percent composition - garzzillo science
TRANSCRIPT
Percent Composition
� The relative amount of the element in a compound � Expressed as a percent
� % Comp will give you either the masses of the elements or the chemical formula � Either way… you will use the following formula:
Percent Composition
Mass from the element
Total Mass (or Molar Mass) 100 %
% Composition Practice
� If a 13.60g sample of a compound contains 8.20g Mg and 5.40g O, what is the % composition of this compound?
� % Mg = (8.20 / 13.60) X 100 = 60.29 %
� % O = (5.40 / 13.60) X 100 = 39.71 %
� Your % values should equal 100%. This is how to check your answer to make sure it makes sense!
% Composition Example #2
� Determine the % composition for glucose, C6H12O6
� Step 1: Determine the Molar Mass:
� 6C = (6 x 12.01) = 72.06g
� 12H = (12 x 1.01) = 12.12g
� 6O = (6 x 16.00) = 96.00g
� Total Molar Mass = 180.18g
% Composition Example #2
� Determine the % composition for glucose, C6H12O6
� Step 2: Calculate the % composition � %C = (72.06 / 180.18) x 100 = 39.99 %
� %H = (12.12 / 180.18) x 100 = 6.73 %
� %O = (96.00 / 180.18) x 100 = 53.28 %
� Remember: the percent should add to 100!
Molecular Formulas
� Are the actual formulas of compounds that are NOT reduced to their lowest whole number ratio
� Example: � Glucose: C6H12O6
� Benzene: C6H6
Empirical Formulas
� Empirical means: achieved through observation or experiments
� The lowest whole number ratio of the atoms of the different elements in a compound
� * May or May NOT be the same as a molecular formula*
� Example: � Glucose: CH2O � Benzene: CH
Empirical, Molecular, or Both?
� N2O5
� Both
� H2SO4 (Sulfuric Acid) � Molecular
� C6H12O6 (Glucose) � Molecular
� H2O (Water) � Both
� H2O2 (Hydrogen peroxide) � Molecular
� C2H7
� Both
� C2H4
� Molecular
� CH4 (Methane) � Both
� HO � Empirical
� C2H8N2
� Empirical
Which have the same Empirical Formula?
� C2H6O2
� No
� C8H14O6
� No
� C7H11O5
� Yes
� C3H9O3
� No
� C2H5O9
� Yes
� C4H6O10
� No
� C4H12O4
� No
Empirical Calculation Example
� Question: A compound is found to contain 25.9% N and 74.1% O, what is the empirical formula?
� Step 1: Assume you have 100.0 g. � Then you will have 25.9g of N and 74.1g of O.
� Step 2: Change masses to moles
25.9 g N
1 mol N
14.0g 1.85 mol N X = Nitrogen:
74.1 g O
1 mol O
16.0g 4.63 mol O X = Oxygen:
� Question: A compound is found to contain 25.9% N and 74.1% O, what is the empirical formula?
� Step 3: � Divide both molar quantities by the smaller number of
moles
Empirical Calculation Example
1.85 mol N
1.85 1 mol N = Nitrogen:
4.63 mol O
1.85 2.5 mol O = Oxygen:
� Question: A compound is found to contain 25.9% N and 74.1% O, what is the empirical formula?
� Step 4: � Multiply by a number to convert the fraction to a whole
number.
Empirical Calculation Example
1 mol N 2 2 mol N X = Nitrogen:
2.5 mol O 2 5 mol O X = Oxygen:
* Must multiply both by the
same number!
� Question: A compound is found to contain 25.9% N and 74.1% O, what is the empirical formula?
� Step 5: � Show the final answer.
Empirical Calculation Example
Empirical Formula
N2O5 =