Which Two Samples Contain The Same Number Of Molecules

You're staring at a multiple-choice question. Four samples. Different masses. Different substances. And the prompt asks: which two contain the same number of molecules?

Your brain freezes. You know it has something to do with moles. Think about it: maybe Avogadro's number. But the numbers on the page — 18 grams here, 44 grams there — they don't match. So how can the molecule count be the same?

Here's the short version: mass doesn't determine molecule count. Moles do. And once you see the pattern, these questions stop being traps and start being free points.

What Is the Mole Concept (And Why It's the Key)

Chemists don't count molecules one by one. So naturally, that would take longer than the age of the universe. Instead, they use the mole — a counting unit, just like a dozen means 12, except a mole means 6.022 × 10²³.

That number is Avogadro's constant. Day to day, it's huge. But the idea is simple: one mole of any substance contains exactly the same number of particles — atoms, molecules, ions, formula units, whatever you're counting.

So if Sample A is 1 mole of water and Sample B is 1 mole of carbon dioxide, they contain the exact same number of molecules*. Even though their masses are completely different (18 g vs 44 g). Even though their volumes, densities, and boiling points have nothing in common.

The mole is the bridge. It translates between the microscopic world (molecules) and the macroscopic world (grams on a balance).

The Formula That Solves Every Version of This Problem

You don't need to memorize Avogadro's number. You need to memorize this:

moles = mass (g) ÷ molar mass (g/mol)

Once you have moles for each sample, you just compare. Same moles = same number of molecules. Different moles = different number of molecules. That said, that's it. The rest is arithmetic.

Why This Question Shows Up Everywhere

High school chemistry. Now, college general chem. And aP Chemistry. The "which two samples" question appears on almost every exam covering the mole concept because it tests the one thing* students consistently misunderstand: mass ≠ molecule count.

Students see "18 g H₂O" and "44 g CO₂" and think: different masses, so different molecule counts.They see "2 g H₂" and "32 g O₂" and think: way different masses, definitely different counts.Now, * Wrong. * Also wrong — both are 1 mole.

The question exists to catch that exact mistake. And it works. A lot.

But once you internalize the logic, you start seeing the pattern in every* version:

• 18 g H₂O vs 44 g CO₂ vs 2 g H₂ vs 32 g O₂ → all 1 mole → all same molecule count
• 9 g H₂O vs 22 g CO₂ vs 1 g H₂ vs 16 g O₂ → all 0.5 mole → all same molecule count
• 36 g H₂O vs 88 g CO₂ → both 2 moles → same molecule count

The substances change. Plus, the masses change. The ratio* stays the same.

How to Solve It Step by Step

Let's walk through a real example. You'll see this exact structure on tests.

Question: Which two of the following samples contain the same number of molecules?
A) 18 g H₂O
B) 44 g CO₂
C) 2 g H₂
D) 16 g O₂

Step 1: Find the molar mass of each substance

This is where periodic tables earn their keep.

• H₂O: 2(1.008) + 16.00 = 18.016 g/mol (call it 18)
• CO₂: 12.01 + 2(16.00) = 44.01 g/mol (call it 44)
• H₂: 2(1.008) = 2.016 g/mol (call it 2)
• O₂: 2(16.00) = 32.00 g/mol

Step 2: Calculate moles for each sample

• A) 18 g ÷ 18 g/mol = 1 mol
• B) 44 g ÷ 44 g/mol = 1 mol
• C) 2 g ÷ 2 g/mol = 1 mol
• D) 16 g ÷ 32 g/mol = 0.5 mol

Step 3: Compare

A, B, and C all have 1 mole. 5 mole.
Which means d has 0. So A, B, and C contain the same number of molecules. D does not.

If the question asks "which two," any pair from A/B/C works. If it asks "which sample is different," it's D.

That's the whole method. No scientific notation. No Avogadro's number required. Just molar mass and division.

Common Mistakes / What Most People Get Wrong

Mistake 1: Comparing Masses Directly

"I see 18 g and 44 g. They're not equal. So the molecule counts aren't equal.

This is the trap. On the flip side, the question wants* you to compare masses. Here's the thing — don't. Compare moles.

Mistake 2: Forgetting Diatomic Elements

Hydrogen is H₂. Nitrogen is N₂. Oxygen is O₂. Fluorine, chlorine, bromine, iodine — all diatomic in their standard state.

If you calculate molar mass of hydrogen as 1 g/mol instead of 2, you'll get 2 moles for 2 g H₂. That's double the real answer. Same for O₂, N₂, Cl₂, etc.

Mistake 3: Using Atomic Mass Instead of Molecular Mass

CO₂ isn't 12 + 16 = 28. It's 12 + 32 = 44.
H₂O isn't 1 + 16 = 17. It's 2 + 16 = 18.

Always count all atoms in the formula. Subscripts matter.

Mistake 4: Rounding Too Early

Molar masses aren't always whole numbers.
That said, cl₂ = 70. 90 g/mol, not 71.
Fe₂O₃ = 159.69 g/mol, not 160.

