7 Quan'ta've Composi'on of Compounds. Chapter Outline. The Mole. The Mole. The Mole. The Mole. Advanced Chemistry

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1 7 Quan'ta've Composi'on of Compounds Chapter Outline 7.1 The Mole Percent Composition of Compounds 7.4 Calculating Empirical Formulas 7.5 Calculating the Molecular Formula from Black pearls are composed of calcium carbonate, CaCO 3. The pearls can be measured by either weighing or counting. Foundations of College Chemistry, 14 th Ed. Morris Hein Susan Arena The Mole Individual atoms are tiny have such a small mass, more convenient units for atoms are needed to be useful on the macroscale. Analogy Fruit in a supermarket is counted by weighing the mass of fruit. If the average mass for a piece of fruit is known, the number of pieces of fruit can be calculated. Example If one orange has a mass of 186 g, then 75 oranges have what mass? 186 g 75 oranges = 13,950 g = kg 1 orange Chemists count atoms in a similar way, by weighing. The stard unit of measurement for chemistry. 1 mole = x objects The number represented by 1 mole, x 10 23, is also called Avogadro s number. Such a large number is useful because even the smallest amount of matter contains extremely large numbers of atoms. The mole is similar to other common units of counting. Example 1 doze 12 objects The Mole The Mole The Mole Moles can be used to describe elements, particles or compounds. Mole is often abbreviated as mol. 1 mol of atoms = x atoms 1 mol of molecules = x molecules 1 mol of electrons = x electrons Avogadro s number can be used as a conversion factor. 1 mol x objects x objects 1 mol How does the mol relate to masses of elements? The atomic mass of 1 mol of any element is defined as the amount of that substance that contains the same number of particles as exactly 12 g of 12 C. 1 mol of any element contains the same number of atoms, but can vary greatly in the overall mass. (Atoms of different elements have different masses) 1

2 Molar Mass: the atomic mass of an element or compound (in grams) which contains Avogadro s number of particles. Molar masses are expressed to 4 significant figures in the text. Determining Molar Mass Convert atomic mass units on the periodic table to grams sum the masses of the total atoms present. Example CaF 2 Molar Mass Molar Mass CaF 2 = g + 2(19.00) g = g Ca 2 F We can use both the mol molar mass as conversion factors. How many moles of lead does 15.0 g of Pb represent? g Pb mol Pb The conversion factor relates g of Pb to moles of Pb g Pb 1 mol Pb g Pb or 1 mol Pb g Pb g Pb 1 mol Pb (Obtain molar mass from the periodic table.) = 7.24 x 10-2 mol Pb How many moles of mercury does 23.0 g of Hg represent? a x 10 3 mol Hg b x 10-1 mol Hg c x 10 1 mol Hg d x 10-3 mol Hg 23.0 g Hg 1 mol Hg g Hg g Hg mol Hg The conversion factor needed: 1 mol Hg g Hg or g Hg 1 mol Hg = 1.15 x 10-1 mol Hg How many Au atoms are contained in 16.0 g of Au? g Au mol Au atoms Au 1 mol Au g Au 1 mol Au 16.0 g Au g Au 1 mol Au x atoms Au x atoms Au 1 mol Au = 4.89 x atoms Au How many Ti atoms are contained in 7.80 g of Ti? a x atoms Ti g Ti mol Ti atoms Ti b x atoms Ti c x atoms Ti d x atoms Ti 1 mol Ti 1 mol Ti g Ti x atoms Ti 1 mol Ti 7.80 g Ti g Ti x atoms Ti 1 mol Ti = 9.81 x atoms Ti What is the mass of 2.13 x atoms of Li? atoms Li mol Li grams Li 2.13 x atoms Li 1 mol Li g Li 1 mol Li x atoms Li 1 mol Li g Li x atoms Li 1 mol Li = 2.46 x 10-5 g Li 2

