Chapter 6 Chemical Composition

Chapter 6 Chemical Composition

Why Is Knowledge of Chemical Composition Important? Everything in nature is either chemically or physically combined with other substances. To know the amount of a material in a sample, you need to know what fraction of the sample it is. Some Applications: The amount of sodium in sodium chloride for diet. The amount of sugar in a can of pop. The amount of hydrogen in water for hydrogen fuel. The amount of chlorine in freon to estimate ozone depletion. 2

Counting Nails by the Pound People usually buy thousands of nails for construction Hardware stores sell nails in the pound. How do I know how many nails I am buying when I buy a pound of nails? 3

Counting Nails by the Pound, Continued A hardware store customer buys 2.60 pounds of nails. A dozen nails has a mass of 0.150 pounds. How many nails did the customer buy? Solution map: 1 dozen nails = 0.150 lbs. 12 nails = 1 dozen nails 4

Counting Nails by the Pound, Continued 1doz.nails 12 nails 2.60 lbs. 0.150 lbs. 1doz. 208 nails The customer bought 2.60 lbs of nails and received 208 nails. He counted the nails by weighing them! 5

Counting Nails by the Pound, Continued What if he bought a different size nail? Would the mass of a dozen be 0.150 lbs? Would there still be 12 nails in a dozen? Would there be 208 nails in 2.60 lbs? How would this effect the conversion factors? 6

Counting Atoms by Moles Just as we can use the dozen to count nails, we can use the mole to count atoms in a sample. 1 mole = 6.022 x 10 23 things. 1 mole of atoms = 6.022 x 10 23 atoms 1 mole of apple = 6.022 x 10 23 apples Note that just like the dozen the mole is a constant number and this number is called the Avogadro s number. 7

Using Moles as Conversion Factors Since 1 mole atoms = 6.022 x10 23 atoms 23 6.022 10 atoms 1mole OR 1mole 6.022 10 23 atoms 8

Example 6.1 A silver ring contains 1.1 x 10 22 silver atoms. How many moles of silver are in the ring? Given: Find: Solution Map: Relationships: Solution: Check: 1.1 10 atoms Ag 22 1.8266 10 1.1 x 10 22 atoms Ag moles Ag 1mol 6.022 10 1 mol = 6.022 x 10 23 atoms atoms Ag -2 mol Since the number of atoms given is less than Avogadro s number, the answer makes sense. Tro's "Introductory Chemistry", Chapter 6 23 Ag atoms mol Ag 1mol 23 6.022 10 1.8 10-2 atoms mol 9 Ag

Example 6.1a Calculate the number of atoms in 2.45 mol of copper. Given: Find: Solution Map: Relationships: mol Cu 2.45 mol Cu atoms Cu 6.022 10 atoms 1mol atoms Cu 1 mol = 6.022 x 10 23 atoms 23 Solution: 6.022 10 2.45 mol Cu 1mol 1.48 10 24 atoms Cu 23 atoms Tro's "Introductory Chemistry", Chapter 6 10

Practice Calculate the number of atoms in 1.45 mol of Sodium. Answer: 8.73 x 10 23 Na atoms 11

Relationship Between Moles and Mass The mass of one mole of a substance is called the molar mass. The molar mass of an element, in grams, is numerically equal to the element s atomic mass, in amu E.g., 1 Carbon atom weighs 12.00 amu, therefore the molar mass of Carbon is 12.00 g The lighter the atom, the less a mole weighs. Carbon (C) is lighter than Sodium (Na); 1 mol C weighs 12.00 g and 1 mol Na weighs 22.99 g. 12

Mole and Mass Relationships Substance Pieces in 1 mole Weight of 1 mole Hydrogen 6.022 x 10 23 atoms 1.008 g Carbon 6.022 x 10 23 atoms 12.01 g Oxygen 6.022 x 10 23 atoms 16.00 g Sulfur 6.022 x 10 23 atoms 32.06 g Calcium 6.022 x 10 23 atoms 40.08 g Chlorine 6.022 x 10 23 atoms 35.45 g Copper 6.022 x 10 23 atoms 63.55 g 1 mole sulfur 32.06 g 1 mole carbon 12.01 g 13

Example 6.2a Calculate the moles of sulfur in 57.8 g of sulfur. Given: Find: 57.8 g S mol S Solution Map: Relationships: Solution: g S 1 mol S = 32.07 g 1mol S 32.07 g 57.8g S 1.80 mols mol S 1mol 32.07 g 14

Example 6.2b Calculate the moles of carbon in 0.0265 g of pencil lead. Given: Find: Solution Map: Relationships: Solution: 0.0265 g C mol C g C 1 mol C = 12.01 g 1mol 12.01g 0.0265 g C 2.21 10-3 mol mol C 1mol 12.01g C 15

