7 Quantitative Composition of Compounds. Chapter Outline. The Mole. Slide 1. Slide 2. Slide 3

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1 1 7 Quantitative Composition of Compounds 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 and Susan Arena Copyright 2 Chapter Outline 7.1 The Mole 7.2 Molar Mass of Compounds 7.3 Percent Composition of Compounds 7.4 Calculating Empirical Formulas 7.5 Calculating the Molecular Formula from the Empirical Formula 3 The Mole Individual atoms are tiny and 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.

2 4 The standard 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 dozen = 12 objects The Mole 5 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 6 The Mole 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)

3 7 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 and 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 8 We can use both the mol and molar mass as conversion factors. How many moles of lead does 15.0 g of Pb represent? Solution Map g Pb mol Pb The conversion factor relates g of Pb to moles of Pb. 1 mol Pb g Pb or g Pb 1 mol Pb (Obtain molar mass from the periodic table.) 15.0 g Pb 1 mol Pb g Pb = 7.24 x 10-2 mol Pb 9 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 Solution Map 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

4 10 How many Au atoms are contained in 16.0 g of Au? Solution Map g Au mol Au atoms Au 16.0 g Au Two conversion factors are needed: 1 mol Au g Au and 1 mol Au g Au 1 mol Au x atoms Au x atoms Au 1 mol Au = 4.89 x atoms Au 11 How many Ti atoms are contained in 7.80 g of Ti? a x atoms Ti b x atoms Ti g Ti Solution Map mol Ti atoms Ti c x atoms Ti Two conversion factors are needed: d x atoms Ti 1 mol Ti 1 mol Ti and 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 12 What is the mass of 2.13 x atoms of Li? Solution Map atoms Li mol Li grams Li 2.13 x atoms Li Two conversion factors are needed: 1 mol Li g Li and 1 mol Li x atoms Li 1 mol Li g Li x atoms Li 1 mol Li = 2.46 x 10-5 g Li

5 13 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 1.28 x 10 8 atoms Ne Solution Map atoms Ne mol Ne grams Ne Two conversion factors are needed: 1 mol Ne g Ne and 1 mol Ne x atoms Ne 1 mol Ne g Ne x atoms Ne 1 mol Ne = 4.29 x g Ne 14 What is the mass of 1.05 mol of Ag? Solution Map mol Ag grams Ag One conversion factor is needed: 1.05 mol Ag 1 mol Ag g Ag g Ag 1 mol Ag = 113. g Ag 15 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 Solution Map mol K grams K One conversion factor is needed: g K 1 mol K 1 mol K g K = 321. g K

6 16 How many hydrogen atoms are in 1.00 moles of H 2 molecules? Solution Map mol H 2 molecules H 2 atoms H 2 Two conversion factors are needed: 1 mol H x molecules H 2 and 1 molecule H 2 2 atoms H 1.00 mol H x molecules H 2 2 atoms H 1 mol H 2 1 molecule H 2 = 1.20 x atoms H 2 17 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 Solution Map mol S 8 molecules S 8 atoms S 8 Two conversion factors are needed: 1 mol S x molecules S 8 and 1 molecule S 8 8atoms S x mol S 23 molecules S 8 8atoms S 8 1 mol S 8 1 molecule S 8 = 4.82 x atoms S 8 18 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 of Compounds Molar Mass = MM O + 2MM H = g + 2(1.008 g) = g

7 19 Molar Mass of Compounds What is the molar mass of aluminum hydroxide, Al(OH) 3? 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 20 Molar Mass of Compounds The molar mass of a compound contains Avogadro s number of formula units/molecules. 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 1molO 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. 21 Molar Mass of Compounds As for elements, we can use both the mol and molar mass of formula units/molecules as conversion factors. How many moles of NaCl are there in 253 g of NaCl? Solution Map g NaCl mol NaCl To convert between g of NaCl and 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

