How Many Carbon Atoms Are Contained In 2.8g Of C2h4

How Many Carbon Atoms Are in 2.8 g of C₂H₄?

Quick chemistry question: you've got 2.It's the kind of problem that looks intimidating at first glance, but it breaks down into a few clean steps once you see the logic. Also, 8 grams of ethylene (C₂H₄), and you need to know how many carbon atoms are sitting inside that sample. Here's the full walkthrough, plus the reasoning behind each move so it actually sticks.

What C₂H₄ Actually Is

Ethylene (C₂H₄) is a small hydrocarbon — two carbons, four hydrogens, double-bonded together. But for this problem, you don't really need its history. It's a gas at room temperature, the simplest alkene*, and a major building block in the chemical industry. You just need its formula and its molar mass.

The molar mass of C₂H₄:

• Carbon: 12.008 g/mol × 4 = 4.02 g/mol
• Hydrogen: 1.01 g/mol × 2 = 24.032 g/mol
• Total: roughly **28.

Most textbooks round this to 28 g/mol, and that's good enough for almost every intro-level problem — including this one.

Why the Question Matters

Moles-to-atoms conversions are the bread and butter of stoichiometry. Miss this, and bigger problems (limiting reactants, empirical formulas, solution chemistry) all start to wobble. The question "how many atoms of X are in Y grams of Z" shows up constantly in general chem, AP chem, and really any lab setting where you're scaling a reaction up or down.

It also trains a specific way of thinking: grams → moles → molecules → atoms. Three conversions, each one a single multiplication. Once you've done it a few times, the pattern becomes reflex.

How to Solve It Step by Step

Here's the path from grams of ethylene all the way to individual carbon atoms.

Step 1: Convert Grams of C₂H₄ to Moles

Take the mass you have and divide by the molar mass.

2.8 g ÷ 28 g/mol = 0.10 mol of C₂H₄

That neat 0.1 is the giveaway that this problem was designed for clean numbers. If you get something messier, double-check your molar mass — easy place to slip.

Step 2: Find Moles of Carbon Atoms

Now look at the formula. Each molecule of C₂H₄ contains 2 carbon atoms. So:

0.10 mol C₂H₄ × 2 mol C / 1 mol C₂H₄ = 0.20 mol of carbon atoms

This is the step people sometimes skip. Practically speaking, they jump straight to atoms and forget that the formula tells you a ratio. The subscripts aren't decoration — they tell you exactly how many of each atom are in one molecule (or one mole of molecules).

Step 3: Convert Moles of Carbon to Actual Atoms

One mole of anything contains Avogadro's number of particles: 6.022 × 10²³.

0.20 mol × 6.022 × 10²³ atoms/mol = 1.2044 × 10²³ carbon atoms

Rounded to a sensible number of sig figs (two, matching the 2.Even so, 8 g starting value), you get about 1. 2 × 10²³ carbon atoms.

That's the answer. But the reasoning is more useful than the number, honestly.

The Shortcut Most People Miss

There's a faster way once you've done this a few times. You can combine steps:

Grams → moles of compound → moles of element → atoms of element

Or, in one line:

(grams ÷ molar mass) × (subscript of element) × Avogadro's number

So: (2.8 ÷ 28) × 2 × 6.022 × 10²³ = 1.

Plug it all in at once, but only after you understand what each piece is doing. Still, the shortcut is for speed, not for learning. If you're shaky on the logic, slow it down.

Common Mistakes to Watch Out For

Forgetting the Subscript

The biggest one. On top of that, a student sees 2. 8 g, divides by 28, gets 0.1 mol of C₂H₄, and then multiplies by Avogadro's number — treating the whole molecule as carbon. That said, that gives 6. 022 × 10²², which is off by a factor of 2. Always check: how many of the atom I'm asked about* are in one molecule* of the compound?

Using the Wrong Molar Mass

Some people accidentally use the mass of carbon alone (12 g/mol) or the mass of CH₂ (14 g/mol) because they misread the formula. C₂H₄ is not C₂H₆ (ethane), and it's not CH₄ (methane). One wrong digit in the formula throws the whole answer off.