3 What is the mass of 1.28 x 10 8 atoms of Ne? a x g Ne b x g Ne c x g Ne d x g Ne atoms Ne mol Ne grams Ne 1 mol Ne g Ne 1 mol Ne x atoms Ne 1 mol Ne 1.28 x g Ne atoms Ne x atoms Ne 1 mol Ne = 4.29 x g Ne What is the mass of 1.05 mol of Ag? mol Ag grams Ag One conversion factor is needed: 1.05 mol Ag 1 mol Ag g Ag g Ag = 113. g Ag 1 mol Ag a g K b x 10-2 g K c g K d x g K What is the mass of 8.21 mol of K? 8.21 mol K mol K = 321. g K grams K One conversion factor is needed: g K 1 mol K 1 mol K g K How many hydrogen atoms are in 1.00 moles of H 2 molecules? mol H 2 molecules H 2 atoms H 2 1 mol H x molecules H 2 1 molecule H 2 2 atoms H x mol H 23 molecules H 2 2 atoms H 2 1 mol H 2 1 molecule H 2 = 1.20 x atoms H 2 How many sulfur atoms are in 2.27 mol of S 8 molecules? a x atoms S 8 b x atoms S 8 c x atoms S 8 d x atoms S 8 mol S 8 molecules S 8 atoms S 8 1 mol S x molecules S 8 1 molecule S 8 8 atoms S x mol S 23 molecules S 8 8 atoms S 8 1 mol S 8 1 molecule S 8 = 1.09 x atoms S 8 Much like an element, molar mass can be defined for a compound. Molar Mass (MM): mass of one mole of the formula unit of a compound. The molar mass of a compound is equal to the sum of the molar masses of all the atoms in the molecule. Example H 2 O Molar Mass = MM O + 2MM H = g + 2(1.008 g) = g 3

4 What is the molar mass of aluminum hydroxide, Al(OH) 3? The molar mass of a compound contains Avogadro s number of formula units/molecules. a g b g c g d g Al(OH) 3 Using the atomic masses of each element: 1 Al = 1(26.98 g) = g 3 O = 3(16.00 g) = g 3 H = 3(1.008 g) = g g H 2 O H 2 O 2 x (6.022 x ) H atoms x O atoms x H 2 O molecules 2 mol H atoms 1 mol O atoms 1 mol H 2 O molecules 2 x g = g H g O g H 2 O Reminder: Pay close attention to whether the desired unit involves atoms or formula units/molecules. Example Cl 2 Contains 2 mol of Cl atoms but only 1 mol of Cl 2 molecules. As for elements, we can use both the mol molar mass of formula units/molecules as conversion factors. How many moles of NaCl are there in 253 g of NaCl? g NaCl mol NaCl To convert between g of NaCl moles, we must first calculate the molar mass of NaCl. MM = g g = 253. g NaCl 1 mol NaCl g NaCl g NaCl = 4.33 mol NaCl a mol TiCl 4 b mol TiCl 4 c x 10 3 mol TiCl 4 d mol TiCl 4 How many moles of TiCl 4 are there in 12.5 g of titanium(iv) chloride? 12.5 g TiCl 4 g TiCl 4 mol TiCl 4 MM TiCl4 = g + 4(35.45) g = g TiCl 4 1 mol TiCl g TiCl 4 = mol TiCl 4 What is the mass of 3.45 mol of Li 2 O? mol Li 2 O g Li 2 O To convert between mol of Li 2 O g, we must first calculate the molar mass of Li 2 O. MM Li2 O = 2(6.941) g g = mol Li 2 O g Li 2 O 1 mol Li 2 O g Li 2 O = 103. g Li 2 O What is the mass of 1.23 mol of PH 3? a x 10-2 g PH 3 b x 10-6 g PH 3 c g PH 3 d g PH mol PH 3 mol PH 3 g PH 3 MM PH3 = g + 3(1.008) g = g PH g PH 3 1 mol PH 3 = 41.8 g PH 3 4

