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Chapter 6 : Chemical Composition

Chapter 6 : Chemical Composition. +. 4 Wood Boards. 6 Nails. 1 Fence Panel. +. For a 12 panels fence…. 4 Dozen Wood Boards. 6 Dozen Nails. 1 Dozen Fence Panels. +. +. 4C. 3H 2. C 4 H 6. +. For a mole of 1,3-Butadiene …. 4 moles of C. 3 moles of H 2. 1 mole of C 4 H 6.

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Chapter 6 : Chemical Composition

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  1. Chapter 6: Chemical Composition + 4 Wood Boards 6 Nails 1 Fence Panel + For a 12 panels fence… 4 Dozen Wood Boards 6 Dozen Nails 1 Dozen Fence Panels +

  2. + 4C 3H2 • C4H6 + • For a mole of 1,3-Butadiene … 4 moles of C 3 moles of • H2 1 mole of • C4H6 +

  3. 1 Dozen = 12 items • The mole is defined (since 1960) as the amount of substance of a system that contains as many entities as there are atoms in 12 g of carbon-12. • Symbol: mol. • Coined by Wilhelm Ostwald in 1893 1 mol = 12 g of carbon-12 • A mole of carbon contains 6.0221415×1023 atoms of carbon, but the same is true for any • other element or molecule; in general: 1 mole = 6.0221415×1023 items # of Molecules = # of moles X 6.022×1023 • 6.0221415×1023 is the Avogadro’s Number

  4. m = number of nails X mass of 1 nail Example: 500g = 100 nails X 5g m = number of moles X mass of 1 mole of the substance m = n X M.M. Where: m = mass (g) n = number of moles (mol) M.M. = Molecular Mass (g/mol) = mass of 1 mole of the substance Sometime the Molecular Mass is measured in Daltons (1Da = 1g/mol)

  5. The molar mass of an element is numerically the same as the atomic weight of the element. They are not however the same; they have different units: • The atomic weight is defined as one twelfth of the mass of an isolated • atom of carbon-12 and is therefore dimensionless • The molar mass is measured in g/mol. • The molar mass of a compound is given by the sum of the atomic weights of the atoms which form the compound. • Example: molar mass of Ca(NO3)2

  6. Write a Solution Map for converting the units : Information Given: 1.1 x 1022 Ag atoms Find: ? moles Conv. Fact.: 1 mole = 6.022 x 1023 Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? atoms Ag moles Ag

  7. Check the Solution: Information Given: 1.1 x 1022 Ag atoms Find: ? moles Conv. Fact.: 1 mole = 6.022 x 1023 Sol’n Map: atoms  mole Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? 1.1 x 1022 Ag atoms = 1.8 x 10-2 moles Ag The units of the answer, moles, are correct. The magnitude of the answer makes sense since 1.1 x 1022 is less than 1 mole.

  8. Write a Solution Map for converting the units : Information Given: 1.75 mol H2O Find: ? g H2O C F: 1 mole H2O = 18.02 g H2O Example:Calculate the mass (in grams) of 1.75 mol of water mol H2O g H2O

  9. Check the Solution: Information Given: 1.75 mol H2O Find: ? g H2O C F: 1 mole H2O = 18.02 g H2O Sol’n Map: mol  g Example:Calculate the mass (in grams) of 1.75 mol of water 1.75 mol H2O = 31.5 g H2O The units of the answer, g, are correct. The magnitude of the answer makes sense since 31.5 g is more than 1 mole.

  10. Chemical Formulas as Conversion Factors • 1 spider  8 legs • 1 chair  4 legs • 1 H2O molecule  2 H atoms  1 O atom

  11. Mole Relationships inChemical Formulas • since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound

  12. Example: Carvone, (C10H14O), is the main component in spearmint oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liquers, soaps and perfumes. Find the mass of carbon in 55.4 g of carvone.

  13. Write a Solution Map for converting the units : Information Given: 55.4 g C10H14O Find: g C CF: 1 mol C10H14O = 150.2 g 1 mol C10H14O  10 mol C 1 mol C = 12.01 g mol C10H14O mol C g C Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). g C10H14O

  14. Check the Solution: Information Given: 55.4 g C10H14O Find: g C CF: 1 mol C10H14O = 150.2 g 1 mol C10H14O  10 mol C 1 mol C = 12.01 g SM: g C10H14O  mol C10H14O  mol C  g C Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O). 55.4 g C10H14O = 44.3 g C The units of the answer, g C, are correct. The magnitude of the answer makes sense since the amount of C is less than the amount of C10H14O.

  15. Percent Composition • Percentage of each element in a compound • By mass • Can be determined from • the formula of the compound • the experimental mass analysis of the compound • The percentages may not always total to 100% due to rounding

  16. Mass Percent as a Conversion Factor • the mass percent tells you the mass of a constituent element in 100 g of the compound • the fact that NaCl is 39% Na by mass means that 100 g of NaCl contains 39 g Na • this can be used as a conversion factor • 100 g NaCl  39 g Na

  17. Example - Percent Composition from the Formula C2H5OH • Determine the mass of each element in 1 mole of the compound 2 moles C = 2(12.01 g) = 24.02 g 6 moles H = 6(1.008 g) = 6.048 g 1 mol O = 1(16.00 g) = 16.00 g • Determine the molar mass of the compound by adding the masses of the elements 1 mole C2H5OH = 46.07 g

  18. Sample - Percent Composition from the Formula C2H5OH • Divide the mass of each element by the molar mass of the compound and multiply by 100%

  19. Empirical Formulas Hydrogen Peroxide Molecular Formula = H2O2 Empirical Formula = HO Benzene Molecular Formula = C6H6 Empirical Formula = CH Glucose Molecular Formula = C6H12O6 Empirical Formula = CH2O

  20. Finding an Empirical Formula • convert the percentages to grams • skip if already grams • convert grams to moles • use molar mass of each element • write a pseudoformula using moles as subscripts • divide all by smallest number of moles • multiply all mole ratios by number to make all whole numbers • if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc. • skip if already whole numbers

  21. All these molecules have the same Empirical Formula. How are the molecules different?

  22. All these molecules have the same Empirical Formula. How are the molecules different?

  23. Molecular Formulas • The molecular formula is a multiple of the empirical formula • To determine the molecular formula you need to know the empirical formula and the molar mass of the compound

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