1. UNIT 9: THE MOLE VOCABULARY: representative particle mole Avogadro’s number molar mass percent composition empirical formula molecular formula hydrate anhydrous dilution molarity

2. Relative Masses Chemists need a convenient method for accurately counting the number of atoms, molecules, or formula units of a substance. The relative masses of atoms are expressed in terms of atomic mass units on an atomic weight scale in which an arbitrary value of exactly 12 atomic mass units is assigned to a carbon atom containing six protons, six neutrons, and six electrons (Carbon- 12). Protons neutrons, and electrons are subatomic particles that we shall discuss in Unit 6. An average hydrogen atom has an atomic weight of 1.0079 atomic mass units and is about 1/12 as heavy as a 12 C atom. An average magnesium atom has an atomic weight of 24.305 atomic mass units and is about twice as heavy as a 12 C atom. atomic mass unit (amu) arbitrary but convenient unit of mass for individual atoms, its use as a unit of mass (weight) in the laboratory is impractical. It is much more convenient to measure mass (weight) in grams. 1 amu = 17 x 10 -24 g

3. Examples: always 3 SF’s H atom = 1.01 amu He atom = 4.00 amu Mg atom = 24.3 amu O atom = 16.0 amu H 2 0 = H2 x 1.00= 2.00 amuO1 x 16.0= 16.0 amu 18.0 amu

4. Atomic masses on the Periodic Table are averages attained by measuring the masses of all of the isotopes in an element and knowing the percentage of each isotope in nature. Cl-37 76% Cl-35 24% 1 averaged Cl atom = 35.5 amu

5. The Mole We need a chemical unit that equals the number of atoms or molecules which has a mass in grams which will be numerically equal to the relative mass. The basic unit is the MOLE. It is the SI base unit used to measure the amount of a substance. The number of atoms of an element in the mass numerically equal to its atomic weight in grams is always the same, 6.02  10 23. There are 6.02  10 23 atoms of hydrogen in 1.0079 grams of hydrogen, 6.02  10 23 atoms in 24.305 grams of magnesium, and 6.02  10 23 atoms in 207.2 grams of lead. 1 mole is the amount of atoms in 12 g of pure carbon-12, or 6.02  10 23 atoms.

6. 1 mole = 6.02  10 23 particles The number is called Avogadro’s number. How large is a mole? 602,213,670,000,000,000,000,000 A mole of pennies would be $602,213,670,000,000,000,000,000 six hundred and two sextillion, two hundred and thirteen quintillion, six hundred and seventy quadrillion dollars six sextillion, twenty-two quintillion, one hundred thirty-six quadrillion, seven hundred trillion dollars 6.02 x 1023 particles of matter

7. 6.02 x 10 23 = Avogadro’s number The number of representative particles in one mole of a pure substance Representative particles = Formula unit (ionic compounds) Molecule (covalent compounds) Atom Ion

8. 1 H atom = 1.01 amu 1 mole of H atom = 1.01 gram = 6.02  10 23 H atoms 1 O 2 molecule = 32.0 amu 1 mole of O 2 = 32.0 g = contains 6.02  10 23 O 2 molecules 1 H 2 O molecule = 18.0 amu 1 mole of H 2 O = 18.0 grams = 6.02  10 23 H 2 O molecules formula unit for ionic compounds 1 formula unit NaCl = 23.0 35.5 58.5 amu 1 mole of NaCl = 58.5 g = 6.02  10 23 NaCl formula units

9. Gram → Mole → Particle Conversions 1 mole = 6.02 x 10 23 particles = molar mass Conversion factors must be used. Example 1. How many atoms are there in 3.60 moles of Na? One mole of Na is 6.02 x 10 23 of Na ? atoms = 3.60 moles 3.60 moles x 6.02 x 10 23 atoms = 2.17 x 10 24 atoms 1 mole

10. Example 2. What is the mass of 5.00 moles of potassium? ? g = 5.00 moles K 5.00 moles K x 39.1 g = 196 g 1 mole Example 3. How many moles are present in 77.5 g of phosphorus? ? moles 77.5 g P 77.5 g P x 1 mole = 2.5 moles S.F. 2.50 moles 31.0 g Example 4. What is the mass of 5.60 moles of CO 2 ? ? g = 5.60 moles CO 2 5.60 moles CO 2 x 44.0 g = 246 g 1 mole add. 12.0 carbon 32.0 oxygen 44.0 g

