Chapter 8 Chemical Composition

8.1 Counting by Weighing 8.2 Atomic Masses: Counting Atoms by Weighing 8.3 The Mole 8.4 Learning to Solve Problems 8.5 Molar Mass 8.6 Percent Composition of Compounds 8.7 Formulas of Compounds 8.8 Calculation of Empirical Formulas 8.9 Calculation of Molecular Formulas

All we need to know is the average mass of the objects. Objects do not need to have identical masses to be counted by weighing. All we need to know is the average mass of the objects. To count the atoms in a sample of a given element by weighing we must know the mass of the sample and the average mass for that element. Copyright © Cengage Learning. All rights reserved

Averaging the Mass of Similar Objects Example: What is the mass of 1000 jelly beans? Not all jelly beans have the same mass. Suppose we weigh 10 jelly beans and find: Now we can find the average mass of a bean. 4. Finally we can multiply to find the mass of 1000 beans! Bean 1 2 3 4 5 6 7 8 9 10 Mass 5.1 g 5.2 g 5.0 g 4.8 g 4.9 g 5.00g Copyright © Cengage Learning. All rights reserved

Averaging the Mass of Different Objects Two samples containing different types of components (A and B), both contain the same number of components if the ratio of the sample masses is the same as the ratio of the masses of the individual components. Copyright © Cengage Learning. All rights reserved

Exercise A pile of marbles weigh 394.80 g. 10 marbles weigh 37.60 g. How many marbles are in the pile? 3.760 g/marble Average mass of one marble = 37.60 g / 10 marbles = 3.76 g / marble 394.80 g / 3.76 g = 105 marbles 10 marbles Or 394.80g ---------------- = 105.0 marbles 37.60 g /marble 3.760 g/marble

1 atomic mass unit (amu) = 1.66 10–24 g Atoms have very tiny masses so scientists made a unit to avoid using very small numbers. 1 atomic mass unit (amu) = 1.66 10–24 g The average atomic mass for an element is the weighted average of the masses of all the isotopes of an element. Copyright © Cengage Learning. All rights reserved

Average Atomic Mass for Carbon Even though natural carbon does not contain a single atom with mass 12.01, for our purposes, we can consider carbon to be composed of only one type of atom with a mass of 12.01. This enables us to count atoms of natural carbon by weighing a sample of carbon. Copyright © Cengage Learning. All rights reserved

Example Using Atomic Mass Units Calculate the mass (in amu) of 431 atoms of carbon. The mass of 1 carbon atom = 12.01 amu. Use the relationship as a conversion factor. 5.18 x 103 amu Copyright © Cengage Learning. All rights reserved

Calculate the mass (in amu) of 75 atoms of aluminum. Exercise Calculate the mass (in amu) of 75 atoms of aluminum. 2.0 x103 amu Al Determine the mass of 1 Al atom. 1 atom of Al = 26.98 amu Use the relationship as a conversion factor. 75 Al atoms × (26.98 amu/1 Al atom) = 2024 amu Copyright © Cengage Learning. All rights reserved

The number equal to the number of carbon atoms in 12 The number equal to the number of carbon atoms in 12.01 grams of carbon. 1 mole of anything = 6.022 x 1023 units of that thing (Avogadro’s number). 1 mole C = 6.022 x 1023 C atoms = 12.01 g C Copyright © Cengage Learning. All rights reserved

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Avogadro’s Number of various elements

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Comparison of 1-Mol Samples of Various Elements A sample of an element with a mass equal to that element’s average atomic mass (expressed in g) contains one mole of atoms (6.022 × 1023 atoms). Comparison of 1-Mol Samples of Various Elements Copyright © Cengage Learning. All rights reserved

Determine the number of copper atoms in a 63.55 g sample of copper. Concept Check Determine the number of copper atoms in a 63.55 g sample of copper. 1 mole 6.022x 1023 atoms Cu 63.55 g Cu ------------ ----------------------------- = 63.55 g Cu 1 mole Cu atoms 6.022×1023 Cu atoms 6.022×1023 Cu atoms; 63.55 g of Cu is 1 mole of Cu, which is equal to Avogadro’s number. Copyright © Cengage Learning. All rights reserved

