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.

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?

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? 1 dozen nails = 0.150 lbs. 12 nails = 1 dozen nails Solution map:

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

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?

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 1023 things. 1 mole of atoms = 6.022 x 1023 atoms 1 mole of apple = 6.022 x 1023 apples Note that just like the “dozen” the mole is a constant number and this number is called the Avogadro’s number.

Using Moles as Conversion Factors Since 1 mole atoms = 6.022 x1023 atoms OR

Tro's "Introductory Chemistry", Chapter 6 Example 6.1—A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? Given: Find: 1.1 x 1022 atoms Ag moles Ag Solution Map: Relationships: 1 mol = 6.022 x 1023 atoms atoms Ag mol Ag Solution: Check: Since the number of atoms given is less than Avogadro’s number, the answer makes sense. Tro's "Introductory Chemistry", Chapter 6

Example 6.1a—Calculate the number of atoms in 2.45 mol of copper. Given: Find: 2.45 mol Cu atoms Cu Solution Map: Relationships: 1 mol = 6.022 x 1023 atoms mol Cu atoms Cu Solution: Tro's "Introductory Chemistry", Chapter 6

Practice—Calculate the number of atoms in 1.45 mol of Sodium. Answer: 8.73 x 1023 Na atoms

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.

Mole and Mass Relationships carbon 12.01 g 1 mole sulfur 32.06 g

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: 1 mol S = 32.07 g g S mol S Solution:

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

Example 6.3—How Many Aluminum Atoms Are in a Can Weighing 16.2 g? Given: Find: 16.2 g Al atoms Al Solution Map: Relationships: 1 mol Al = 26.98 g, 1 mol = 6.022 x 1023 g Al mol Al atoms Al Solution:

Tro's "Introductory Chemistry", Chapter 6 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

Practice 1) Calculate the grams of carbon in 2 Practice 1) Calculate the grams of carbon in 2.21 x 10-3 mol of lead pencil. Answer: 0.0265 g Carbon

Practice 2) How many copper atoms are in a penny weighing 3.10 g? Answer: 2.94 x 1022 Copper atoms

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 (H2O) Since 1 mole of H2O contains 2 moles of H and 1 mole of O. Molar mass = 2(1.01 g H) + 16.00 g O = 18.02 g.

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

Example 6. 4a—How many moles are in 50. 0 g of PbO2. (Pb = 207 Example 6.4a—How many moles are in 50.0 g of PbO2? (Pb = 207.2, O = 16.00) Given: Find: 50.0 g mol PbO2 moles PbO2 Solution Map: Relationships: 1 mol PbO2 = 239.2 g g PbO2 mol PbO2 Solution: Check: Since the given amount is less than 239.2 g, the moles being < 1 makes sense.

Practice—What is the mass of 4.78 x 1024 NO2 molecules? Answer: 365 g

Example 6.5—How many formula units are in 50.0 g of PbO2? (PbO2 = 239.2) Given: Find: 50.0 g PbO2 formula units PbO2 Solution Map: Relationships: 239.2 g = 6.022 x 1023 g PbO2 units PbO2 Solution:

Mole Relationships in Chemical Formulas Examples Moles of compound Moles of constituents 1 mol NaCl 1 mol Na, 1 mol Cl 1 mol H2O 2 mol H, 1 mol O 1 mol CaCO3 1 mol Ca, 1 mol C, 3 mol O 1 mol C6H12O6 6 mol C, 12 mol H, 6 mol O Fact check: How many moles of H atoms are in 1 mole of CH3OCH2CH3 Tro's "Introductory Chemistry", Chapter 6

Example 6.6—Calculate the moles of oxygen in 1.7 moles of CaCO3. Given: Find: 1.7 mol CaCO3 mol O Solution Map: Relationships: 1 mol CaCO3 = 3 mol O mol CaCO3 mol O Solution: Tro's "Introductory Chemistry", Chapter 6

Example 6.7—Find the mass of carbon in 55.4 g C10H14O. Given: Find: 55.4 g C10H14O g C Solution Map: Relationships: 1 mol C10H14O = 150.2 g, 1 mol C = 12.01 g, 10 mol C : 1 mol C10H14O g C10H14O mol C10H14O mol C g C Solution:

Practice—Find the Mass of Sodium in 6. 2 g of NaCl (MM: Na = 22 Practice—Find the Mass of Sodium in 6.2 g of NaCl (MM: Na = 22.99, Cl = 35.45) Answer: 2.4 g

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. Tro's "Introductory Chemistry", Chapter 6

Example 6.9—Find the mass percent of Cl in C2Cl4F2. Given: Find: C2Cl4F2 % Cl by mass Solution Map: Relationships: Solution:

Practice—Determine the mass percent composition of Ca and Cl in CaCl2: Answer: 36.11 % Ca and 63.86 % Cl

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 Tro's "Introductory Chemistry", Chapter 6

Example. : The FDA recommends that you consume 2. 4 g of sodium a day **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 Solve this in class Answer: 6.2 g NaCl

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.

