Question

An ionic compound is composed of the following elements: Nitrogen – 31.57% Hydrogen – 9.10% Phosphorus – 23.27% Oxygen – 36.06% What is the empirical formula for this compound?

Answer 1

Answer:

N3H12PO3

Explanation:

Step 1:

Data obtained from the question.

Nitrogen (N) = 31.57%

Hydrogen (H) = 9.10%

Phosphorus (P) = 23.27%

Oxygen (O) = 36.06%

Step 2:

Divide by their molar masses.

This is illustrated below:

N = 31.57/14 = 2.255

H = 9.10/ 1 = 9.1

P = 23.27/31 = 0.751

O = 36.06/16 = 2.254

Step 3:

Divide by the smallest number. This is illustrated below:

N = 2.255/0.751 = 3

H = 9.1 /0.751 = 12

P = 0.751 /0.751 = 1

O = 2.254/0.751 = 3

The empirical formula is N3H12PO3

Answer 2

Answer:

The empirical formula is: N3H12PO3

Explanation:

To calculate the empirical formula, the values ​​of each element must first be taken in 100gr of the compound:

• 31.57% Nitrogen = 31.57gr N

• 9.10% Hydrogen = 9.10 gr H

• 23.27% Phosphorus = 23.27gr P

• 36.06% Oxygen = 36.06gr O

Then the grams of each element are divided by their atomic weight:

• 31.57grN / 14U = 2,255 N

• 9.10grH / 1.001U = 9.091 H

• 23.27grP / 31U = 0.75 P

• 36.06grO / 16U = 2.253 O

Then each data is divided by the smallest number that is that of Phosphorus:

• 2,255 N / 0.75 = 3N

• 9,091 H / 0.75 = 12H

• 0.75 P / 0.75 = 1 P

• 2,253 O / 0.75 = 3 O

And the data we use is put into a formula of the compound:

• NHPO

And its values ​​are assigned:

• N3H12PO3