11. Example 5. How many moles of Fe 3+ ions and SO 4 2- ions are present in 4.68 moles of iron(III) 2 sulfate 3 ? 2 Fe and 3 SO 4 3(32.1) 1 formula unit 1 formula unit 12(16.0) + 2(55.8) 399.9 2 moles of Fe and 3 moles of SO 4 1 mole Fe 2 (SO 4 ) 3 1 mole Fe 2 (SO 4 ) 3 1

12. Example 6. ? moles Fe 3+ = 4.68 moles Fe 2 (SO 4 ) 3 x 2 moles of Fe 1 mole Fe 2 (SO 4 ) 3 9.36 moles of Fe 3+ Example 7. ? moles SO 4 2- = 4.68 moles Fe 2 (SO 4 ) 3 x 3 moles of SO 4 2- 1 mole Fe 2 (SO 4 ) 3 14.0 moles SO 4 2-

13. Converting Between Moles and Particles Particles to moles Use the inverse of Avogadro’s number as the conversion factor.

14. The Mass of a Mole 1 mol of copper and 1 mol of carbon have different masses. One copper atom has a different mass than 1 carbon atom. Molar mass is the mass in grams of one mole of any pure substance. The molar mass of any element is numerically equivalent to its atomic mass and has the units g/mol.

16. Moles to mass 3.00 moles of copper has a mass of 191 g.

17. Using Molar Mass Convert mass to moles with the inverse molar mass conversion factor. Convert moles to atoms with Avogadro’s number as the conversion factor.

18. Using Molar Mass This figure shows the steps to complete conversions between mass and atoms. A mole always contains the same number of particles; however, moles of different substances have different masses.

19. Chemical Formulas and the Mole Chemical formulas indicate the numbers and types of atoms contained in one unit of the compound. One mole of CCl 2 F 2 contains one mole of C atoms, two moles of Cl atoms, and two moles of F atoms. One mole of CCl 2 F 2 contains one carbon atom, two Chlorine atoms, and two Fluorine atoms.

20. Mass of atoms is found on the periodic table K = 39.1 g/mol For elements, the conversion factor is the molar mass of the element. Mass of compounds must be calculated H 2 0 = 2(H) + O = 2(1.0) + 16.0 = 18.0 g/mol CO 2 = C + 2(O) = 12.0 + 2(16.0) = 44.0 g/mol Given 3 moles of Li, what is the mass? 3 mol Li x 6.941 g = 20.823 g Li 1 mol

21. The procedure is the same for compounds, except that you must first calculate the molar mass of the compound. Mass of compounds must be calculated The molar mass of a compound equals the molar mass of each element, multiplied by the moles of that element in the chemical formula, added together. “ADD ‘M UP” CO 2 = C + 2(O) C1 x 12.0= 12.0 O2 x 16.0= 32.0 44.0 g/mole

22. Converting Mass of an element to Moles Given 20.823 g of Li, calculate the number of moles? 20.823 g Li x 1mol = 3 mol Li 6.941 g

23. Converting the Mass of a Compound to Moles The conversion factor is the inverse of the molar mass of the compound. How many moles are there in 27 g H 2 0? 27 g H 2 0 X 1 mol H 2 0 = 1.5 mol H 2 0 18.0 g H 2 0

24. Converting the Mass of a Compound to Number of Particles Convert mass to moles of compound with the inverse of molar mass. Convert moles to particles with Avogadro’s number. 27 g H 2 0 x 1 mol H 2 0 x 6.02 x 10 23 molec H 2 0 = 9.0 x 10 23 molec H 2 0 18.0 g H 2 O 1 mol H 2 O How many molecules are in 27 g H 2 O? Step 1: 27 g H 2 0 x 1 mol H 2 0 = 1.5 mol H 2 0 18.0 g H 2 0 Step 2: 1.5 mol H 2 0 x 6.02 x 10 23 molec H 2 0 = 9.0 x 10 23 molec H 2 0 1 mol H 2 0

25. CONVERTING THE MASS OF A COMPOUND TO PARTICLES (More Practice) Elements: Calculate the number of atoms from 1.00 g K. 1.00 g K x (1 mol K/39.1 g K) x (6.02 x 10 23 atoms K/1 mol K) = 1.54 x 10 22 atoms K Step 1: Convert mass to moles of compound with the inverse of molar mass. Step 2: Convert moles to particles with Avogadro’s number. Compounds: (Remember to calculate molar mass) 54 g H 2 SO 4 to Particles. 54 g H 2 SO 4 x 1 mol H 2 SO 4 / 98.08 g H 2 SO 4 x 6.02 x 10 23 molec H 2 SO 4 / 1 mol H 2 SO 4 = 3.3 x 10 23 molec H 2 SO 4

26. Converting the Mass of a Compound to Number of Particles (cont.) This figure summarizes the conversions between mass, moles, and particles.

