Best examples of empirical formula examples from percent composition

Starting with simple examples of empirical formula from percent composition

Textbook problems may feel artificial, but they’re still the cleanest way to see the logic. Let’s start with one of the classic examples of empirical formula examples from percent composition and then build up to more realistic chemistry.

Example 1: A binary compound of carbon and hydrogen

Suppose an unknown compound is found to be 80.0% carbon and 20.0% hydrogen by mass.

Chem teachers love this one because the math is tidy, but the method you use here is exactly the same method you’ll use on messier, real examples.

Step 1 – Assume a 100 g sample
If the compound is 80.0% C and 20.0% H, then in 100 g you have:

• 80.0 g C
• 20.0 g H

Assuming 100 g keeps the numbers aligned with the percentages.

Step 2 – Convert grams to moles

Using approximate molar masses (from any standard periodic table, for example the one maintained by NIST at nist.gov):

• Carbon: 80.0 g ÷ 12.01 g/mol ≈ 6.66 mol C
• Hydrogen: 20.0 g ÷ 1.008 g/mol ≈ 19.84 mol H

Step 3 – Divide by the smallest mole value

Smallest is about 6.66 mol.

• C: 6.66 ÷ 6.66 ≈ 1.00
• H: 19.84 ÷ 6.66 ≈ 2.98 ≈ 3

Empirical formula: CH₃

That’s your first example of empirical formula from percent composition: a simple hydrocarbon with a 1:3 C:H ratio.

Examples of empirical formula examples from percent composition with three elements

Binary compounds are nice, but most real chemistry involves at least three elements. Let’s look at examples of empirical formula from percent composition that include oxygen and nitrogen, the workhorses of biochemical and environmental chemistry.

Example 2: A nitrogen–oxygen compound (air pollution style)

Imagine a gas sample analyzed from an industrial stack is 30.4% nitrogen and 69.6% oxygen by mass. Regulatory agencies like the U.S. EPA track nitrogen oxides because they contribute to smog and acid rain (see background at epa.gov).

Step 1 – 100 g basis

• 30.4 g N
• 69.6 g O

Step 2 – Convert to moles

• N: 30.4 g ÷ 14.01 g/mol ≈ 2.17 mol
• O: 69.6 g ÷ 16.00 g/mol = 4.35 mol

Step 3 – Divide by the smallest

• N: 2.17 ÷ 2.17 ≈ 1.00
• O: 4.35 ÷ 2.17 ≈ 2.00

Empirical formula: NO₂

This matches nitrogen dioxide, a real-world air pollutant. This is one of the best examples because it connects the math to something you might actually see in an environmental report.

Example 3: A compound containing C, H, and O (food chemistry flavor)

Suppose a carbohydrate-like compound has the following percent composition:

• 40.0% carbon
• 6.7% hydrogen
• 53.3% oxygen

These numbers are similar to some simple sugars and sugar alcohols discussed in nutrition chemistry (for background on carbohydrates, see introductory material from Harvard’s nutrition resources at hsph.harvard.edu).

Step 1 – 100 g sample

• 40.0 g C
• 6.7 g H
• 53.3 g O

Step 2 – Convert to moles

• C: 40.0 ÷ 12.01 ≈ 3.33 mol
• H: 6.7 ÷ 1.008 ≈ 6.65 mol
• O: 53.3 ÷ 16.00 ≈ 3.33 mol

Step 3 – Divide by the smallest

Smallest is about 3.33.

• C: 3.33 ÷ 3.33 ≈ 1.00
• H: 6.65 ÷ 3.33 ≈ 2.00
• O: 3.33 ÷ 3.33 ≈ 1.00

Empirical formula: CH₂O

This is a classic result. A lot of simple carbohydrates have the empirical formula CH₂O, even though their molecular formulas (like C₆H₁₂O₆ for glucose) are multiples of that. It’s one of the most common examples of empirical formula from percent composition you’ll see in general chemistry.

Real examples: empirical formula from percent composition in the lab

So far, these have been clean, rounded numbers. Let’s move into real examples that look more like what you’d get from a lab instrument or a data sheet.

Example 4: Hydrated salt from percent water

Hydrated ionic compounds are a favorite lab exercise. Assume you heat a sample of a copper sulfate hydrate and measure the mass before and after heating:

• Mass of CuSO₄·xH₂O before heating: 2.50 g
• Mass of CuSO₄ after heating: 1.60 g

The lost mass is water:

• Water lost = 2.50 g − 1.60 g = 0.90 g H₂O

Step 1 – Moles of anhydrous CuSO₄

Molar mass of CuSO₄ ≈ 159.6 g/mol.

