Question

What is the name of the empirical formula for a compound that is 36.86% N and 63.14% O by mass?

Answer 1

Given total percent = 36.86% + 63.14% = 100% hence ther are no other elements present in the given compound.

1. Divide each percent by its corresponding molar mass of the element.

2. Further divide the ratio comes from the point 1 by one of the least value to get nearest integer value.

Step 1:

Oxygen = 63.14 / 15.999 = 3.94649665604

Nitrogen = 36.86 / 14.0067 = 2.63159773537

Step 1:

now divide both ratio value by 2.63159773537 to get an integer value.

O:N = 3.94649665604 / 2.63159773537 : 2.63159773537 / 2.63159773537

=                        1.49965802258 : 1

still we have non-integer values, hence multiply by an integer 2 to get integer values

= 3:2 = O:N ratio of elements menas 3 oxygen atoms and 2 nitrogen atoms are present in the given compound.

This is the required coefficients as subscripts to the elements

Answer: empirical formula is    N₂O₃         and the name of the compound is      Dinitrogen trioxide

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