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Percent Comp

Percent Comp. Percentage composition. Indicates the relative amount of each element present in a compound. Calculating percentage composition. Step 1 : Calculate molar mass Step 2 : Divide the subtotal for each element’s mass by the molar mass.

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Percent Comp

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  1. Percent Comp

  2. Percentage composition Indicates the relative amount of each element present in a compound.

  3. Calculating percentage composition Step 1 : Calculate molar mass Step 2 : Divide the subtotal for each element’s mass by the molar mass. Step 3: Multiply by 100 to convert to a percentage.

  4. Example 1 1. Calculate the molar mass of water, H2O. Step 1 – H- 2(1.01)= 2.02 O – 1(16.0)= 16.0 ________ 18.02

  5. Step 2 2.02/18.02 X 100 = 11.21% 16.0 /18.02 X 100 = 88.79 % Water is composed of 11.21 % hydrogen and 88.79 % oxygen.

  6. Example 2 Calculate the percentage composition of sucrose , C12H22O11.

  7. Example 3 Find the percentage composition of hydrogen in sulfuric acid.

  8. Empirical Formula From percentage to formula

  9. Empirical Formula • Theempirical formula gives the ratio of the number of atoms of each element in a compound. CompoundFormulaEmpirical Formula Hydrogen peroxide H2O2 OH Benzene C6H6 CH Ethylene C2H4 CH2 Propane C3H8 C3H8

  10. Calculating Empirical • Pretend that you have a 100 gram sample of the compound. • That is, change the % to grams. • Convert the grams to mols for each element. • Write the number of mols as a subscript in a chemical formula. • Divide each number by the least number. • Multiply the result to get rid of any fractions.

  11. A compound was analyzed to be 82.67% carbon and 17.33% hydrogen by mass. What is the empirical formula for the compound? Assume 100 g of sample, then 82.67 g are C and 17.33 g are H EXAMPLE 1

  12. Convert masses to moles: 82.67 g C x mole/12.011 g = 6.88 moles C 17.33 g H x mole/1.008 g = 17.19 mole H Find relative # of moles (divide by smallest number) EXAMPLE 1

  13. Convert moles to ratios: 6.88/6.88 = 1 C 17.19/6.88 = 2.50 H Or 2 carbons for every 5 hydrogens C2H5 EXAMPLE 1

  14. Example 2 • Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N. • Assume 100 g so • 38.67 g C x 1mol C = 3.220 mole C 12.01 gC • 16.22 g H x 1mol H = 16.09 mole H 1.01 gH • 45.11 g N x 1mol N = 3.219 mole N 14.01 gN

  15. Example 2 • 3.220 mole C • 16.09 mole H • 3.219 mole N • C3.22H16.09N3.219 If we divide all of these by the smallest one It will give us the empirical formula

  16. Example 2 • The ratio is 3.220 mol C = 1 mol C 3.219 molN 1 mol N • The ratio is 16.09 mol H = 5 mol H 3.219 molN 1 mol N • C1H5N1 is the empirical formula

  17. Molecular Formula • Is the actual Formula • Only applies to covalent compounds • Find the Mass of the Empirical Formula • Divide the actual molar mass by the mass of one mole of the empirical formula. • Caffeine has a molar mass of 194 g. what is its molecular formula? Empirical Formula C4H5N2O1

  18. Caffeine Example • Find the mass of the empirical formula Empirical Formula C4H5N2O1 C= 4 X 12.01 H = 5 X 1.01 N = 2 X 14.01 O = 1 X 15.99 Total Mass = 97g

  19. Caffeine Example • Find x if • 194 g • 97 g = 2 C4H5N2O1 2 X C8H10N4O2.

  20. Example 2 • A compound is known to be composed of 71.65 % Cl, 24.27% C and 4.07% H. Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

  21. Example 2 = 2.0mol • 71.65 Cl 24.27C 4.07 H. = 2.0mol = 4.0mol

  22. Example 2 • Cl2C2H4 We divide by lowest (2mol ) • Cl1C1H2 would give an empirical wt of 48.5g/mol Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula?

  23. would give an empirical wt of 48.5g/mol Its molar mass is known (from gas density) is known to be 98.96 g. What is its molecular formula? = 2 =

  24. 2 X Cl1C1H2 = Cl2C2H4

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