2. Chapter 10 The Mole “Making Measurements in Chemistry” T. Witherup 2006

3. Chapt. 10 OBJECTIVES Define the “mole” & describe its importance. Define the “mole” & describe its importance. Identify & use Avogadro’s number. Identify & use Avogadro’s number. Define “molar mass” & explain how it relates the mass of a substance to its number of particles. Define “molar mass” & explain how it relates the mass of a substance to its number of particles. Convert among the number of particles, moles & mass of a substance. Convert among the number of particles, moles & mass of a substance. Describe “molar volume” & use it to solve problems. Describe “molar volume” & use it to solve problems. Find the percentage composition of a formula. Find the percentage composition of a formula. Use percentage composition to find the formula of an unknown sample. Use percentage composition to find the formula of an unknown sample. Find empirical & molecular formulas. Find empirical & molecular formulas.

4. 10-1 Chemical Measurements Atomic Mass: the mass of an atom expressed relative to the mass assigned to the carbon-12 isotope, in amu (atomic mass units). Atomic Mass: the mass of an atom expressed relative to the mass assigned to the carbon-12 isotope, in amu (atomic mass units). 1 amu = 1/12 of the mass of Carbon-12. 1 amu = 1/12 of the mass of Carbon-12. Since C-12 has 6p + 6n = 12 particles, its mass = 12 amu. Since C-12 has 6p + 6n = 12 particles, its mass = 12 amu. Since C-13 has 6p + 7n = 13 particles, its mass = 13 amu. Since C-13 has 6p + 7n = 13 particles, its mass = 13 amu. And C-14 has a mass of ??? amu? And C-14 has a mass of ??? amu? (Why don’t we consider the mass of the electrons in these atoms?) (Why don’t we consider the mass of the electrons in these atoms?)

5. Average Atomic Mass Since “natural” carbon has 1.1% the C-13 isotope and only a trace amount of the C- 14 isotope, its average atomic mass is dominated by C-12, or 12.011 amu. The average atomic mass accounts for all (natural) isotopes of an element. The average atomic mass of each element may be found on the Periodic Table. NOTE: In our course, when using average atomic masses, always round to the ‘hundredths’ place.

6. Average Atomic Mass of Elements ELEMENT AVERAGE ATOMIC MASS (amu) AVERAGE ATOMIC MASS, rounded (amu) Hydrogen1.007941.01 Carbon12.011112.01 Oxygen15.999416.00 Chlorine35.45335.45 Iron55.84755.85

7. Formula Mass Formula Mass: the sum of the atomic masses of all atoms in a compound. Water (H2O) has 2 Hydrogen atoms and 1 Oxygen atom. Formula Mass of H2O is 2 X (1.01) amu + 16.00 amu = 18.02 amu. What is the formula mass of methane, CH4? O Of NaCl? (For ionic compounds we refer to a “formula unit” rather than a “formula mass,” but it is essentially the same idea.) f ammonia, NH3? f glucose, C6H12O6?

8. But what is a “mole”? That furry creature who burrows in the yard? The dark pigmentation on our skin? A massive stone structure used as a breakwater or pier? An undercover agent? NO! A mole is a special chemical term used to count atoms!

9. The MOLE (mol) A mole of any element is defined as the amount of the element that contains as many atoms as there are in exactly 12 g of the carbon-12 isotope. A mole of any element is defined as the amount of the element that contains as many atoms as there are in exactly 12 g of the carbon-12 isotope. A mole is found experimentally to be equal to 6.022 X 10 23 atoms of C-12, which is called Avogadro’s Number (N A ). A mole is found experimentally to be equal to 6.022 X 10 23 atoms of C-12, which is called Avogadro’s Number (N A ). 12 g Carbon-12 = 1 mol Carbon 12 atoms = 6.022 X 10 23 Carbon 12 atoms 12 g Carbon-12 = 1 mol Carbon 12 atoms = 6.022 X 10 23 Carbon 12 atoms The molar mass of any substance is the mass of one mole of that substance. Molar mass is numerically equal to the atomic mass, molecular mass, or formula mass of the substance. A mole of any substance contains Avogadro’s number of units of that substance (6.022 X 10 23 units). A mole of any substance contains Avogadro’s number of units of that substance (6.022 X 10 23 units).

