Remember… Law of Constant Composition a compound contains elements in a certain fixed proportions...
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![Page 1: Remember… Law of Constant Composition a compound contains elements in a certain fixed proportions (ratios), regardless of how the compound is prepared.](https://reader034.fdocuments.us/reader034/viewer/2022051215/5697bfbe1a28abf838ca28ef/html5/thumbnails/1.jpg)
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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.
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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
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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)
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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
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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
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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
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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.
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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
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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
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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
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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
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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
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Molecular Formula- shows the actual number of atoms
Example: C6H12O6
Empirical Formula - shows the ratio between atoms
Example: CH2O
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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
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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
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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
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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
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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
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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
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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
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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