What is the empirical formula for a compound that contains 79.86 % iodine and 20.14 % oxygen by mass?

IO₂

Explanation:

We have been given the mass percentages of the elements that makes up the compound:

Mass percentage given are:

Iodine = 79.86%

Oxygen = 20.14%

To calculate the empirical formula which is the simplest formula of the compound, we follow these steps:

> Express the mass percentages as the mass of the elements of the compound.

> Find the number of moles by dividing through by the atomic masses

> Divide by the smallest and either approximate to nearest whole number or multiply through by a factor.

> The ratio is the empirical formula of the compound.

Solution:

I O

% of elements 79.86 20.14

Mass (in g) 79.86 20.14

Moles(divide by

Atomic mass) 79.86/127 20.14/16

Moles 0.634 1.259

Dividing by

Smallest 0.634/0.634 1.259/0.634

1 2

The empirical formula is IO₂

i think it is d. e because the next one is igneous.

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).