10. What’s in a Mole? Atoms – one mole of an element contains 6.022 X 10 23 atoms of the element. Atoms – one mole of an element contains 6.022 X 10 23 atoms of the element. Molecules – one mole of a molecular (covalent) compound contains 6.022 X 10 23 molecules. Molecules – one mole of a molecular (covalent) compound contains 6.022 X 10 23 molecules. Formula Units – one mole of an ionic compound contains 6.022 X 10 23 formula units. Formula Units – one mole of an ionic compound contains 6.022 X 10 23 formula units. Gizmos – one mole of gizmos contains 6.022 X 10 23 gizmos. Gizmos – one mole of gizmos contains 6.022 X 10 23 gizmos. Anything – one mole of anything contains 6.022 X 10 23 anythings! Anything – one mole of anything contains 6.022 X 10 23 anythings! N A (6.022 X 10 23 ) is very practical for counting small particles, especially things like atoms, ions and molecules. N A (6.022 X 10 23 ) is very practical for counting small particles, especially things like atoms, ions and molecules.

11. EXAMPLES How many pens in 1 mole of pens? How many pens in 1 mole of pens? How many atoms in 63.546g of Cu? How many atoms in 63.546g of Cu? How many atoms in 6.3546g of Cu? How many atoms in 6.3546g of Cu? How many molecules in 1 mole of sugar (C 6 H 12 O 6 )? How many molecules in 1 mole of sugar (C 6 H 12 O 6 )? How many molecules in 10 moles of sugar? How many molecules in 10 moles of sugar? How many carbon atoms in 1 mole of sugar? How many oxygen atoms? How many carbon atoms in 1 mole of sugar? How many oxygen atoms? How many formula units in 1 mole of CaCl 2 ? How many formula units in 1 mole of CaCl 2 ? How many calcium ions in 1 mole of CaCl 2 ? How many calcium ions in 1 mole of CaCl 2 ? How many chloride ions in 1 mole of CaCl 2 ? How many chloride ions in 1 mole of CaCl 2 ? How many chloride ions in 0.1 mole of CaCl 2 ? How many chloride ions in 0.1 mole of CaCl 2 ?

12. 10-2 Mole Conversions by Factor Label Method (1) Mass and Moles Mass and Moles Use the molar mass of the substance. Use the molar mass of the substance. 1 Mole = n grams of the substance, so 1 = n grams/Mole, but also 1 = Mole/n grams of the substance. 1 Mole = n grams of the substance, so 1 = n grams/Mole, but also 1 = Mole/n grams of the substance. Use these conversions by setting up Factor Labels to cancel the units! Use these conversions by setting up Factor Labels to cancel the units! See examples on next slide. See examples on next slide.

13. 10-2 Mole Conversions by Factor Label Method (1) Examples How many moles in 75.0 g of iron? Example A: How many moles in 75.0 g of iron? How many grams in 0.250 mol Na? Example B: How many grams in 0.250 mol Na? 75.0g Fe X 1 mol Fe 55.85g Fe = 1.34 mol Fe 0.250 mol Na X 22.99g Na 1 mol Na = 5.75g Na

14. Solving ‘Mole Problems” (1) MOLES MASS Molar mass mol/g g/mol

15. 10-2 Mole Conversions by Factor Label Method (2) Particles and Moles Particles and Moles Use Avogadro’s Number (6.022 X 10 23 ) of particles. Use Avogadro’s Number (6.022 X 10 23 ) of particles. (6.022 X 10 23 )particles/mol  1 mol/(6.022 X 10 23 )particles (6.022 X 10 23 )particles/mol  1 mol/(6.022 X 10 23 )particles Again, set up Factor Labels to cancel units! Again, set up Factor Labels to cancel units! See examples on next slide. See examples on next slide.

16. 10-2 Mole Conversions by Factor Label Method (2) Examples = 0.697 mol CO 2 How many atoms in 0.25 mol Na? Example C: How many atoms in 0.25 mol Na? 0.250 mol Na X 6.022 X10 23 atoms Na mol Na 1 mol Na = 1.51 X 10 23 atoms Na Example D: How many moles in 4.20 X 10 23 molecules of CO 2 ? 4.20 X 10 23 Molecules CO 2 X 1 mol CO 2 6.022 X 10 23 molecules CO 2

17. Solving ‘Mole Problems” (2) MOLES PARTICLES Number of Particles in 1 mole (6.02 X 10 23 )

18. 10-2 Mole Conversions by Factor Label Method (3) Gases and Moles Gases and Moles Avogadro proposed that equal volumes of gases contain the same number of gas particles at a given temperature & pressure. Avogadro proposed that equal volumes of gases contain the same number of gas particles at a given temperature & pressure. Therefore one mole of gas #1 would have the same volume as one mole of gas #2. Therefore one mole of gas #1 would have the same volume as one mole of gas #2. It is observed that one mole of any gas occupies 22.4 liters @ STP (molar volume). It is observed that one mole of any gas occupies 22.4 liters @ STP (molar volume). STP = Standard Temperature and Pressure, 0° C and 1 atmosphere. STP = Standard Temperature and Pressure, 0° C and 1 atmosphere. Once more, set up Factor Labels to cancel the units! Once more, set up Factor Labels to cancel the units!

