A compound was analyzed and found to have the following percent of composition: Nitrogen 11.64% and Chlorine 88.36 %. Determine the empirical formula for the compound ( in order to solve this problem consider the percentages as if they were masses of the respective elements and follow the process to find the empirical formula for the compound).

Nitrogen - moles: 11.64/14 = 0.83

Chlorine - moles: 88.36/35.5 = 2.489

Ratio - Chlorine (0.83/0.83 = 1) - Nitrogen (2.489/0.83 = 2.998)

Empirical formula NCl₃

hope that helps

Answer Link

kumaripoojavt kumaripoojavt · Answer 2 · 2022-07-23T11:18:32+02:00

NCl₃ is the empirical formula for the compound.

What are moles?

A mole is defined as 6.02214076 × [tex]10^{23}[/tex]of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Moles of nitrogen : [tex]\frac{mass}{molar \;mass}[/tex]

Nitrogen - moles: 11.64 ÷ 14 = 0.83

Moles of Chlorine: [tex]\frac{mass}{molar \;mass}[/tex]

Chlorine - moles: 88.36 ÷ 35.5 = 2.489

Ratio:

Chlorine (0.83 ÷ 0.83 = 1) : Nitrogen (2.489 ÷ 0.83 = 2.998)

The whole number would be 1:3

Empirical formula NCl₃

Learn more about moles here:

https://brainly.com/question/8455949

#SPJ2