Our professor taught us the right process of getting the Empirical and Molecular Formula of the Compounds. Here is the sample problem that she showed us.

The problem is:

Given the following percent composition 26.50% K, 35.37% Cr and 39.10% O, determine the empirical and molecular formula of the compound if the molar mass of the compound is 294 6/mol.

Step 1: State the given.

% mass of K = 26.50%

% mass of Cr = 35.37%

% mass of O = 39.10%

Step 2: Convert percent to grams by copying the percent value and write grams.

% mass of K = 26.50% = 26.50 g.

% mass of Cr = 35.37% = 35.37 g.

% mass of O = 39.10% = 39.10 g.

Step 3: Calculate the number of moles by multiplying the atomic weights to the individual masses.

mol of K = 26.50 g x 1 mol/40.00 g = 0.66 mol

mol of Cr = 35.37 g x 1 mol/52.00 g = 0.68 mol

mol of O = 39.10 g x 1 mol/16.00 g = 2.44 mol

Step 4: Look for the smallest number of mole and use it as a divisor to obtain a whole number ratio.

In the problem, the smallest number of mole is of K which is 0.66 mol.

K = 0.66 mol/ 0.66 mol = 1

Cr = 0.68 mol/0.66 = 1.03

O = 2.44 mol/0.66 mol = 3.69

Empirical Formula is KCrO4

Step 4: Compute for the empirical mass based on the empirical formula, then add the answer.

K = 1 x 40.00 g/mol = 40.00 g/mol

Cr = 1 x 52.00 g/mol = 52.00 g/mol

O = 4 x 16.00 g/mol = 64.00 g/mol

Empirical Mass = 156 g/mol

Step 5: Compute for "n", to solve for Molecular Formula.

n = molar mass/empirical mass

n = 294 g/mol/156 g/mol

n = 1.88

n = 2

Molecular Formula = n(E.F.)

Molecular Formula = 2 (KCrO4)

M. F. = K2Cr2O8

There you go. I hope that I have helped you in some other way in getting the empirical and molecular mass of compounds.

The problem is:

Given the following percent composition 26.50% K, 35.37% Cr and 39.10% O, determine the empirical and molecular formula of the compound if the molar mass of the compound is 294 6/mol.

Step 1: State the given.

% mass of K = 26.50%

% mass of Cr = 35.37%

% mass of O = 39.10%

Step 2: Convert percent to grams by copying the percent value and write grams.

% mass of K = 26.50% = 26.50 g.

% mass of Cr = 35.37% = 35.37 g.

% mass of O = 39.10% = 39.10 g.

Step 3: Calculate the number of moles by multiplying the atomic weights to the individual masses.

mol of K = 26.50 g x 1 mol/40.00 g = 0.66 mol

mol of Cr = 35.37 g x 1 mol/52.00 g = 0.68 mol

mol of O = 39.10 g x 1 mol/16.00 g = 2.44 mol

Step 4: Look for the smallest number of mole and use it as a divisor to obtain a whole number ratio.

In the problem, the smallest number of mole is of K which is 0.66 mol.

K = 0.66 mol/ 0.66 mol = 1

Cr = 0.68 mol/0.66 = 1.03

O = 2.44 mol/0.66 mol = 3.69

Empirical Formula is KCrO4

Step 4: Compute for the empirical mass based on the empirical formula, then add the answer.

K = 1 x 40.00 g/mol = 40.00 g/mol

Cr = 1 x 52.00 g/mol = 52.00 g/mol

O = 4 x 16.00 g/mol = 64.00 g/mol

Empirical Mass = 156 g/mol

Step 5: Compute for "n", to solve for Molecular Formula.

n = molar mass/empirical mass

n = 294 g/mol/156 g/mol

n = 1.88

n = 2

Molecular Formula = n(E.F.)

Molecular Formula = 2 (KCrO4)

M. F. = K2Cr2O8

There you go. I hope that I have helped you in some other way in getting the empirical and molecular mass of compounds.

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