Example 6.3 How Many Aluminum Atoms Are in a Can Weighing 16.2 g? Given: Find: Solution Map: Relationships: Solution: 16.2 g Al atoms Al 1mol 26.98 g 1 mol Al = 26.98 g, 1 mol = 6.022 x 10 23 16.2 g Al g Al mol Al atoms Al 3.62 10 1mol Al 26.98g Al 23 atoms Al 23 6.022 10 atoms 1mol 23 6.022 10 1mol atoms 16

Practice 1) Calculate the grams of carbon in 2.21 x 10-3 mol of lead pencil. 2) How many copper atoms are in a penny weighing 3.10 g? Tro's "Introductory Chemistry", Chapter 6 17

Practice 1) Calculate the grams of carbon in 2.21 x 10-3 mol of lead pencil. Answer: 0.0265 g Carbon 18

Practice 2) How many copper atoms are in a penny weighing 3.10 g? Answer: 2.94 x 10 22 Copper atoms 19

Molar Mass of Compounds The molar mass of a compound can be calculated by adding up the molar mass of all the atoms that the compound is compose of. Example: Calculate the molar mass of water (H 2 O) Since 1 mole of H 2 O contains 2 moles of H and 1 mole of O. Molar mass = 2(1.01 g H) + 16.00 g O = 18.02 g. 20

Example 6.4 Calculate the mass of 1.75 mol of H 2 O. Given: Find: Solution Map: Relationships: Solution: Check: 1.75 mol H 2 O g H 2 O mol H 2 O 18.02 g 1mol H O 1 mol H 2 O = 18.02 g 2 g H 2 O 18.02 g 1.75 mol H2O 1mol 31.535 g 31.5 g H O Since the given amount is more than 1 mol, the mass being > 18 g makes sense. 2 21

Example 6.4a How many moles are in 50.0 g of PbO 2? (Pb = 207.2, O = 16.00) Given: Find: Solution Map: Relationships: Solution: Check: 50.0 g mol PbO 2 moles PbO 2 g PbO 2 mol PbO 2 1 2 mol PbO 239.2 g 1 mol PbO 2 = 239.2 g 50.0 g PbO 2 0.20903 mol 1mol 239.2 g 0.209 mol PbO Since the given amount is less than 239.2 g, the moles being < 1 makes sense. 2 22

Practice What is the mass of 4.78 x 10 24 NO 2 molecules? Answer: 365 g 23

Example 6.5 How many formula units are in 50.0 g of PbO 2? (PbO 2 = 239.2) Given: Find: Solution Map: Relationships: 50.0 g PbO 2 formula units PbO 2 g PbO 2 units PbO 2 23 6.022 10 molecules 239.2 g 239.2 g = 6.022 x 10 23 Solution: 50.0 g PbO 1.26 10 2 23 23 6.022 10 units PbO 239.2 g PbO units PbO 2 2 2 24

Mole Relationships in Chemical Formulas Examples Moles of compound Moles of constituents 1 mol NaCl 1 mol Na, 1 mol Cl 1 mol H 2 O 2 mol H, 1 mol O 1 mol CaCO 3 1 mol Ca, 1 mol C, 3 mol O 1 mol C 6 H 12 O 6 6 mol C, 12 mol H, 6 mol O Fact check: How many moles of H atoms are in 1 mole of CH 3 OCH 2 CH 3 Tro's "Introductory Chemistry", Chapter 6 25

Example 6.6 Calculate the moles of oxygen in 1.7 moles of CaCO 3. Given: Find: Solution Map: Relationships: 1.7 mol CaCO 3 mol O mol CaCO 3 1 mol CaCO 3 = 3 mol O 3 mol O 1mol CaCO 3 mol O Solution: 1.7 mol CaCO 5.1 mol O 3 3 mol O 1mol CaCO 3 Tro's "Introductory Chemistry", Chapter 6 26

Example 6.7 Find the mass of carbon in 55.4 g C 10 H 14 O. Given: Find: Solution Map: Relationships: 55.4 g C 10 H 14 O g C Solution: 1mol C10H 55.4 g C H14O 150.2 g g C 10 H 14 O mol C 10 H 14 O mol C g C 1mol 150.2 g 1 mol C 10 H 14 O = 150.2 g, 1 mol C = 12.01 g, 10 mol C : 1 mol C 10 H 14 O O 10 mol C 1mol C H 10 mol C 1mol C H 12.01g C O 1mol C 14 10 10 14 10 14 O 12.01g 1mol 44.3 g C 27

