Best examples of identifying unknown substances with percent composition

Real examples of identifying unknown substances with percent composition

Before any theory, let’s start with situations where percent composition really decides what a substance is. These are the kinds of examples of identifying unknown substances with percent composition that show up in classrooms, quality control labs, and forensic work.

Imagine you’re given a white powder with no label. It could be baking soda, washing soda, table salt, or something less friendly. You burn it, dissolve it, weigh it, and from that data you calculate what percent of the mass is carbon, hydrogen, oxygen, sodium, or chlorine. From those percentages, you build an empirical formula. From the empirical formula, you match it to a real compound. That basic storyline repeats constantly in analytical chemistry.

Let’s walk through several real examples and build up the method as we go.

Classic lab example of identifying an unknown hydrate

A favorite classroom example of using percent composition involves hydrates—ionic salts that trap water molecules in their crystal structure, like CuSO₄·5H₂O.

You’re handed a pale blue crystal labeled only as “Unknown Hydrate of Copper(II) Sulfate.” Your job is to figure out how many water molecules are attached.

You heat the sample to drive off the water and track the mass.

• Mass of crucible + hydrate before heating: 25.63 g
• Mass of crucible + anhydrous salt after heating: 24.79 g
• Mass of water lost: 0.84 g

Suppose the empty crucible has a mass of 23.79 g.

• Mass of hydrate: 25.63 − 23.79 = 1.84 g
• Mass of anhydrous CuSO₄: 24.79 − 23.79 = 1.00 g
• Mass of water: 0.84 g

Now you calculate percent composition by mass:

• % H₂O = (0.84 g ÷ 1.84 g) × 100 ≈ 45.7%
• % CuSO₄ = (1.00 g ÷ 1.84 g) × 100 ≈ 54.3%

Convert masses to moles:

• Moles of CuSO₄ = 1.00 g ÷ 159.6 g/mol ≈ 0.00627 mol
• Moles of H₂O = 0.84 g ÷ 18.0 g/mol ≈ 0.0467 mol

Divide by the smaller value:

• CuSO₄: 0.00627 ÷ 0.00627 ≈ 1
• H₂O: 0.0467 ÷ 0.00627 ≈ 7.45 ≈ 7 (within experimental error)

So the empirical formula is roughly CuSO₄·7H₂O. Your percent composition data has identified the unknown hydrate.

This is one of the best examples of identifying unknown substances with percent composition because you can see every step: raw mass data → percent composition → mole ratio → identity.

Food chemistry: checking a “mystery sweetener” with percent composition

Percent composition isn’t just for inorganic salts. It’s heavily used in food chemistry, especially when verifying that an ingredient is what the supplier claims.

Say a manufacturer receives a white crystalline sweetener that’s supposed to be sucrose (C₁₂H₂₂O₁₁). A quality control chemist combusts a small sample and measures the carbon dioxide and water produced. From those values, they back-calculate the percent composition of carbon, hydrogen, and oxygen in the compound.

Suppose the analysis of an unknown sweetener gives:

• 42.1% carbon
• 6.4% hydrogen
• 51.5% oxygen (by difference)

Now compare to theoretical percent composition of sucrose:

• Molar mass of C₁₂H₂₂O₁₁ ≈ 342.3 g/mol
• Carbon: (12 × 12.01) ÷ 342.3 ≈ 42.1%
• Hydrogen: (22 × 1.008) ÷ 342.3 ≈ 6.5%
• Oxygen: (11 × 16.00) ÷ 342.3 ≈ 51.4%

The numbers line up almost perfectly. This is a textbook example of identifying an unknown substance with percent composition: the experimental percentages match the theoretical ones within normal lab error, so the sweetener is almost certainly sucrose and not, say, lactose or a sugar alcohol.

For context, food and nutrition labels often rely on related analytical methods. The FDA’s resources on food analysis and labeling give a sense of how seriously this is taken in industry: https://www.fda.gov/food.

Environmental chemistry: identifying a pollutant salt in water

Environmental labs routinely receive water samples with “something dissolved” in them. One example of identifying unknown substances with percent composition comes from analyzing an unknown sulfate salt in polluted groundwater.

A sample is evaporated to dryness, leaving a white solid. Further tests show it contains only a metal cation and sulfate (SO₄²⁻). The lab measures the percent composition of the solid and finds:

• 46.6% metal (M)
• 53.4% sulfate (SO₄)

You assume the formula is MSO₄. The molar mass of SO₄ is about 96.1 g/mol. Let the molar mass of M be x.

