more distinct forms known as allotropes. Ø For example, oxygen (O 2) and ozone (O 3) are allotropes of oxygen. A structural formula shows not only the elemental composition, but also the general arrangements.
Worked Example 5. 5 Write the molecular formula of ethanol based on its ball-and-stick model, shown here. Strategy Refer to the labels on the atoms (or see Table 5. 3). There are two carbon atoms, six hydrogen atoms, and one oxygen atom, so the subscript on C will be 2 and the subscript on H will be 6, and there will be no subscript on O. Solution C 2 H 6 O Think About It Often the molecular formula for a compound such as ethanol (consisting of carbon, hydrogen, and oxygen) is written so that the formula more closely resembles the actual arrangement of atoms in the molecule. Thus, the molecular formula for ethanol is commonly written as C 2 H 5 OH.
Covalent Bonding and Molecules Molecular substances can also be represented using empirical formulas, the whole-number ratio of elements. While, the molecular formulas tell us the actual number of atoms (the true formula), the empirical formula gives the simplest formula. Molecular formula: N 2 H 4 Empirical formula: NH 2 The molecular and empirical formulas are often the same.
Covalent Bonding and Molecules
Worked Example 5. 6 Write the empirical formulas for the following molecules: (a) glucose (C 6 H 12 O 6), a substance known as blood sugar; (b) adenine (C 5 H 5 N 5), also known as vitamin B 4; and (c) nitrous oxide (N 2 O), a gas that is used as an anesthetic (“laughing gas”) and as an aerosol propellant for whipped cream. Strategy To write the empirical formula, the subscripts in the molecular formula must be reduced to the smallest possible whole numbers (without altering the relative numbers of atoms).
Worked Example 5. 6 (cont. ) Solution (a) Dividing each of the subscripts in the molecular formula for glucose by 6, we obtain the empirical formula CH 2 O. If we had divided the subscripts by 2 or 3, we would have obtained the formulas C 3 H 6 O 3 and C 2 H 4 O 2, respectively. Although the ratio of carbon to hydrogen to oxygen atoms in each of these formulas is correct (1: 2: 1), neither is the simplest formula because the subscripts are not in the smallest possible whole-number ratio. (b) Dividing each subscript by 5, we get the empirical formula CHN. (c) Because the subscripts in the formula for nitrous oxide are already the smallest possible whole numbers, its empirical formula is the same as its molecular formula N 2 O. Think About It Make sure that the ratio in each empirical formula is the same as that in the corresponding molecular formula and that the subscripts are the smallest possible whole numbers. In part (a), for example, the ratio of C: H: O in the molecular formula is 6: 12: 6, which is equal to 1: 2: 1, the ratio expressed in the empirical formula.
5. 6 Naming Molecular Compounds Remember that binary molecular compounds are substances that consist of just two different elements. Nomenclature: 1) Name the first element that appears in the formula. 2) Name the second element that appears in the formula, changing its ending to –ide. Examples: HCl hydrogen chloride HI hydrogen iodide
Naming Molecular Compounds Greek prefixes are used to denote the number of atoms of each element present.
Naming Molecular Compounds The prefix mono- is generally omitted for the first element. For ease of pronunciation, we usually eliminate the last letter of a prefix that ends in “o” or “a” when naming an oxide. Example: N 2 O 5 is dinitrogen pentoxide not dinitrogen pentaoxide
Worked Example 5. 7 Name the following binary molecular compounds: (a) NF 3 and (b) N 2 O 4. Strategy Each compound will be named using the systematic nomenclature including, where necessary, appropriate Greek prefixes. Solution (a) nitrogen trifluoride (b) dinitrogen tetroxide Think About It Make sure that the prefixes match the subscripts in the molecular formulas and that the word oxide is not preceded immediately by an “a” or an “o”.