• Moles CuSO₄ = 1.60 g ÷ 159.6 g/mol ≈ 0.0100 mol

Step 2 – Moles of water

Molar mass H₂O ≈ 18.02 g/mol.

• Moles H₂O = 0.90 g ÷ 18.02 g/mol ≈ 0.0499 mol

Step 3 – Ratio of water to salt

• H₂O : CuSO₄ = 0.0499 : 0.0100 ≈ 4.99 : 1 ≈ 5 : 1

Empirical formula: CuSO₄·5H₂O

Here, the empirical formula applies to the whole hydrate, not just the anhydrous salt. This is one of the best examples of empirical formula examples from percent composition because it matches a real, named compound: copper(II) sulfate pentahydrate, the bright blue solid you’ll see in many teaching labs.

Example 5: Combustion analysis of an organic compound

Modern organic analysis still uses the same logic behind these examples of empirical formula from percent composition, especially in combustion analysis.

Suppose burning 0.500 g of an unknown compound containing C, H, and O produces:

• 0.733 g CO₂
• 0.300 g H₂O

From these masses, labs calculate the percent composition and then the empirical formula.

Step 1 – Moles of C from CO₂

Molar mass CO₂ ≈ 44.01 g/mol.

• Moles CO₂ = 0.733 ÷ 44.01 ≈ 0.0167 mol
• Each mole of CO₂ has 1 mole of C → moles C = 0.0167 mol

Mass of C:

• 0.0167 mol × 12.01 g/mol ≈ 0.200 g C

Step 2 – Moles of H from H₂O

Molar mass H₂O ≈ 18.02 g/mol.

• Moles H₂O = 0.300 ÷ 18.02 ≈ 0.0167 mol
• Each mole of H₂O has 2 moles of H → moles H = 2 × 0.0167 ≈ 0.0334 mol

Mass of H:

• 0.0334 mol × 1.008 g/mol ≈ 0.0337 g H

Step 3 – Mass and moles of O by difference

Total mass of sample = 0.500 g.

Mass of O = 0.500 − (0.200 + 0.0337) ≈ 0.266 g O

Moles of O:

• 0.266 ÷ 16.00 ≈ 0.0166 mol

Step 4 – Mole ratios

• C: 0.0167 mol
• H: 0.0334 mol
• O: 0.0166 mol

Divide by the smallest (≈0.0166):

• C: 0.0167 ÷ 0.0166 ≈ 1.0
• H: 0.0334 ÷ 0.0166 ≈ 2.0
• O: 0.0166 ÷ 0.0166 ≈ 1.0

Empirical formula: CH₂O

Once again, a CH₂O empirical formula pops up, reinforcing why it’s one of the most common examples of empirical formula from percent composition in organic and biochemistry.

More complex examples include halogens and nitrogen

To really test your understanding, you want examples that don’t fall into the usual “carbohydrate” pattern. These next real examples bring in halogens and nitrogen, which show up constantly in pharmaceuticals and industrial chemicals.

Example 6: A chlorine-containing compound

An unknown compound is found to be:

• 24.3% carbon
• 4.1% hydrogen
• 71.6% chlorine

Step 1 – 100 g basis

• C: 24.3 g
• H: 4.1 g
• Cl: 71.6 g

Step 2 – Convert to moles

• C: 24.3 ÷ 12.01 ≈ 2.02 mol
• H: 4.1 ÷ 1.008 ≈ 4.07 mol
• Cl: 71.6 ÷ 35.45 ≈ 2.02 mol

Step 3 – Divide by smallest

Smallest is about 2.02.

• C: 2.02 ÷ 2.02 ≈ 1.0
• H: 4.07 ÷ 2.02 ≈ 2.0
• Cl: 2.02 ÷ 2.02 ≈ 1.0

Empirical formula: CH₂Cl

This is the empirical formula of chloromethane and related compounds. It’s a clear example of empirical formula from percent composition where a halogen dominates the mass percent even though there’s only one Cl atom in the formula.

Example 7: A nitrogen-containing organic compound

Now let’s consider a compound used as a model for amino-containing molecules. Analysis shows:

• 46.7% nitrogen
• 13.3% hydrogen
• 40.0% carbon

Step 1 – 100 g sample

• N: 46.7 g
• H: 13.3 g
• C: 40.0 g

Step 2 – Convert to moles

• N: 46.7 ÷ 14.01 ≈ 3.33 mol
• H: 13.3 ÷ 1.008 ≈ 13.2 mol
• C: 40.0 ÷ 12.01 ≈ 3.33 mol

Step 3 – Divide by the smallest

Smallest is about 3.33.

