Chapter 8 The Mole Concept by Christopher G. Hamaker Illinois State University © 2014 Pearson Education, Inc. 1

Avogadro’s Number In a lab, we cannot work with individual atoms or molecules. They are too small. Avogadro’s number (symbol N) is the number of atoms in 12.01 grams of carbon. Its numerical value is 6.02 x 1023. It is called 1 mole Therefore, a 12.01 g sample of carbon contains 6.02 x 1023 carbon atoms and it is 1 mole. Similarly, 18.02 g sample of water contains 6.02 x 1023 H2O molecules (1 mole water).

Analogies for Avogadro’s Number The volume occupied by one mole of softballs would be about the size of Earth. One mole of Olympic shot put balls has about the same mass as that of Earth. One mole of hydrogen atoms laid side by side would circle Earth about 1 million times?.

1 mol = Avogadro’s number = 6.02 x 1023 units Mole Calculations I The mole (mol) is a unit of measure for an amount of a chemical substance. A mole is Avogadro’s number of particles, which is 6.02 x 1023 particles (any particles- atoms, molecules, ions, protons, neutrons, electrons, etc) 1 mol = Avogadro’s number = 6.02 x 1023 units We can use the mole relationship to convert between the number of particles and the mass of a substance.

Mole Calculations I, Continued How many sodium atoms are in 0.240 mol Na? We want atoms of Na. We have 0.240 mol Na. 1 mole Na = 6.02 x 1023 atoms Na. = 1.44 x 1023 atoms Na 0.240 mol Na x 1 mol Na 6.02 x 1023 atoms Na

Mole Calculations I, Continued How many moles of aluminum are in 3.42 x 1021 atoms Al? We want moles Al. We have 3.42 x 1021 atoms Al. 1 mol Al = 6.02 x 1023 atoms Al. = 5.68 x 10–3 mol Al 3.42 x 1021 atoms Al x 1 mol Al 6.02 x 1023 atoms Al

Mole Calculations II We will be using the unit analysis method again. Recall the following steps:

Molar Mass The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance. The atomic mass of carbon is 12.01 amu. Therefore, the molar mass of carbon is 12.01 g/mol. Since nitrogen occurs naturally as a diatomic, N2, the molar mass of nitrogen gas is two times 14.01 g or 28.02 g/mol. Similarly, the molar mass of H2O is 18.02 g/mol (sum of masses of 2H and one O) 1 amu = 1.6605e-24 g

One Mole of Several Substances C12H22O11 H2O mercury sulfur NaCl copper lead K2Cr2O7 Pb =207.20, Cu = 63.55 S = 32.06, Hg = 200.59 C = 12.01, K = 39.10, H = 1.01, Cr = 51.99, O = 16.00, N = 14.01

Calculating Molar Mass The molar mass of a substance is the sum of the molar masses of each element. What is the molar mass of copper(II) nitrite, Cu(NO2)2? The sum of the atomic masses is as follows: 63.55 + 2(14.01 + 16.00 + 16.00) = 63.55 + 2(46.01) = 155.57 amu The molar mass for Cu(NO2)2 is 155.57 g/mol.

6.02 x 1023 particles = 1 mol = molar mass Mole Calculations II Now we will use the molar mass of a compound to convert between grams of a substance and moles or particles of a substance. 6.02 x 1023 particles = 1 mol = molar mass If we want to convert particles to mass, we must first convert particles to moles, and then we can convert moles to mass.

Mole–Mass Calculation What is the mass of 1.33 moles of titanium, Ti? We want grams. We have 1.33 moles of titanium. Use the molar mass of Ti: 1 mol Ti = 47.88 g Ti. = 63.7 g Ti 1.33 mole Ti x 47.88 g Ti 1 mole Ti

Atoms–Mass Calculation What is the mass of 2.55 x 1023 atoms of lead? We want grams. We have atoms of lead. Use Avogadro’s number to convert atoms to moles and then molar mass of Pb. 2.55 × 1023 atoms Pb x 1 mol Pb 6.02×1023 atoms Pb 207.2 g Pb 1 mole Pb x = 87.9 g Pb

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 u, Zn = 65.38, Ag = 107.9). Since magnesium is lighter than zinc and silver, it will have more moles in a 100.0 g substance than Zn and Silver. (Mg = 24.31 u, Zn = 65.38, Ag = 107.9)

