# 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.

Versi interaktif: https://www.calcopenly.com/id/science/moles-stoichiometry-calculator
Topik: Kalkulator sains

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.

## Masukan

- **Calculate** (pilihan: Grams ↔ moles ↔ particles, Limiting reagent)
- **Molar mass from** (pilihan: Rumus, Typed value)
- **Rumus**
- **Massa molar**
- **I have** (pilihan: Massa, Moles, Particles)
- **Massa**
- **Amount**
- **Amount unit**: Used for the amount you type and the amount in the result (pilihan: µmol, mmol, mol, kmol)
- **Number of particles**: Molecules, atoms or formula units; 3e23 works
- **Coefficient of reactant A**
- **Reactant A formula**
- **Mass of A**
- **Coefficient of reactant B**
- **Reactant B formula**
- **Mass of B**
- **Coefficient of the product**
- **Product formula**

## Hasil

- Amount (mol) — hasil utama
- Massa (g)
- Particles
- Molar mass used (g/mol)
- Limiting reagent
- Theoretical yield of product (g)
- Product formed (mol)
- Excess reactant left over (g)
- Extent of reaction (mol)

## Rumus

$$
n = \frac{m}{M},\quad N = nN_A;\qquad \xi = \min\!\left(\frac{n_A}{a}, \frac{n_B}{b}\right),\quad m_{\text{product}} = c\,\xi\,M_{\text{product}}
$$

## Contoh penyelesaian

### 36.03 g of water is 2 mol

- Calculate: Grams ↔ moles ↔ particles
- Molar mass from: Rumus
- Rumus: H2O
- I have: Massa
- Massa: 36.03 g
- Amount unit: mol
- **Amount: 2 mol**
- **Particles: 1.20443 × 10²⁴**
- Sumber pemeriksaan: 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: Rumus
- Rumus: C
- I have: Moles
- Amount: 1
- Amount unit: mol
- **Particles: 6.02214 × 10²³**
- **Massa: 12.011 g**
- Sumber pemeriksaan: 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
- Massa molar: 58.44 g/mol
- I have: Moles
- Amount: 0.25
- Amount unit: mol
- **Massa: 14.61 g**
- Sumber pemeriksaan: Python 3.8 decimal: 0.25 × 58.44

### 3.011 × 10²³ molecules of CO₂

- Calculate: Grams ↔ moles ↔ particles
- Molar mass from: Rumus
- Rumus: CO2
- I have: Particles
- Amount unit: mol
- Number of particles: 3.011e23
- **Amount: 0.499988 mol**
- **Massa: 22.004 g**
- Sumber pemeriksaan: Python 3.8 decimal: 3.011e23 / N_A = 0.4999883; × 44.009 g/mol

### 2 H₂ + O₂ → 2 H₂O with 10 g H₂ and 64 g O₂

- Calculate: Limiting reagent
- Coefficient of reactant A: 2
- Reactant A formula: H2
- Mass of A: 10 g
- Coefficient of reactant B: 1
- Reactant B formula: O2
- Mass of B: 64 g
- Coefficient of the product: 2
- Product formula: H2O
- **Limiting reagent: O2**
- **Theoretical yield of product: 72.0645 g**
- **Excess reactant left over: 1.9355 g**
- Sumber pemeriksaan: Python 3.8 decimal: ξ = min(10/2.016/2, 64/31.998) = 2.000125; 2ξ × 18.015 = 72.0645 g

### N₂ + 3 H₂ → 2 NH₃ with 28 g N₂ and 6 g H₂

- Calculate: Limiting reagent
- Coefficient of reactant A: 1
- Reactant A formula: N2
- Mass of A: 28 g
- Coefficient of reactant B: 3
- Reactant B formula: H2
- Mass of B: 6 g
- Coefficient of the product: 2
- Product formula: NH3
- **Limiting reagent: H2**
- **Theoretical yield of product: 33.7917 g**
- **Excess reactant left over: 0.208333 g**
- Sumber pemeriksaan: Python 3.8 decimal: ξ = min(28/28.014, 6/2.016/3) = 0.992063; 2ξ × 17.031 = 33.7917 g

## Pertanyaan

### 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 %.

### Seberapa akurat “Grams to moles and stoichiometry calculator”?

Akurasi bergantung pada masukan dan asumsi metode. Aritmetika desimal memakai 50 digit signifikan, tetapi perkiraan, metode numerik dan data sumber bisa kurang presisi; pembulatan yang ditampilkan tidak menghilangkan batasan itu. Contoh penyelesaian yang diperiksa dengan sumber independen: 7. Misalnya, “36.03 g of water is 2 mol” diperiksa dengan Python 3.8 decimal: 36.03 / 18.015 = 2; × 6.02214076e23 (IUPAC 2021 weights).

### Dari mana metode ini berasal?

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).

## Sumber

- [CODATA 2022 / SI 2019 — Avogadro constant N_A = 6.022 140 76 × 10²³ mol⁻¹ (exact)](https://physics.nist.gov/cgi-bin/cuu/Value?na)
- [OpenStax Chemistry 2e, §4.4 Reaction yields (limiting reactant, theoretical yield)](https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields)