The compound X2(Y Z3)3 has a molar mass of 282.23 g and a...

The compound $X_2\left(Y Z_3\right)_3$ has a molar mass of 282.23 g and a percent composition (by mass) of $19.12 \% \mathrm{X}$, $29.86 \% \mathrm{Y}$, and $51.02 \% \mathrm{Z}$. What is the compound?

Key Concepts

Percent Composition

Percent composition represents the mass percentage of each element within a compound. It is calculated by dividing the mass of the element in one mole of the compound by the molar mass of the compound, then multiplying by 100. This concept allows chemists to relate the masses of the different elements to each other, which is essential when deducing the empirical or molecular formula from experimental data.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole. It is obtained by summing the atomic masses of all atoms in the compound’s formula. Knowing the molar mass of a compound, along with its percent composition, is crucial when reconciling the experimental data with the theoretical formula, particularly in scaling the empirical formula to the molecular formula.

Empirical Formula Determination

The empirical formula of a compound is the simplest whole-number ratio of the elements present. It is determined by converting the percent composition to masses (typically using an assumed 100 g sample), then converting these masses to moles using the atomic masses, and finally dividing by the smallest number of moles present. This process highlights the fundamental stoichiometric relationships between the atoms in a compound.

Molecular Formula Calculation

The molecular formula of a compound shows the actual number of atoms of each element in a molecule. It can be determined by comparing the empirical formula mass to the compound’s molar mass. By finding a multiplying factor that relates the empirical mass to the molar mass, the empirical formula is scaled up to the molecular formula. This method bridges the gap between the simplest ratio of elements and the actual composition of the molecule.

Transcript

Please give Ace some feedback