The combustion of 40.5 mg of a compound containing C, H, and O, and extracted from the bark of t

step 1 So in the question here, we have a combustion of a C -H -O compound, C -X -H -Y -O -Z, that has

The combustion of 40.5 mg of a compound containing C, H, and...

The combustion of 40.5 mg of a compound containing $C$, $\mathrm{H},$ and $\mathrm{O},$ and extracted from the bark of the sassafras tree, produces $110.0 \mathrm{mg}$ of $\mathrm{CO}_{2}$ and $22.5 \mathrm{mg}$ of $\mathrm{H}_{2} \mathrm{O} .$ The molar mass of the compound is $162 \mathrm{g} / \mathrm{mol}$. Determine its empirical and molecular formulas.

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Key Concepts

Molecular Formula Determination

After determining the empirical formula, its molar mass is compared to the compound's actual molar mass. The actual molecular formula is found by multiplying the subscripts in the empirical formula by an appropriate integer factor, ensuring that the total mass corresponds to the known molar mass of the compound.

Combustion Analysis

This method involves burning an organic compound to produce CO? and H?O, which release carbon and hydrogen respectively. By measuring the amounts of these products, one can calculate the number of moles of carbon and hydrogen that were originally present in the compound. Oxygen content is then determined indirectly, typically by difference, providing a complete basis for analysis.

Empirical Formula Determination

This concept refers to finding the simplest integer ratio of the elements in a compound. In combustion analysis, the masses of CO? and H?O produced are converted to moles of carbon and hydrogen, and any remaining mass is attributed to oxygen. The resulting mole ratios are then simplified to their smallest whole numbers to obtain the empirical formula.

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