Empirical and Molecular Formulas. Formaldehyde CH 2 O Acetic acid C 2 H 4 O 2 Gylceradehyde C 3 H 6 O 3 40% C; 6.7% H; 53.3% O.

Empirical and Molecular Formulas

Formaldehyde CH2O

Acetic acid C2H4O2

Gylceradehyde C3H6O3

40% C; 6.7% H; 53.3% O

Empirical Formula

• Analysis of chemical compounds gives the % composition of each element.

• From this we can determine the Empirical Formula.

• The empirical formula shows the smallest whole number

mole ratio of the atoms in a compound.

CH2OCH3OOCH = C2H4O2

CH3OCH3O

Empirical FormulaFor ionic compounds, the formula unit is usually the compound’s empirical formula.

For molecular compounds, the molecular formula and the empirical formula can be different.

Molecular Formula

Empirical Formula

H2O2 HO

C6H12O6 CH2O

Determine the empirical formula for a compound containing 2.128 g Cl and 1.203 g Ca.Steps1. Find mole amounts for each element.2. Divide each mole amount by the smallest mole amount.

Example:

Example:1. Find mole amounts for each element (i.e., convert mass mole).

2.128 g Cl x 1 mol Cl = 0.06003 mol Cl 35.45 g Cl

1.203 g Ca x 1 mol Ca = 0.03001 mol Ca

40.08 g Ca

Example:

2. Divide each mole by the smallest mole amount.

Cl = 0.0600 mol Cl = 2.00 mol Cl 0.0300

Ca = 0.0300 mol Ca = 1.00 mol Ca

0.0300Ratio is 1 Ca: 2 ClEmpirical Formula = CaCl2

Example 2:

A compound is analyzed and found to contain 36.70% potassium, 33.27% chlorine, and 30.03% oxygen. What is the empirical formula of the compound?

KClO2

A compound weighing 298.12 g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula?

Hint:

Percent to mass

Mass to mole

Divide by small

Multiply ‘til whole

A compound weighing 298.12 g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula?

Percent to mass: Mg – (72.2%/100)*298.12 g = 215.24 g N – (27.8%/100)*298.12 g = 82.88 g

Mass to mole: Mg – 215.24 g * ( 1 mole ) = 8.86 mole 24.3

gN – 82.88 g * ( 1 mole ) = 5.92 mole 14.01

gDivide by small: Mg - 8.86 mole/5.92 mole = 1.50 N - 5.92 mole/5.92 mole = 1.00

mole

Multiply ‘til whole: Mg – 1.50 x 2 = 3.00N – 1.00 x 2 = 2.00 Mg3N2

Molecular Formula

The molecular formula gives the actual number of atoms of each element in a molecular compound.

Steps1. Find the empirical formula.2. Calculate the Empirical Formula Mass (EFM).3. Divide the molar mass by the “EFM” to get the “factor”.4. Multiply empirical formula by factor to get molecular formula.

Molecular Formula

Find the molecular formula for a compound whose molar mass is ~124.06 and empirical formula is CH2O3. 2. “EFM” = 62.03 g

3. 124.06/62.03 = 2

4. 2(CH2O3) = C2H4O6

Factor

Factor EmpiricalFormula

MolecularFormula

Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol.

Steps1. Find the empirical formula.2. Calculate the Empirical Formula Mass.3. Divide the molar mass by the “EFM”.4. Multiply empirical formula by factor.

Example:

Empirical formula.A. Find mole amounts.4.90 g N x 1 mol N = 0.350 mol

N14.01 g N

11.2 g O x 1 mol O = 0.700 mol O

16.00 g O

Example

B. Divide each mole by the smallest mole.

N = 0.350 = 1.00 mol N0.350

O = 0.700 = 2.00 mol O0.350

Empirical Formula = NO2

Empirical Formula Mass = 46.01 g/mol

Example:

Molecular formulaMolar Mass = 92.0 g/mol = 2.00Emp. Formula Mass 46.01 g/mol

Molecular Formula = 2 x Emp. Formula = N2O4

A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?

A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?

g C – (48.38/100)*528.39 g = 255.64 g

g H – (8.12/100)*528.39 g = 42.91 g

g O – (43.5/100)*528.39 g = 229.85 g

mole C - 255.64 g * ( 1 mole ) = 21.29 mol 12.01

gmole H – 42.91 g * ( 1 mole ) = 42.49 mol 1.01 g

mole O – 229.85 g * ( 1 mole ) = 14.37 mol16.00 g

A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?

From last slide: 21.29 mol C, 42.49 mol H, 14.27 mol O

C – 21.29/14.27 = 1.49

H – 42.49/14.27 = 2.98 (esentially 3)

O – 14.27/14.27 = 1.00

C – 1.49 x 2 = 3

H – 3 x 2 = 6

O – 1 x 2 = 2

C3H6O2

A 528.39 g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~222.25 g/mol. What is its molecular formula?

From last slide: Empirical formula = C3H6O2

“EFM” = 74.09

Molar mass = 222.24 = ~3

EFM 74.09

3(C3H6O2) = C9H18O6