If the given masses are precise (like 70.And 90 g Cl₂), use precise molar masses. If they're rounded (71 g Cl₂), rounded molar masses are fine. Match the precision of the problem.

Mistake 5: Confusing "Number of Molecules" with "Number of Atoms"

1 mole of H₂O has 6.022 × 10²³ molecules*.
But it has 3 ×

Mistake 5 – Mixing Up Molecules and Atoms

When a question asks for the number of molecules, you stop after counting whole formula units.
When it asks for the number of atoms, you must multiply the molecule count by the number of atoms that each molecule contains.

Substance	Atoms per molecule	Atoms in 1 mol of substance
H₂O	3 (2 H + 1 O)	3 × 6.022 × 10²³ = 1.So naturally, 022 × 10²³ = 1. That's why 807 × 10²⁴
O₂	2 (2 O)	2 × 6. Also, 807 × 10²⁴
CO₂	3 (1 C + 2 O)	3 × 6. In practice, 022 × 10²³ = 1. 204 × 10²⁴
H₂	2 (2 H)	2 × 6.022 × 10²³ = 1.

So, 1 mol of H₂O contains 6.022 × 10²³ molecules but 1.807 × 10²⁴ atoms**.
If a test asks which sample has the same number of oxygen atoms, you would compare the product of (moles × atoms‑per‑molecule) for each option, not just the mole count. Worth keeping that in mind.

---

Practice Problem

Question: Which two of the following samples contain the same number of oxygen atoms?

A) 16 g O₂  B) 44 g CO₂  C) 18 g H₂O  D) 8 g O₂

Hint:* First find moles of each sample, then multiply by the number of O atoms per molecule.

---

Solution

Sample	Molar mass (g mol⁻¹)	Mass (g)	Moles	O atoms per molecule	Total O atoms (relative)
A) O₂	32.Practically speaking, 00	16	0. 50	2	0.On the flip side, 50 × 2 = 1. 0
B) CO₂	44.01	44	1.00	2	1.00 × 2 = 2.0
C) H₂O	18.In practice, 02	18	1. 00	1	1.00 × 1 = 1.

| D) 8 g O₂ | 32.00 | 8 | 0.25 | 2 | 0.25 × 2 = 0.

Comparison of total oxygen atoms

• A) 1.0  C) 1.0  → Equal
• B) 2.0  D) 0.5  → Not equal

Hence, samples A (16 g O₂) and C (18 g H₂O) contain the same number of oxygen atoms.

---

Why the answer matters

The trick is to remember that atoms* are the fundamental units you’re counting, whereas molecules* are the entities you convert between mass and moles. Even if two samples contain the same number of moles, the number of atoms they hold can differ dramatically because each molecule may house a different count of the element in question.

Continue exploring with our guides on what does the roman numeral c mean and volume of a cone with diameter.

---

Quick‑reference cheat sheet

Common pitfall	What to do instead
Assuming 1 g = 1 mol	Divide the mass by the molecular* mass of the compound.
Using atomic mass for a molecule	Add the masses of all atoms in the formula. But
Rounding too early	Keep the significant figures that match the given data until the final step.
Confusing molecules with atoms	Multiply the mole number by the number of atoms per molecule when the problem asks for atoms.

---

Take‑away

1. Always convert mass → moles using the correct molecular mass.
2. Count every atom in the formula; subscripts matter.
3. Keep track of what the question actually requests—moles, molecules, or atoms.
4. Maintain precision throughout the calculation.

By internalizing these habits, the “double‑mole” and “half‑mole” errors vanish, and you’ll find that stoichiometry becomes a matter of plugging numbers into well‑defined relationships rather than a source of frustration.

Happy calculating!

Test Your Understanding

Before moving on, try these quick variations without looking back at the worked example. The goal is to build the reflex of mass → moles → atoms* automatically.

1. Which sample contains the greatest number of oxygen atoms?

   • 32 g O₂
   • 44 g CO₂
   • 36 g H₂O

2. How many grams of CO would contain the same number of oxygen atoms as 24 g of O₃?
   (Molar masses: CO = 28.01 g mol⁻¹, O₃ = 48.00 g mol⁻¹)

3. True or False: “Because 18 g of H₂O and 16 g of O₂ both represent 1 mol of oxygen atoms*, they also contain the same number of hydrogen atoms.”

Answers: 1) 44 g CO₂ (2 mol O atoms); 2) 56 g CO; 3) False—water has 2 mol H atoms, O₂ has none.*

---

Connecting to the Bigger Picture

This problem is a microcosm of every stoichiometry calculation you will meet:

Step in this problem	General stoichiometry equivalent
Mass → moles (divide by molar mass)	Reactant mass → moles of reactant
Moles × subscript → atom count	Mole ratio from balanced equation → moles of product
Compare atom counts	Compare actual yield to theoretical yield

Mastering the “atom-accounting” mindset here means you are already doing limiting-reagent problems, percent-yield calculations, and gas-law conversions—you’re just changing the units at the end.

