Empirical & Molecular FormulasLaw 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.

Empirical FormulaEmpirical Formula is the formula that gives the lowest ratio of atoms in a compound. It does not necessarily tell you the exact number of each type of atom. Example 1: The percent composition of a compound is 69.9 % iron and 30.1% oxygen. What is the empirical formula of a compound?

Example 1: 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 100mO= 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 FenO= m/M = 30.1g/16.00g/mol = 1.88 mol OStep 4: State the Amount RationFe : nO1.25mol : 1.88 mol

Step 5: Calculate lowest whole number ratio1.25mol : 1.88 mol1.25mol 1.25 mol1 : 1.52 : 3

Empirical Formula is Fe2O3When you dont get a whole number, multiply entire ratio by 2, 3, 4 etc. until you get a whole number

Example 2: The percent composition of a compound is 21.6% sodium, 33.3% chlorine, and 45.1% oxygen. What is the empirical formula of the compound?

Step 1: List the given valuesCl=33.3%, Na = 21.6% and O = 45.1%

Step 2: Calculate the mass (m) of each element in a 100g sample.mCl= 33.3 x 100g = 33.3g Cl 100mNa= 21.6 x 100g = 21.6g Na 100mO= 45.1 x 100g = 45.1g O 100

Step 3: Convert Mass (m) into moles (n)nCl= m/M = 33.3g/35.5g/mol = 0.94 mol ClnNa= m/M = 21.6g/23.0g/mol = 0.94 mol NanO= m/M = 45.1g/16.00g/mol = 2.82 mol OStep 4: State the Amount RationFe : nNa : nO0.94mol : 0.94mol : 2.82 mol

Step 5: Calculate lowest whole number ratio0.94mol : 0.94mol : 2.82 mol0.94mol : 0.94mol : 0.94 mol1 : 1: 3

Empirical Formula is NaClO3

Molecular FormulaMolecular 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 formulaThe molar mass of the compound

Molecular Formula- shows the actual number of atomsExample: C6H12O6

Empirical Formula - shows the ratio between atomsExample: 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=CH3OMcompound = 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 CH3OTherefore, the molecular formula is C3H9O3

Molecular formula molar massEmpirical formula molar mass93.12 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 formulaStep 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 samplemC=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 CnH= m/M = 6.67g/1.01g/mol = 6.60 mol HnO= m/M = 53.30g/16.00g/mol = 3.33 mol OStep 4: State the Amount RationC : nH : nO3.33mol : 6.60mol : 3.33 mol

Step 5: Calculate lowest whole number ratio3.33mol : 6.60mol : 3.33 mol3.33mol : 3.33mol : 3.33 mol

1 : 2: 1

Empirical Formula is CH2O

Step 6: Determine the molar mass for the empirical formulaMEmpirical = 12.01g/mol + 2(1.01g/mol) + 16.00g/mol = 30.03 g/molStep 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 mass180.18 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 CnH= m/M = 6.70g/1.01g/mol = 6.65 mol HnO= m/M = 42.6g/16.00g/mol = 2.66 mol OnN= m/M = 18.7g/14.01g/mol = 1.33 mol N

State the Amount RationC : nH : nO : nN2.66mol : 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 formulaMEmpirical = 75.08gDivide 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 mass75.08 g/mol75.08 g/mol

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