Worked Example 5. 8 Write the chemical formulas for the following binary molecular compounds: (a) sulfur tetrafluoride and (b) tetraphosphorus decasulfide. Strategy The formula for each compound will be deduced using the systematic nomenclature guidelines. Solution (a) SF 4 (b) P 4 S 10 Think About It Double-check that the subscripts in the formulas match the prefixes in the compound names: (a) 4 = tetra and (b) 4 = tetra and 10 = deca.
Compounds Containing Hydrogen The names of molecular compounds containing hydrogen do not usually conform to the systematic nomenclature guidelines. Many are called by the common, nonsystematic names or by names that do not indicate explicitly the number of H atoms present. Examples: B 2 H 6 Si. H 4 NH 3 PH 3 H 2 O H 2 S Diborane Silane Ammonia Phosphine Water Hydrogen sulfide
Compounds Containing Hydrogen One definition of an acid is a substance that produces hydrogen ions (H+) when dissolved in water. HCl is an example of a binary compound that is an acid when dissolved in water. To name these types of acids: 1) remove the –gen ending from hydrogen 2) change the –ide ending on the second element to –ic. hydrogen chloride → hydrochloric acid
Compounds Containing Hydrogen A compound must contain at least one ionizable hydrogen atom to be an acid upon dissolving.
Organic Compounds Our nomenclature discussion so far has focused on inorganic compounds, generally defined as those without carbon. Organic compounds contain carbon and hydrogen, sometimes in combination with other atoms. Hydrocarbons contain only carbon and hydrogen. The simplest hydrocarbons are called alkanes.
Organic Compounds
Organic Compounds
Organic Compounds Many organic compounds contain groups of atoms known as functional groups, which often determine a molecule’s reactivity.
5. 7 Covalent Bonding in Ionic Species Polyatomic ions consist of a combination of two or more atoms. Formulas are determined following the same rule as for ionic compounds containing only monatomic ions: ions must combine in a ratio that give a neutral formula overall. Calcium phosphate: Ca 2+ PO 43– Ca 3(PO 4)2 Sum of charges: 3(+2) + 2(– 3) = 0
Covalent Bonding in Ionic Species
Covalent Bonding in Ionic Species
Worked Example 5. 9 Name the following ionic compounds: (a) Fe 2(SO 4)3, (b) Al(OH)3, and (c) Hg 2 O. Strategy Begin by identifying the cation and anion in each compound, and then combine the names for each, eliminating the word ion. Solution (a) Fe 2(SO 4)3 is iron(III) sulfate. (b) Al(OH)3 is aluminum hydroxide. (c) Hg 2 O is mercury(I) oxide. Think About It Be careful not to confuse the subscript in the formula with the charge in the metal ion. In part (a), for example, the subscript on Fe is 2, but this is an iron(III) compound.
Covalent Bonding in Ionic Species Oxoanions are polyatomic anions that contain one or more oxygen atoms and one atom (the “central atom”) of another element. Starting with the oxoanions that end in –ate, we can name these ions as follows: 1) The ion with one more O atom than the –ate ion is called the per…ate ion. Thus, Cl. O 3 - is the chlorate ion, so Cl. O 4 - is the perchlorate ion. 2) The ion with one less O atom than the –ate ion is called the –ite ion. Thus, Cl. O 2 - is the chlorite ion. 3) The ion with two fewer O atom than the –ate ion is called the hypo…ite ion. Thus, Cl. O- is the hypochlorite ion. At minimum, memorize the oxoanions that end in –ate so you can apply these guidelines when necessary.