Concept Check Which of the following is closest to the average mass of one atom of copper? a) 63.55 g b) 52.00 g c) 58.93 g d) 65.38 g e) 1.055 x 10-22 g The correct answer is e. The mass of one atom of copper is going to be extremely small, so only letter “e” makes sense. The calculated solution is (1 Cu atom) × (1 mol Cu/6.022×1023 Cu atoms) × (63.55 g Cu/1 mol Cu). Copyright © Cengage Learning. All rights reserved

Calculate the number of iron atoms in a 4.48 mole sample of iron. Concept Check Calculate the number of iron atoms in a 4.48 mole sample of iron. 2.70 × 1024 Fe atoms (4.48 mol Fe) × (6.022×1023 Fe atoms / 1 mol Fe) = 2.70×1024 Fe atoms. Copyright © Cengage Learning. All rights reserved

Concept Check A sample of 26.98 grams of Al has the same number of atoms as _______ grams of Au. a) 26.98 g b) 13.49 g c) 197.0 g d) 256.5 g 1 mole Al 1 mole Au 196.97g Au 26.98 g Al ----------- --------------- ---------------- = 197.0 g Au 26.98 g Al 1 mole Al 1 mole Au The correct answer is c. That is the mass of 1 mole of gold. The 26.98 g of Al represents 1 mole of Al. Thus both samples contain Avogadro’s number, 6.022 × 1023 atoms. Copyright © Cengage Learning. All rights reserved

Concept Check Which of the following 100.0 g samples contains the greatest number of atoms? Magnesium Zinc Silver The correct answer is a. Magnesium has the smallest average atomic mass (24.31 amu). Since magnesium is lighter than zinc and silver, it will take more atoms to reach 100.0 g. Copyright © Cengage Learning. All rights reserved

Rank the following according to number of atoms (greatest to least): Exercise Rank the following according to number of atoms (greatest to least): 107.9 g of silver (= 1 mole) 70.0 g of zinc (> 1 mole) 21.0 g of magnesium (< 1 mole) b) a) c) b, a, c; The greater the number of moles, the greater the number of atoms present. Zinc contains 1.07 mol, Ag contains 1.000 mol, and Mg contains 0.864 mol. Copyright © Cengage Learning. All rights reserved

Conceptual Problem Solving Where are we going? Read the problem and decide on the final goal. How do we get there? Work backwards from the final goal to decide where to start. Reality check. Does my answer make sense? Is it reasonable? Copyright © Cengage Learning. All rights reserved

Mass in grams of one mole of the substance: Molar Mass of N = 14.01 g/mol Molar Mass of H2O = 18.02 g/mol (2 × 1.008 g) + 16.00 g Molar Mass of Ba(NO3)2 = 261.35 g/mol 137.33 g + (2 × 14.01 g) + (6 × 16.00 g) Copyright © Cengage Learning. All rights reserved

Calculations Using Molar Mass Moles of a compound = Mass of a sample (g) = (moles of sample)(molar mass of compound) Copyright © Cengage Learning. All rights reserved

What is the molar mass of nickel(II) carbonate? Exercise What is the molar mass of nickel(II) carbonate? a) 118.7 g/mol b) 134.7 g/mol c) 178.71 g/mol d) 296.09 g/mol The formula for nickel(II) carbonate is NiCO3. Therefore the molar mass is: 58.69 + 12.01 + 3(16.00) = 118.70 g/mol. The correct answer is a. The formula for nickel(II) carbonate is NiCO3. Therefore the molar mass is: 58.69 + 12.01 + 3(16.00) = 118.7 g/mol.