Percent by mass of M in MSO₄:

• %M = x ÷ (x + 96.1) × 100

Set this equal to the experimental value:

• x ÷ (x + 96.1) × 100 ≈ 46.6

Solve for x:

• x ≈ 0.466(x + 96.1)
• x ≈ 0.466x + 44.8
• x − 0.466x ≈ 44.8
• 0.534x ≈ 44.8
• x ≈ 83.9 g/mol

A metal with atomic mass near 84 g/mol points you toward strontium (Sr, about 87.6 g/mol) or krypton (a gas, so not relevant here). Considering typical groundwater contaminants, SrSO₄ is a reasonable match.

In real environmental work, this kind of percent composition calculation is paired with instrumental methods like ICP-MS or ion chromatography, but the logic is the same: use the mass percentages to narrow down the identity. Agencies like the U.S. Geological Survey describe broader water analysis strategies here: https://www.usgs.gov/special-topics/water-science-school.

Forensic-style example: matching a mystery white powder

Forensic labs love percent composition as a first-pass screening tool. Picture this realistic example of identifying an unknown substance with percent composition during a mock crime lab exercise.

You’re given an unlabeled white powder seized at a scene. Preliminary tests suggest it’s an ionic compound containing sodium and possibly carbonate or chloride. After careful analysis, you determine the mass percent composition:

• 27.4% sodium
• 14.3% carbon
• 58.3% oxygen

Convert to moles (assume a 100 g sample for easy math):

• Na: 27.4 g ÷ 22.99 g/mol ≈ 1.19 mol
• C: 14.3 g ÷ 12.01 g/mol ≈ 1.19 mol
• O: 58.3 g ÷ 16.00 g/mol ≈ 3.64 mol

Divide all by the smallest (1.19):

• Na: 1.19 ÷ 1.19 ≈ 1
• C: 1.19 ÷ 1.19 ≈ 1
• O: 3.64 ÷ 1.19 ≈ 3.06 ≈ 3

Empirical formula: NaCO₃. In reality, sodium carbonate is Na₂CO₃, so you suspect rounding or slight experimental error. If you scale the ratio to whole numbers, you get 2:1:3, which matches Na₂CO₃.

You compare this to theoretical percent composition of Na₂CO₃:

• Molar mass Na₂CO₃ ≈ 105.99 g/mol
• %Na: (2 × 22.99) ÷ 105.99 ≈ 43.4%
• %C: 12.01 ÷ 105.99 ≈ 11.3%
• %O: (3 × 16.00) ÷ 105.99 ≈ 45.3%

If your measured values are reasonably close after correcting for moisture or impurities, you can argue the unknown is sodium carbonate. This is one of the best examples of identifying unknown substances with percent composition for teaching, because students can easily compare it to alternatives like NaCl or NaHCO₃ and see how the percentages differ.

Pharmaceutical quality control: verifying a generic drug salt

Drug manufacturers must confirm that the active ingredient in a generic medication matches the branded version. Percent composition is part of that toolkit.

Consider an unknown sample claimed to be lidocaine hydrochloride, an anesthetic often used in creams and injections. Its formula is C₁₄H₂₂N₂O·HCl. A lab determines the percent composition of chlorine in the sample.

Suppose the experimental result is 10.8% Cl by mass.

Theoretical calculation:

• Molar mass of C₁₄H₂₂N₂O ≈ 234.34 g/mol
• Add HCl: +36.46 g/mol → total ≈ 270.80 g/mol
• Percent Cl = 35.45 ÷ 270.80 × 100 ≈ 13.1%

If the unknown sample’s chlorine percentage is significantly lower (10.8% vs. 13.1%), that suggests either:

• It’s not lidocaine hydrochloride, or
• The sample is impure or mixed with filler.

In real pharmaceutical analysis, labs combine percent composition with sophisticated instruments (HPLC, mass spectrometry). But the core logic is the same as in simpler examples of identifying unknown substances with percent composition you see in class: compare experimental percentages to theoretical values and decide whether they match.

Organizations like the U.S. Pharmacopeia (https://www.usp.org) set detailed standards for this kind of testing, and the FDA monitors compliance.

Combustion analysis: organic unknowns and empirical formulas

If you’ve taken a college-level general chemistry course, you’ve almost certainly seen examples of identifying unknown substances with percent composition using combustion analysis of organic compounds.

Picture an unknown organic compound containing only C, H, and O. A 0.250 g sample is burned completely in oxygen, producing 0.550 g of CO₂ and 0.225 g of H₂O.