Covalent Bonding in Ionic Species perchlorate chlorite hypochlorite Cl. O 4 Cl. O 3 Cl. O 2 Cl. O- nitrate nitrite NO 3 NO 2 - phosphate phosphite PO 43 PO 33 - sulfate sulfite SO 42 SO 32 -
Worked Example 5. 10 Name the following species: (a) Br. O 4 -, (b) HCO 3 -, and (c) H 2 CO 3. Strategy Each species is either an oxoanion or an oxoacid. Identify the “reference oxidation” (the one with the –ate ending) for each, and apply the rules to determine appropriate names. Think It Remembering all these names and formulas Solution (a)About Br. O 4 - has one more O atom than the bromate ion (Br. Ois 3 ), so Br. O 4 greatly facilitated is the perbromate ion. by memorizing the common ions that end in –ate. chlorate Cl. O 3 nitrate NO 3 - Because HCO 2 carbonate CO 3 hydrogen (b) CO 32 - isiodate the carbonate. IOion. atom, 3 3 has one ionizable bromate C 2 O 42 it is called the hydrogen Br. O carbonate ion. oxalate 3 sulfate SO 42 chromate Cr. O 42 phosphate PO 43 - atoms andpermanganate (c) With two ionizable hydrogen no charge on the. Mn. O compound, H 2 CO 3 is 4 carbonic acid. Think About It Make sure that the charges sum to zero in each compound formula. In part (a), for example, Hg 2+ + 2 Cl- = (+2) + 2(-1) = 0; in part (b), (+2) + 2(-1) = 0; and in part (c), 3(+1) + (-3) = 0.
Worked Example 5. 11 Determine the formula of sulfurous acid. Strategy The –ous ending in the name of an acid indicates that the acid is derived from an oxoanion ending in –ite. The oxoanion must be sulfite, SO 32 -, so add enough hydrogen ions to make a neutral formula. Solution The formula of sulfurous acid is H 2 SO 3. Think About It Remembering all these names and formulas is greatly facilitated by memorizing the common ions that end in -ate.
Hydrates A hydrate is a compound that has a specific number of water molecules within its solid structure. For example, in its normal state, copper(II) sulfate has five water molecules associated with it. Systematic name: copper(II) sulfate pentahydrate Formula: Cu(SO)4 ∙ 5 H 2 O Some other hydrates are Ba. Cl 2 ∙ 2 H 2 O Li. Cl ∙ H 2 O Mg. SO 4 ∙ 7 H 2 O Sr(NO 3)2 ∙ 4 H 2 O
Hydrates When the water molecules are driven off by heating, the resulting compound, Cu(SO)4, is sometimes called anhydrous copper(II) sulfate. Anhydrous means the compound no longer has water molecules associated with it.
Familiar Inorganic Compounds
5. 8 Molecular and Formula Mass The molecular mass is the mass in atomic mass units (amu) of an individual molecule. To calculate molecular mass, multiply the atomic mass for each element in a molecule by the number of atoms of that element and then total the masses Molecular mass of H 2 O = 2(atomic mass of H) + atomic mass of O = 2(1. 008 amu) + 16. 00 amu = 18. 02 amu Because the atomic masses on the periodic table are average atomic masses, the result of such a determination is an average molecular mass, sometimes referred to as the molecular weight.
Molecular and Formula Mass Although an ionic compound does not have a molecular mass, we can use its empirical formula to determine its formula mass (the mass of a “formula unit”), sometimes called the formula weight.
Worked Example 5. 12 Calculate the molecular mass or the formula mass, as appropriate, for each of the following corresponds: (a) propane, C 3 H 8, (b) lithium hydroxide, Li. OH, and (c) barium acetate, Ba(C 2 H 3 O 2)2. Strategy Determine the molecular mass (for each molecular compound) or formula mass (for each ionic compound) by summing all the atomic masses. Solution For. About each compound, multiply theyou number atomsthe by number the atomic mass Think It Double-check that have of counted of eachofelement and then for sumeach the calculated atoms correctly compound values. and that you have used the proper atomic masses from the periodic table. (a) The molecular mass of propane is 3(12. 01 amu) + 8(1. 008 amu) = 44. 09 amu (b) The formula mass of lithium hydroxide is 6. 941 amu + 16. 00 amu + 1. 008 amu = 23. 95 amu. (c) The formula mass of barium acetate is 137. 3 amu + 4(12. 01 amu) + 6(1. 008 amu) + 4(16. 00 amu) = 255. 4 amu.
