Remember… Law of Constant Composition a compound contains elements in a certain fixed proportions (ratios), regardless of how the compound is prepared.

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


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


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


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


75.08 g/mol75.08 g/mol

Remember… Law of Constant Composition a compound contains elements in a certain fixed proportions (ratios), regardless of how the compound is prepared.

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