• N: 3.33 ÷ 3.33 ≈ 1.0
• H: 13.2 ÷ 3.33 ≈ 4.0
• C: 3.33 ÷ 3.33 ≈ 1.0

Empirical formula: CH₄N

This pattern appears in simple amines. Again, the method is identical to all the earlier examples of empirical formula examples from percent composition: assume 100 g, convert to moles, divide by the smallest, and clean up the ratios.

Modern context: why percent composition still matters in 2024–2025

With high-resolution mass spectrometry and advanced NMR everywhere, you might think percent composition is old-fashioned. But even in 2024–2025, the logic behind these examples of empirical formula from percent composition shows up in:

• Pharmaceutical analysis – Early characterization of new drug candidates often starts with elemental analysis (CHN analysis) to check purity and verify empirical formula before moving to more complex structural tools. Many academic labs still send samples to elemental analysis services that report percent C, H, N.
• Environmental monitoring – Regulatory bodies like the U.S. EPA and international agencies publish data on pollutant compositions. Converting reported mass percentages into empirical formulas helps compare field data to known compounds (see EPA resources on air pollutants at epa.gov).
• Material science – Researchers developing new battery materials, catalysts, or polymers often report empirical formulas based on elemental analysis, especially in early-stage work.

If you’re planning to move into any lab-heavy field, being comfortable with these real examples of percent-composition-to-empirical-formula conversions is not just a school exercise; it’s part of how you sanity-check data.

Common pitfalls when working through examples of empirical formula from percent composition

Looking back at all these examples of empirical formula examples from percent composition, a few recurring issues stand out:

Rounding too early
If you round mole values too aggressively, you can miss ratios like 1.5 or 2.5 that need to be multiplied by 2 to get whole numbers. When you see something like 1.49 or 2.51, don’t force it to 1 or 3 instantly. Consider whether multiplying all ratios by 2 or 3 gives a cleaner set of integers.

Forgetting oxygen by difference
In combustion analysis (like Example 5), students often forget that oxygen’s mass is found by subtracting the masses of C and H from the total. This “by difference” approach is standard in real labs and is described in many general chemistry textbooks hosted by universities (for example, open-course materials from MIT or other .edu sources).

Confusing empirical and molecular formulas
Several of the best examples here (CH₂O especially) are empirical formulas that correspond to multiple possible molecular formulas. To go from empirical to molecular, you need the molar mass from experimental data (such as mass spectrometry). The empirical formula alone gives ratios, not exact atom counts.

FAQ: Short answers built around real examples

Q: Can you give a simple example of finding an empirical formula from percent composition?
Yes. A classic example of this process is a compound that is 40.0% C, 6.7% H, and 53.3% O by mass. Assuming 100 g, converting to moles, and normalizing the ratios leads to an empirical formula of CH₂O. This is one of the most widely used examples of empirical formula from percent composition in introductory chemistry.

Q: How do I know if my mole ratios are correct in these examples of empirical formula calculations?
After dividing by the smallest mole value, check whether the numbers are very close to whole numbers or simple fractions like 1.5, 2.0, 2.5, or 3.0. If you get 1.50, 2.00, and 3.00, multiply all by 2 to clear the half. Comparing your result to known patterns (such as CH₂O for carbohydrates or NO₂ for nitrogen oxides) can also help you spot unrealistic answers.

Q: Are there real-world examples of empirical formula from percent composition outside the classroom?
Absolutely. Environmental chemists analyzing air pollutants, pharmaceutical chemists checking the composition of new drug candidates, and materials scientists characterizing new compounds all use percent composition data to confirm empirical formulas. These are not just made-up textbook problems.

Q: What’s the difference between an example of empirical formula and an example of molecular formula?
An empirical formula shows the simplest whole-number ratio of atoms (like CH₂O), while a molecular formula shows the actual number of atoms in a molecule (like C₆H₁₂O₆). Multiple molecular formulas can share the same empirical formula, which is why percent composition alone can’t tell you everything about structure.

Q: Where can I find reliable reference data for atomic masses when working through these examples?
For accurate atomic weights and related data, chemists often use resources from the National Institute of Standards and Technology at nist.gov. University general chemistry pages (.edu) are also reliable for quick periodic tables and worked examples.

If you can comfortably walk through each of these examples of empirical formula examples from percent composition, you’re in good shape for both exams and real lab work. The math is repetitive by design; the more you practice on real examples, the more automatic the process becomes.