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CH 9: The Mole

CH 9: The Mole. Renee Y. Becker CHM 1025 Valencia Community College. Avogadro’s Number. 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.

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CH 9: The Mole

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

  2. Avogadro’s Number • Avogadro’s number (symbol N) is the number of atoms in 12.01 grams of carbon. • Its numerical value is 6.02 × 1023. • Therefore, a 12.01 g sample of carbon contains 6.02 × 1023 carbon atoms.

  3. The Mole • 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 × 1023 particles. 1 mol = Avogadro’s number = 6.02 × 1023 units • We can use the mole relationship to convert between the number of particles and the mass of a substance.

  4. Analogies for Avogadro’s Number • 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.

  5. One Mole of Several Substances C12H22O11 H2O lead mercury K2Cr2O7 sulfur copper NaCl

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

  7. Example 2 • How many moles of potassium are in 1.25 × 1021 atoms K?

  8. Molar Mass • 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, O2, the molar mass of oxygen gas is 2 times 16.00 g or 32.00 g/mol.

  9. 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 magnesium nitrate, Mg(NO3)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(NO3)2 is 148.33 g/mol.

  10. Example 3 Calc. the molar mass: • Fe2O3 • C6H8O7 • C16H18N2O4

  11. Mole Calculations • 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 × 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.

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

  13. Example 5 • What is the mass of 2.55 × 1023 atoms of lead?

  14. Example 6 • How many O2 molecules are present in 0.470 g of oxygen gas?

  15. Example 7 • What is the mass of a single molecule of sulfur dioxide? The molar mass of SO2 is 64.07 g/mol.

  16. Molar Volume • 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.

  17. Molar Volume of Gases • We now have a new unit factor equation: 1 mole gas = 6.02 × 1023 molecules gas = 22.4 L gas

  18. Gas Density molar mass in grams/mol = density, g/L molar volume in liters/mol • 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:

  19. Example 8 • What is the density of ammonia gas, NH3, at STP?

  20. Example 9 • 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.

  21. Mole Unit Factors • We now have three interpretations for the mole: • 1 mol = 6.02 × 1023 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.

  22. Example 10 • A sample of methane, CH4, occupies 4.50 L at STP. How many moles of methane are present?

  23. Example 11 • What is the mass of 3.36 L of ozone gas, O3, at STP?

  24. Example 12 • How many molecules of hydrogen gas, H2, occupy 0.500 L at STP?

  25. Percent Composition • The percent compositionof 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.

  26. 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 mole of the compound. • One mole of H2O contains 2 mol of hydrogen and 1 mol of oxygen. • 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

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

  28. Example 13 • TNT (trinitrotoluene) is a white crystalline substance that explodes at 240 °C. Calculate the percent composition of TNT, C7H5(NO2)3.

  29. Percent Composition of TNT

  30. Empirical Formulas • The empirical formulaof 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.

  31. 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 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.

  32. Calculating Empirical Formulas 1 mol Ra 1.640 g Ra × = 0.00726 mol Ra 226.03 g Ra 1 mol O 0.115 g O × = 0.00719 mol O 16.00 g O • The molar mass of radium is 226.03 g/mol, and the molar mass of oxygen is 16.00 g/mol. • 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.

  33. Example 14 • Determine the empirical formula of sodium sulfate given the following data A 3.00 g sample of NaxSyOz is comprised of 0.9722 g of sodium, 0.678 g of sulfur and 1.35 g of oxygen

  34. Empirical Formulas from Percent 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.

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

  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

  37. Molecular Formulas (CH)n 26 g/mol CH 13 g/mol = • 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: n = 2 and the molecular formula is C2H2.

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

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

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