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Empirical and Molecular Formulas

Empirical and Molecular Formulas. Review. We learned how to calculate the molar mass of compounds. Calculate the molar mass of Ca(CN) 2 . 1 x Ca = 1 x 40.08 g/mol = 40.08 g/mol 2 x C = 2 x 12.01 g/mol = 24.02 g/mol 2 x N = 2 x 14.01 g/mol = 28.02 g/mol TOTAL = 92.12 g/mol. Review.

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Empirical and Molecular Formulas

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  1. Empirical and Molecular Formulas

  2. Review • We learned how to calculate the molar mass of compounds. • Calculate the molar mass of Ca(CN)2. • 1 x Ca = 1 x 40.08 g/mol = 40.08 g/mol • 2 x C = 2 x 12.01 g/mol = 24.02 g/mol • 2 x N = 2 x 14.01 g/mol = 28.02 g/mol • TOTAL = 92.12 g/mol

  3. Review • We also learned how to determine the percentage composition of a compound. • Calculate the % composition of Ca(CN)2. • %Ca = (40.08)/(92.12) x 100% = 43.51% Ca • %C = (24.02)/(92.12) x 100% =26.07% C • %N = (28.02)/(92.12) x 100% = 30.42% N

  4. O H H Empirical Formulas • Molecular Formula (MF) - shows how many atoms are actually in a molecule. • EXAMPLE: Glucose has the MF C6H12O6. • EXAMPLE: Water has the MF H2O.

  5. Empirical Formulas • Empirical Formula (EF) - shows the lowest whole-number ratio of atoms in a compound. • EXAMPLE: Glucose has the EF CH2O. • EXAMPLE: Water has the EF H2O. • Different cmpds can have different MFs but have the same EF. • EXAMPLE: NO2 and N2O4 have different MFs but the same EF (NO2).

  6. Empirical Formulas • You can discover the empirical formula of a compound if you know the % composition. • Will NOT tell you which molecular formula is correct!

  7. Empirical Formulas • An unknown compound is analyzed: • 15.77% carbon • 84.23% sulfur • Calculate the EF.

  8. Empirical Formulas • First, assume you have exactly 100 grams of the sample. • Why 100 grams? • Because percents become grams. • In 100 grams of this compound you would have: • 15.77 g C • 84.23 g S

  9. x x Empirical Formulas • Next, change grams to moles. • 15.77 g C • 84.23 g S 1 mol C = 1.313 mol C 12.01 g C 1 mol S = 2.626 mol S 32.07 g S

  10. Empirical Formulas • The formula so far: • C1.313S2.626 • Divide all subscripts by the lowest one. • CS2 • This is the empirical formula of our mystery compound. • We don’t know if it’s the correct molecular formula. • Could be CS2, C2S4, C3S6, C4S8, etc...

  11. Empirical Formulas • A mystery compound has the following composition: • 3.0856% hydrogen • 31.604% phosphorus • 65.310% oxygen • What is this compound’s empirical formula?

  12. Empirical Formulas • Convert grams to moles. • 3.0856 g H x = 3.0613 mol H • 31.604 g P x = 1.0203 mol P • 65.310 g O x = 4.0820 mol O • Formula so far: H3.0613P1.0203O4.0820 • Reduced: H3PO4

  13. Empirical Formulas • A mystery compound has the following composition: • 25.940% nitrogen • 74.060% oxygen • What is this compound’s empirical formula?

  14. Empirical Formulas • Convert grams to moles. • 25.940 g N x = 1.8520 mol N • 74.060 g O x = 4.6289 mol O • Formula so far: N1.8520O4.6289 • Reduced: NO2.5 • Double subscripts to eliminate fractions. • N2O5

  15. Molecular Formulas • If we know a compound’s empirical formula and its molar mass, we can work out its molecular formula. • Benzene, a common non-polar solvent, has the empirical formula CH and a molar mass of 78.12 g/mol • Formula mass of CH: 13.02 g/mol • How many times does this go into 78.12 g/mol? • 78.12 g/mol  13.02 g/mol = 6 times • Molecular formula of benzene = (CH)6 = C6H6

  16. Molecular Formulas • The empirical formula of uracil (a base found in RNA) is C2H2NO. If the molar mass of uracil is 122.09 g/mol, what is the molecular formula of uracil? • Formula mass of C2H2NO: 56.05 g/mol • How many times does this go into 122.09 g/mol? • 122.09 g/mol  56.05 g/mol ≈ 2 • Molecular formula of uracil = (C2H2NO)2 = C4H4N2O2

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