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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.

更新于 已验证的示例：7

Used for the amount you type and the amount in the result
试一试
Amount
mol
Amount: 2 mol
有效数字：6；取最近值，等距时远离零
质量
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

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

  1. Amount in moles

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

  2. 质量

    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.

关于Grams to moles and stoichiometry calculator

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.

计算示例

36.03 g of water is 2 mol

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

核验来源：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
公式
公式
C
I have
Moles
Amount
1
Amount unit
mol
Particles
6.02214 × 10²³
质量
12.011 g

核验来源：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
摩尔质量
58.44 g/mol
I have
Moles
Amount
0.25
Amount unit
mol
质量
14.61 g

核验来源：Python 3.8 decimal: 0.25 × 58.44

3.011 × 10²³ molecules of CO₂

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

核验来源：Python 3.8 decimal: 3.011e23 / N_A = 0.4999883; × 44.009 g/mol

常见问题

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”有多准确？

准确性取决于输入值和方法的假设。十进制运算使用50位有效数字，但估算、数值方法和源数据的精度可能较低；显示时的舍入并不能消除这些限制。 已按独立来源核验的计算示例：7。 例如，“36.03 g of water is 2 mol”根据Python 3.8 decimal: 36.03 / 18.015 = 2; × 6.02214076e23 (IUPAC 2021 weights)进行核验。

这种方法出自哪里？

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).

关于此计算器

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}}

来源

  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)

已对照来源验证

此计算器包含 7 个已解示例，答案来自独立来源。这些示例会在测试套件中运行，你也可以在此运行验证。

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