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Chapter 10

Chapter 10. Chemical Quantities, the Mole, and Conversions. The Mole. Measuring Matter -The amount of something is usually determined one of three ways; by counting, by mass, or by volume. -Groups are made to make counting easier. (Remember the bean lab?) A bunch, a dozen, a ream, a can.

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Chapter 10

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  1. Chapter 10 Chemical Quantities, the Mole, and Conversions

  2. The Mole Measuring Matter -The amount of something is usually determined one of three ways; by counting, by mass, or by volume. -Groups are made to make counting easier. (Remember the bean lab?) A bunch, a dozen, a ream, a can

  3. What is a mole? 6.022 x 1023 of anything! Known as Avogadro’s number Representative Particles are the smallest division of a substance that maintains the characteristics of that substance -For molecular compounds R.Ps are molecules -For ionic compounds R.Ps are formula units -For elements R.Ps are atoms

  4. Mass and the Mole 1 mole of any element is equal to the atomic mass on the periodic table AKA molar mass For example: 18g of H2O = 6.022 x 1023 molecules

  5. Converting Number of particles to moles… RP x 1 mole =Moles 6.02 x 1023 RPs How many moles are 2.80 x 1024 atoms of silicon? 4.65 moles

  6. Moles to number of particles… moles x 6.02 x 1023 RPs. = RPs 1 mole How many atoms are in 1.14 mol Co? 6.86 x 1023 atoms Co

  7. The mass of a mole of an element The atomic mass is the mass of one atom of an element (in AMU) Based on the mass of one atom of Carbon-12 AKA – relative mass since each element is compared to C-12 Not very practical when working with chemicals in real life

  8. Molar Mass is the atomic mass of an element expressed in grams or, the mass of one mole of atoms of a particular element expressed in grams Numerically equal to the atomic mass in amu’s because the Mole is a constant number!

  9. Formula Mass is the sum of the average atomic masses of all the elements represented in the formula (in amu’s) • Example: Use the atomic masses in the periodic table to calculate the formula mass of SrCl2 FM = 87.62 + 2(35.453) = 158.526 amu • Importance – the formula mass of a compound gives us useful information about the quantities of reactants needed to make new products (must follow the Law of Conservation of Mass)

  10. Molecular Mass is the same as formula mass, but this term in used specifically for molecular compounds Not used for ionic compounds Example – water: H2O = 2(1.00797) + 15.9994 = 18.01534 amu

  11. Molar Mass is the mass in GRAMS of one mole (6.02 x 10 23 particles) numerically equal to the formula mass of a compound (units = g/mol)

  12. Mole-Mass and Mole-Volume Relationships • Mole-Mass Relationship Use the molar mass of an element or compound to convert between the mass of a substance and the moles of a substance Moles to Mass (grams): Multiply by the Molar Mass Example – Find the mass, in grams, of 4.52 x 10-3 mol C20H42

  13. C20H42 Step 1: Find the Molar Mass: 20 (12.01115) + 42 (1.00797) = 282.55774 g/mol Step 2: Convert: 4.52 x 10-3 mol x 282.55774 g C20H42 = 1 mol C20H42 Answer: 1.28 g C20H42

  14. Grams to moles: Divide by the Molar Mass Example – Calculate the number of moles in 75.0 g of Dinitrogen trioxide. Step 1: Write the correct chemical formula: N2O3 Step 2: Find the Molar Mass: 2( 14.0067) + 3(15.9994) = 76.0116 g/mol Step 3: Convert: 75.0 g N2O3 x 1 mole N2O3 = 76.0116 g N2O3 Answer: 0.987 mole N2O3

  15. Remember This? # of RPs # of Moles Mass (g) x Avogadro’s # x molar mass

  16. The Mole Volume Relationship Avogadro’s Hypothesis states equal volumes of gases at the same temperature and pressure contain equal numbers of molecules Standard Temperature and Pressure (STP) = 0 ˚C (273 ˚K) and 101.325 kPa (1 atm, 760 mmHg, 760 torr)

  17. Molar Volume = the 22.4 L of space occupied by 6.02 x 1023 representative particles of any gas at STP Example – What is the volume of 3.20 x 10-3 mol CO2 at STP? 3.20 x 10-3 mol CO2 x 22.4 L = ? 1 mole Answer: 0.0717 L CO2

  18. Example: – How many moles of CO2 are in 57.0 L of CO2? 57.0 L CO2 x 1 mole = 22.4 L Answer: 2.54 mol CO2

  19. Calculating Molar Mass from Density Background – the density of gases is measured in g/L at a specific temperature, and, the density of a gas at STP is a constant value (characteristic property) To find Molar Mass of a gas from density – multiply the density at STP by the molar volume at STP  

  20. Example – A gaseous compound composed of sulfur and oxygen, which is linked to the formation of acid rain, has a density of 3.58 g/L at STP. What is the molar mass of this gas? Molar Mass = 3.58 g of gas x 22.4 L = 1 L 1 mol Answer: 80.2 g/mol