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Sig Fig Confusion

The starting mass is 2.2044 × 10²³ is technically more "precise" but technically wrong for the problem. So stick to 1. 8 g — two sig figs. So your final answer should have two sig figs too. Reporting 1.2 × 10²³.

Mixing Up Avogadro's Number

It's 6.So 022 × 10²⁴. Even so, 022 × 10²² (a real mix-up I've seen) and not 6. 022 × 10²³, not 6.If your answer comes out off by exactly a factor of 10, that's probably why.

Variations of the Same Problem Worth Knowing

The same logic works for almost any "how many atoms" question. A few quick examples:

• How many hydrogen atoms in 2.8 g of C₂H₄? Same setup, but multiply by 4 instead of 2. Answer: about 2.4 × 10²³.
• How many carbon atoms in 5.6 g of C₂H₄? Double the sample, double the answer: about 2.4 × 10²³.
• How many total atoms in 2.8 g of C₂H₄? Add carbons and hydrogens: 1.2 × 10²³ + 2.4 × 10²³ = 3.6 × 10²³ atoms.

Notice the pattern: every variation follows the same three steps, just with different numbers plugged in. Master the method once and you can solve dozens of problems.

Quick Mental Check

Here's a sanity check that works without doing the full math. A mole of C₂H₄ weighs about 28 g. You've got 2.8 g, which is exactly one-tenth of a mole. Also, one mole of C₂H₄ has 2 moles of carbon in it (because of the subscript). So one-tenth of a mole of C₂H₄ has 0.Here's the thing — 2 moles of carbon. Even so, multiply by Avogadro's number and you're done. Because of that, if your answer is anywhere near 10²³, you're on the right track. Still, if it's near 10²², you probably forgot the ×2. If it's near 10²⁴, you flipped a decimal somewhere.

FAQ

What is the molar mass of C₂H₄?

About 28.05 g/mol. Most problems let you round to 28.

Why do we multiply by 2?

Because each molecule of C₂H₄ contains 2 carbon atoms. The subscript in the formula tells you the atom count per molecule.

What is Avogadro's number?

6.022 × 10²³. It defines how many particles (atoms, molecules, ions, whatever) are in one mole of a substance.

Could I do this with 2.8 g of CO₂ instead?

Same method, different numbers. In real terms, cO₂ has a molar mass of about 44 g/mol and 1 carbon per molecule. So: (2.Plus, 8 ÷ 44) × 1 × 6. 022 × 10²³ ≈ 3.8 × 10²² carbon atoms. The logic doesn't change.

How many significant figures should my answer have?

Match the least precise number in your problem. Here, that's 2.8 g (two sig figs), so the final answer is 1.2 × 10²³.

Wrapping Up

So — 2.Even so, 8 grams of C₂H₄ contains about 1. Still, 2 × 10²³ carbon atoms. The whole thing comes down to three moves: grams to moles, moles of compound to moles of the atom you care about, and moles of atoms to actual atom count using Avogadro's number.

Once you've done it a few times, the steps stop feeling like steps and start feeling like one continuous motion. And honestly, that's the goal. Chemistry is full of conversions like this — pressure to moles, volume to moles, concentration to moles — and the atoms-from-mass problem is probably the friendliest introduction to the genre.

A few parting tips:

• Always write out the units. They cancel like numbers and tell you whether your setup makes sense before you even crunch the digits.
• Don't skip the formula-to-moles step. Jumping straight from grams to atoms is where most errors creep in.
• Round only at the end. Keep one or two extra digits during your calculation and trim to sig figs in the final answer.

If you can do this problem cleanly, you can do molarity problems, stoichiometry problems, and gas law problems. The skeleton is the same: identify what you have, identify what you want, and find the bridge between them. Practically speaking, the bridge here was the molar mass and Avogadro's number. In other problems, it'll be a different constant — but the thinking is identical.

So next time you see a question like this, don't panic. Break out your three steps, grab Avogadro's number, and let the units do the heavy lifting. You've got this.