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Empirical and Molecular Formulas

Empirical and Molecular Formulas. 10.4. Empirical formulas give the lowest whole number ratio of atoms in the compound. You’ve been using these for ionic compounds all along. NaBr Fe 2 O 3 H 2 SO 4 . Molecular formulas show the exact composition of 1 molecule.

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Empirical and Molecular Formulas

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  1. Empirical and Molecular Formulas 10.4

  2. Empirical formulas give the lowest whole number ratio of atoms in the compound. You’ve been using these for ionic compounds all along. NaBr Fe2O3 H2SO4

  3. Molecular formulas show the exact composition of 1 molecule. N2 CH4 C2H4 methane ethene It may not be in the lowest whole number ratio because it shows exactly what is in one molecule.

  4. Percent Composition Percent composition describes the masses of the atoms in a compound; all of the other formulas describe the number of atoms (or moles).

  5. The empirical formula for hydrogen chloride is HCl; it’s a 1:1 ratio. But by mass, the composition looks like this: 97 % Cl 3 % H

  6. Remember, generally speaking: Calculating percents: part * 100 whole The percent means ‘per 100’.

  7. Empirical formulas contain the information necessary to find the percent composition. Converting from an empirical formula to percent composition means converting from a mole ratio (empirical formula) to a mass ratio (percent composition).

  8. What is the percent composition of Fe2O3? The empirical formula indicates a ratio of 2:3 atoms or moles. In 1 mole of Fe2O3, there are 2 moles of Fe and 3 moles of O.

  9. Calculate the masses of the moles: Total mass = 159.7 g

  10. Calculate percents, (part/whole)*100: The percent composition for Fe2O3 is 69.94% Fe and 30.06 % O by mass.

  11. Percent composition Empirical Formula Suppose a chemist had a compound that was 27.00 % carbon and 73.00 % oxygen. What is the empirical formula for this compound?

  12. First, convert all % into moles: 27.00 % C and 73.00 % O % means ‘per 100’, so assume a 100 g sample. 27% would be 27 g

  13. So our mystery compound has this ratio: 2.248 moles C : 4.563 moles O Our empirical formula begins to take shape as C2.248O4.563 Empirical formulas are always in the lowest whole number ratio

  14. Dividing both by the smaller gives a ratio of C1O2.03 which can be rounded to CO2 1 is understood and not written The compound is carbon dioxide.

  15. Sometimes the division doesn’t come out in a decimal that can reasonably be rounded to the nearest whole number. 1:3.97 can be rounded to 1:4 1: 2.5 cannot be rounded, but still needs to be reported as whole numbers. Multiply both by some factor that will result in the lowest whole number ratio. X 2 2:5 which works X 4 4:10 which can be reduced

  16. Once you have the empirical formula, you can determine the molecular formula if necessary – but you must have the molar mass of the compound in question.

  17. There is a great flow chart on page 347 (Figure 10-15) that condenses the steps – put it in your notes for a quick reference.

  18. The empirical formula for calcium chloride is CaCl2. What is the percent composition? • Start with what you know: • 1 mole of Ca for every 2 moles of Cl • 1 mole of Ca has a mass of 40.08 g • 1 mole of Cl has a mass of 35.45 g

  19. Figure out the masses involved for CaCl2. 1 mole Ca = 40.08 g Total mass for a formula unit of CaCl2 is 40.08 g + 70.90 g = 110.98 g

  20. Now the percentage composition calculations: CaCl2 is 36.11 % Ca and 63.89 % Cl by mass

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