---

Final Thoughts

Chemistry rewards systematic thinking over memorization. When a question asks for atoms*, ions*, or formula units*, the workflow never changes:

1. Write what you know (mass, volume, concentration).
2. Convert to moles using the appropriate constant (molar mass, molar volume, Avogadro’s number).
3. Use the chemical formula or balanced equation as the bridge to the requested particle.
4. Check that your answer’s units match the question—if they don’t, you’ve missed a conversion factor.

The four samples in this exercise were deliberately chosen to look similar (round numbers, familiar compounds) but to test whether you respect the subscripts. That discipline—pausing to count the atoms inside* each molecule—is what separates a correct answer from a plausible guess.

Keep a periodic table handy, keep your units visible, and keep practicing. The patterns will soon feel as natural as reading a sentence.

Happy calculating—and may your significant figures always be on your side!

Extending the Workflow: From Simple Atoms to Complex Molecules

So far we’ve focused on counting individual atoms within a single compound. In real laboratory work, the same three‑step rhythm appears, but the “bridge” (the chemical formula or balanced equation) is often far more involved. Here are a few ways to keep the process smooth when you encounter those more elaborate scenarios.

1. Multi‑component Species

When the target particle is a polyatomic ion or a coordination complex, the subscript pattern can be hidden inside parentheses.

• Example: How many oxygen atoms are present in 5.0 g of calcium carbonate, CaCO₃?
  1. Mass → moles: (5.0\ \text{g} ÷ 100.09\ \text{g mol}^{-1} = 0.050\ \text{mol}).
  2. Moles × subscript: Each formula unit contains 3 O atoms, so (0.050\ \text{mol} × 3 = 0.150\ \text{mol O}).
  3. Moles → atoms: (0.150\ \text{mol} × 6.022×10^{23} = 9.0×10^{22}) O atoms.

The key is to treat the entire formula as a “mini‑equation” that tells you how many of each atom are bundled together.

2. Balanced‑Equation Stoichiometry

When a reaction is involved, the bridge is the balanced chemical equation. The same conversion factors (molar mass, Avogadro’s number) still apply, but you must first translate masses of reactants into moles of product using the mole ratios.

• Example: From the combustion of methane (CH₄ + 2 O₂ → CO₂ + 2 H₂O), how many water molecules are produced from 8.0 g of CH₄?
  1. Mass → moles of CH₄: (8.0\ \text{g} ÷ 16.04\ \text{g mol}^{-1} = 0.50\ \text{mol}).
  2. Mole ratio: 1 mol CH₄ gives 2 mol H₂O, so (0.50\ \text{mol} × 2 = 1.0\ \text{mol H₂O}).
  3. Moles → molecules: (1.0\ \text{mol} × 6.022×10^{23} = 6.0×10^{23}) water molecules.

Notice how the subscript counting from the previous section is now replaced by the stoichiometric coefficients from the balanced equation.

3. Gas‑Phase Conversions

If the problem involves gases, the molar volume (22.4 L mol⁻¹ at STP) can be inserted between mass and moles, or between moles and particles, depending on what the question asks.

• Example: How many nitrogen atoms are present in 5.6 L of N₂ gas at STP?
  1. Volume → moles: (5.6\ \text{L} ÷ 22.4\ \text{L mol}^{-1} = 0.25\ \text{mol N₂}).
  2. Moles × subscript: Each N₂ molecule holds 2 N atoms, so (0.25\ \text{mol} × 2 = 0.50\ \text{mol N}).
  3. Moles → atoms: (0.50\ \text{mol} × 6.022×10^{23} = 3.0×10^{23}) nitrogen atoms.

4. Practical Tips to Avoid Common Slip‑Ups

Pitfall	Quick Check
Mis‑reading subscripts (e.g., thinking CO₂ has 2 O atoms per molecule)	Write the formula, circle each subscript, count them before moving on.
Forgetting Avogadro’s number when the answer is expected in particles	Ask yourself: “Do I need atoms, molecules, or formula units?” If yes, multiply by (6.022×10^{23}).
Unit mismatches (grams vs. moles vs. liters)	Keep a small conversion table handy: molar mass (g mol⁻¹), molar volume (L mol⁻¹),

and Avogadro’s number ((6.Consider this: 022 \times 10^{23}) particles/mol). | Always perform a "sanity check" on your units to ensure they cancel out correctly in your calculation.

Summary of the Conversion Workflow

To master stoichiometry, view every problem as a series of connected bridges. You are rarely converting directly from one unit to another; instead, you are moving through the "Mole Bridge." Whether you are starting with mass, volume, or particle count, your first objective should almost always be to convert that value into moles. Once you have the moles, you can use the chemical formula (subscripts) or the balanced equation (coefficients) to pivot to the desired substance, and finally, use Avogadro's number to reach the final count of atoms or molecules.

Conclusion

Stoichiometry is the mathematical backbone of chemistry. While the individual steps—calculating molar mass, applying mole ratios, and using Avogadro’s number—may seem simple in isolation, their true power lies in their ability to link the macroscopic world we can weigh on a scale to the microscopic world of individual atoms. By mastering these conversion pathways and maintaining strict attention to unit consistency, you can predict the outcomes of chemical reactions with precision, turning abstract formulas into tangible, measurable results.