5. 9 Percent Composition of Compounds A list of the percent by mass of each element in a compound is known as the compound’s percent composition by mass. where n is the number of atoms of the element in a molecule or formula unit of the compound
Percent Composition of Compounds For a molecule of H 2 O 2:
Percent Composition of Compounds We could also have used the empirical formula of hydrogen peroxide (HO) for the calculation. In this case, we could have used the empirical formula mass, the mass in amu of one empirical formula, in place of the molecular formula. The empirical formula mass of H 2 O 2 (the mass of HO) is 17. 01 amu. %H = 1. 008 amu H × 100% = 5. 926% 17. 01 amu H 2 O 2 %O = 16. 00 amu O × 100% = 94. 06% 17. 01 amu H 2 O 2
Worked Example 5. 13 Lithium carbonate, Li 2 CO 3, was the first “mood-stabilizing” drug approved by the FDA for the treatment of mania and manic-depressive illness, also known as bipolar disorder. Calculate the percent composition by mass of lithium carbonate. Strategy Use Equation 5. 1 to determine the percent by mass contributed by each element in the compound. Think About It Make sure that the percent composition results for Solution For eachsum element, multiply the 100. number of atoms atomic a compound to approximately (In this case, by thethe results summass, divide to byexactly the formula mass, and multiply by 100 percent–– 18. 79% + 16. 25% + 64. 96% = 100. 00%–– but remember that because 2× 6. 941 amu Li of rounding, the percentages may sum to %Li = × 100%less. = 18. 79% very slightly more slightly 73. 89 amuor. Livery CO 2 3 %C = 12. 01 amu C × 100% = 16. 25% 73. 89 amu Li 2 CO 3 %O = 3× 16. 00 amu O × 100% = 64. 96% 73. 89 amu Li 2 CO 3
5. 10 Molar Mass The molar mass (M ) of a substance is the mass in grams of one mole of the substance. The molar mass of an element is numerically equal to its atomic mass. 1 mol C = 12. 01 g 1 C atom = 12. 01 amu The molar mass of a compound is the sum of molar masses of the elements it contains. 1 mol H 2 O = 2 × 1. 008 g + 16. 00 g = 18. 02 g 1 mol Na. Cl = 22. 99 g + 35. 45 g = 58. 44 g
Molar Mass When expressing the molar mass of elements such as oxygen and hydrogen, we have to be careful to specify which form of the element we mean. For instance, the element oxygen exists predominantly as O 2. But we might also mean atomic oxygen (O). We must consider the context to tell which form of the element, O 2 or O, is intended. Context Oxygen means Molar mass How many moles of oxygen react with 2 moles of hydrogen to produce water? O 2 32. 00 g How many moles of oxygen are there in 1 mole of water? O 16. 00 g Air is approximately 21 percent oxygen. O 2 32. 00 g Many organic compounds contain oxygen. O 16. 00 g
Interconverting Mass, Moles, and Numbers of Particles
Worked Example 5. 14 Determine (a) the number of moles of CO 2 in 10. 00 g of carbon dioxide and (b) the mass of 0. 905 mole of sodium chloride. Strategy Use molar mass to convert from mass to moles and to convert from moles to mass. The molar mass of carbon dioxide (CO 2) is 44. 01 g/mol and the molar mass of sodium chloride (Na. Cl) is 58. 44 g/mol. Solution (a) 10. 00 g CO 2 × 1 mol CO 2 = 0. 2272 mol CO 2 44. 01 g CO 2 (b) 0. 905 mol Na. Cl × 58. 44 g Na. Cl = 52. 9 g Na. Cl 1 mol Na. Cl Think About It Always double-check unit cancellations in problems such as these–errors are common when molar mass is used as a conversion factor. Also make sure that the results make sense. In both cases, a mass smaller than the molar mass corresponds to less than a mole of substance.
