Molecular Weight Calculator - Calculate Molar Mass & Composition
Molecular Weight Calculator
Examples: H2O (water), C6H12O6 (glucose), Ca(OH)2 (calcium hydroxide)
Common Compounds
Enter a molecular formula and click Calculate
Examples: H2O, C6H12O6, Ca(OH)2, H2SO4
Understanding Molecular Weight Calculation
Quick Summary: Calculate molecular weight for any chemical formula by summing atomic weights of all constituent atoms. Our free calculator supports complex formulas with parentheses, provides detailed elemental composition analysis, and includes a database of common compounds. Essential for chemistry students, researchers, and laboratory professionals.
What is Molecular Weight?
Molecular weight (also called molecular mass or relative molecular mass) represents the sum of atomic weights of all atoms in a molecule. It is expressed in atomic mass units (amu) or unified atomic mass units (u), but more commonly in grams per mole (g/mol) when referred to as molar mass. These terms are numerically identical and often used interchangeably in chemistry.
Fundamental relationship: One mole of any substance contains Avogadro's number (6.022 × 10²³) of molecules. The molecular weight in grams equals the mass of one mole of that substance. For example, water (H2O) has a molecular weight of 18.016, meaning 18.016 grams of water contains exactly 6.022 × 10²³ water molecules.
Molecular Weight = Σ (Number of atoms × Atomic weight)
Molecular Weight: Sum of all atomic weights in formula
Atomic Weight: Average mass of element's isotopes (from periodic table)
Number of atoms: Count from subscripts in chemical formula
Units: amu (atomic mass units) or g/mol (grams per mole)
Step-by-Step Calculation Method
Simple Formulas Without Parentheses
For straightforward formulas like H2O or CO2, identify each element and its subscript (number of atoms), then multiply by the atomic weight:
Example: Water (H2O)
- Hydrogen (H): 2 atoms × 1.008 amu = 2.016 amu
- Oxygen (O): 1 atom × 16.00 amu = 16.00 amu
- Total Molecular Weight: 2.016 + 16.00 = 18.016 amu (or g/mol)
Example: Glucose (C6H12O6)
- Carbon (C): 6 atoms × 12.01 amu = 72.06 amu
- Hydrogen (H): 12 atoms × 1.008 amu = 12.096 amu
- Oxygen (O): 6 atoms × 16.00 amu = 96.00 amu
- Total: 72.06 + 12.096 + 96.00 = 180.156 amu
Complex Formulas With Parentheses
Parentheses indicate groups of atoms that appear multiple times. Multiply everything inside parentheses by the subscript outside:
Example: Calcium Hydroxide Ca(OH)2
Step 1: Expand the parentheses: Ca + (OH) × 2 = Ca + 2O + 2H
Step 2: Count atoms: 1 Ca, 2 O, 2 H
Step 3: Calculate:
- Ca: 1 × 40.08 = 40.08
- O: 2 × 16.00 = 32.00
- H: 2 × 1.008 = 2.016
Total: 40.08 + 32.00 + 2.016 = 74.096 g/mol
Example: Aluminum Sulfate Al2(SO4)3
Step 1: Expand: 2 Al + (SO4) × 3 = 2 Al + 3 S + 12 O
Step 2: Calculate:
- Al: 2 × 26.98 = 53.96
- S: 3 × 32.07 = 96.21
- O: 12 × 16.00 = 192.00
Total: 53.96 + 96.21 + 192.00 = 342.17 g/mol
Elemental Composition and Mass Percent
Elemental composition describes the percentage by mass of each element in a compound. This information is crucial for analytical chemistry, identifying unknown substances, and quality control in manufacturing.
Mass Percent = (Element's Total Mass ÷ Molecular Weight) × 100%
Example: Glucose (C6H12O6) with molecular weight 180.156 g/mol:
| Element | Atoms | Total Mass (amu) | Mass Percent |
|---|---|---|---|
| Carbon (C) | 6 | 72.06 | 40.00% |
| Hydrogen (H) | 12 | 12.096 | 6.71% |
| Oxygen (O) | 6 | 96.00 | 53.29% |
Interpretation: Glucose is 40% carbon by mass, 6.71% hydrogen, and 53.29% oxygen. The percentages must sum to 100%. This composition is characteristic of carbohydrates, which are generally composed of carbon, hydrogen, and oxygen in approximately 1:2:1 atomic ratio, resulting in oxygen dominating the mass percentage due to its higher atomic weight.
Applications in Chemistry
Stoichiometry and Chemical Equations
Molecular weight is the bridge between the molecular world and the macroscopic world. It allows conversion between moles (counting molecules) and mass (weighing substances):
moles = mass (g) ÷ molecular weight (g/mol)
mass (g) = moles × molecular weight (g/mol)
Example: Combustion of methane (CH4): CH4 + 2O2 → CO2 + 2H2O
If you burn 16 grams of methane (molecular weight 16.04 g/mol), you have 16 ÷ 16.04 ≈ 1 mole of CH4. According to the balanced equation, 1 mole of CH4 produces 1 mole of CO2 (44.01 g/mol) and 2 moles of H2O (18.016 g/mol each). Therefore, you will produce 44.01 g of CO2 and 36.032 g of H2O. Total mass is conserved: 16 g CH4 + 64 g O2 = 44.01 g CO2 + 36.032 g H2O = 80.042 g products (close to 80 g reactants, with rounding).