5 How many molecules of H 2 S are present in 7.53 g of H 2 S? How many atoms of H are present in the sample? g H 2 S mol H 2 S molecules H 2 S atoms H MM H2S = 2(1.008) g g = g H 2 S 7.53 g H 2 S 1 mol H 2 S g H 2 S 1.33 x molecules H 2 S x molecules H 2 S 1 mol H 2 S = 1.33 x molecules H 2 S 2 atoms H = 2.66 x atoms H 1 molecule H 2 S How many molecules of H 2 O 2 are there in g of the compound? a x molecules H 2 O 2 b x molecules H 2 O 2 g H c x O 2 mol H 2 O 2 molecules H 2 O 2 molecules H 2 O 2 d x molecules H 2 O 2 MM H2O2 = 2(1.008) g + 2(16.00) g = g H 2 O g H 2 O 2 1 mol H 2 O x molecules H 2 O g H 2 O 2 1 mol H 2 O 2 = 1.34 x molecules H 2 O 2 Percent Composi'on of Compounds Percent = parts per 100 parts Percent composition: mass percent of each element in a compound Molar mass: total mass (100%) of a compound % composition is independent of sample size % composition can be determined by: 1. Knowing the compound s formula or 2. Using experimental data Percent Composi'on from the Compound s Formula Two Step Strategy 1. the molar mass of the compound. 2. Divide the total mass of each element by the compound s molar mass multiply by 100. Total element mass % of the element = 100 Compound molar mass Percent Composi'on of Compounds the percent composition of K 2 S. Step 1 compound molar mass MM K2S = 2(39.10) g g = g Step 2 % composition of each element. % K = 2(39.10) g K g 100 = % K % S = g S g 100 = % S Notice the sum of the percentages must equal 100%. This provides another way of determining the % composition of a specific element, if the other %s are known. Percent Composi'on of Compounds a % b % c % d. 11.1% the percent composition of O in H 2 O 2. Step 1 molar mass MM H2O2 = 2(1.008) g + 2(16.00)g = g Step 2 % composition O 2(16.00) g O % O = 100 = % O g 5

6 Percent Composi'on of Compounds the % composition of K 2 CrO 4. Step 1 compound molar mass MM K2CrO4 = 2(39.10) g g + 4(16.00) g = g Step 2 % composition 2(39.10) g K % K = 100 = % K g g Cr % Cr = 100 = % Cr g 4(16.00) g O % O = 100 = % O g Percent Composi'on from Experimental Data Two Step Strategy 1. the mass of the compound formed. 2. Divide the mass of each element by the total mass multiply by 100. Total element mass % of the element = 100 Total compound mass Percent Composi'on from Experimental Data When heated in air, 1.63 g of Zn reacts with 0.40 g of oxygen to give ZnO. the percent composition of the compound formed. Step 1 the mass of the compound formed. Mass compound = Mass Zn + Mass O = 1.63 g g = 2.03 g compound Step 2 % composition % Z 1.63 g Zn 2.03 g 100 = 80.3 % Zn % O = 0.40 g O 2.03 g 100 = 20. % O 100.3% Total should be +/0.5% of 100 Aluminum chloride forms by reaction of g of Al with g of chlorine. What is the percent composition of Cl in the compound? a % b % c % d % Percent Composi'on from Experimental Data Step 1 the mass of the compound formed. Mass compound = Mass Al + Mass Cl = g g = g compound Step 2 % composition g Cl % Cl = 100 = 79.8 % g Empirical Molecular Formula Empirical Molecular Formula Empirical Formula: smallest whole number ratio of atoms in a compound Molecular Formula: actual formula of a compound. Represents the total number of atoms in one formula unit of the compound. Whole number multiple of the empirical formula Example Acetylene (C 2 H 2 ) Benzene (C 6 H 6 ) Both have the same empirical formula CH. Each compound is a multiple of CH. Acetylene C 2 H 2 = (CH) 2 Benzene C 6 H 6 = (CH) 6 Formula Composition % C %H Molar Mass (g/mol) CH (empirical formula) C 2 H 2 (acetylene) (2 x 13.02) C 6 H 6 (benzene) (6 x 13.02) Each compound has very different chemical physical properties even though they share the same empirical formula. Compounds with the same empirical formula have the same percent composition. Molar mass = molar mass of the empirical unit multiple of the unit 6