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 70.0 g of zinc 21.0 g of magnesium b, a, c; The greater the number of moles, the greater the number of atoms present. Zinc contains 1.07 mol, Ag contains 1.00 mol, and Mg contains 0.864 mol. (Mg = 24.31 u, Zn = 65.38, Ag = 107.9)

Mass–Molecule Calculation How many F2 molecules are present in 2.25 g of fluorine gas? We want molecules F2. We have grams F2. Convert g to moles using molar mass F2 Use Avogadro’s number for converting moles to molecules 2.25 g F2 x 1 mol F2 38.00 g F2 6.02 x 1023 molecules F2 x 3.56 x 1022 molecules F2

Mass of an Atom or Molecule What is the mass of a single molecule of sulfur dioxide? The molar mass of SO2 is 64.07 g/mol. We want mass of a molecule of SO2, we have the molar mass of sulfur dioxide. Use Avogadro’s number and the molar mass of SO2 as follows: 64.07 g SO2 1 mol SO2 6.02 x 1023 molecules SO2 x 1.06 x 10–22 g/molecule

Molar Volume At standard temperature and pressure, 1 mol of any gas occupies 22.4 L. The volume occupied by 1 mol of gas (22.4 L) is called the molar volume. Standard temperature and pressure are 0 C and 1 atm. STP = 0 0C, 1 atm, NTP = 20 0C, 1 atm

Molar Volume of Gases We now have a new unit factor equation: 1 mol gas = 6.02 x 1023 molecules gas = 22.4 L gas

One Mole of a Gas at STP The box below has a volume of 22.4 L, which is the volume occupied by 1 mol of a gas at STP.

Gas Density The density of gases is much less than that of liquids. We can easily calculate the density of any gas at STP. The formula for gas density at STP is as follows: = density, g/L molar mass in grams molar volume in liters

Calculating Gas Density What is the density of methane gas, CH4, at STP? First we need the molar mass for methane. 12.01 + 4(1.01) = 16.05 g/mol The molar volume CH4 at STP is 22.4 L/mol. Density is mass/volume. What is the density of CO2 = 0.717 g/L 16.05 g/mol 22.4 L/mol

Molar Mass of a Gas We can also use molar volume to calculate the molar mass of an unknown gas. An unknown gas has 4.29 g and occupies 1.50 L at STP. What is the molar mass? We want g/mol; we have g/L. 4.29 g 1.50 L 22.4 L 1 mol x = 64.1 g/mol

Mole Calculations III We now have three interpretations for the mole: 1 mol = 6.02 x 1023 particles 1 mol = molar mass 1 mol = 22.4 L at STP This gives us three unit factors to use to convert among moles, particles, mass, and volume.

Calculating Molar Volume A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present? We want moles; we have volume. Use molar volume of a gas: 1 mol = 22.4 L. 4.50 L CH4 x = 0.201 mol CH4 1 mol CH4 22.4 L CH4

Calculating Mass Volume What is the mass of 3.36 L of ozone gas, O3, at STP? We want mass O3; we have 3.36 L O3. Convert volume to moles, then moles to mass. 3.36 L O3 x x 22.4 L O3 1 mol O3 48.00 g O3 = 7.20 g O3

Calculating Molecule Volume How many molecules of argon gas, Ar, occupy 0.430 L at STP? We want molecules Ar; we have 0.430 L Ar. Convert volume to moles, and then moles to molecules. 0.430 L Ar x 1 mol Ar 22.4 L Ar 6.02 x 1023 molecules Ar x = 1.16 x 1022 molecules Ar

Percent Composition The percent composition of a compound lists the mass percent of each element. For example, the percent composition of water, H2O, is 11% hydrogen and 89% oxygen. All water contains 11% hydrogen and 89% oxygen by mass.

Calculating Percent Composition There are a few steps to calculating the percent composition of a compound. Let’s practice using H2O. Assume you have 1 mol of the compound. One mole of H2O contains 2 mol of hydrogen and 1 mol of oxygen. Therefore, 2(1.01 g H) + 1(16.00 g O) = molar mass H2O 2.02 g H + 16.00 g O = 18.02 g H2O

More Calculating Percent Composition Next, find the percent composition of water by comparing the masses of hydrogen and oxygen in water to the molar mass of water. 2.02 g H 18.02 g H2O x 100% = 11.2% H 16.00 g O 18.02 g H2O x 100% = 88.79% O

Percent Composition Problem TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 °C. Calculate the percent composition of TNT, C7H5(NO2)3. 7(12.01 g C) + 5(1.01 g H) + 3 (14.01 g N + 32.00 g O) = g C7H5(NO2)3 84.07 g C + 5.05 g H + 42.03 g N + 96.00 g O = 227.15 g C7H5(NO2)3

