Stoichiometric Mass Converter
Convert stoichiometric mass between units instantly. Includes conversion tables, common equivalents, and calculation formulas.
Stoichiometric Mass Converter
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Formula: Mass Product = (Mass Reactant / MW Reactant) × (coeff Product / coeff Reactant) × MW Product
Worked example — 244.30 g CO₂
Formula
Mass Product = (Mass Reactant / MW Reactant) × (coeff Product / coeff Reactant) × MW Product
Converts mass of one substance to the equivalent mass of another using molar masses and stoichiometric coefficients from a balanced equation.
Worked Examples
Example 1: Water to CO₂
Problem:100g H₂O, MW=18.015. Find equivalent CO₂ (MW=44.01), 1:1 ratio.
Solution:Moles = 100/18.015 = 5.551. Mass CO₂ = 5.551 × 44.01 = 244.30 g
Result:244.30 g CO₂
Frequently Asked Questions
What is stoichiometry?
Stoichiometry is the calculation of reactants and products in chemical reactions using balanced equations and molar ratios.
How do I find molar mass?
Add the atomic masses of all atoms in the molecular formula. For example, H₂O = 2(1.008) + 16.00 = 18.015 g/mol.
What is a limiting reactant and how does it affect yield calculations?
The limiting reactant is the reagent that runs out first and therefore determines the maximum amount of product that can form. To identify it, convert each reactant's mass to moles, divide by its stoichiometric coefficient, and the smallest value indicates the limiting reactant. Yield calculations based on the non-limiting reactant will overestimate the actual product mass.
What is percent yield and how is it calculated?
Percent yield measures the efficiency of a chemical reaction: % yield = (actual yield / theoretical yield) × 100%. The theoretical yield is the maximum mass of product predicted by stoichiometry (what Stoichiometric Mass Converter computes). The actual yield is what is measured in the lab after the reaction. Losses from incomplete reactions, side reactions, and handling typically result in yields of 60-95% in laboratory settings.
How do the coefficients of a balanced equation enter a mass conversion?
The coefficients set the mole ratio, and mass converts only through moles. You divide the reactant mass by its molar mass to get moles, multiply by the product coefficient divided by the reactant coefficient, then multiply by the product molar mass. Skipping the coefficient ratio is the single most common error: it silently assumes a 1:1 reaction, which is wrong for anything like 2 H2 + O2 giving 2 H2O.
Why must the equation be balanced before any mass conversion?
A balanced equation is the statement that atoms are conserved, and the mole ratio it encodes is the only valid bridge between two substances. An unbalanced equation gives a ratio that does not conserve atoms, so the predicted product mass will not match the reactant mass consumed plus any co-products. Balancing first, then converting, is what makes the arithmetic obey conservation of mass.
What is the difference between empirical formula and molecular formula for molar mass?
The empirical formula gives the simplest whole-number atom ratio; the molecular formula gives the actual atom count per molecule. Glucose is CH2O empirically but C6H12O6 molecularly, and the molar masses differ by the same factor of six. Stoichiometric mass conversions must use the molecular formula, because the balanced equation is written in terms of real molecules, not reduced ratios.
How do moles connect mass to particle counts?
Moles are the bridge: n = m / M, where m is mass in grams and M is molar mass in g/mol. Multiplying moles by the Avogadro constant, about 6.022 x 10^23 per mole, gives the number of molecules or formula units. That is why molar mass is expressed in g/mol; it is numerically the same as the relative molecular mass in atomic mass units, which is what makes the bookkeeping convenient.
How do hydrates and reagent purity change the molar mass you should use?
A hydrate such as copper sulfate pentahydrate carries water in its formula, so its molar mass includes five water molecules and a given mass contains proportionally less of the anhydrous salt. Similarly, a reagent quoted at 95 percent purity delivers only that fraction as the intended substance. Use the molar mass of the exact solid you weighed, then scale by purity, or the conversion will overstate the moles available.
Why do procedures deliberately use one reactant in excess?
Adding excess of a cheap or easily removed reagent drives a reversible reaction further toward products, in line with Le Chatelier's principle, and ensures the expensive or hazardous reactant is consumed completely. The trade-off is that the excess must be separated later. When a reagent is in excess it is by definition not the limiting reactant, so theoretical yield must be computed from the other one.
Reviewed for accuracy by Manoj Kumar, Mathematics Educator · Editorial policy
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