Limiting Reagent Determination
In chemical reactions with multiple reactants, molecular weight helps identify which reactant runs out first (the limiting reagent), controlling how much product forms:
Example: React 10 g of hydrogen (H2, MW = 2.016) with 80 g of oxygen (O2, MW = 32.00) to form water: 2H2 + O2 → 2H2O
- Moles of H2: 10 g ÷ 2.016 g/mol = 4.96 moles
- Moles of O2: 80 g ÷ 32.00 g/mol = 2.50 moles
- Stoichiometry: Need 2 moles H2 per 1 mole O2. With 4.96 moles H2, we need 4.96 ÷ 2 = 2.48 moles O2
- Conclusion: O2 is in slight excess; H2 is the limiting reagent. Water produced = 4.96 moles H2O × 18.016 g/mol = 89.36 g
Empirical and Molecular Formula Determination
Analytical chemistry often determines percent composition experimentally (e.g., combustion analysis). From this data, chemists calculate the empirical formula (simplest whole-number ratio) and use molecular weight to find the molecular formula:
Example: A compound contains 40.00% C, 6.71% H, 53.29% O by mass. Molecular weight from mass spectrometry is approximately 180 g/mol.
- Convert percentages to moles: Assume 100 g sample: 40.00 g C ÷ 12.01 = 3.33 mol C; 6.71 g H ÷ 1.008 = 6.66 mol H; 53.29 g O ÷ 16.00 = 3.33 mol O
- Find smallest ratio: Divide all by 3.33: C = 1, H = 2, O = 1. Empirical formula: CH2O (empirical weight = 30 g/mol)
- Determine molecular formula: 180 ÷ 30 = 6. Molecular formula = (CH2O)6 = C6H12O6 (glucose)
Atomic Weights and Isotopes
Atomic weights listed on the periodic table are weighted averages of all naturally occurring isotopes of each element. For example, natural chlorine is 75.8% chlorine-35 (34.969 amu) and 24.2% chlorine-37 (36.966 amu):
Atomic weight of Cl = (0.758 × 34.969) + (0.242 × 36.966) = 35.45 amu
For most laboratory calculations, standard atomic weights from IUPAC (International Union of Pure and Applied Chemistry) are sufficient. However, for precise isotope studies, nuclear chemistry, or mass spectrometry analysis, specific isotope masses must be used instead of average atomic weights.
Common Calculation Mistakes and How to Avoid Them
- Forgetting subscripts: H2O has 2 hydrogens, not 1. Always check subscripts carefully. If no subscript appears, there is 1 atom of that element.
- Mishandling parentheses: Ca(OH)2 has 2 oxygens and 2 hydrogens, not 1 of each. Multiply everything inside parentheses by the outside subscript.
- Nested parentheses: Expand from innermost to outermost. For Fe4[Fe(CN)6]3, first expand Fe(CN)6, then multiply by the outside subscript.
- Wrong atomic weights: Use current values from reliable sources. Older periodic tables may have outdated weights. Our calculator uses IUPAC 2021 values.
- Rounding too early: Keep full precision during intermediate calculations, then round the final answer to appropriate significant figures (typically 2-4 decimal places for molecular weight).
- Confusing molecular weight with mass: Molecular weight is 18.016 g/mol for water; this is NOT the mass of one molecule (which would be 18.016 ÷ Avogadro's number = 2.99 × 10⁻²³ g).
Practical Laboratory Applications
Solution Preparation
To prepare solutions of specific molarity, molecular weight converts desired moles into grams to weigh:
Example: Prepare 500 mL of 0.1 M NaCl solution (NaCl MW = 58.44 g/mol)
- Moles needed = 0.1 mol/L × 0.5 L = 0.05 mol
- Mass needed = 0.05 mol × 58.44 g/mol = 2.922 g
- Weigh 2.922 g NaCl, dissolve in water, dilute to exactly 500 mL total volume
Quality Control and Purity Assessment
If measured elemental composition deviates from calculated values, the sample may contain impurities, wrong compound, or experimental error. For example, if supposed pure glucose shows 38% carbon instead of 40%, it likely contains impurities or degradation products.
Drug Dosage Calculations
Pharmaceutical chemistry uses molecular weight to convert between different dose units. For example, converting a drug dose from milligrams to millimoles (mmol) for physiological calculations, or calculating equivalent doses of different salt forms of the same drug (e.g., comparing calcium carbonate to calcium citrate supplements).
Additional Resources
For more information on molecular weight, chemical formulas, and stoichiometry:
- IUPAC Periodic Table of Elements - Official atomic weights from IUPAC
- NIST Chemistry WebBook - Comprehensive chemical and physical property data
- ChemSpider - Chemical structure database with molecular weights
- PubChem Database - Open chemistry database from NIH