Exercise Consider equal mole samples of N2O, Al(NO3)3, and KCN. Rank these from least to most number of nitrogen atoms in each sample. a) KCN, Al(NO3)3, N2O b) Al(NO3)3, N2O, KCN c) KCN, N2O, Al(NO3)3 d) Al(NO3)3, KCN, N2O The correct answer is c. Assume you have one mole of each sample. This means you have 1 mole of N in KCN, 2 moles of N in N2O and 3 moles of N in Al(NO3)3. Moles are directly related through Avogadro’s number. Copyright © Cengage Learning. All rights reserved

Consider separate 100.0 gram samples of each of the following: Exercise Consider separate 100.0 gram samples of each of the following:  H2O, N2O, C3H6O2, CO2 Rank them from greatest to least number of oxygen atoms. H2O, CO2, C3H6O2, N2O 1/18 2/44 2/74 1/44 1/18 1/22 1/37 1/44 18gpm 44gpm 74gpm 44gpm H2O, CO2, C3H6O2, N2O; The number of oxygen atoms in each is: H2O = 3.343×1024 CO2 = 2.737×1024 C3H6O2 = 1.626×1024 N2O = 1.368×1024 100 gw(1 mole/18gw)(6.022x1023 molec/1mole)(# Oxygen atoms) =

Exercise How many grams of fluorine are contained in one molecule of boron trifluoride? a) 3.155 × 10–23 g b) 9.465 × 10–23 g c) 6.022 × 1023 g d) 3.433 × 1025 g The correct answer is b. 1 molecule BF3 × (3 atoms F/1 molecule BF3) × (1 mol F/6.022×1023 atoms F) × (19.00 g F/1 mol F) = 9.465 × 10–23 g F

Mass percent of an element: For iron in iron(III) oxide, (Fe2O3): % Copyright © Cengage Learning. All rights reserved

Percent Composition What is the % composition of C6H12O6? 12.011g C 72.066 180.155 x100%=40.0% C 6 moles C -------------- = 72.066 g C 1 mole C 1.0079g H 12.0948 180.155 12 moles H -------------- = 12.0948 g H x100%=6.7% H 1 mole H 15.999g O 95.994 180.155 6 moles O -------------- = 95.994g O x100%=53.3% O 1 mole O 180.155 % composition is 40.0% C, 6.7% H, and 53.3% O

Exercise Morphine, derived from opium plants, has the potential for use and abuse. It’s formula is C17H19NO3. What percent, by mass, is the carbon in this compound? a) 12.0 % b) 54.8 % c) 67.9 % d) 71.6 % The correct answer is d. First determine the molar mass of the compound, then divide that into the total mass of carbon present, and finally multiply that by 100 to convert from a fraction into a percentage. (17  12.01) / [(17  12.01) + (19  1.01) + (1  14.01) + (3  16.00)] = 0.716 0.716  100 = 71.6 %

Consider separate 100.0 gram samples of each of the following: Exercise Consider separate 100.0 gram samples of each of the following:  H2O, N2O, C3H6O2, CO2 Rank them from highest to lowest percent oxygen by mass. H2O, CO2, C3H6O2, N2O 16/18 32/44 32/74 16/44 0.89 0.73 0.43 0.36 18gpm 44gpm 74gpm 44gpm H2O, CO2, C3H6O2, N2O; The percent oxygen by mass in each is: H2O = 88.81% CO2 = 72.71% C3H6O2 = 43.20% N2O = 36.35%

Empirical Formulas The empirical formula of a compound is the simplest whole number ratio of the atoms present in the compound. The empirical formula can be found from the percent composition of the compound.

Simplest whole-number ratio Formulas Empirical formula = CH Simplest whole-number ratio Molecular formula = (empirical formula)n [n = integer] Molecular formula = C6H6 = (CH)6 Actual formula of the compound Copyright © Cengage Learning. All rights reserved

Steps for Determining the Empirical Formula of a Compound A gaseous compound containing carbon and hydrogen was analyzed and found to consist of 83.65% carbon by mass. Determine the empirical formula of the compound. Obtain the mass of each element present (in grams). Assume you have 100 g of the compound. 83.65% C = 83.65 g C (100.00 – 83.65) 16.35% H = 16.35 g H Copyright © Cengage Learning. All rights reserved

Steps for Determining the Empirical Formula of a Compound 2. Determine the number of moles of each type of atom present. Copyright © Cengage Learning. All rights reserved