  21. Percent Composition and Chemical Formulas • Definition – relative amounts of substances in a compound expressed as percents by mass • Percentage Composition from Mass Data • Equation: Mass of Element x 100 = % Mass Compound

  22. It has been measured in lab that 1.000g of water contains 0.112g Hydrogen. Find the mass percentage of hydrogen • Mass %H = 0.112g x 100 = 11.2 %H 1.000g

  23. Percentage Composition from the Chemical Formula (Atomic mass of elem.) x (# atoms of elem.) x 100 Molar mass of the compound

  24. What is the percent composition by mass of hydrogen in water? • Mass %H = (1.00797g) x 2 x 100 = 18.01534g • Answer: 11.19013 %H

  25. Percentage Composition as a Conversion Factor What is the mass of hydrogen in 30.0g of water? Mass H = (Mass %  100) x mass sample = (11.2%  100) x 30.0g = Answer: 3.36gH

  26. Hydrates (compounds containing water): You must include the water of hydration in the formula mass • What are the mass percentages of the elements in Na2CO310 H2O? • MM = 2(22.98977) + 12.01115 + 3(15.9994) + 10(18.01534) = 286.14229g

  27. %Na = 2(22.98977) x 100 = 16.06877 % Na 286.14229 • %C = 12.01115 x 100 = 4.197614 % C 286.14229 • %O = 3(15.9994) x 100 = 16.77424 % O 286.14229 • %H2O = 10(18.01534) x 100 = 62.95938 % H2O 286.14229

  28. Empirical Formulas • A chemical formula showing the smallest (simplest) whole-number ratio of atoms in a compound • Steps for determining Empirical Formula from % composition Step 1: Determine the masses of the elements in a 100g (since its based on a percent by mass) sample Step 2: Convert these masses in grams to moles by dividing by the atomic mass

  29. Steps for empirical formulas (contd) Step 3: Determine the simplest mole ratio between the elements by dividing each by the smallest number of moles Step 4: Use these mole ratios to write the subscripts in the formula

  30. Example! • Calcium Fluoride occurs as the mineral Fluorite. It contains 51.3% by mass of Ca and 48.7% by mass of F. Determine the Empirical formula. • Step 1: Start with a 100g sample of calcium fluoride 51.3% of 100g = 51.3 g Ca 48.7% of 100g = 48.7 g F • Step 2: Convert to Moles 51.3 g Ca x 1 mole Ca = 1.28 mole Ca 40.08 g Ca 48.7 g F x 1 mole F = 2.56 mole F 18.998403 g F • Step 3: Ratio mole F = 2.56 mole F = 2 - ratio is 1:2 mole Ca 1.28 mole Ca

  31. CaF2 is the empirical formula • Another Example! Substance X and Y. Suppose you found the number of moles of X and Y in 100.0g of a compound to be 1.96 mole X and 2.94 mole Y. What is the empirical formula? Answer: X2Y3

  32. How are these similar? C2H2 C8H8

  33. Molecular Formulas • The actual formula of a molecular compound which is often a multiple of the empirical formula (E.g. C2H6 instead of CH3)

  34. Steps for Determining Molecular Formula from Empirical Formula The Molecular Mass must be determined by experimental analysis or be given in the problem Divide the Molecular Mass by the Formula Mass (Molar Mass) and adjust the subscripts in the Empirical Formula by multiplying by this resulting number

  35. Example Water – the simplest formula is H2O, but could the molecular formula be H4O2 or H8O4? • Molecular Mass = 18.01 g by analysis • Formula Mass(Molar Mass) = 2(1.00797) + 15.9994 = 18.01534g • Ratio of MM/FM = 18.01  18.0534 or 1:1 so this is also the molecular formula

  36. Example 2 Hydrogen Peroxide – the simplest formula is HO • Molecular Mass = 34.0g by analysis (given) • Formula Mass (Molar Mass) = 1.00797 g + 15.9994 g = 17.00737 g • Ratio of MM/FM = 34.0  17.00737 = 2:1 so the formula is doubled to H2O2

  37. Example 3 A compound is found by analysis to consist of 40.1% S and 59.9% O. The Molecular Mass is 80.1g. Determine the Empirical Formula, the Molecular Formula, and name the compound • Step 1: in a 100.0g sample 40.1% of 100.0 g = 40.1 g S 59.9% of 100.0 g = 59.0 g O • Step 2: convert to moles 40.1 g S x 1 mole S = 1.25 mole S 32.06g S 59.9 g O x 1 mole O = 3.74 mole O 15.9994 g O

  38. Step 3: ratio 3.74 mole O  1.25 mole S = 2.992:1 or 3:1 • Step 4: empirical formula SO3 • Step 5: mass ratios FM* = 32.06 g S + 3(15.9994 g)O = 80.0582 g SO3 MM = 80.1g (given in problem) 1:1 so the molecular formula = the empirical

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