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Remember…
• Law of Constant Composition a compound contains elements in a certain fixed proportions (ratios), regardless of how the compound is prepared or where it is found in nature.
• If you have one molecule of methane gas, you will always have 1 carbon atoms and 4 hydrogen atoms.
1. Mass Spectrometer
• This machine measure the molar mass of a compound.
• A small sample of the compound is vaporized and hit with a beam of electrons
•The fragments are put through an electric field and the amount of deflection determines molar mass
2. Combustion Analyzer• Is an instrument that can determine the
percentages of carbon, hydrogen, oxygen & nitrogen in a compounds
• A combustion reaction occurs and the individual parts of the products are captured and measured
• Using mass of products and individual atom mass, one can determine the percent composition
CxHyOz + O2 (g) CO2 (g) + H2O (g)
From Thursday…
What is the percent composition of formaldehyde, CH2O
M = 30.03 g mol
C = 12.01 g mol = 39.99%
30.03 g mol
H = 2.02 g mol = 6.73%
30.03 g mol
O = 16.00 g mol = 53.28%
30.03 g mol
What is the percent composition of acetic acid, C2H4O2
M = 60.06 g mol
C = 24.02 g mol = 39.99%
60.06 g mol
H = 4.04 g mol = 6.73%
60.06 g mol
O = 32.00 g mol = 53.28%
60.06 g mol
What`s the differenceMolecular Formula
Empirical Formula
Ratio
Benzene C6H6 CH
Acetylene C2H2 CH
Glucose C6H12O6 CH2O
Hydrogen peroxide
H2O2 HO
Water H2O H2O
Ammonia NH3 NH3
1:1
1:1
1:2:1
1:1
2:1
1:3
What does a molecular formula show
What does a Empirical formula show
Exact number and types of atoms in the molecule
Gives the lowest ratio of atoms in a compound. It does not necessarily tell you the exact number of each type of atom.
A compound was found to be composed of 85.6% carbon and 14.4% hydrogen. What is the empirical formula
Step 1: List the given valuesC=85.6% and H = 14.4%
Step 2: Calculate the mass (m) of each element in a 100g sample.
mC= 85.6% x 100g = 85.6g 100
mH= 14.4% x 100g = 14.4g 100
Step 3: Convert Mass (m) into moles (n)
nC= m/M = 85.6g/12.01g/mol = 7.1274 mol C
nH= m/M = 14.4g/1.008g/mol = 14.257 mol H
Step 4: State the Amount RationC : nH
7.1274 mol : 14.257 mol
Step 5: Calculate lowest whole number ratio, by dividing by the lowest amount of moles.
C = 7.1274 = 1 H = 14.257 = 2 7.1274 7.1274
Empirical Formula
CH2
The percent composition of a compound is 69.9 % iron and 30.1% oxygen. What is the empirical formula of a compound?
Step 1: List the given valuesFe=69.9% and O = 30.1%
Step 2: Calculate the mass (m) of each element in a 100g sample.
mFe= 69.9 x 100g = 69.9g 100
mO= 30.1 x 100g = 30.1g 100
Step 3: Convert Mass (m) into moles (n)
nFe= m/M = 69.9g/55.86g/mol = 1.25 mol Fe
nO= m/M = 30.1g/16.00g/mol = 1.88 mol OStep 4: State the Amount RationFe : nO
1.25mol : 1.88 mol
Step 5: Calculate lowest whole number ratio1.25mol : 1.88 mol1.25mol 1.25 mol1 : 1.52 : 3 Empirical Formula
is Fe2O3
When you don’t get a whole number, multiply entire ratio by 2, 3, 4 etc. until you get a whole number
Molecular Formula
• Molecular Formula of a compound tells you exact number of atoms in one molecule of a compound. This formula may be equal to the empirical formula or may be a multiple of this formula.