8 22 Molar Mass of Compounds How many moles of TiCl 4 are there in 12.5 g of titanium(iv) chloride? a mol TiCl 4 b mol TiCl 4 c x 10 3 mol TiCl 4 d mol TiCl 4 Solution Map g TiCl 4 mol TiCl 4 MM TiCl4 = g + 4(35.45) g = g TiCl g TiCl 4 1 mol TiCl g TiCl 4 = mol TiCl 4 23 Molar Mass of Compounds What is the mass of 3.45 mol of Li 2 O? Solution Map mol Li 2 O g Li 2 O To convert between mol of Li 2 O and g, we must first calculate the molar mass of Li 2 O. MM Li2 O = 2(6.941) g g = g Li 2 O 3.45 mol Li 2 O g Li 2 O 1 mol Li 2 O = 103. g Li 2 O 24 Molar Mass of Compounds What is the mass of 1.23 mol of PH 3? a x 10-2 g PH 3 Solution Map b x 10-6 g PH 3 mol PH 3 gph 3 c g PH 3 d g PH 3 MM PH3 = g + 3(1.008) g = g PH mol PH g PH 3 1 mol PH 3 = 41.8 g PH 3

9 25 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? Solution Map g H 2 S mol H 2 S molecules H 2 S atoms H Molar Mass of Compounds MM H2 S = 2(1.008) g g = g H 2 S 1 mol H 7.53 g H 2 S 2 S g H 2 S x molecules H 2 S 1 mol H 2 S = 1.33 x molecules H 2 S 2 atoms H 1.33 x molecules H 2 S = 2.66 x atoms H 1 molecule H 2 S 26 Molar Mass of Compounds 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 Solution Map 2 c x g H molecules H 2 O 2 O 2 mol H 2 O 2 molecules H 2 O 2 2 d x molecules H 2 O 2 MM H2 O 2 = 2(1.008) g + 2(16.00) g = g H 2 O g H 2 O 2 1 mol H 2O x molecules H 2 O g H 2 O 2 1 mol H 2 O 2 = 1.34 x molecules H 2 O 2 27 Percent Composition 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

10 28 Percent Composition 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 and multiply by 100. Total element mass % of the element = 100 Compound molar mass 29 Percent Composition of Compounds the percent composition of K 2 S. Step 1 compound molar mass MM K2 S = 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. 30 Percent Composition of Compounds the percent composition of O in H 2 O 2. a % b % c % d. 11.1% Step 1 molar mass MM H2 O 2 = 2(1.008) g + 2(16.00)g = g Step 2 % composition O 2(16.00) g O % O = 100 = % O g

11 31 Percent Composition of Compounds the % composition of K 2 CrO 4. Step 1 compound molar mass MM K2 CrO 4 = 2(39.10) g g + 4(16.00) g = g Step 2 % composition % K = 2(39.10) g K g 100 = % K % Cr = g Cr g 100 = % Cr % O = 4(16.00) g O g 100 = % O 32 Percent Composition from Experimental Data Two Step Strategy 1. the mass of the compound formed. 2. Divide the mass of each element by the total mass and multiply by 100. Total element mass % of the element = 100 Total compound mass 33 Percent Composition 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 1.63 g Zn % Zn = 100 = 80.3 % Zn 2.03 g 0.40 g O % O = 100 = 20. % O 2.03 g 100.3% Total should be +/0.5% of 100

12 34 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 Composition 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 35 Empirical and 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 ) and 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 36 Empirical and Molecular Formula 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 and 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

13 37 Calculating Empirical Formulas 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 38 Calculating Empirical Formulas Strategy to an Empirical Formula: 1. Assume a starting mass of the compound (usually g) and 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 smallest mole amount. 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. 39 Calculating Empirical Formulas the empirical formula for a compound that contains 11.19% H and 88.79% O. Step 1 Find amounts of each element In a g sample, there are g H and g O Step 2 Convert g to moles using element molar masses g H 1 mol H g H = mol H g O 1 mol O g O = mol O

