Question

a compound is found to contain 7.522 % phosphorus and 92.47 % iodine by mass. to answer the question, enter the elements in the order presented above. question 1: the empirical formula for this compound is. question 2: the molar mass for this compound is 411.7 g/mol. the molecular formula for this compound is.

Explanation:

Step1: Assume 100 g of the compound

The mass of phosphorus ($P$) is $7.522$ g and the mass of iodine ($I$) is $92.47$ g.

Step2: Calculate the moles of each element

The molar - mass of $P$ is $M_{P}=30.97$ g/mol, so the moles of $P$, $n_{P}=\frac{7.522\ g}{30.97\ g/mol}\approx0.243$ mol. The molar - mass of $I$ is $M_{I}=126.9$ g/mol, so the moles of $I$, $n_{I}=\frac{92.47\ g}{126.9\ g/mol}\approx0.729$ mol.

Step3: Find the mole - ratio

Divide each number of moles by the smaller number of moles ($n_{P} = 0.243$ mol). For $P$, $\frac{0.243\ mol}{0.243\ mol}=1$. For $I$, $\frac{0.729\ mol}{0.243\ mol}=3$.

Step1: Calculate the empirical - formula mass

The empirical formula is $PI_{3}$. The empirical - formula mass $M_{empirical}=30.97\ g/mol+3\times126.9\ g/mol=30.97\ g/mol + 380.7\ g/mol=411.67\ g/mol$.

Step2: Find the ratio of molar mass to empirical - formula mass

The molar mass $M = 411.7$ g/mol. The ratio $n=\frac{M}{M_{empirical}}=\frac{411.7\ g/mol}{411.67\ g/mol}\approx1$.

Step3: Determine the molecular formula

Since $n = 1$, the molecular formula is the same as the empirical formula.

Answer:

$PI_{3}$