Empirical and Molecular Formula

Empirical and Molecular Formula

High School

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Learning Objectives

• Define and differentiate between empirical and molecular formulas.

• Determine the empirical formula from percent composition or mass data.

• Calculate the molecular formula from the empirical formula and molar mass.

• Explain the relationship between empirical and molecular formulas.

Key Vocabulary

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Molecular Formula

A formula showing the actual number and kinds of atoms in one molecule of a compound.

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Empirical Formula

A formula with the lowest whole-number ratio of elements that are present in a compound.

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Ionic Compound Formula

These formulas are always empirical, representing the simplest whole-number ratio of ions in the compound.

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Molar Mass

The mass of one mole of a substance, which is typically expressed in grams per mole (g/mol).

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Empirical vs. Molecular Formulas

• A molecular formula shows the actual atoms in a compound, like C₆H₆.

• An empirical formula is the simplest ratio, so for benzene (C₆H₆) it is CH.

• Ionic compounds like NaCl and MgCl₂ always use empirical formulas.

• Molecular formulas can be empirical (H₂O) or multiples like glucose (C₆H₁₂O₆).

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Solved Example 1
The molecular formula for octane, a component of gasoline, is C₈H₁₈. What is the empirical formula for octane?

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Step 1: Analyze and Sketch the Problem

• Goal: Find the empirical formula for octane.

• Knowns: The molecular formula is C₈H₁₈.

• Unknown: The empirical formula.

• Formula: Find the greatest common factor (GCF) of the subscripts and divide them by it to get the simplest whole-number ratio.

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Solved Example 1
The molecular formula for octane, a component of gasoline, is C₈H₁₈. What is the empirical formula for octane?

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Step 2: Solve for the Unknown

• The subscripts in the molecular formula C₈H₁₈ are 8 for carbon and 18 for hydrogen. The greatest common factor (GCF) of these subscripts is 2.

• Divide each subscript by the GCF (2) to get the simplest whole-number ratio. Carbon: 8 ÷ 2 = 4. Hydrogen: 18 ÷ 2 = 9. The empirical formula is C₄H₉.

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Solved Example 1
The molecular formula for octane, a component of gasoline, is C₈H₁₈. What is the empirical formula for octane?

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Step 3: Evaluate the Answer

• The new subscripts are 4 and 9, and their greatest common factor (GCF) is 1.

• Since the GCF is 1, the formula C₄H₉ is in its simplest form, which confirms the answer is correct.

Multiple Choice

The molecular formula for glucose is C₆H₁₂O₆. What is its empirical formula?

• CH₂O

• C₂H₄O₂

• C₃H₆O₃

• C₆H₁₂O₆

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Calculating the Empirical Formula

• Follow the rhyme: percent to mass, mass to moles, divide by small, multiply 'til whole.

• ​Assume a 100g sample to convert each element's percentage directly into grams.

• Divide the moles of each element by the smallest mole value for the ratio.

• Multiply non-whole number ratios to get whole numbers for the final subscripts.

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Solved Example 2
A compound is found to contain 25.9% nitrogen and 74.1% oxygen by mass. What is the empirical formula of the compound?

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Step 1: Analyze and Sketch the Problem

• Goal: Determine the empirical formula.

• Knowns: The compound is 25.9% Nitrogen (N) and 74.1% Oxygen (O).

• Unknown: The simplest whole number ratio of atoms in the compound.

• Process: Percent to mass, mass to moles, divide by small, multiply 'til whole.

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Solved Example 2
A compound is found to contain 25.9% nitrogen and 74.1% oxygen by mass. What is the empirical formula of the compound?

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Step 2: Solve for the Unknown

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Solved Example 2
A compound is found to contain 25.9% nitrogen and 74.1% oxygen by mass. What is the empirical formula of the compound?

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Step 3: Evaluate the Answer

• The final subscripts (2 and 5) are whole numbers, and the process correctly converted percentages to a whole number mole ratio.

• The answer N₂O₅ is a valid empirical formula.

Multiple Choice

After converting the masses of elements in a compound to moles, you find there are 2.5 moles of element X and 5.0 moles of element Y. What is the correct next step?

• Multiply both mole values by 2 to get whole numbers

• Write the formula as X_2.5Y_5.0

• Divide both mole values by the smaller value, 2.5

• Convert the moles back to grams to verify the masses

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Finding the Molecular Formula

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Solved Example 3
A compound has an empirical formula of CH₂O and a molecular mass of 180.18 g/mol. What is its molecular formula? (Atomic masses: C=12.01, H=1.01, O=16.00)

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Step 1: Analyze and Sketch the Problem

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Solved Example 3
A compound has an empirical formula of CH₂O and a molecular mass of 180.18 g/mol. What is its molecular formula? (Atomic masses: C=12.01, H=1.01, O=16.00)

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Step 2: Solve for the Unknown

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Solved Example 3
A compound has an empirical formula of CH₂O and a molecular mass of 180.18 g/mol. What is its molecular formula? (Atomic masses: C=12.01, H=1.01, O=16.00)

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Step 3: Evaluate the Answer

• Calculate the molar mass of the resulting molecular formula, C₆H₁₂O₆: (6 × 12.01) + (12 × 1.01) + (6 × 16.00) = 180.18 g/mol.

• The calculated molar mass matches the given molecular mass. The answer is correct and logical.

Multiple Choice

A compound has an empirical formula of P₂O₅ and an empirical formula mass of ~142 g/mol. If its molecular mass is ~284 g/mol, what is the multiplier 'n' used to find the molecular formula?

• 4

• 1

• 2

• 3

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Common Misconceptions

Misconception	Correction
The empirical and molecular formulas are always different.	They can be the same, like for water (H2O).
A molecular formula is always a multiple of its empirical formula.	The multiplier can be 1, making the formulas identical.
Molecular formula can be found using only percent composition.	Molar mass is also needed to find the molecular formula.

Multiple Choice

A compound is analyzed and found to contain 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. What is its empirical formula?

• CHO

• CHO₂

• C₂H₄O₂

• CH₂O

Multiple Choice

Which of the following pairs of chemical formulas share the same empirical formula?

• H₂O and H₂O₂

• N₂O₄ and N₂O₅

• CO and CO₂

• C₂H₂ and C₆H₆

Multiple Choice

A compound's empirical formula is C₂H₃, and its molar mass is approximately 54 g/mol. Analyze this information to determine its molecular formula.

• C₄H₆

• C₈H₁₂

• C₂H₃

• C₆H₉

Multiple Choice

A sample contains 72.2g of Magnesium (Mg) and 27.8g of Nitrogen (N), and its formula mass is found to be ~100.9 g/mol. Determine both the empirical and molecular formulas.

• EF=MgN, MF=Mg₂N₂

• EF=MgN₂, MF=MgN₂

• EF=Mg₂N₃, MF=Mg₂N₃

• EF=Mg₃N₂, MF=Mg₃N₂

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Summary

• Molecular formulas show true atoms; empirical formulas show the simplest ratio.

• To find the empirical formula: percent to mass, mass to moles, and divide by small.

• The molecular formula is a whole-number multiple of the empirical formula.

• Find the molecular formula using the empirical formula and the compound's molar mass.

Poll

On a scale of 1-4, how confident are you about calculating empirical and molecular formulas?

• 1

• 2

• 3

• 4