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Percent Composition

Percent Composition. The total amount or % of an element in its compound. Percent Composition. Three STEPS Formula determination  using Criss-cross determine subscripts 2) Formula mass  using periodic table, total all elements 3) % Calculation

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Percent Composition

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  1. Percent Composition The total amount or % of an element in its compound.

  2. Percent Composition Three STEPS • Formula determination  using Criss-cross determine subscripts 2) Formula mass  using periodic table, total all elements 3) % Calculation  element mass/formula mass x 100

  3. Percent Composition Find the percent composition of Iron(II)Nitrate Step 1 Determine the formula  Fe(NO3)2 Step 2 Determine the formula mass  1 Fe atom 1 x 55.8 amu = 55.8 amu 2 N atoms 2 x 14 amu = 28 amu 6 O atoms 6 x 16 amu = 96 amu179.8 amu

  4. Percent Composition Step 3. Calculate % of total mass 1 Fe atom = 55.8 amu Fe x 100 = 31.0 % 179.8 amu Fe(NO3)2 2 N atoms = 28 amu N x 100 = 15.6% 179.8 amu Fe(NO3)2 6 O atom = 96 amu O x 100 = 53.4 % 179.8 amu Fe(NO3)2

  5. Percent Composition Problems: • Determine the % composition of Na and C in Sodium Carbonate.

  6. Percent Composition 2. Determine the % composition of all the elements in Aluminum Sulfate.

  7. Percent Composition 3. Determine the % composition of all the elements in Sodium Hydroxide.

  8. Percent Composition • Determine the % composition of C in Carbon Dioxide.

  9. Percent Composition • In a sample of Pitchblende, Uranium ore, it is determined that there are 2.5 g of Uranium in a 500.0 g sample. What % of the ore is Uranium? Little different problem but still % comp. Total Ore = 500.0 g Uranium = 2.5 g % = U / Ore x 100  2.5 g/ 500.0 g x100 U = 0.50% of the Ore

  10. % Composition Practice Problem What is the % composition of all the elements in Ammonium Phosphite. Step 1  Formula (NH4)3PO3 Step 2  Molar Mass 3(14.0) + 12(1.0) + 31.0 + 3 (16.0) = 133.0 g Step 3  % Comp N = 42/133 x 100 = 31.6 % H = 12/133 x 100 = 9.0 % P = 31/133 x 100 = 23.3 % O = 48/133 x100 = %36.1 %

  11. Empirical Formula What if you were given the % and asked to find the formula. A compound contains 32.38 % Na 22.65 % S 44.99 % O What is the formula of this compound?

  12. Empirical Formula Step 1 Calculate the number of moles for all elements in compound. a) If given as mass convert using molar mass. b) If given as a %, treat as if it is a 100 g Step 2 Determine the smallest mole value. Step 3 Determine mole ratio of elements by dividing by lowest mole value.

  13. Empirical Formula Step 4 If a mole ratio develops that is not a whole # or easily rounded off, multiply all ratios by a small integer (2, 3 or 4) to convert the fraction into a whole #. Step 5 Apply mole ratio to determine the Empirical formula. (The whole # ratio become the subscripts!) Bonus Step If asked to find the Molecular formula, divide the empirical formula mass into the molecular mass to get an integer. Multiply that integer through the empirical formulas subscripts. The result will be the molecular formula.

  14. Empirical Formula Step 1 Convert all values to moles Start with a 100 g sample if in % 32.38 % Na  32.38 g of Na / 23.0 g Na 22.65 % S  22.65 g of S / 32.1 g S 44.99 % O  44.99 g of O / 16.0 g O

  15. Empirical Formula Step 2:: Determine the smallest value. 32.38 g of Na = 1.41 mol Na 22.65 g of S = 0.706 mol S  smallest 44.99 g of O = 2.81 mol O

  16. Empirical Formula Step 3:: Divide through to get mole ratio. Na  1.41 mol Na /0.706  1.997 mol  2 S  0.706 mol S /0.706  1 mol  1 O  2.81 mol O/0.706  3.98 mol  4 0.706 mol

  17. Empirical Formula Step 4:: Ratio are whole #  Skip step Step 5 :: whole # ratios convert to subscripts Na  2 S  1 O  4 Na2SO4 Sodium Sulfate

  18. Empirical Formula Homework From your books:: Go to page 246. Review sample problems L, M and N Complete practice problems 1-3  Pg. 247 Complete practice problems 1-2  Pg. 249

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