Silberberg Chapter 3
Silberberg Chapter 3 CHEM 211
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This 5 page Class Notes was uploaded by Taylor Swifty on Sunday September 25, 2016. The Class Notes belongs to CHEM 211 at University of Missouri - Kansas City taught by Todor Gounev in Fall 2016. Since its upload, it has received 6 views. For similar materials see General Chemistry I in Chemistry at University of Missouri - Kansas City.
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Date Created: 09/25/16
3.1 • The Mole All the ideas in this chapter rely on understanding a key concept related to a unit called the mole. Daily life in the laboratory involves measuring substances: We want to know numbers of chemical entities — atoms, ions, molecules, or formula units — that react with each other How can we possibly count or weigh such minute objects? Chemists have devised a unit, called the mole, to count chemical entities by weighing them. Defining the Mole Mole (abbreviated mol) is the SI unit for amount of substance defined as amount of a substance that contains same number of entities as the number of atoms in 12 g of carbon12 Avogadro’s number (in honor of the 19thcentury Italian physicist Amedeo Avogadro) 23 One mole (1 mol) contains 6.022 x 10 entities (to four significant figures) Thus, 23 1 mol of carbon12 contains 6.022 x 10 carbon12 atoms 1 mol of H 2 contains 6.022 x 10 H O 2olecules 23 1 mol of NaCl contains 6.022 x 10 NaCl formula units A counting unit, like dozen, tells you number of objects but not mass a mass unit, like kilogram, tells you mass of objects but not number The mole tells you both—the number of objects in a given mass of substance: 1 mol of carbon12 contains 6.022 x 10 carbon12 atoms and has a mass of 12 g What does it mean that the mole unit allows you to count entities by weighing the sample? Suppose you have a sample of carbon12 and want to know the number of atoms present. You find that the sample weighs 6 g, so it is 0.5 mol of carbon12 and, thus, contains 3.011 x 10 atoms: 6 g of carbon12 is 0.5 mol of carbon12 and contains 3.01131023 atoms Knowing the amount (in moles), the mass (in grams), and the number of entities becomes very important when we mix different substances to run a reaction. The central relationship between masses on the atomic scale and on the macroscopic scale is the same for elements and compounds: Elements. The mass in atomic mass units (amu) of one atom of an element is the same numerically as the mass in grams (g) of 1 mole of atoms of the element. Recall from Chapter 2 that each atom of an element is considered to have the atomic mass given in the periodic table (see margin). Thus, 1 atom of S has a mass of 32.07 amu and 1 mol (6.02231023 atoms) of S has a mass of 32.07 g 1 atom of Fe has a mass of 55.85 amu and 1 mol (6.02231023 atoms) of Fe has a mass of 55.85 g Note, also, that since atomic masses are relative, 1 Fe atom weighs 55.85/32.07 as much as 1 S atom, and 1 mol of Fe weighs 55.85/32.07 as much as 1 mol of S. Compounds. The mass in atomic mass units (amu) of one molecule (or formula unit) of a compound is the same numerically as the mass in grams (g) of 1 mole of the compound. Thus, for example, 1 molecule of H O2has a mass of 18.02 amu and 1 mol (6.02231023 molecules) of H O has a ma2s of 18.02 g 1 formula unit of NaCl has a mass of 58.44 amu and 1 mol (6.02231023 formula units) of NaCl has a mass of 58.44 g Here, too, because masses are relative, 1 H O 2olecule weighs 18.02/58.44 as much as 1 NaCl formula unit, and 1 mol of H O2weighs 18.02/58.44 as much as 1 mol of NaCl. The two key points to remember about the importance of the mole unit are: • The mole lets us relate the number of entities to the mass of a sample of those entities. • The mole maintains the same numerical relationship between mass on the atomic scale (atomic mass units, amu) and mass on the macroscopic scale (grams, g). In everyday terms, a grocer does not know there are 1 dozen eggs from their weight or that there is 1 kilogram of beans from their count, because eggs and beans do not have fixed masses. But, by weighing out 63.55 g (1 mol) of copper, a chemist does know that there are 6.02231023 copper atoms, because all copper atoms have an atomic mass of 63.55 amu. Figure 3.1 shows 1 mole of some familiar elements and compounds. Determining Molar Mass The molar mass (Μ) of a substance is the mass per mole of its entities (atoms, molecules, or formula units) and has units of grams per mole (g/mol). The periodic table is indispensable for calculating molar mass: 1. Elements. To find the molar mass, look up the atomic mass and note whether the element is monatomic or molecular. • Monatomic elements. The molar mass is the periodictable value in grams per mole.