Steps for Determining the Empirical Formula of a Compound 3. Divide the number of moles of each element by the smallest number of moles to convert the smallest number to 1. If all of the numbers so obtained are integers, these are the subscripts in the empirical formula. If one or more of these numbers are not integers, go on to step 4. Copyright © Cengage Learning. All rights reserved

Steps for Determining the Empirical Formula of a Compound 4. Multiply the numbers you derived in step 3 by the smallest integer that will convert all of them to whole numbers. This set of whole numbers represents the subscripts in the empirical formula. Copyright © Cengage Learning. All rights reserved

Empirical Formula from the Percent Composition What is the Empirical Formula if the % composition is 43.2% K, 39.1% Cl, and some O? 1 mole K 1.10 1.10 43.2 g K -------------- = 1.10 moles K =1.0 mole K 39.098g K 1 mole Cl 1.10 1.10 39.1 g Cl -------------- = 1.10 moles Cl = 1.0 mole Cl 35.453 g Cl 1 mole O 1.10 1.10 17.7 g O ------------- = 1.10 moles O = 1.0 mole O 15.999g O The Empirical Formula is KClO (MW =90.550g/mole) If MW of the real formula is 90.550, what is the actual formula? (90.550)/(90.550) = 1 KClO x 1 = KClO

Empirical Formula from the Percent Composition What is the Empirical Formula if the % composition is 40.0% C, 6.7% H, and 53.3% O? 1 mole C 3.33 3.33 40.0 g C -------------- = 3.33 moles C =1.0 mole C 12.011g C 1 mole H 6.64 3.33 6.7 g H -------------- = 6.64 moles H = 2.0 mole H 1.0079g H 1 mole O 3.33 3.33 53.3 g O ------------- = 3.33 moles O = 1.0 mole O 15.999g O The Empirical Formula is CH2O (MW =30.026) If MW of the real formula is 180.155, what is the actual formula? (180.155)/(30.026) = 6 CH2O x 6 = C6H12O6

Exercise The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). What is the empirical formula? 49.3 g C x 1 mole C/12g = 4.11 mole C 6.9 g H x 1 mole H/1g = 6.9 mole H 43.8 g O x 1 mole O/16g = 2.74 mole O 4.11/2.74 = 1.5 mole C x 2 = 3 mole C 6.9/2.74 = 2.5 mole H x 2 = 5 mole H 2.74/2.74 = 1 mole O x 2 = 2 mole O C3H5O2 Assume 100.0g. 49.3 g of carbon is 4.105 mol C (49.3/12.01). 6.9 g of hydrogen is 6.845 mol H (6.9/1.008). 43.8 g of oxygen is 2.7375 mol O (43.8/16.00). The ratio of C:H:O is 1.5:2.5:1 (C: 4.105/2.7375 = 1.5; H: 6.845/2.7375 = 2.5). Multiplying each by 2 to get a whole number becomes a ratio of 3:5:2. The empirical formula is therefore C3H5O2. Copyright © Cengage Learning. All rights reserved

where n is a whole number The molecular formula is the exact formula of the molecules present in a substance. The molecular formula is always an integer multiple of the empirical formula. Molecular formula = (empirical formula)n where n is a whole number Copyright © Cengage Learning. All rights reserved

Molar mass of C3H7 = 43.086 g/mol Continued Example… A gaseous compound containing carbon and hydrogen was analyzed and found to consist of 83.65% carbon by mass. The molar mass of the compound is 86.2 g/mol. You determined the empirical formula to be C3H7. What is the molecular formula of the compound? Molar mass of C3H7 = 43.086 g/mol C3H7 × 2 = C6H14 Copyright © Cengage Learning. All rights reserved

Flow Chart of the Relationships of Calculations Copyright © Cengage Learning. All rights reserved

Exercise The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. The empirical formula is C3H5O2. What is the molecular formula? C6H10O4 The molar mass of the empirical formula is 73.07 g/mol, which goes into the molar mass of the molecular formula 2 times (146/73.07). The molecular formula is therefore C6H10O4. Mw C3H5O2 = 3x12 + 5x1 + 2x16 =73 g/mole 146/73 = 2 2 x C3H5O2 = Copyright © Cengage Learning. All rights reserved