• To determine, you need:– The empirical formula– The molar mass of the compound
Molecular Formula- shows the actual number of atoms
Example: C6H12O6
Empirical Formula - shows the ratio between atoms
Example: CH2O
The empirical formula of a compound is CH3O and its molar mass is 93.12g/mol. What is the
molecular formula?Step 1: List given valuesEmpirical Formula=CH3O
Mcompound = 93.12 g/mol
Step 2: Determine the molar mass for the empirical formula, CH3O.
MEmpirical = 12.01g/mol + 3(1.01g/mol) + 16.00g/mol
= 31.04 g/mol
Step 3. Divide the molar mass by the empirical formula molar mass.
= = 3Step 4. Calculate Molecular Formula by
multiplying this number by the empirical formula.
Molecular formula = x (empirical formula)3 x CH3O
Therefore, the molecular formula is C3H9O3
Molar massEmpirical formula molar mass
93.12 g/mol31.04 g/mol
Example 2: The percent composition of a compound is determined by a combustion and analyzer is a 40.03% carbon, 6.67% hydrogen,
& 53.30% oxygen. The molar mass is 180.18g/mol. What is the molecular formula
Step 1: List given valuesC= 40.03%, O=53.30%, H=6.67%Mcompound = 180.18 g/mol
Step 2: Calculate the mass of each element in a 100g sample
mC=40.03g mO=53.30g mH=6.67g
Step 3: Convert Mass (m) into moles (n)
nC= m/M = 40.03g/12.01g/mol = 3.33 mol C
nH= m/M = 6.67g/1.01g/mol = 6.60 mol H
nO= m/M = 53.30g/16.00g/mol = 3.33 mol O
Step 4: State the Amount RationC : nH : nO
3.33mol : 6.60mol : 3.33 mol
Step 5: Calculate lowest whole number ratio3.33mol : 6.60mol : 3.33 mol3.33mol : 6.60mol : 3.33 mol
1 : 2: 1Empirical Formula is CH2O
Step 6: Determine the molar mass for the empirical formula
MEmpirical = 12.01g/mol + 2(1.01g/mol) + 16.00g/mol
= 30.03 g/mol
Step 7. Divide the molar mass by the empirical formula molar mass.
= = 6Step 8. Calculate Molecular Formula by
multiplying this number by the empirical formula.
Molecular formula = x (empirical formula)6 x (CH2O)
Therefore, the molecular formula is C6H12O6
Molar massEmpirical formula molar mass
180.18 g/mol30.03 g/mol
Example 3: The percent composition of a compound is determined by a combustion and analyzer is a 32.0% carbon, 6.70% hydrogen, 42.6% oxygen & 18.7% nitrogen. The molar mass is 75.08g/mol. What is the molecular
formula?Calculate the mass of each element in a 100g samplemC=32.0g mO=42.6g mH=6.70g mN=18.7g
Convert Mass (m) into moles (n)
nC= m/M = 32.0g/12.01g/mol = 2.66 mol C
nH= m/M = 6.70g/1.01g/mol = 6.65 mol H
nO= m/M = 42.6g/16.00g/mol = 2.66 mol O
nN= m/M = 18.7g/14.01g/mol = 1.33 mol N
State the Amount RationC : nH : nO : nN
2.66mol : 6.65mol : 2.6 mol: 1.33mol
Step 5: Calculate lowest whole number ratio2.66mol : 6.65mol : 2.6 mol: 1.33mol1.33mol : 1.33mol : 1.33 mol: 1.33mol
2 : 5: 2: 1
Empirical Formula is C2H5O2N
Determine the molar mass for the empirical formula
MEmpirical = 75.08g
Divide the molar mass by the empirical formula molar mass.
=
= 1Calculate Molecular Formula by multiplying
this number by the empirical formula.Molecular formula = x (empirical formula)1 x (C2H5O2N)
Therefore, the molecular formula is C2H5O2N
Molar massEmpirical formula molar mass
75.08 g/mol75.08 g/mol
