Mole Calculator with Steps

Compute moles from mass, volume, particles, or concentration with clarity fast. See every algebraic step, unit handling, and rounding explained live for you. Export results and worked steps to CSV or PDF for records easily. Trust accurate chemistry math with guided, verifiable solutions always.

Calculator

Avogadro’s constant: 6.02214076×1023 mol−1
Result rounding; steps always retain unrounded values.
g/mol
Enter directly or compute from a chemical formula.
mol
mol/L
Accepts plain numbers or scientific notation like 6.022e23.
g
mol/L
mol/L
Leave exactly one of C1, V1, C2, V2 blank to solve it.

Result

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Steps

Choose a calculation type, enter inputs, then click Calculate.

Saved results

# Type Inputs Result

Example data

Use these to test and compare your results.

#TypeInputsExpected
1 Mass → Moles m = 18 g, formula = H2O (M ≈ 18.015) n ≈ 0.9992 mol
2 Volume & Molarity → Moles V = 250 mL, C = 0.200 mol/L n = 0.0500 mol
3 Particles → Moles N = 3.011×1023 n = 0.5000 mol
4 Moles → Particles n = 0.125 mol N ≈ 7.5277×1022
5 Moles → Mass n = 0.0500 mol, NaCl (M ≈ 58.44) m ≈ 2.922 g
6 % Composition Formula = H2O; sample mass = 10 g H ≈ 11.19% (1.12 g), O ≈ 88.81% (8.88 g)
7 Dilution C1=1.0 M, V1=25 mL, C2=0.250 M, V2=? V2 = 100 mL

how many atoms are contained in 12.5 moles of nitrogen calculator with step?

Clarify meaning of “nitrogen”: single atoms (N) or diatomic molecules (N2). Results differ by a factor of two.

Case A — nitrogen atoms (N)

  1. Relation: N = n × NA.
  2. Substitute: n = 12.5 mol, NA = 6.02214076×1023 mol−1.
  3. Compute: N = 12.5 × 6.02214076×1023 = 7.52767595e+24 atoms ≈ 7.53 ×10^24 atoms.

Case B — nitrogen gas molecules (N2)

  1. Molecules first: Nmolecules = n × NA = 12.5 × 6.02214076×1023 = 7.52767595e+24 molecules.
  2. Each N2 has 2 atoms ⇒ total atoms = 2 × Nmolecules = 2 × 7.52767595e+24 = 1.50553519e+25.
  3. Rounded: ≈ 1.51 ×10^25 atoms.
InterpretationExpressionValue
N atoms (N)12.5 × NA7.53 ×10^24 atoms
N2 gas → atoms2 × 12.5 × NA1.51 ×10^25 atoms

Tip: To reproduce above, choose “Moles → Particles.” For N2, multiply the particle count by 2 to convert molecules to atoms.

what is a mole in chemistry? step-by-step guide

Definition: A mole is 6.02214076×1023 elementary entities (Avogadro’s constant). It links macroscopic mass to particle counts.

Worked example — 5.00 g of glucose (C6H12O6)

  1. Compute molar mass: M = 6·12.0107 + 12·1.00794 + 6·15.9994 = 180.156 g/mol.
  2. Convert to moles: n = m / M = 5.00 g / 180.156 g/mol = 0.027754 mol.
  3. Particles (molecules): N = n·NA = 0.027754 × 6.02214076×1023 = 1.671369e+22 ≈ 1.671 ×10^22 molecules.
  4. Total atoms: each molecule has 24 atoms ⇒ atoms = 24 × molecules = 4.011287e+23 ≈ 4.011 ×10^23 atoms.

convert grams to moles: worked example with steps

Example — 5.85 g of NaCl

  1. Molar mass: M(NaCl) = 22.98977 + 35.453 = 58.443 g/mol.
  2. Formula: n = m / M.
  3. Substitute: n = 5.85 g / 58.443 g/mol = 0.100098 mol.
  4. Report to four significant figures: 0.1001 mol.

Tip: If the sample is in mg or kg, convert to grams before dividing by M.

calculate molarity from moles and volume: steps

Example — n = 0.0500 mol KCl in 250 mL solution

  1. Convert volume: 250 mL = 0.250 L.
  2. Use: C = n / V.
  3. Substitute: C = 0.0500 mol / 0.250 L = 0.200000 mol/L.
  4. Rounded to three significant figures: 0.200 M.

Remember: volumes must be in liters for C in mol/L.

limiting reagent & theoretical yield — step-by-step walkthrough

Balanced equation: N2 + 3 H2 → 2 NH3

Given

  • m(N2) = 28.00 g, M(N2) = 28.01340 g/mol ⇒ n(N2) = 0.99952 mol
  • m(H2) = 6.00 g, M(H2) = 2.01588 g/mol ⇒ n(H2) = 2.97637 mol

1) Identify the limiting reagent

  1. Compare n/coefficient: N2: 0.99952/1 = 0.99952; H2: 2.97637/3 = 0.99212.
  2. Smaller value limits ⇒ H2 is the limiting reagent.

2) Theoretical yield of NH3

  1. Use stoichiometric ratio: n(NH3) = (coeff NH3 / coeff limiting) × n(limiting).
  2. n(NH3) = 1.98425 mol; M(NH3) = 17.03052 g/mol.
  3. m(NH3) = n × M = 33.79 g (theoretical).

3) Excess reagent left over

n(N2) required = 0.99212 mol ⇒ leftover n(N2) = 0.00740 mol ⇒ leftover m(N2) ≈ 0.207 g.

4) Optional — percent yield

If actual NH3 collected is 30.0 g: % yield = (actual/theoretical) ×100 = 30.0 / 33.79 ×100 = 88.8%.

QuantityValue
n(N2)0.99952 mol
n(H2)2.97637 mol
Limiting reagentH2
n(NH3) theoretical1.98425 mol
m(NH3) theoretical33.79 g
Leftover N20.00740 mol (0.207 g)

Recreate this in the tool: compute each reactant’s moles with “Mass → Moles”, divide by coefficients to find the limiter, then get product moles by ratio and convert to grams.

Formula used

  • n = m / M — moles from mass and molar mass.
  • m = n × M — mass from moles and molar mass.
  • n = C × V — moles from molarity and volume.
  • C = n / V — molarity from moles and volume.
  • n = N / NA — moles from particles.
  • N = n × NA — particles from moles.
  • wi(%) = (ai Mi / Mtotal) × 100 — percent by mass of element i.
  • C1V1 = C2V2 — dilution relation.

Units: m in g, M in g/mol, V in L (or mL), C in mol/L. Convert inputs to base units where required.

How to use this calculator

  1. Select a Calculation type based on the data you have.
  2. Provide required inputs. To auto-get M, type the chemical formula then click Compute molar mass.
  3. Click Calculate. Review Steps to see the algebra and unit handling.
  4. Use Add to results table to store a row. Export via CSV or PDF.

Tip: Choose an appropriate number of significant figures for reporting.

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Important Note: All the Calculators listed in this site are for educational purpose only and we do not guarentee the accuracy of results. Please do consult with other sources as well.