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Empirical and Molecular Formulas

Empirical and Molecular Formulas. This presentation has been brought to you by the CSI Lab located in N. Haven, Ct. Empirical and Molecular. Molecular Formula: The formula for a compound that exist as the nbr. of atoms. Ex. H 2 0 2  Hydrogen Peroxide Empirical Formula:

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Empirical and Molecular Formulas

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  1. Empirical and Molecular Formulas This presentation has been brought to you by the CSI Lab located in N. Haven, Ct.

  2. Empirical and Molecular • Molecular Formula: • The formula for a compound that exist as the nbr. of atoms. • Ex. H202 Hydrogen Peroxide • Empirical Formula: • Indicate the lowest whole number ratio of atoms of each element in a compound. • Ex. HO  Hydrogen Peroxide • Ho, Ho, Ho!!!

  3. The molecular formula of glucose is… • C6H12O6, , its empirical formula is… • CH2O … • So what’s my point? • Empirical formulas are lowest whole number ratios of atoms of each type …to determine them experimentally we need numbers of atoms!

  4. Consider the following: • Analysis of a compound at CSI North Haven Labs indicates that it contains • 32.38 g Na • 22.65 g S • 44.99 g O • How can we use this data to determine the formula?...Think… • Mass  number of atoms!

  5. Remember doing mole conversions? • Use the molar mass of element convert to moles. • A mole is a number of atoms … you now have a ratio of atoms in the compound. • Divide each mole amount by the smallest mole amount… whole # ratio??? • Use ratio to write subscripts

  6. 1.408 mol Na : 0.7063 mol S : 2.812mol O • Divide each by smallest  • 1.993 mol Na : 1 mol S : 3.981 mol O • Na2SO4

  7. Try this.. • Analysis indicates that a compound contains 78.1% Boron and 21.9% hydrogen determine the empirical formula. • Why is this one any different from the last example? • Assume you have a 100 gram sample…now how many grams of boron do you have?

  8. 7.22 mol B : 21.7 mol H • Divide through by smallest  • 1 mol B : 3.01 mol H • BH3

  9. Try this one … • Analysis of a 10.150 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.433 g. What is the empirical formula? • Grams of P = 4.433g Grams of O = ? • Now try it…

  10. 0.1431 mol P : 0.3573 mol O • Divide by smallest  • 1 mol P : 2.497 mol O … • Now what?...Multiply through by two to obtain whole numbers… 2 : 5 P2O5

  11. HW • Study for quiz next class • Read pgs 229-233 • Answer questions #1-3 pg 233

  12. Find empirical formulas and the empirical masses… 1. 36.48% Na, 25.41% S, 38.11 % O… Na2SO3 126 g/mol • 53.70 % iron and 46.30% sulfur… Fe2S3 208 g/mol • 1.04 g K, 0.70 g Cr and 0.86 g O… K2CrO4 194 g/mol 4. 36.51% C, 6.20% H and 57.29 % O … CH2O 30 g/mol

  13. Molecular Formulas • You have already solved for the Empirical Formula • Now find the mass of the Empirical Formula • Ex. CH2O  ~30 g/mol • You will be given a molecular weight to start with (~184.62 g/mol) • Now divide the molecular mass by the Empirical mass

  14. This will tell you the ratio, or the amount of times that the Empricial Formula goes into the Molecular formula • (184.62 g/mol)/(30 g/mol) • Answer should be close to a whole number • 6.154  6 • Multiply the Empirical Formula by that whole nbr. and you have the molecular formula • C6H12O6

  15. Quantitative Analysis • Various methods are used to determine the mass or percent composition of each element in a given compound…more about those later…

  16. Analysis by Combustion

  17. Percent Composition Believe it or not you’ve already done it.

  18. % comp. • Tells you the percent of the mass made up by each element in the compound • The mass of each element in a compound compared to the entire mass of the compound and multiplied by 100 %

  19. Find the percent composition of copper (I) sulfide, Cu2S

  20. A sample of an unkn. Compound with a mass of 0.2370 g is extracted from the roots of a plant. Decomposition of the sample produces 0.09480 g of carbon, 0.1264 g of oxgyen, and 0.0158 g of hydrogen. What is the percentage comp. of the compound?

  21. Hydrates • inorganic salts which contain a specific number of water molecules that are loosely attached • Written as: CuSO4• 5H2O • Copper Sulfate with 5 water molecules

  22. Formula for a hydrate • Finding the empirical formula for a hydrate involves finding the mass difference of the hydrate and anhydrous salt by driving off the excess water: • DEMO:http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/stoichiometry/empirical.html

  23. Example you conduct an experiment using CuWO4• xH2O • Mass of hydrate = 2.286 g • Mass of anhydrous salt = 2.050 g • Mass of H2O driven off = 0.236 g • CuWO4 = 311.4 g/mol • Mol salt = 2.050g (1mol/311.4g) = 0.00658 mol • Mol H2O = 0.236g (1mol/18g) = 0.0131 mol • Divide by the smallest mol amount • 0.0131 mol/0.00658 mol =~2 ; 2 water molecules • CuWO4 • 2H2O

  24. One interesting note! • A compound with a molecular mass of 59.7 g is determined to be made up of 40% Na and 60% Cl. What is the Empirical Formula? What is the Molecular Formula? • Empirical Formula – NaCl • Molecular Formula – NaCl • What are the charges (oxidation nbrs) on the Na cation and Cl anion? • +1 and -1 • What is the overall charge of the cmpd? • ZERO

  25. Oxidation Numbers by Group 0 1+ 2+ 3+ 3- 2- 1- varies Tend to have more than one oxidation number (+1 to +7) 3+ 3+ or 4+

  26. The sum of the oxidation nbrs in an ionic cmpd must be zero. • Ex. Magnesium Iodide • Mg+2 I-1 • Goes to MgI2 • Again the charge on the cmpd is 0

  27. Aluminum Oxide Al2O3?

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