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Calculate Molar Mass Calculator

Molar Mass Formula:

\[ \text{Molar Mass} = \sum (\text{Atomic Masses} \times \text{Counts}) \]

g/mol
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1. What is Molar Mass?

Molar mass is the mass of a given substance (chemical element or chemical compound) divided by its amount of substance. It is expressed in grams per mole (g/mol) and is calculated by summing the atomic masses of all atoms in the molecule.

2. How Does the Calculator Work?

The calculator uses the formula:

\[ \text{Molar Mass} = \sum (\text{Atomic Masses} \times \text{Counts}) \]

Where:

Explanation: The calculator multiplies each atomic mass by its count in the molecule, then sums all these values to get the total molar mass.

3. Importance of Molar Mass Calculation

Details: Molar mass is essential for converting between grams and moles, which is fundamental in stoichiometric calculations, preparing solutions, and determining empirical and molecular formulas.

4. Using the Calculator

Tips: Enter atomic masses (comma separated) in g/mol and their corresponding counts (comma separated). The number of values in both fields must match.

5. Frequently Asked Questions (FAQ)

Q1: Where can I find atomic masses?
A: Atomic masses are listed in the periodic table. For most calculations, you can use the standard atomic weight rounded to two decimal places.

Q2: How do I calculate molar mass for a compound?
A: Identify all atoms in the compound's formula, look up their atomic masses, multiply each by its count in the formula, then sum all these values.

Q3: What's the difference between molecular mass and molar mass?
A: Molecular mass is the mass of one molecule (in atomic mass units), while molar mass is the mass of one mole of molecules (in grams per mole).

Q4: How precise should my atomic masses be?
A: For most practical purposes, 2 decimal places is sufficient. For very precise work, use more decimal places or isotopic masses.

Q5: Does this work for ionic compounds?
A: Yes, the same principle applies to ionic compounds. Use the formula unit (the simplest ratio of ions in the compound) for your calculation.

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