7 To calculate an empirical formula, you need to know: 1. The elements present in the compound 2. The atomic masses of each element (from the Periodic Table) 3. The ratio (by mass or %) of the combined elements Strategy to an Empirical Formula: 1. Assume a starting mass of the compound (usually g) express the mass of each element in grams. 2. Convert g of each element to mol using molar mass. (These numbers may or may not be whole numbers.) 3. Divide each of the mole amounts from Step 2 by the The new numbers are the subscripts in the empirical formula. Special Case: If fractions are encountered, multiply by a common factor to provide whole numbers for each subscript. the empirical formula for a compound that contains 11.19% H 88.79% O. Step 1 Find amounts of each element In a g sample, there are g H g O Step 2 Convert g to moles using element molar masses g H g O 1 mol H g H 1 mol O g O = mol H = mol O mol H mol O mol O mol O = = Empirical formula is H 2 O the empirical formula for a compound that contains 56.68% K, 8.68% C 34.73% O. a. K 3 C 2 O 3 b. K 4 C 2 O 6 c. K 2 CO 3 Step 1 Find amounts of each element In a g sample, there are g K, 8.68 g C g O Step 2 Convert g to moles d. KCO 2 1 mol K g K = mol K g K 8.68 g C g O 1 mol C g C 1 mol O g O = mol C = mol O the empirical formula for a compound that contains 56.68% K, 8.68% C 34.73% O. a. K 3 C 2 O 3 b. K 4 C 2 O 6 c. K 2 CO 3 d. KCO mol K mol mol C mol mol O mol = = = Empirical formula is: K 2 CO 3 7

8 the empirical formula for a compound that contains g Fe g S? a. FeS 2 b.fe 3 S 2 c. FeS d. Fe 2 S 3 Step 1 Find amounts of each element Already provided in problem g Fe S Step 2 Convert g to moles g Fe g S 1 mol Fe g Fe 1 mol S g S = mol Fe = mol S Common Fractions mol Fe = = Decimal Fraction mol / mol S mol = = Empirical formula is Fe 2 S / / / /4 To get a whole number, multiply the decimal by the corresponding number in the denominator of the fraction. Example After dividing, you get 0.75 (=3/4) Multiply by the denominator 4(0.75) = 4(3/4) = 3 If molar mass is known, the molecular formula can be calculated from the empirical formula. Molecular formula is a multiple of the empirical formula. Need to determine the value of n. Solving for n The molecular formula can be calculated from the empirical formula if the compound s molar mass is known. Molecular formula = multiple of the empirical formula (EF) MF Determining the multiple n gives the molecular formula Molar mass Mass of empirical formula = number of empirical units in the molecular formula Molar mass Mass of empirical formula = number of empirical units in the molecular formula A compound with the empirical formula NH 2 was found to have a molar mass of g. What is the molecular formula? Molar mass Mass of empirical formula = (1.008) = number of empirical units in the molecular formula A compound with the empirical formula NO 2 was found to have a molar mass of g. What is the molecular formula? a) NO 2 b)n 2 O 4 c) N 3 O6 d) N 4 O g = (16.00) g Molecular formula = (NO 2 ) 2 = N 2 O 4 Molecular formula = (NH 2 ) 2 = N 2 H 4 8

9 Propylene contains 14.3 % H 85.7 % C has a molar mass of g. What is its molecular formula? Plan empirical formula then determine the molecular formula Step 1 Find compound masses In g of compound, 14.3 g H 85.7 g C Step 2 Convert g to moles 1 mol H 14.3 g H = 14.2 mol H g H 85.7 g C 1 mol C g C = 7.14 mol C 14.2 mol H 7.14 mol 7.14 mol C 7.14 mol = 1.99 = 1.00 Empirical formula = CH 2 With EF, calculate the molecular formula g = (1.008) g Molecular formula = (CH 2 ) 3 = C 3 H 6 the molecular formula for a compound that contains 80.0% C 20.0% H with a molar mass of g. a. CH 3 b. CH 2 c. C 2 H 6 d. C 2 H 4 Plan empirical then molecular formula Step 1 Find compound masses In g of compound, 20.0 g H 80.0 g C Step 2 Convert g to moles 1 mol H 20.0 g H g H 80.0 g C 1 mol C g C = 19.8 mol H = 6.66 mol C 19.8 mol H 6.66 mol 6.66 mol C 6.66 mol = 2.97 = 1.00 Empirical formula = CH 3 From empirical formula, calculate the molecular formula g = (1.008) g Molecular formula = (CH 3 ) 2 = C 2 H 6 Learning Objec'ves Learning Objec'ves 7.1 The Mole Apply the concept of the mole, molar mass, Avogadro s number to solve chemistry problems. 7.2 the molar mass of a compound. 7.3 Percent Composition of Compounds the percent composition of a compound from its chemical composition from experimental data. 7.4 Calculating Empirical Formulas Determine the empirical formula for a compound from its percent composition. 7.5 Calculating the Molecular Formula from Compare an empirical formula to a molecular formula calculate a molecular formula from an empirical formula, using the molar mass. 9