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Determining Chemical Formulas

Understand how to determine empirical formulas, calculate mole ratios, and find molecular formulas in chemistry with step-by-step examples and practice questions.

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Determining Chemical Formulas

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  1. Determining Chemical Formulas 7-4

  2. Beaker Breaker • Name or write the following formulas: • AlN =___________________ • CO = ___________________ • Al2S3 =___________________ • Copper (II) Oxide =______________

  3. Empirical formula • formula that shows the smallest whole-number MOLE ratio of the different atoms in a compound • In ionic cmpds, the formula unit usually is the empirical formula

  4. What is the empirical formula of the following molecular formulas? • Ethyne, C2H2 • Benzene, C6H6 • Glucose, C6H12O6 • Acetic Acid

  5. Ethyne, C2H2 • CH • Benzene, C6H6 • CH • Glucose, C6H12O6 • CH2O • Acetic Acid • CH3COOH = C2H4O2 = CH2O

  6. Calculation of an empirical formula • Empirical formula: smallest whole-number MOLE ratio of the different atoms in a compound • ∴ 1)determine moles of each element • 2)determine the simplest whole # ratio of these moles

  7. What would the formula be of a cmpd that was composed of 2 mol hydrogen and 1 mol oxygen? • H:O mole ratio = 2:1 • ∴ H2O1…but always written H2O

  8. What would the formula be of a cmpd composed of 0.78 mol Al & 2.34 mol Br? • Mole ratio of Al: Br • = 0.78 : 2.34 • smallest whole # ratio????? • 0.78 mole Al = 1.0 Al 0.78 • 2.34 mol Br = 3.0 Br 0.78 • Al:Br = 1:3 AlBr3

  9. What is the empirical formula for a cmpd that contains 0.900 g calcium and 1.60 g chlorine? • Determine moles of Ca and Cl • Determine the simplest whole # mole ratio • 0.900 g Ca x 1 mol Ca = 0.0224 mole Ca 40.08 g Ca • 1.60 g Cl x 1 mol Cl = 0.0451 mol Cl 35.45 g Cl

  10. Simplest whole-# mole ratio? • 0.0224 mole Ca • 0.0451 mol Cl • 0.0224 mole Ca = 1.00 = 1 0.0224 mole • 0.0451 mol Cl = 2.01 = 2 0.0224 mole ∴ Ca:Cl = 1:2 CaCl2

  11. Beaker Breaker • A sample of indium chloride weighing 0.5000 g is found to contain 0.2404 g of chlorine. What is the empirical formula of the indium compound?

  12. Determine the empirical formula in a cmpd that is 40.0% C, 6.71% H, and 53.3% O. • Assume a 100.0 g sample. • 40.0 g C = 3.33 mol C • 6.71 g H = 6.64 mol H • 53.3 g O = 3.33 mol O • 3.33 mole C/3.33 = 1.00 = 1 • 6.64 mole H/3.33 = 1.99 = 2 • 3.33 mol O/3.33 = 1.00 = 1 • C:H:O = 1:2:1 CH2O

  13. If the empirical formula is CH2 and the molar mass is 42 g/mol, what is the molecular formula? • mass of empirical formula = 14 g/mol • mass of “real” formula = 42 g/mol • ∴ “real” is 42/14 or 3 times heavier! • C: H • 1: 2 (from empirical formula) • is really, 3( 1:2) = 3:6 C3H6 • (check: C3H6 does have a mass of 42 g/mol)

  14. If the molecular mass for the prev. problem (cmpd that is 40.0% C, 6.71% H, and 53.3% O)is 90 g/mol, determine the molecular formula. • The actual formula is how many times heavier than the empirical? • 3 times • ∴ molecular formula is 3xs bigger • 3 (C:H:O = 1:2:1) =3:6:3 C3H6O3

  15. Beaker Breaker • Write the symbol and oxidation # (incl +/-) for the following ions: • Calcium ____________________ • Chloride ___________________ • Oxide ____________________ • Sulfate ____________________

  16. Practice • Complete section review on page 233 #1-5

  17. Do Section Review p. 233 • Na2SO3 • Fe2S3 • K2CrO4 • N2O5 • N2O4

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