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

Percent Composition Three STEPS 1)Formula determination  using Criss-cross determine subscripts Formula mass 2) Formula mass  using periodic table, total all elements % Calculation 3) % Calculation element mass/formula mass x 100  element mass/formula mass x 100

Percent Composition Find the percent composition of Iron(II)Nitrate Step 1 Determine the formula  Fe(NO 3 ) 2 Step 2 Determine the formula mass  1 Fe atom 1 x 55.8 amu = 55.8 amu 2 N atoms2 x 14 amu = 28 amu 6 O atoms6 x 16 amu = 96 amu amu

Percent Composition Step 3. Calculate % of total mass Fe atom31.0 % 1 Fe atom = 55.8 amu Fe x 100 = 31.0 % amu Fe(NO 3 ) 2 N atoms15.6% 2 N atoms = 28 amu N x 100 = 15.6% amu Fe(NO 3 ) 2 O atom53.4 % 6 O atom = 96 amu O x 100 = 53.4 % amu Fe(NO 3 ) 2

Percent Composition Problems: 1.Determine the % composition of Na and C in Sodium Carbonate. FormulaTotal mass Math% Na 2 CO amu Na =46/106 x100 C =12/106 x100 O =48/106 x100 Na= 43.4% C= 11.3% O= 45.3%

Percent Composition 2. Determine the % composition of all the elements in Aluminum Sulfate. FormulaTotal mass Math% Al 2 (SO 4 ) amu Al =54/342.3 x100 S =96.3/342.3 x100 O =192/342.3 x100 Al= 15.8% S= 28.1% O= 56.1%

Percent Composition 3. Determine the % composition of all the elements in Sodium Hydroxide. FormulaTotal mass Math% NaOH 40.0 amu Na = 23.0/40.0 x100 H = 1/40.0 x100 O = 16/40.0 x100 Na= 57.5% H= 2.5% O= 40.0%

Percent Composition 4.Determine the % composition of C in Carbon Dioxide. FormulaTotal mass Math% CO amu C =12.0/44.0 x100 O =32.0/44.0 x100 C= 27.3% O= 72.7%

Percent Composition 5.In a sample of Pitchblende, Uranium ore, it is determined that there are 2.5 g of Uranium in a g sample. What % of the ore is Uranium? Little different problem but still % comp. Total Ore = g Uranium = 2.5 g % = U / Ore x 100  2.5 g/ g x100 U = 0.50% of the Ore

% Composition Practice Problem What is the % composition of all the elements in Ammonium Phosphite. Step 1  Formula (NH 4 ) 3 PO 3 Step 2  Molar Mass 3(14.0) + 12(1.0) (16.0) = g Step 3  % CompN = 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 %

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

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.

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 #. The whole # ratio become the subscripts Step 5  Apply mole ratio to determine the Empirical formula. (The whole # ratio become the subscripts!) Bonus StepMolecular formula 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.

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

Empirical Formula Step 2 Step 2:: Determine the smallest value g of Na = 1.41 mol Na g of S = mol S  smallest g of O = 2.81 mol O

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

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

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