Question

A compound is found to contain 10.37 % phosphorus, 4.689 % nitrogen, and 84.95 % iodine by weight. The molecular weight for this compound is 298.8 g/mol. What is the molecular formula for this compound

Answer 1

P₂N₂I₄

Step-by-step explanation:

Firstly, we calculate the empirical formula of the compound. To do this we find the moles of each element. Treat the percentages as masses.

The formula for moles is:

`moles = (mass)/(relative \: atomic \: mass)`

The masses of each element are:

Phosphorus = 10.37g

Nitrogen = 4.689g

Iodine = 84.95g

The relative atomic masses of each element are:

Phosphorus = 31.0

Nitrogen = 14.0

Iodine = 126.9

Using these values, we can calculate the moles of each element:

`moles \: of \: phosphorus = (10.37)/(31.0) = 0.335`

`moles \: of \: nitrogen \: = \: (4.689)/(14.0) = 0.335`

`moles \: of \: iodine \: = \: (84.95)/(126.9) = 0.669`

Using the moles, write them in a ratio to find the ratio of moles:

Phosphorus : nitrogen: iodine

0.335 : 0.335: 0.669

The smallest number here is 0.335, so divide each number by 0.335

0.335÷0.335=1

0.335÷0.335=1

0.669÷0.335=1.99... = 2 (round to the nearest whole number)

These numbers are used as the subscripts of each element.

This means that the empirical formula is:

PNI₂

To calculate the molecular formula:

`(molar \: mass \: of \: molecular \: formula)/(molar \: mass \: of \: empirical \: formula)`

The molar mass of empirical formula is adding the molar masses of compound PNI2:

31.0 + 14.0 + 126.9 = 171.9

The molar mass of molecular weight or mass = 298.8 (given from the question)

Substitute these in the formula above:

`(298.8)/(171.9) = 1.738...`

1.738 = 2 rounded to the nearest whole number.

Multiply all the subscripts seen in the empirical formula by the number 2:

PNI₂ × 2 = P₂N₂I₄