* For example, the molar mass of neon is 20.18 g/mol, and the molar mass of gold is 197.0 g/mol. • Molecular elements. You must know the formula to determine the molar mass (see Figure 2.13). For example, in air, oxygen exists most commonly as diatomic molecules, so the molar mass of O2 is twice that of O: Molar mass (Μ) of O = 2 x Μ of O = 2 x 16.00 g/mol = 32.00 g/mol The most common form of sulfur exists as octatomic molecules, S : 8 Μ of S 8 8 x Μ of S = 8 x 32.07 g/mol = 256.6 g/mol 2. Compounds. The molar mass is the sum of the molar masses of the atoms in the formula. Thus, from the formula of sulfur dioxide, SO2, we know that 1 mol of SO2 molecules contains 1 mol of S atoms and 2 mol of O atoms: Μ of SO =2Μ of S + (2 x Μ of O) = 32.07 g/mol + (2 x 16.00 g/mol) = 64.07 g/mol Similarly, for ionic compounds, such as potassium sulfide (K S), 2e have Μ of K S2= (2 x Μ of K) + Μ of S = (2 x 39.10 g/mol) + 32.07 g/mol = 110.27 g/mol 1 atom of S has a mass of 32.07 amu and 1 mol (6.02231023 atoms) of S has a mass of 32.07 g 1 atom of Fe has a mass of 55.85 amu and 1 mol (6.02231023 atoms) of Fe has a mass of 55.85 g 1 molecule of H2O has a mass of 18.02 amu and 1 mol (6.02231023 molecules) of H2O has a mass of 18.02 g 1 formula unit of NaCl has a mass of 58.44 amu and 1 mol (6.02231023 formula units) of NaCl has a mass of 58.44 g *The mass value in the periodic table has no units because it is a relative atomic mass, given by the atomic mass (in amu) divided by 1 amu ( 1 12 mass of one 12C atom in amu): Relative atomic mass 5 atomic mass 1amu2 1 12 mass of 12C 1amu2 Therefore, you use the same number for the atomic mass and for the molar mass. Thus, subscripts in a formula refer to individual atoms (or ions) as well as to moles of atoms (or ions). Table 3.1 summarizes these ideas for glucose, C6H12O6 (see margin), the essential sugar in energy metabolism. Converting Between Amount, Mass, and Number of Chemical Entities One of the most common skills in the lab— and on exams—is converting between amount (mol), mass (g), and number of entities of a substance. 1. Converting between amount and mass. If you know the amount of a substance, you can find its mass, and vice versa. The molar mass (m), which expresses the equivalence between 1 mole of a substance and its mass in grams, is the conversion factor. • From amount (mol) to mass (g), multiply by the molar mass: Mass (g) 5 amount (mol) 3 no. of grams 1 mol (3.2) • From mass (g) to amount (mol), divide by the molar mass (multiply by 1/m): Amount (mol) 5 mass (g) 3 1 mol no. of grams (3.3) 2. Converting between amount and number. Similarly, if you know the amount (mol), you can find the number of entities, and vice versa. Avogadro’s number, which expresses the equivalence between 1 mole of a substance and the number of entities it contains, is the conversion factor. • From amount (mol) to number of entities, multiply by Avogadro’s number: No. of entities 5 amount (mol) 3 6.02231023 entities 1 mol (3.4) • From number of entities to amount (mol), divide by Avogadro’s number: Amount (mol) 5 no. of entities 3 1 mol 6.02231023 entities (3.5) AmountMassNumber Conversions Involving Elements We begin with amountmassnumber relationships of elements. As Figure 3.2 shows, convert mass (in grams) or number of entities (atoms or molecules) to amount (mol) first. For molecular elements, Avogadro’s number gives molecules per mole. Let’s work through a series of sample problems that show these conversions for both elements and compounds. AmountMassNumber Conversions Involving Compounds Only one new step is needed to solve amountmassnumber problems involving compounds: we need the chemical formula to find the molar mass and the amount of each element in the compound. The relationships are shown in Figure 3.3, and Sample Problems 3.4 and 3.5 apply them. The Importance of Mass Percent For many purposes, it is important to know how much of an element is present in a given amount of compound. In this section, we find the composition of a compound in terms of mass percent and use it to find the mass of each element in the compound. Determining Mass Percent from a Chemical Formula Each element contributes a fraction of a compound’s mass, and that fraction multiplied by 100 gives the element’s mass percent. Finding the mass percent is similar on the molecular and molar scales: • For a molecule (or formula unit) of compound, use the molecular (or formula) mass and chemical formula to find the mass percent of any element X in the compound: Mass % of element X = [atoms of X in formula x atomic mass of X(amu)]/molecular (or formula) mass of compound (amu) x 100 • For a mole of compound, use the molar mass and formula to find the mass percent of each element on a mole basis: Mass % of element X = [moles of X in formula x molar mass of X (g/mol)] / [mass (g) of 1 mol of compound] x 100 (3.6) As always, the individual mass percents add up to 100% (within rounding). In Sample Problem 3.6, we determine the mass percent of each element in a compound. Determining the Mass of an Element from Its Mass Percent Sample Problem 3.6 shows that an element always constitutes the same fraction of the mass of a given compound (see Equation 3.6). We can use that fraction to find the mass of element in any mass of a compound: Mass of element = mass of compound x mass of element in 1 mol of compound / mass of 1 mol of compound (3.7) For example, to find the mass of oxygen in 15.5 g of nitrogen dioxide, we have Mass (g) of O = 15.5 g NO x 2 m2l x M of O (g/mol) / mass (g) of 1 mol NO 2 = 15.5 g NO x 232.00 g O / 46.01 g NO = 10.8 2 O
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