Interactive Periodic Table

Search / filter; laid out with real period & group positions.

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* Lanthanides and Actinides shown as detached rows.

Molar Mass Calculator

Enter a chemical formula (e.g., H₂O, C₆H₁₂O₆, NaCl)

Chemical Formula

Result:

Molar Mass = 18.015 g/mol

Formula

M = \sum (\text{atomic weight} \times \text{number of atoms})

Show Work (with your inputs)

Step 1 — Contribution from H

Multiply atomic weight of H by number of atoms:

1.008 \times 2 = 2.016

Step 2 — Contribution from O

Multiply atomic weight of O by number of atoms:

15.999 \times 1 = 15.999

Final Step — Total Molar Mass

M = 2.016 + 15.999 = 18.015\ \text{g/mol}

M = 18.015 g/mol

Equation Balancer

Paste an unbalanced equation like Fe + O₂ → Fe₂O₃. Use + for species, → or -> for arrows.

Molarity & Dilution Calculator

Calculate solution concentration and dilution values.

Molarity

Moles of solute (mol)Solution volume

Result:

M = 1.000 M

Formulas

$M = \frac{n}{V}$
$C_1 V_1 = C_2 V_2$

Dilution

Leave one field blank; it will be solved for you.

C₁ (M)V₁C₂ (M)V₂

Solution Concentration Converter

Compute molality, mass %, ppm, and ppb from solute/solvent data.

Solute mass (g)Solvent mass (g)Solute molar mass (g/mol)

Formulas

$n = \frac{m_{\text{solute}}}{M}$
$m = \frac{n}{m_{\text{solvent}}\ (\text{kg})}$
$\text{Mass\%} = \frac{m_{\text{solute}}}{m_{\text{solute}} + m_{\text{solvent}}} \times 100$
$\text{ppm} = \frac{m_{\text{solute}}}{m_{\text{solute}} + m_{\text{solvent}}} \times 10^6$
$\text{ppb} = \frac{m_{\text{solute}}}{m_{\text{solute}} + m_{\text{solvent}}} \times 10^9$

Isotope Finder

Search isotopes by element name/symbol or mass number. Examples: C-14, 14C, Carbon-14, O-18, 14.

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Stoichiometry

Mass ↔ Moles ↔ Particles

Formula

Molar mass

180.2 g/mol

I have…

Mass (g)Moles (mol)Particles

Mass

180 g

Moles

0.99913 mol

Particles

6.0169 × 10^23

Formulas

$M = \sum n_i \times \text{AW}_i$
$n = \tfrac{m}{M}$
$m = n \times M$
$N = n \times N_a$
$n = \tfrac{N}{N_a}$

Show Work (with your inputs)

Step 1 — Molar Mass

M = 180.16\ \text{g/mol}

Step 2 — Mass → Moles

n = \frac{180}{180.16} = 0.99913\ \text{mol}

n = 0.99913 mol

Step 3 — Moles → Particles

N = n \times N_a = 0.99913 \times 6.0221 \times 10^{23} = 6.0169 \times 10^{23}

N = 6.0169 × 10^23

Atom counts

C × 6
H × 12
O × 6

Reaction Stoichiometry

Convert between reactants and products using mole ratios

Reaction (balanced)

Given

Target

Result: 20 mol

Formulas

$n_{target} = n_{given} \times \tfrac{\text{coeff}_{target}}{\text{coeff}_{given}}$
$m = n \times M$
$N = n \times N_a$
$V = n \times 22.4\ \text{L (STP)}$

Show Work (with your inputs)

Step 1 — Convert Given to Moles

n_{given} = 10\ \text{mol}

n₍given₎ = 10 mol

Step 2 — Mole Ratio

n_{target} = 10 \times \tfrac{2}{1} = 20\ \text{mol}

n₍target₎ = 20 mol

Step 3 — Convert Target to Desired Unit

n_{target} = 20\ \text{mol}

20 mol

Limiting Reagent & Percent Yield

Enter a balanced equation, choose reactant inputs, pick a product, and (optionally) enter the actual product mass.