19. 10-2 Mole Conversions by Factor Label Method (3) Examples (Gases) Example E: What is the volume of 13.0 moles of hydrogen gas at STP? 13.0 mol H 2 X 22.4 L H 2 1 mol H 2 = 291 L H 2 Example F: How many moles are in 250. mL of oxygen at STP? at STP? 250. mL O 2 X 1 L O 2 1000 mL O 2 1 mol O 2 22.4 L O 2 = 0.0112 mol O 2 X

20. Solving ‘Mole Problems” (3) MOLES VOLUME of gas @ STP Molar volume (22.4L/mol @ STP)

21. 10-2 Mole Conversions by Factor Label Method (Summary) (See p 330.) Mass and Moles Mass and Moles Use the molar mass of the substance. Use the molar mass of the substance. Particles and Moles Particles and Moles Use Avogadro’s Number (6.02 X 10 23 ) of particles. Use Avogadro’s Number (6.02 X 10 23 ) of particles. Gases and Moles Gases and Moles Use the Molar Volume (22.4 liters @ STP). Use the Molar Volume (22.4 liters @ STP). STP = Standard Temperature and Pressure, 0° C and 1 atmosphere. STP = Standard Temperature and Pressure, 0° C and 1 atmosphere. Set up Factor Labels to cancel units! Set up Factor Labels to cancel units! DON’T GET LAZY! Include labels to ensure that ALL units cancel correctly. DON’T GET LAZY! Include labels to ensure that ALL units cancel correctly. Multi-step conversions are easily done if you are careful with the labels! Multi-step conversions are easily done if you are careful with the labels!

22. Summary: Solving ‘Mole Problems” MOLES MASS Molar mass PARTICLES Number of Particles in 1 mole (6.02 X 10 23 ) VOLUME of gas @STP Molar volume (22.4L/mol @ STP)

23. 10-3 Empirical & Molecular Formulas Percentage Composition Percentage Composition The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100%. The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100%. Example 1 Example 1 2.45 g aluminum oxide decomposes into 1.30 g aluminum & 1.15 g oxygen. What is the percentage composition? 2.45 g aluminum oxide decomposes into 1.30 g aluminum & 1.15 g oxygen. What is the percentage composition? %O = (1.15g O/2.45g Aluminum Oxide) X 100% = 46.9% Oxygen (O, not O 2 ) %O = (1.15g O/2.45g Aluminum Oxide) X 100% = 46.9% Oxygen (O, not O 2 ) %Al = (1.30g Al/2.45g Aluminum Oxide) X 100% = 53.1% Aluminum %Al = (1.30g Al/2.45g Aluminum Oxide) X 100% = 53.1% Aluminum As a check, note that 46.9% + 53.1% = 100.0%. As a check, note that 46.9% + 53.1% = 100.0%.

24. 10-3 Empirical & Molecular Formulas (cont’d) Percentage Composition (cont’d) Percentage Composition (cont’d) Example 2 Example 2 Determine the percent composition of CaCO 3. Determine the percent composition of CaCO 3. Molar Mass = 40.08 + 12.01 +3(16.00) = 100.09g/mol Molar Mass = 40.08 + 12.01 +3(16.00) = 100.09g/mol %Ca = (40.08g Ca/100.09g CaCO 3 ) X 100% = 40.04% Ca %Ca = (40.08g Ca/100.09g CaCO 3 ) X 100% = 40.04% Ca % C = (12.01g C/100.09g CaCO 3 ) X 100% = 12.00% C % C = (12.01g C/100.09g CaCO 3 ) X 100% = 12.00% C % O = (48.00g O/100.09g CaCO 3 ) X 100% = 47.96% O % O = (48.00g O/100.09g CaCO 3 ) X 100% = 47.96% O Check: 40.04% + 12.00% + 47.96% = 100.00% Check: 40.04% + 12.00% + 47.96% = 100.00%

25. Determining Empirical Formulas Empirical Formula Empirical Formula The formula that gives the simplest whole number ratio of the atoms of the elements in the formula. The formula that gives the simplest whole number ratio of the atoms of the elements in the formula. Example 1 Example 1 What is the empirical formula of a compound containing 2.644g of gold and 0.476g of chlorine? What is the empirical formula of a compound containing 2.644g of gold and 0.476g of chlorine? 0.476g Cl X (1 mol Cl/35.45g Cl) = 0.0134 mol Cl 0.476g Cl X (1 mol Cl/35.45g Cl) = 0.0134 mol Cl 2.644g of Au X (1 mol Au/196.97g Au) = 0.0134 mol Au 2.644g of Au X (1 mol Au/196.97g Au) = 0.0134 mol Au Empirical formula = Au 0.0134 Cl 0.0134 or simply AuCl Empirical formula = Au 0.0134 Cl 0.0134 or simply AuCl

26. Determining Empirical Formulas Example 2 What is the empirical formula of a compound with 5.75 g Na, 3.50 g N & 12.00 g O? What is the empirical formula of a compound with 5.75 g Na, 3.50 g N & 12.00 g O? First, find the mole amounts. First, find the mole amounts. 5.75g Na X (1 mol Na/22.99g Na) = 0.250 mol Na 5.75g Na X (1 mol Na/22.99g Na) = 0.250 mol Na 3.50 g N X (1 mol N/14.01g N) = 0.250 mol N 3.50 g N X (1 mol N/14.01g N) = 0.250 mol N 12.00g O X (1 mol O/16.00g O) = 0.750 mol O 12.00g O X (1 mol O/16.00g O) = 0.750 mol O Empirical formula = Na 0.250 N 0.250 O 0.750 Empirical formula = Na 0.250 N 0.250 O 0.750 Divide each mole quantity by the smallest to get whole numbers. (0.250 in this case) Divide each mole quantity by the smallest to get whole numbers. (0.250 in this case) Empirical formula = NaNO 3 Empirical formula = NaNO 3

27. Determining Molecular Formulas Molecular Formula Molecular Formula The formula that gives the actual number of atoms of each element in a molecular compound. The formula that gives the actual number of atoms of each element in a molecular compound. Example 3 Example 3 Hydrogen peroxide has a molar mass of 34.00 g/mol and a chemical composition of 5.90% H & 94.1% O. What is its molecular formula? Hydrogen peroxide has a molar mass of 34.00 g/mol and a chemical composition of 5.90% H & 94.1% O. What is its molecular formula? First, find the empirical formula, assuming the percents are mass. First, find the empirical formula, assuming the percents are mass. For 5.90% H: 5.90gH X (1 mol H/1.01g H) = 5.84 mol H For 5.90% H: 5.90gH X (1 mol H/1.01g H) = 5.84 mol H For 94.1% O: 94.1g O X (1 mol O/16.00g O) = 5.88 mol O For 94.1% O: 94.1g O X (1 mol O/16.00g O) = 5.88 mol O Empirical formula = H 5.84 O 5.88 or HO. Empirical formula = H 5.84 O 5.88 or HO. But molar mass = 34.00g/mol and HO is only 17.01g/mol. But molar mass = 34.00g/mol and HO is only 17.01g/mol. Therefore, Molecular Formula = 2(HO) or H 2 O 2. Therefore, Molecular Formula = 2(HO) or H 2 O 2.

28. Determining Molecular Formulas Example 4 A compound contains 42.56 g Pd and 0.8000 g H. If its molar mass is 216.8 g/mol, find the molecular formula. A compound contains 42.56 g Pd and 0.8000 g H. If its molar mass is 216.8 g/mol, find the molecular formula. First, find the empirical formula. First, find the empirical formula. 42.56g Pd X (1 mol Pd/106.42 g Pd) = 0.4000 mol Pd 42.56g Pd X (1 mol Pd/106.42 g Pd) = 0.4000 mol Pd 0.8000g H X (1 mol H/1.00g H) = 0.800 mol H 0.8000g H X (1 mol H/1.00g H) = 0.800 mol H Empirical formula: Pd 0.4000 H 0.800 or PdH 2 Empirical formula: Pd 0.4000 H 0.800 or PdH 2 PdH 2 has a mass = 108.44 g/mol. PdH 2 has a mass = 108.44 g/mol. Since molar mass = 216.8 and the empirical mass = 108.4, the Molecular Formula is twice the empirical formula [2(PdH 2 )] or simply Pd 2 H 4. Since molar mass = 216.8 and the empirical mass = 108.4, the Molecular Formula is twice the empirical formula [2(PdH 2 )] or simply Pd 2 H 4.

29. Did we cover the OBJECTIVES? Define the “mole” & describe its importance. Identify & use Avogadro’s number. Define “molar mass” & explain how it relates the mass of a substance to its number of particles. Convert among the number of particles, moles & mass of a substance. Describe “molar volume” & use it to solve problems. Find the percentage composition of a formula. Use percentage composition to find the formula of an unknown sample. Find empirical & molecular formulas.

30. How to be successful at solving Mole Problems: USE FACTOR LABELS! PRACTICE! PRACTICE! PRACTICE!