From CO₂:

• Moles CO₂ = 0.550 g ÷ 44.01 g/mol ≈ 0.0125 mol
• Each mole of CO₂ contains 1 mol of C → 0.0125 mol C
• Mass of C = 0.0125 mol × 12.01 g/mol ≈ 0.150 g

From H₂O:

• Moles H₂O = 0.225 g ÷ 18.02 g/mol ≈ 0.0125 mol
• Each mole of H₂O contains 2 mol H → 0.0250 mol H
• Mass of H = 0.0250 mol × 1.008 g/mol ≈ 0.0252 g

Mass of O in the original compound is by difference:

• Total mass = 0.250 g
• Mass O = 0.250 − (0.150 + 0.0252) ≈ 0.0748 g

Convert to moles:

• C: 0.150 g ÷ 12.01 ≈ 0.0125 mol
• H: 0.0252 g ÷ 1.008 ≈ 0.0250 mol
• O: 0.0748 g ÷ 16.00 ≈ 0.00468 mol

Divide by the smallest (0.00468):

• C: 0.0125 ÷ 0.00468 ≈ 2.67
• H: 0.0250 ÷ 0.00468 ≈ 5.34
• O: 0.00468 ÷ 0.00468 ≈ 1

Multiply all by 3 to get whole numbers:

• C: 8
• H: 16
• O: 3

Empirical formula: C₈H₁₆O₃.

At this point, you’d compare to known compounds with that empirical formula and use other data (boiling point, NMR, IR) to finish the identification. But the percent composition step—turning mass data into an empirical formula—is the backbone of many real examples in organic analysis.

For a deeper dive into combustion and elemental analysis in research labs, the National Institute of Standards and Technology (NIST) has technical resources at https://www.nist.gov.

Modern trends (2024–2025): where percent composition still matters

In 2024–2025, chemistry labs are full of high-end instruments: mass spectrometers, NMR machines, and high-resolution chromatographs. So why are students still drilled on examples of identifying unknown substances with percent composition?

Because the logic behind percent composition is baked into modern methods:

• Elemental analyzers still report mass percentages of C, H, N, S, and sometimes O. Chemists check whether those match the theoretical percent composition of the compound they think they made.
• Environmental and materials labs use percent composition to validate sensor data and calibrate instruments.
• Battery, semiconductor, and catalyst research often starts with verifying that a synthesized material has the correct elemental ratios—again, a percent composition problem at its core.

So when you practice a classroom example of identifying an unknown substance with percent composition, you’re basically training the same reasoning skills that chemists use before they trust any fancy instrument.

FAQ: examples of identifying unknown substances with percent composition

Q: Can you give a simple classroom example of identifying an unknown substance with percent composition?
Yes. A classic one is heating an unknown hydrate like “MgSO₄·xH₂O,” measuring how much mass is lost as water, and then using percent composition to find x. Once you know x, you can name the compound (for example, magnesium sulfate heptahydrate) and say you’ve identified the unknown.

Q: How many elements do I need data for to identify an unknown using percent composition?
You usually need percent composition for all elements present in significant amounts. In many examples of identifying unknown substances with percent composition, that means three elements (like C, H, O) or four (like C, H, N, O). If you’re missing one, you can sometimes get it by difference (100% minus the others), but that only works cleanly if you know which elements are present.

Q: What’s the difference between empirical formula and molecular formula in these examples?
The empirical formula shows the simplest whole-number ratio of atoms. The molecular formula shows the actual number of each atom in a molecule. Percent composition only gives you the empirical formula. To get the molecular formula, you also need the molar mass. Many real examples of identifying unknown substances with percent composition stop at the empirical formula, then use molar mass from another experiment to scale up to the molecular formula.

Q: Are there medical or biological examples of identifying unknown substances with percent composition?
Yes. In biochemistry and medical research, percent composition is used to check the identity and purity of small molecules and to verify the elemental makeup of new drug candidates. While hospitals rely on faster clinical tests for day-to-day work, research labs at places like the NIH (https://www.nih.gov) still use elemental analysis as a supporting tool.

Q: Why do textbooks keep using the same few examples of identifying unknown substances with percent composition?
Because they work. Hydrates, simple ionic salts, and small organic molecules keep the math manageable while still mirroring real lab situations. Once you’re comfortable with those, you can tackle more complicated cases—like mixed samples, trace metals, or materials with large unit cells—using the same reasoning.

Percent composition might look like just another formula problem on paper, but the examples of identifying unknown substances with percent composition you’ve seen here are exactly how chemists think when the label is missing and the stakes are real.