Question

Determine the empirical formula for a compound that is 36.86% n and 63.14% o by mass.

Answer 1

Answer:

$NO_{1.499}$

Explanation:

Let assume that 100 kg of the compound is tested. The quantity of kilomoles for each element are, respectively:

$n_{N} = \frac{36.86\,kg}{14.006\,\frac{kg}{kmol} }$

$n_{N} = 2.632\,kmol$

$n_{O} = \frac{63.14\,kg}{15.999\,\frac{kg}{kmol} }$

$n_{O} = 3.946\,kmol$

Ratio of kilomoles oxygen to kilomole nitrogen is:

$n^{*} = \frac{3.946\,kmol}{2.632\,kmol}$

$n^{*}= 1.499$

It means that exists 1.499 kilomole oxygen for each kilomole nitrogen.

The empirical formula for the compound is:

$NO_{1.499}$

Answer 2

Answer:

N₂O₃

Explanation:

Empirical Formula: The Empirical formula of a compound is the smallest formula of that compound that tells us the component element in the molecule of that compound and the ratio in which these elements are combined together.

From the question,

Given: N = 36.86 %, 0 = 63.14 %

Step 1: divide by their relative atomic mass.

N                 :              O

36.86/14                   63.14/16

2.633           :              3.946

Step 2: Divide by the smallest

N                       :          O

2.633/2.633                3.946/2.633

1                       :            1.5

1                       :           3/2

2                       :            3

Hence the empirical formula = N₂O₃