Combustion Analysis Calculator

How to use the combustion analysis calculator

What this calculator does. It converts measured carbon dioxide and water from a complete-combustion experiment into the amounts of carbon and hydrogen that were present in an unknown sample. For a C, H, O compound, it obtains oxygen by subtracting the carbon and hydrogen masses from the original sample mass. It then reduces the mole amounts to a whole-number empirical formula. When you also provide the compound's molar mass, it estimates the integer multiple needed to report a molecular formula. It does not identify structural isomers, functional groups, reaction completeness, or experimental contamination.

When to use it. Use this tool to check an organic chemistry lab calculation, prepare a pre-lab example, verify homework involving empirical formulas, or compare how measurement changes affect a proposed formula. The method assumes complete combustion and that the selected element set matches the sample. The IUPAC definition of an empirical formula explains why the result is the simplest whole-number atomic ratio rather than necessarily the true molecular composition.

How to calculate. The calculator opens with a complete demonstration and an immediately available XLSX workbook. Follow these steps:

1. Choose Substance. Select “C, H, O compound” when oxygen may be present, or “Hydrocarbon” when the sample contains only carbon and hydrogen.
2. Replace the demonstration values with your measured Carbon dioxide (CO₂) mass and Water (H₂O) mass, both in grams. For a C – H – O compound, also enter Sample mass.
3. Optionally enter Sample's molar mass in g/mol. Leave it blank when you only need an empirical formula.
4. Read the empirical formula, empirical formula mass, elemental detail table, and – when molar mass is available – the molecular formula and multiplier.
5. Select Download Excel to create a workbook from the current validated state. Reset clears the demonstration data and results; export remains unavailable until a complete valid data set is entered again.

Input guide. Substance is required and controls whether oxygen is calculated. Sample's molar mass is an optional positive decimal in g/mol; 90.0779 g/mol is a realistic example, and increasing it can increase the molecular multiplier while leaving the empirical formula unchanged. Sample mass is a positive decimal in grams and is required for C – H – O mode; 12.915 g is the demonstration value. It must exceed the calculated carbon-plus-hydrogen mass, otherwise oxygen by difference would be negative. Carbon dioxide (CO₂) mass and Water (H₂O) mass are required positive decimal masses in grams; the demonstration uses 18.942 g and 7.749 g. Enter a period as the decimal separator, do not use scientific notation, and do not attach unit text inside a field.

Output guide. Empirical formula is the reduced atomic ratio and is an exact identity only to the extent that the measured data and rounding support it. Empirical formula mass is the molar mass of one empirical-formula unit. Molecular formula multiplies all empirical subscripts by a near-integer ratio of sample molar mass to empirical formula mass; “Not available” means no molar mass was supplied or the ratio is not acceptably close to an integer. Molecular multiplier is that integer. Oxygen by difference is meaningful only in C – H – O mode and becomes zero for a hydrocarbon. In the detail table, mass is the recovered elemental mass, amount is moles, normalized ratio divides by the smallest mole amount, and formula count is the final whole-number subscript.

Worked example. Burning the 12.915 g demonstration sample yields 18.942 g CO₂ and 7.749 g H₂O. Using standard atomic masses, the products correspond to about 5.169 g carbon and 0.867 g hydrogen. Oxygen by difference is about 6.879 g. Dividing each mass by its atomic mass gives mole amounts close to 0.430, 0.860, and 0.430 mol, which reduce to 1:2:1. The empirical formula is therefore CH₂O, with an empirical formula mass of about 30.026 g/mol. Dividing 90.0779 g/mol by that mass gives approximately 3, so the molecular formula is C₃H₆O₃.