Worked Example 5. 15 (a) Determine the number of water molecules and the numbers of H and O atoms in 3. 26 g of water. (b) Determine the mass of 7. 92× 1019 carbon dioxide molecules. Strategy Use molar mass and Avogadro’s number to convert from mass to molecules, and vice versa. Use the molecular formula of water to determine the numbers of H and O atoms. Solution 1 mol H 2 O 6. 022× 1023 H 2 O molecules = 1. 09× 1023 H 2 O molecules (a) 3. 26 g H 2 O× 18. 02 g H O × 1 mol H 2 O 2 Using the molecular formula, we can determine the number of H and O atoms in 3. 26 g of H 2 O as follows: 2 H atoms 1. 09× 1023 H 2 O molecules × 1 H O molecule = 2. 18× 1023 H atoms 2 1. 09× 1023 H 2 O molecules × 1 O atom 23 1 H 2 O molecule = 1. 09× 10 H atoms
Worked Example 5. 15 (cont. ) Solution (b) 7. 92× 1019 CO 2 molecules × 44. 01 g CO 2 1 mol CO 2 × 1 mol CO 2 6. 022× 1023 CO 2 molecules = 5. 79× 10 -3 g CO 2 Think About It Again, check the cancellation of units carefully and make sure that the magnitudes of your results are reasonable.
Determination of Empirical Formula and Molecular Formula from Percent Composition Using the concepts of the mole and molar mass, we can now use an experimentally determined percent composition to determine the empirical and/or molecular formula. The empirical formula gives only the ratio of atoms in a molecule, so there may be multiple compounds with the same empirical formula. If we know the approximate molar mass of the compound, we can determine the molecular formula by dividing the molar mass by the empirical formula mass. We then multiply the empirical formula by this number to obtain the molecular formula.
Worked Example 5. 16 Determine the empirical formula of a compound that is 30. 45 percent nitrogen and 69. 55 percent oxygen by mass. Given that the molar mass of the compound is approximately 92 g/mol, determine the molecular formula of the compound. Strategy Assume a 100 -g sample so that the mass percentages of nitrogen and oxygen given in the problem statement correspond to the masses of N and O in the compound. Then, using the appropriate molar masses, convert the grams of each element to moles. Use the resulting numbers as subscripts in the empirical formula, reducing them to the lowest possible whole numbers for the final answer. To calculate the molecular formula, first divide the molar mass given in the problem statement by the empirical formula mass. Then, multiply the subscripts in the empirical formula by the resulting number to obtain the subscripts in the molecular formula. The molar masses of N and O are 14. 01 and 16. 00 g/mol, respectively. One hundred grams of a compound that is 30. 45 percent nitrogen and 69. 35 percent oxygen by contains 30. 45 g N and 69. 55 g O.
Worked Example 5. 16 (cont. ) Solution 1 mol N = 2. 173 mol N 14. 01 g N 1 mol O 69. 55 g O × = 4. 347 mol O 16. 00 g O 30. 45 g N × The gives a formula of N 2. 173 O 4. 347. Dividing both subscripts by the smaller of the two to get the smallest possible whole numbers (2. 173/2. 173 = 1, 4. 347/2. 173 ≈ 2) gives an empirical formula of NO 2. Finally, dividing the approximate molar mass (92 g/mol) by the empirical formula mass [14. 01 g/mol + 2(16. 00 g/mol) = 46. 01 g/mol] gives 92/46. 01 ≈ 2. Then, multiplying both subscripts in the empirical formula by 2 gives the molecular formula, N 2 O 4. Think About It Use the method described in Worked Example 5. 13 to calculate the percent composition of the molecular formula N 2 O 4 and verify that it is the same as that given in this problem.
5 Chapter Summary: Key Points Compounds Lewis Dot Symbols Ionic Compounds and Ionic Bonding Chemical Formulas Lattice Energy Ionic Compound Nomenclature Lewis Theory of Bonding Covalent Bonding Laws of Definite Proportions and Multiple Proportions Molecular, Structural, and Empirical Formulas Molecular Compound Nomenclature Inorganic and Organic Compounds Polyatomic Ions Oxyanions Oxoacids Hydrates Molecular Mass and Formula Mass Empirical Formula Mass Molar Mass
- Slides: 78