Practice Find the Mass of Sodium in 6.2 g of NaCl (MM: Na = 22.99, Cl = 35.45) Answer: 2.4 g 28

Percent Composition of Compounds Percentage of each element in a compound by mass can be determined from: The formula of the compound. The experimental mass analysis of the compound. Percentage mass of element X in 1mol mass of 1mol of the compound 100% Tro's "Introductory Chemistry", Chapter 6 29

Example 6.9 Find the mass percent of Cl in C 2 Cl 4 F 2. Given: Find: Solution Map: Relationships: Solution: C 2 Cl 4 F 2 % Cl by mass Mass% Cl Mass% element molar mass C Cl F 4 molar mass Cl molar mass C Cl F X 4 molar mass Cl 4(35.45 g/mol) 2 4 2 2 mass element mass1mol of 141.8 g/mol Mass% Cl 100% 69.58% 203.8 g/mol 4 2 141.8 g/mol 100% X in 1mol 100% compound 2(12.01) 4(35.45) 2(19.00) 203.8 g/mol

Practice Determine the mass percent composition of Ca and Cl in CaCl 2 : (Ca = 40.08, Cl = 35.45) Answer: 36.11 % Ca and 63.86 % Cl 31

Mass Percent as a Conversion Factor The mass percent tells you the mass of a constituent element in 100 g of the compound. The fact that NaCl is 39% Na by mass means that 100 g of NaCl contains 39 g Na. This can be used as a conversion factor. 100 g NaCl = 39 g Na g NaCl 39 g Na 100 g NaCl g Na g 100 g NaCl Na 39 g Na g NaCl Tro's "Introductory Chemistry", Chapter 6 32

**Example**: The FDA recommends that you consume 2.4 g of sodium a day. How many grams of NaCl must you consume a day in order to keep up with your sodium intake. NaCl is 39 % sodium by mass Answer: 6.2 g NaCl 33

Empirical Formulas The simplest, whole-number ratio of atoms in a molecule is called the empirical formula. Can be determined from percent composition or combining masses. The molecular formula is a multiple of the empirical formula. 34

Empirical Formulas, Continued Hydrogen Peroxide Molecular formula = H 2 O 2 Empirical formula = HO Benzene Molecular formula = C 6 H 6 Empirical formula = CH Glucose Molecular formula = C 6 H 12 O 6 Empirical formula = CH 2 O 35

Example Determine the Empirical Formula of Benzopyrene, C 20 H 12, Find the greatest common factor (GCF) of the subscripts. 20 and 12 GCF = 4 Divide each subscript by the GCF to get the empirical formula. C 20/4 H 12/4 = C 5 H 3 Empirical formula = C 5 H 3 36

Finding an Empirical Formula From % Mass 1. Convert the percentages to grams. a. Skip if already grams. 2. Convert grams to moles. a. Use molar mass of each element. 3. Write a pseudoformula using moles as subscripts. 4. Divide all by smallest number of moles. 5. Multiply all mole ratios by number to make all whole numbers, if necessary. a. -If ratio is 0.5, multiply all by 2 -if ratio is 0.33 or 0.67, multiply all by 3 -if ratio is 0.25 multiply by 4 b. Skip if already whole numbers after Step 4. 37

Example 6.11 Finding an empirical formula from experimental data A laboratory analysis of aspirin determined the following mass percent composition. Find the empirical formula. C = 60.00% H = 4.48% O = 35.53% Tro's "Introductory Chemistry", Chapter 6 38

Example: Find the empirical formula of aspirin with the given mass percent composition. Step 1: Convert percentages to grams Given: C = 60.00% H = 4.48% O = 35.53% Therefore, in 100 g of aspirin there are 60.00 g C, 4.48 g H, and 35.53 g O. 39

Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: 60.00 g C, 4.48 g H, 35.53 g O Find: empirical formula, C x H y O z Conversion Factors: 1 mol C = 12.01 g 1 mol H = 1.01 g 1 mol O = 16.00 g Step 2: Convert masses to moles: 1mol C 60. 00 g C 4. 996 mol C 12.01g C 1mol H 4.48 g H 4. 44 mol H 1.01g H 1mol O 35.53 g O 2. 221mol O 16.00 g O 40

Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: 4.996 mol C, 4.44 mol H, 2.221 mol O Write the empirical formula, C x H y O z Step 3: Write a pseudoformula in the form C x H y O z. C 4.996 H 4.44 O 2.221 41

Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: C 4.996 H 4.44 O 2.221 Find: empirical formula, C x H y O z Solution Map: g C,H,O mol C,H,O mol ratio empirical formula Step 4: Find the mole ratio by dividing by the smallest number of moles in C 4.996 H 4.44 O 2.221 C 4.996 2.221 C H 2.25 4.44 2.221 H 2 O O 1 2.221 2.221 42

Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: C 2.25 H 2 O 1 Find: empirical formula, C x H y O z Solution Map: g C,H,O mol C,H,O mol ratio empirical formula Step 5: Multiply subscripts by factor to give whole number. C 2.25H 2O1 { } x 4 C 9 H 8 O 4 43

Practice 1 Determine the Empirical Formula of Tin Fluoride, which Contains 75.7% Sn and the Rest Fluorine. Practice 2 Determine the Empirical Formula of a compound which Contains 69.94% Fe and the Rest Oxygen. 47

Practice 1 Determine the Empirical Formula of Stannous Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00). Answer: SnF 2 48

Answer: SnF 2 Tro's "Introductory Chemistry", Chapter 6 49

Practice 2 Determine the Empirical Formula of a compound which Contains 69.94 % Fe (55.85) and the Rest Oxygen (16.00). Answer: Fe 2 O 3 50

Answer: Fe 2 O 3 51

Empirical Formula and Molecular Formula. Name Empirical Molecular Molar Formula Formula Mass, g Glyceraldehyde CH 2 O C 3 H 6 O 3 90 Erythrose CH 2 O C 4 H 8 O 4 120 Arabinose CH 2 O C 5 H 10 O 5 150 Glucose CH 2 O C 6 H 12 O 6 180 52

Molecular Formulas The molecular formula is a multiple of the empirical formula. To determine the molecular formula, you need to know the empirical formula and the molar mass of the compound. Molar mass real formula = Factor used to multiply subscripts Molar mass empirical formula 53

Example 6.13 Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C 5 H 8. 1. Determine the empirical formula. May need to calculate it as previous. C 5 H 8 2. Determine the molar mass of the empirical formula. 5 C = 60.05, 8 H = 8.064 C 5 H 8 = 68.11 g/mol 54

Example 6.13 Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C 5 H 8, Continued. 3. Divide the given molar mass of the compound by the molar mass of the empirical formula. Round to the nearest whole number. 204 g/mol 68.11g/mol 3 Tro's "Introductory Chemistry", Chapter 6 55

Example 6.13 Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C 5 H 8, Continued. 4. Multiply the empirical formula by the factor above to give the molecular formula. (C 5 H 8 ) 3 = C 15 H 24 Tro's "Introductory Chemistry", Chapter 6 56

Practice Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C 5 H 3. What is its Molecular Formula? (C = 12.01, H=1.01) Answer: C 20 H 12 57

Example 6.14 Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N. (C=12.01, H=1.01, N=14.01) Tro's "Introductory Chemistry", Chapter 6 58

Example 6.14 Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N, Continued Given: 74.0% C, 8.7% H, {100 (74.0+8.7)} = 17.3% N in 100 g nicotine there are 74.0 g C, 8.7 g H, and 17.3 g N. Find: C x H y N z Conversion Factors: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol N = 14.01 g Solution Map: g C g H mol C mol H pseudoformula mole ratio whole number ratio empirical formula g N mol N 59

Example 6.14 Determine the Molecular Formula of Nicotine, which has a Molar Mass of 162 g and is 74.0% C, 8.7% H, and the Rest N, Continued. Apply solution map: 1mol C 74. 0 g C 6.16 mol C 12.01g 1mol H 8.7 g H 8.6 mol H 1.01g 1mol N 17.3 g N 1.23 mol N 14.01g mol. mass nicotine mol. mass emp. form. C 6.16 H 8.6 N 1.23 C6.16H 8.6 N1.23 C5H7N 1.23 1.23 1.23 C 5 = 5(12.01 g) = 60.05 g N 1 = 1(14.01 g) = 14.01 g H 7 = 7(1.01 g) = 7.07 g C 5 H 7 N = 81.13 g 162 g 81.13 g {C 5 H 7 N} x 2 = C 10 H 14 N 2 2 60

Recommended Study Problems Chapter 6 NB: Study problems are used to check the student s understanding of the lecture material. Students are EXPECTED TO BE ABLE TO SOLVE ALL THE SUGGESTED STUDY PROBLEMS. If you encounter any problems, please talk to your professor or seek help at the HACC-Gettysburg learning center. Questions from text book Chapter 6, p 189 3, 7, 11, 17, 19, 25, 27, 37, 51, 57, 59, 69, 71, 73, 79, 85, 87, 89, 97, 102, 117, 116 ANSWERS -The answers to the odd-numbered study problems are found at the back of your textbook 61