Empirical Formulas, Continued Hydrogen Peroxide Molecular formula = H2O2 Empirical formula = HO Benzene Molecular formula = C6H6 Empirical formula = CH Glucose Molecular formula = C6H12O6 Empirical formula = CH2O

Example—Determine the Empirical Formula of Benzopyrene, C20H12, 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. C20/4H12/4 = C5H3 Empirical formula = C5H3

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.

Tro's "Introductory Chemistry", Chapter 6 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

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.

Step 2: Convert masses to moles: 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, CxHyOz 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:

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, CxHyOz Step 3: Write a pseudoformula in the form CxHyOz . C4.996H4.44O2.221

Find: empirical formula, CxHyOz Solution Map: g C,H,O  mol C,H,O  Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: C4.996H4.44O2.221 Find: empirical formula, CxHyOz 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 C4.996H4.44O2.221

Step 5: Multiply subscripts by factor to give whole number. Example: Find the empirical formula of aspirin with the given mass percent composition. Information: Given: C2.25H2O1 Find: empirical formula, CxHyOz Solution Map: g C,H,O  mol C,H,O  mol ratio  empirical formula Step 5: Multiply subscripts by factor to give whole number. { } x 4 C9H8O4

Example 6.12—Finding an Empirical Formula from Experimental Data A 3.24-g sample of titanium reacts with oxygen to form 5.40 g of the metal oxide. What is the formula of the oxide?

Example 6.12 Solution: Step 1: Convert percentages to grams Skip, since masses are given Given: Ti = 3.24 g compound = 5.40 g Therefore mass O = 5.40 g – 3.24 g = 2.16 g Step 2: Convert masses to moles

Example 6.12 Cont’d Step 3: Write a pseudoformula in the form TixOy Step 4: Find the mole ratio by dividing by the smallest number of moles in Ti0.0677O0.135 Ti0.0677O0.135

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.

Practice 1—Determine the Empirical Formula of Stannous Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00). Answer: SnF2

Tro's "Introductory Chemistry", Chapter 6 Answer: SnF2

Practice 2—Determine the Empirical Formula of a compound which Contains 69.94 % Fe (55.85) and the Rest Oxygen (16.00). Answer: Fe2O3

Answer: Fe2O3

Empirical Formula and Molecular Formula. Name Empirical Formula Molecular Molar Mass, g Glyceraldehyde CH2O C3H6O3 90 Erythrose C4H8O4 120 Arabinose C5H10O5 150 Glucose C6H12O6 180

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 massreal formula Molar massempirical formula = Factor used to multiply subscripts

Example 6.13—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8. Determine the empirical formula. May need to calculate it as previous. C5H8 Determine the molar mass of the empirical formula. 5 C = 60.05, 8 H = 8.064 C5H8 = 68.11 g/mol

Tro's "Introductory Chemistry", Chapter 6 Example 6.13—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued. Divide the given molar mass of the compound by the molar mass of the empirical formula. Round to the nearest whole number. Tro's "Introductory Chemistry", Chapter 6

Tro's "Introductory Chemistry", Chapter 6 Example 6.13—Determine the Molecular Formula of Cadinene if it has a Molar Mass of 204 g and an Empirical Formula of C5H8, Continued. Multiply the empirical formula by the factor above to give the molecular formula. (C5H8)3 = C15H24 Tro's "Introductory Chemistry", Chapter 6

Practice—Benzopyrene has a Molar Mass of 252 g and an Empirical Formula of C5H3. What is its Molecular Formula? (C = 12.01, H=1.01) Answer: C20H12

Tro's "Introductory Chemistry", Chapter 6 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

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: CxHyNz Conversion Factors: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol N = 14.01 g Solution Map: g C mol C whole number ratio mole ratio pseudo- formula empirical formula g H mol H g N mol N

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: C5 = 5(12.01 g) = 60.05 g N1 = 1(14.01 g) = 14.01 g H7 = 7(1.01 g) = 7.07 g C5H7N = 81.13 g C6.16H8.6N1.23 {C5H7N} x 2 = C10H14N2

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