14 40 Calculating Empirical Formulas Step 3 Convert to whole numbers by dividing by the smallest mole amount mol H mol O mol O mol O = = Empirical formula is H 2 O 41 Calculating Empirical Formulas the empirical formula for a compound that contains 56.68% K, 8.68% C and 34.73% O. Step 1 Find amounts of each element a. K 3 C 2 O 3 In a g sample, there are b. K 4 C 2 O g K, 8.68 g C and g O c. K 2 CO 3 Step 2 Convert g to moles d. KCO 2 1 mol K g K = mol K 8.68 g C g K 1 mol C g C = mol C g O 1 mol O g O = mol O 42 the empirical formula for a compound that contains 56.68% K, 8.68% C and 34.73% O. Step 3 Convert to whole numbers by dividing by the smallest mole amount. a. K 3 C 2 O 3 b. K 4 C 2 O 6 c. K 2 CO 3 d. KCO 2 Calculating Empirical Formulas mol K mol mol C mol mol O mol = = = Empirical formula is: K 2 CO 3

15 43 the empirical formula for a compound that contains g Fe and g S? a. FeS 2 b.fe 3 S 2 c. FeS d. Fe 2 S 3 Calculating Empirical Formulas Step 1 Find amounts of each element Already provided in problem g Fe and S Step 2 Convert g to moles g Fe 1 mol Fe g Fe = mol Fe g S 1 mol S g S = mol S 44 Calculating Empirical Formulas Step 3 Convert to whole numbers by dividing by the smallest mole amount. 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 45 Calculating the Molecular Formula from the Empirical Formula 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 Molar mass n = Mass of empirical formula = number of empirical units in the molecular formula

16 46 Calculating the Molecular Formula from the Empirical Formula 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) n = MF Determining the multiple n gives the molecular formula Molar mass n = Mass of empirical formula = number of empirical units in the molecular formula 47 Calculating the Molecular Formula from the Empirical Formula n = 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 n = = (1.008) = number of empirical units in the molecular formula Molecular formula = (NH 2 ) 2 = N 2 H 4 48 Calculating the Molecular Formula from the Empirical 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 n = = (16.00) g Molecular formula = (NO 2 ) 2 = N 2 O 4

17 49 Calculating the Molecular Formula from the Empirical Formula Propylene contains 14.3 % H and 85.7 % C and has a molar mass of g. What is its molecular formula? Plan empirical formula and then determine the molecular formula Step 1 Find compound masses In g of compound, 14.3 g H and 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 50 Calculating the Molecular Formula from the Empirical Formula Step 3 Convert to whole numbers by dividing by the smallest mole amount 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 n = = (1.008) g Molecular formula = (CH 2 ) 3 = C 3 H 6 51 Calculating the Molecular Formula from the Empirical Formula the molecular formula for a compound that contains 80.0% C and 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 and then molecular formula Step 1 Find compound masses In g of compound, 20.0 g H and 80.0 g C Step 2 Convert g to moles 20.0 g H 1 mol H g H = 19.8 mol H 80.0 g C 1 mol C g C = 6.66 mol C

18 52 Calculating the Molecular Formula from the Empirical Formula Step 3 Convert to whole numbers by dividing by the smallest mole amount mol H 6.66 mol = mol C = mol Empirical formula = CH 3 From empirical formula, calculate the molecular formula g n = = (1.008) g Molecular formula = (CH 3 ) 2 = C 2 H 6 53 Learning Objectives 7.1 The Mole Apply the concept of the mole, molar mass, and Avogadro s number to solve chemistry problems. 7.2 Molar Mass of Compounds the molar mass of a compound. 7.3 Percent Composition of Compounds the percent composition of a compound from its chemical composition and from experimental data. 54 Learning Objectives 7.4 Calculating Empirical Formulas Determine the empirical formula for a compound from its percent composition. 7.5 Calculating the Molecular Formula from the Empirical Formula Compare an empirical formula to a molecular formula and calculate a molecular formula from an empirical formula, using the molar mass.