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Grams to moles and stoichiometry calculator

Convert grams to moles and particles for any chemical formula, and find the limiting reagent, theoretical yield and leftover excess of a balanced reaction.

更新日 検証済みの例：7

Used for the amount you type and the amount in the result
試す
Amount
mol
Amount: 2 mol
有効桁数：6；最も近い値へ、等距離ならゼロから遠い値へ
質量
36.03g
Particles
1.20443 × 10²⁴
Molar mass used
18.015g/mol

36.03 g of H₂O is 2 mol, or 1.204 × 10²⁴ particles.

Mass, amount and particles of H₂O

質量36.03 gAmount2 molParticles1.204 × 10²⁴÷ 18.015 g/mol× 18.015 g/mol× 6.022 × 10²³÷ 6.022 × 10²³
計算方法 S

  1. Amount in moles

    n=mM=36.0318.015=2 moln = \frac{m}{M} = \frac{36.03}{18.015} = 2\ \mathrm{mol}

  2. 質量

    m=nM=(2)(18.015)=36.03 gm = nM = (2)(18.015) = 36.03\ \mathrm{g}

  3. Particles

    N=nNA=(2)(6.02214076×1023)=1.20443×1024N = nN_A = (2)(6.02214076\times10^{23}) = 1.20443 \times 10^{24}

    N_A is exact in the SI since 2019.

Grams to moles and stoichiometry calculatorについて

The amount of substance links mass to particle count: n = m/M, where M is the molar mass worked out from the formula, and N = n × N_A, where the Avogadro constant N_A is 6.022 140 76 × 10²³ per mole. In limiting-reagent mode, each reactant's moles are divided by its coefficient in the balanced equation; the smallest quotient is how far the reaction can run, and it sets the theoretical yield of product.

Typical uses are weighing out reagents and predicting product mass before an experiment. The default, 36.03 g of water, is 2 mol, or 1.204 × 10²⁴ molecules. For 2 H₂ + O₂ → 2 H₂O with 10 g of hydrogen and 64 g of oxygen, oxygen runs out first and the yield is 72.06 g of water.

The yield assumes the equation is balanced and the reaction goes to completion, so measured yields come out lower.

計算例

36.03 g of water is 2 mol

Calculate
Grams ↔ moles ↔ particles
Molar mass from
計算式
計算式
H2O
I have
質量
質量
36.03 g
Amount unit
mol
Amount
2 mol
Particles
1.20443 × 10²⁴

照合元：Python 3.8 decimal: 36.03 / 18.015 = 2; × 6.02214076e23 (IUPAC 2021 weights)

One mole of carbon atoms

Calculate
Grams ↔ moles ↔ particles
Molar mass from
計算式
計算式
C
I have
Moles
Amount
1
Amount unit
mol
Particles
6.02214 × 10²³
質量
12.011 g

照合元：SI 2019 definition of the mole (Avogadro constant exact); standard atomic weight of carbon 12.011 (IUPAC 2021)

0.25 mol NaCl with a typed molar mass

Calculate
Grams ↔ moles ↔ particles
Molar mass from
Typed value
モル質量
58.44 g/mol
I have
Moles
Amount
0.25
Amount unit
mol
質量
14.61 g

照合元：Python 3.8 decimal: 0.25 × 58.44

3.011 × 10²³ molecules of CO₂

Calculate
Grams ↔ moles ↔ particles
Molar mass from
計算式
計算式
CO2
I have
Particles
Amount unit
mol
Number of particles
3.011e23
Amount
0.499988 mol
質量
22.004 g

照合元：Python 3.8 decimal: 3.011e23 / N_A = 0.4999883; × 44.009 g/mol

よくある質問

How do you convert grams to moles?

Divide the mass in grams by the molar mass in g/mol: n = m/M. Water has a molar mass of 18.015 g/mol, so 36.03 g of water is 36.03 ÷ 18.015 = 2 mol. To go back, multiply moles by the molar mass. The molar mass is the sum of the atomic weights in the formula, which the calculator adds up from IUPAC 2021 values.

What is Avogadro's number?

It is the number of particles in one mole: exactly 6.022 140 76 × 10²³ per mole since the 2019 revision of the SI, which defines the mole by fixing this constant. One mole of carbon therefore contains 6.022 140 76 × 10²³ atoms and weighs 12.011 g. Multiply an amount in moles by this number to get molecules, atoms or formula units.

How do you find the limiting reagent?

Convert each reactant's mass to moles, then divide by its coefficient in the balanced equation; the reactant with the smaller result runs out first. For N₂ + 3 H₂ → 2 NH₃ with 28 g of N₂ and 6 g of H₂, nitrogen gives 0.9995 mol ÷ 1 = 0.9995 and hydrogen gives 2.976 mol ÷ 3 = 0.992, so hydrogen limits the yield to 33.79 g of ammonia.

What is the difference between theoretical yield and percent yield?

Theoretical yield is the product mass if the limiting reagent reacted completely, which is what this calculator reports. Percent yield compares what you actually collected: actual ÷ theoretical × 100. With a theoretical yield of 72.06 g of water and 65.0 g collected, the percent yield is 90.2 %. Transfer losses, side reactions and incomplete reaction keep real yields below 100 %.

「Grams to moles and stoichiometry calculator」の精度はどのくらいですか？

精度は入力値と計算方法の前提に依存します。十進演算には有効数字50桁を使いますが、推定、数値計算手法、元データの精度はそれより低い場合があります。表示の丸め処理でこれらの制約がなくなるわけではありません。 独立した出典の解答と照合した計算例：7。 例えば、「36.03 g of water is 2 mol」はPython 3.8 decimal: 36.03 / 18.015 = 2; × 6.02214076e23 (IUPAC 2021 weights)と照合しています。

この計算方法の出典は何ですか？

CODATA 2022 / SI 2019 — Avogadro constant N_A = 6.022 140 76 × 10²³ mol⁻¹ (exact); OpenStax Chemistry 2e, §4.4 Reaction yields (limiting reactant, theoretical yield).

この計算機について

n=mM,N=nNA;ξ=min⁡ ⁣(nAa,nBb),mproduct=c ξ Mproductn = \frac{m}{M},\quad N = nN_A;\qquad \xi = \min\!\left(\frac{n_A}{a}, \frac{n_B}{b}\right),\quad m_{\text{product}} = c\,\xi\,M_{\text{product}}

出典

  1. CODATA 2022 / SI 2019 — Avogadro constant N_A = 6.022 140 76 × 10²³ mol⁻¹ (exact)
  2. OpenStax Chemistry 2e, §4.4 Reaction yields (limiting reactant, theoretical yield)

出典と照合済み

この計算機には、独立した出典の解答を使った計算例が 7 件あります。テストに組み込まれており、ここでも実行できます。

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