CHEMISTRY LAB
A small chemistry toolkit for high-school students. Choose a topic, then use Calculator or Practice mode.
Open a tool below. Everything runs in your browser; nothing is uploaded.
Atoms & formulas
01 · Build an atom
Z = protons; A = protons + neutrons; charge = protons − electrons.
Worked example
Example: oxygen-16 has 8 protons and 8 neutrons. A neutral atom has 8 electrons; O²⁻ has 10.
Explore the first 20 elements. Change particles to see the element, mass number and electric charge.
Particle counts are a schematic, not orbital positions. Mass number is not molar mass.
How many neutrons does oxygen-16 contain? Z = 8.
Hint
Neutrons = A − Z.
Worked solution
16 − 8 = 8 neutrons.
02 · Molar mass
Add the atomic weights, multiplied by their atom counts.
Worked example
Example: M(H₂O) = 2 × 1.008 + 15.999 = 18.015 g/mol.
Examples:
| Element | Atoms | Atomic weight | g/mol |
|---|
Approximate classroom values from CIAAW’s abridged atomic weights (2024). Natural isotopic variation and uncertainty are not included in the result.
Supported symbols:
Supports nested parentheses and Unicode subscripts. No hydrates, ionic charges or isotope notation.
CIAAW · Atomic-weight referenceCalculate M(H₂O), using H = 1.008 and O = 15.999.
Hint
There are two hydrogen atoms.
Worked solution
2 × 1.008 + 15.999 = 18.015 g/mol.
03 · Mass and moles
n = m/M; m = n × M. Use grams and g/mol.
Worked example
Example: 18.015 g of H₂O ÷ 18.015 g/mol = 1 mol.
How many moles are in 36 g water? For this exercise, M = 18 g/mol.
Hint
Use n = m/M.
Worked solution
36/18 = 2 mol.
04 · Percentage composition
Element percentage = element contribution / molar mass × 100%.
Worked example
Example: using C = 12 and O = 16, CO₂ has M = 44 g/mol; carbon is 12/44 × 100% ≈ 27.27%.
Find the oxygen mass percentage in CaCO₃. Use Ca = 40, C = 12, O = 16.
Hint
M = 40 + 12 + 3 × 16.
Worked solution
Oxygen contributes 48 out of 100: 48%.
Solutions
05 · Concentration
Molar: c = n/V. Mass percentage: cₚ = mₛ/mᵣ × 100%. Solution mass includes solute and solvent.
Worked example
Example: 0.25 mol in 500 mL = 0.25/0.5 = 0.5 mol/dm³. Separately, 10 g solute in 200 g solution gives 5%.
What is the mass percentage of 10 g solute in 200 g total solution?
Hint
Divide by solution mass, not solvent mass.
Worked solution
10/200 × 100% = 5%.
06 · Dilution
c₁V₁ = c₂V₂. This model assumes no reaction and no loss of solute.
Worked example
Example: V₁ = 0.2 × 250 / 2 = 25 mL. Dilute to a final volume of 250 mL.
A 2 mol/dm³ stock is diluted to 250 mL at 0.2 mol/dm³. Find V₁.
Hint
V₁ = c₂V₂/c₁.
Worked solution
0.2 × 250 / 2 = 25 mL.
Reactions
07 · Balance equations
Change coefficients, not the subscripts in chemical formulas.
Worked example
Example: 2 H₂ + O₂ → 2 H₂O. Both sides contain 4 H atoms and 2 O atoms.
Adjust the numbers before each formula. The number of each kind of atom must match on both sides.
Challenge
| Element | Left | Right |
|---|
In H₂ + O₂ → H₂O, what is the smallest whole-number coefficient before H₂?
Hint
Start with oxygen: use 2 H₂O.
Worked solution
2 H₂ + O₂ → 2 H₂O. The coefficient is 2.
08 · Reaction stoichiometry
Balanced equation → moles → coefficient ratio → product mass.
Worked example
Example: 2.016 g H₂ = 1 mol H₂. In 2 H₂ + O₂ → 2 H₂O, the H₂:H₂O ratio is 1:1, giving 18.015 g water.
2 H₂ + O₂ → 2 H₂O. From 4 g H₂, how many grams of H₂O form? Use M(H₂)=2 and M(H₂O)=18 g/mol; oxygen is in excess.
Hint
The H₂:H₂O mole ratio is 1:1.
Worked solution
4/2 = 2 mol H₂ → 2 mol H₂O; 2 × 18 = 36 g.