1. CH 9: The Mole Renee Y. Becker CHM 1025 Valencia Community College 1

2. 2 Avogadro’s number (symbol N) is the number of atoms in 12.01 grams of carbon. Its numerical value is 6.02 × 10 23. Therefore, a 12.01 g sample of carbon contains 6.02 × 10 23 carbon atoms. Avogadro’s Number

3. 3 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 × 10 23 particles. 1 mol = Avogadro’s number = 6.02 × 10 23 units We can use the mole relationship to convert between the number of particles and the mass of a substance. The Mole

4. 4 The volume occupied by one mole of softballs would be about the size of the Earth. One mole of Olympic shotput balls has about the same mass as the Earth. Analogies for Avogadro’s Number

5. 5 One Mole of Several Substances C 12 H 22 O 11 H2OH2O mercury sulfur NaCl copper lead K 2 Cr 2 O 7

6. 6 How many sodium atoms are in 0.120 mol Na? Example 1

7. 7 How many moles of potassium are in 1.25 × 10 21 atoms K? Example 2

8. 8 The atomic mass of any substance expressed in grams is the molar mass (MM) of that substance. The atomic mass of iron is 55.85 amu. Therefore, the molar mass of iron is 55.85 g/mol. Since oxygen occurs naturally as a diatomic, O 2, the molar mass of oxygen gas is 2 times 16.00 g or 32.00 g/mol. Molar Mass

9. 9 The molar mass of a substance is the sum of the molar masses of each element. What is the molar mass of magnesium nitrate, Mg(NO 3 ) 2 ? The sum of the atomic masses is: 24.31 + 2(14.01 + 16.00 + 16.00 + 16.00) = 24.31 + 2(62.01) = 148.33 amu The molar mass for Mg(NO 3 ) 2 is 148.33 g/mol. Calculating Molar Mass

10. Example 3 Calc. the molar mass: 1.Fe 2 O 3 2.C 6 H 8 O 7 3.C 16 H 18 N 2 O 4 10

11. 11 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 × 10 23 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 Calculations

12. 12 What is the mass of 1.33 moles of titanium, Ti? Example 4

13. 13 What is the mass of 2.55 × 10 23 atoms of lead? Example 5

14. 14 How many O 2 molecules are present in 0.470 g of oxygen gas? Example 6

15. 15 What is the mass of a single molecule of sulfur dioxide? The molar mass of SO 2 is 64.07 g/mol. Example 7

16. 16 At standard temperature and pressure, 1 mole of any gas occupies 22.4 L. The volume occupied by 1 mole of gas (22.4 L) is called the molar volume. Standard temperature and pressure are 0  C and 1 atm. Molar Volume

17. 17 We now have a new unit factor equation: 1 mole gas = 6.02 × 10 23 molecules gas = 22.4 L gas Molar Volume of Gases

18. 18 The density of gases is much less than that of liquids. We can calculate the density of any gas at STP easily. The formula for gas density at STP is: = density, g/L molar mass in grams/mol molar volume in liters/mol Gas Density

19. 19 What is the density of ammonia gas, NH 3, at STP? Example 8

20. 20 We can also use molar volume to calculate the molar mass of an unknown gas. 1.96 g of an unknown gas occupies 1.00 L at STP. What is the molar mass? We want g/mol; we have g/L. Example 9

21. 21 We now have three interpretations for the mole: – 1 mol = 6.02 × 10 23 particles – 1 mol = molar mass – 1 mol = 22.4 L at STP for a gas This gives us 3 unit factors to use to convert between moles, particles, mass, and volume. Mole Unit Factors

22. 22 A sample of methane, CH 4, occupies 4.50 L at STP. How many moles of methane are present? Example 10

23. 23 What is the mass of 3.36 L of ozone gas, O 3, at STP? Example 11

24. 24 How many molecules of hydrogen gas, H 2, occupy 0.500 L at STP? Example 12

25. 25 The percent composition of a compound lists the mass percent of each element. For example, the percent composition of water, H 2 O is: – 11% hydrogen and 89% oxygen All water contains 11% hydrogen and 89% oxygen by mass. Percent Composition

26. 26 There are a few steps to calculating the percent composition of a compound. Let’s practice using H 2 O. – Assume you have 1 mole of the compound. – One mole of H 2 O contains 2 mol of hydrogen and 1 mol of oxygen. – 2(1.01 g H) + 1(16.00 g O) = molar mass H 2 O – 2.02 g H + 16.00 g O = 18.02 g H 2 O Calculating Percent Composition

27. 27 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 H 2 O × 100% = 11.2% H 16.00 g O 18.02 g H 2 O × 100% = 88.79% O Calculating Percent Composition % H = % O =

28. 28 TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 °C. Calculate the percent composition of TNT, C 7 H 5 (NO 2 ) 3. Example 13

29. 29 Percent Composition of TNT

30. 30 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 C 6 H 6. – The empirical formula of benzene is CH. The molecular formula of octane is C 8 H 18. – The empirical formula of octane is C 4 H 9. Empirical Formulas

31. 31 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 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

32. 32 The molar mass of radium is 226.03 g/mol, and the molar mass of oxygen is 16.00 g/mol. We get Ra 0.00726 O 0.00719. Simplify the mole ratio by dividing by the smallest number. We get Ra 1.01 O 1.00 = RaO is the empirical formula. Calculating Empirical Formulas 1 mol Ra 226.03 g Ra 1.640 g Ra ×= 0.00726 mol Ra 1 mol O 16.00 g O 0.115 g O ×= 0.00719 mol O

33. Example 14 Determine the empirical formula of sodium sulfate given the following data A 3.00 g sample of Na x S y O z is comprised of 0.9722 g of sodium, 0.678 g of sulfur and 1.35 g of oxygen 33

34. 34 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 Formulas from Percent Composition

35. 35 Calculate the moles of each element: 1 mol C 12.01 g C 92.2 g C ×= 7.68 mol C 1 mol H 1.01 g H 7.83 g H ×= 7.75 mol H The ratio of elements in acetylene is C 7.68 H 7.75. Divide by the smallest number to get the formula: 7.68 C = C 1.00 H 1.01 = CH 7.75 7.68 H Empirical Formula for Acetylene

36. Example 15 Find the empirical formula of Iron(III) chloride from the following data In Iron(III) chloride – Fe 34.43 % by mass – Cl 65.57 % by mass 36

37. 37 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 C 2 H 2. Molecular Formulas

38. Example 16 Find the empirical and molecular formula for hydrazine, N x H y The molar mass of hydrazine is 32.046 g/mol In hydrazine – N is 87.42% by mass – H is 12.58 % by mass 38

39. 39 We can use the following flow chart for mole calculations: Chapter Summary