Balanced equation

Reactants

BaCl2

M = 208.2280 g/mol

Treat as excess

Amount (mol)

2 AgNO3

M = 169.8720 g/mol

Treat as excess

Amount (mol)

Products

2 AgCl

M = 143.3180 g/mol

Ba(NO3)2

M = 261.3360 g/mol

Target product for yieldActual mass of product (g)

Formulas

$n = \frac{m}{M}$
$n = M \times V$
$\xi = \frac{n}{\nu}$
$n_{\text{product}} = \xi \times \nu_{\text{product}}$
$m_{\text{product}} = n_{\text{product}} \times M_{\text{product}}$
$\%\ \text{yield} = \frac{m_{\text{actual}}}{m_{\text{theoretical}}} \times 100$

Examples

Theoretical Yield & Percent Yield

Enter a balanced equation, input reactants (mol, g, or solution), select a product, and (optionally) enter actual yield.

Balanced equation

Reactants

BaCl2

M = 208.2280 g/mol

Treat as excess

Amount (mol)

2 AgNO3

M = 169.8720 g/mol

Treat as excess

Amount (mol)

Products

2 AgCl

M = 143.3180 g/mol

Ba(NO3)2

M = 261.3360 g/mol

Target product for yieldActual mass of product (g)

Formulas

$n = \tfrac{m}{M} \quad ; \quad m = n \times M$
$n = M \times V \quad (\text{for solutions})$
$\xi = \tfrac{n}{\nu} \quad (\text{reaction extent})$
$n_{prod} = \xi \times \nu_{prod}$
$m_{prod} = n_{prod} \times M_{prod}$
$\%\,yield = \tfrac{m_{actual}}{m_{theoretical}} \times 100$

Examples

Ideal Gas Law Solver

PV = nRT

Solve for:

Pressure:

Volume:

Moles (n):Temperature:

Result:

P = 1.0006 atm

Show Work

Step 1: Conversion

\text{Convert volume: } 22.4\,L \to 22.4\,L

Step 2: Conversion

\text{Convert temperature: } 273.15\,K \to 273.15\,K

Final Step — Solve

P = \frac{nRT}{V} = \frac{1\times 0.082057\times 273.15}{22.4}

P = 1.0006 atm

Gas Law Variations

Boyle • Charles • Avogadro • Dalton

Choose a gas law:

Solve for:

P₁V₁P₂V₂

Result:

P2 = —

Formulas

$P_1 V_1 = P_2 V_2 \quad \text{(Boyle)}$
$\tfrac{V_1}{T_1} = \tfrac{V_2}{T_2} \quad \text{(Charles)}$
$\tfrac{V_1}{n_1} = \tfrac{V_2}{n_2} \quad \text{(Avogadro)}$
$P_\text{total} = \sum_i P_i \quad \text{(Dalton)}$

Show Work

Step 1 — Apply Boyle’s Law

P_1 V_1 = P_2 V_2

Final Step — Solve for P₂

P_2 = \frac{P_1 V_1}{V_2} = \frac{1\times 1}{NaN} \;\Rightarrow\; \mathbf{—}

Bond Enthalpies

Explore bond dissociation energies (BDEs) for different bond types and molecules.

Average BDE by Bond Type

Bond	Average BDE (kcal/mol)	# Samples
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Solubility Rules

Reference guide for predicting solubility in aqueous solutions.

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Chemistry Tools

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Interactive Periodic Table

Molar Mass Calculator

Formula

Equation Balancer

Molarity & Dilution Calculator

Molarity

Formulas

Dilution

Solution Concentration Converter

Formulas

Isotope Finder

Stoichiometry

Formulas

Reaction Stoichiometry

Given

Target

Formulas

Limiting Reagent & Percent Yield

Reactants

Products

Formulas

Theoretical Yield & Percent Yield

Reactants

Products

Formulas

Ideal Gas Law Solver

Gas Law Variations

Formulas

Bond Enthalpies

Average BDE by Bond Type

Solubility Rules