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Empirical and Molecular Formula
Tuesday, January 7, 2014
Steps Used to Determine Empirical Formula
Step 1: List Given Values
Tuesday, January 7, 2014
Steps Used to Determine Empirical Formula
Step 1: List Given Values
Step 2: Calculate Mass(m) of Each Element in a 100g sample
Step 3: Convert mass (m) into Amount (n)
n= m / M
Tuesday, January 7, 2014
Steps Used to Determine Empirical Formula
Step 1: List Given Values
Step 2: Calculate Mass(m) of Each Element in a 100g sample
Step 3: Convert mass (m) into Amount (n)
n= m / M
Step 4: State Amount Ratio
Step 5: Calculate the Lowest Whole-Number Amount Ratio
Tuesday, January 7, 2014
Example
A certain compound contains 5.9 % hydrogen and 94.1% oxygen. Determine the empirical formula of the compound.
Tuesday, January 7, 2014
Solution
Consider 100g of the compound
Element Mass in 100 g sample
Amount (n)n= m / M Ratio
Hydrogen
Oxygen
5.9 n= 5.9/ 1=5.9
5.9/5.9=1
94.1 n=94.1/16=5.9
5.9/5.9=1
Tuesday, January 7, 2014
Example
The empirical formula of a compound is HCO2. If the compound has a molecular mass of 90 g/mol, determine it’s molecular formula.
Tuesday, January 7, 2014
Example
Step 1: Empirical formula is HCO2.
Step 2:
Given: Empirical Formula HCO2
Given: M = 90.0 g/mol
Step 3: Determine the Molar Mass of the empirical formula
HCO2:
H = 1 x 1.0 g = 1 g
C= 1 x 12.0 g = 12 g
O = 2 x 16 g = 32 g
=45.0g/mol
Tuesday, January 7, 2014
AnswerStep 4: Determine Ratio of Molar Mass of Compound to Molar Mass of Empirical Formula
Mcompound / M HCO2= 90 g/mol / 45 g/mol
=2Step 5: Calculate the Molecular Formula
Molecular Formula = 2(empirical formula)
2 x HCO2
C2H2O4
Tuesday, January 7, 2014
The analysis of a compound shows that it is made up of 21.9 % Na, 45.7 % C, 1.9% H and 30.5% O
What is the molecular formula of a compound if its molecular mass is 210.0g/mol?
Tuesday, January 7, 2014
Percent Composition
Tuesday, January 7, 2014
Steps to Calculating Percentage Composition
Step 1: Calculate Total Mass of Each Element in the Compound.
Step 2: Calculate Molecular Mass of the Compound
Step 3: Calculate Percentage Composition by Mass of Compound
Tuesday, January 7, 2014
Example
Calculate the percentage composition by mass of alanine, C3H7NO2.
Tuesday, January 7, 2014
ExampleStep 1:
C3H7NO2
C= 3 x 12.0 u = 36.0 u
H = 7 x 1.0 u = 7 u
N= 1 x 14.0 u = 14 u
O = 2 x 16.0 u = 32.0 u
Step 2:
Total above = 89.0 u
Tuesday, January 7, 2014
Example%C = (mass of C in 1 molecule / mass of 1 molecule) x 100
=(36 u/89 u) x 100 = 40.4 %
%H = mass of H in 1 molecule / mass of 1 molecule x 100
=(7 u/89 u) x 100 = 7.9 %
%N = mass of N in 1 molecule / mass of 1 molecule x 100
=(14 u/89 u) x 100 = 15.7 %
%O = mass of O in 1 molecule / mass of 1 molecule x 100
=(32 u/89 u) x 100 = 36.0 %
Tuesday, January 7, 2014
Example
What is the percent composition of glucose? C6H12O6
Tuesday, January 7, 2014