27. Percent Composition The percent by mass of any element in a compound can be found by dividing the mass of the element by the mass of the compound and multiplying by 100.

28. Percent Composition The percent by mass of each element in a compound is the percent composition of a compound. Percent composition of a compound can also be determined from its chemical formula.

29. % composition of C 3 F 6 Total mass of C = total mass of F = Mass of compound C 3 F 6 = % of C = % of F = Check your work!

30. Empirical Formula The empirical formula for a compound is the smallest whole-number mole ratio of the elements. You can calculate the empirical formula from percent by mass by assuming you have 100.00 g of the compound. Then, convert the mass of each element to moles. The empirical formula may or may not be the same as the molecular formula. Molecular formula of hydrogen peroxide = H 2 O 2 Empirical formula of hydrogen peroxide = HO

31. Molecular Formula The molecular formula specifies the actual number of atoms of each element in one molecule or formula unit of the substance. Molecular formula is always a whole-number multiple of the empirical formula. To find the molecular formula you must know -Empirical formula -Molar mass of compound Can be the same as the empirical formula but often is not

32. Molecular Formula (cont.)

33. Calculating Empirical Formulas 1) Change % to g (assume 100 g of compound, so 30% = 30 g) 2) Convert each element from g to moles 3) Divide each mole amount by the smallest number from step 2 4) Change to a whole number = subscript in empirical formula

34. Example: 48.64% C, 8.16% H, 43.20% O 48.64g C x 1 mol C 12.01 g C 8.16g H x 1 mol H 1.008 g H 43.20g O x 1 mol O 16.00 g O

35. 4.050 mol C = 1.5 mol C 2.700 8.10 mol H = 3 mol H 2.700 2.700 mol O = 1 mol O 2.700

36. The simplest mole ratio is (1.5 mol C), (3 mol H), (1 mol O) Multiply each number in the ratio-in this case 2-that yields a ratio of whole numbers. 1.5 mol C x 2 = 3 mol C 3 mol H x 2 = 6 mol H 1 mol O x 2 = 2 mol O The simplest whole number ratio is 3 atoms C, 6 atoms H, 2 atoms O. Thus, the formula is C 3 H 6 O 2

37. Naming Hydrates A hydrate is a compound that has a specific number of water molecules bound to its atoms. The number of water molecules associated with each formula unit of the compound is written following a dot. Sodium carbonate decahydrate = Na 2 CO 3 10H 2 O

38. Naming Hydrates (cont.)

39. Analyzing a Hydrate When heated, water molecules are released from a hydrate leaving an anhydrous compound. To determine the formula of a hydrate, find the number of moles of water associated with 1 mole of hydrate.

40. Analyzing a Hydrate Weigh hydrate. Heat to drive off the water. Weigh the anhydrous compound. Subtract and convert the difference to moles. The ratio of moles of water to moles of anhydrous compound is the coefficient for water in the hydrate.

41. Use of Hydrates Anhydrous forms of hydrates are often used to absorb water, particularly during shipment of electronic and optical equipment. In chemistry labs, anhydrous forms of hydrates are used to remove moisture from the air and keep other substances dry.

42. Molarity One way to represent concentration (how much solute is present in a given amount of solution). Molarity (M) = the moles of solute present in a liter of solution 6 moles HCl moles of solute 6 moles HCl 0.500 L of solution L of solution 1 L solution = 12 M = 6 M (6 molar)

43. Dilution and Volumes of Solutions Required for Reactions Dilution is the process of making a solution less concentrated by adding some of the solution to more solvent. In all dilutions, the number of moles of solute remains the same. After dilution, the new molarity is simply the number of moles of solute present divided by the total volume of solution.

44. The number of moles of solute in a given volume of a solution of known concentration can always be determined by multiplying the volume times the molarity. liters x moles solute = moles solute liter Volume x Molarity = amount of solute

45. Problems using Molarity 1) What is the Molarity of a solution which contains 29.3 g NaCl in 250 mL of solution? ?M = moles of solute = 29.3 g NaCl x 1 mole = 2.00 M L of solution.250 L solution 58.5 g 2) How many moles of KNO 3 are contained in 300 mL in 0.500 M solution? ? moles = 0.300 L x 0.500 moles = 0.150 mole 1 L 3) How many grams of KNO 3 are needed to prepare 5.00 mL of a 0.200 M solution? ? g = 0.500 L x 0.200 mole x 101.1 g = 10.1 grams KNO 3 1 L 1 mole