Percent Composition of TNT 84.07 g C 227.15 g TNT x 100% = 37.01% C 5.05 g H 227.15 g TNT x 100% = 2.22% H 42.03 g N 227.15 g TNT x 100% = 18.50% N 96.00 g O 227.15 g TNT x 100% = 42.26% O

Empirical Formula The empirical formula of a compound is the simplest whole number ratio of ions in a formula unit or atoms of each element in a molecule. The molecular formula of benzene is C6H6. The empirical formula of benzene is CH. The molecular formula of octane is C8H18. The empirical formula of octane is C4H9. Na2O AlF3 Ca2O2

Calculating Empirical Formulas We can calculate the empirical formula of a compound from its composition data. We can determine the mole ratio of each element from the mass to determine the empirical formula of radium oxide, Ra?O?. – A 1.640 g sample of radium metal was heated to produce 1.755 g of radium oxide. What is the empirical formula? – We have 1.640 g Ra and 1.755 – 1.640 = 0.115 g O.

Calculating Empirical Formulas, Continued – The molar mass of radium is 226.03 g/mol, and the molar mass of oxygen is 16.00 g/mol. 1 mol Ra 226.03 g Ra 1.640 g Ra x = 0.00726 mol Ra 1 mol O 16.00 g O 0.115 g O x = 0.00719 mol O – We get Ra0.00726O0.00719. Simplify the mole ratio by dividing by the smallest number. – We get Ra1.01O1.00 = RaO is the empirical formula.

Empirical Formulas from Mass Composition We can also use percent composition data to calculate empirical formulas. Assume that you have 100 grams of sample. Acetylene is 92.2% carbon and 7.83% hydrogen. What is the empirical formula? – If we assume 100 grams of sample, we have 92.2 g carbon and 7.83 g hydrogen.

Empirical Formula for Acetylene Calculate the moles of each element. 1 mol C 12.01 g C 92.2 g C x = 7.68 mol C 1 mol H 1.01 g H 7.83 g H x = 7.75 mol H The ratio of elements in acetylene is C7.68H7.75. Divide by the smallest number to get the following formula: 7.68 C = C1.00H1.01 = CH 7.75 H

Molecular Formula The empirical formula for acetylene is CH. This represents the ratio of C to H atoms on acetylene. The actual molecular formula is some multiple of the empirical formula, (CH)n. Acetylene has a molar mass of 26 g/mol. Find n to find the molecular formula: = CH (CH)n 26 g/mol 13 g/mol n = 2 and the molecular formula is C2H2.

What is the empirical formula? What is the molecular formula? 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. What is the empirical formula? What is the molecular formula? 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. 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.

From % composition get the empirical formula: mol of C = 49. 3g/12 From % composition get the empirical formula: mol of C = 49.3g/12.01g mol−1 = 4.11 mol mol of H = 6.9g/1.008g mol−1 = 6.85 mol mol of O = 43.8g/16.00g mol−1 = 2.74 mol C: 4.11/2.74 = 1.5 H: 6.85/2.74 = 2.5 O: 1 (x 2) No fractions, simplest whole number is EF = C3H5O2 Its (EF) molar mass is 73g/mol. Adipic acid molar mass is 146g/mol, twice that of EF. Therefore molecular formula is 2 x C3H5O2 = C6H10O4

Critical Thinking: Avogadro’s Number In 1911, Ernest Rutherford estimated the value of Avogadro’s number as 6.11 x 1023 using alpha particles from radium. The most recent measurements of Avogadro’s number were made using X-ray diffraction of silicon crystals. Currently, the most accurate value for Avogadro’s number is 6.0221415 x 1023.

Chapter Summary Avogadro’s number is 6.02 x 1023, and is 1 mole of any substance. The molar mass of a substance is the sum of the atomic masses of each element in the formula. At STP, 1 mole of any gas occupies 22.4 L.

Chapter Summary, Continued We can convert between the number of particles and moles of a substance using Avogadro’s number (1 mol = 6.02 x 1023 particles). We can convert between mass of a substance and moles of a substance using the molar mass (g/mol). We can convert between the volume of a gas at STP and moles of a gas using molar volume at STP (1 mol = 22.4 L).

Chapter Summary, Continued The percent composition of a substance is the mass percent of each element in that substance. The empirical formula of a substance is the simplest whole number ratio of the elements in the formula. The molecular formula is a multiple of the empirical formula.