Question

(10 points) What is the empirical formula for a compound that is 54.48% carbon, 13.75% hydrogen, and 31.77 % nitrogen?

Answer 1

The Empirical formula will be for given compound that is 54.48% carbon, 13.75% hydrogen, and 31.77 % nitrogen is  $C_2H_6N$ .

Explanation:

The formula which gives the simple whole number ratio of the atoms of various elements present in one molecule of the compound is understood as empirical formula. This can be more appropriately understood by solving the given problem as follows:

The % of elements given as:

C = 54.48%

H = 13.75%

N = 31.77%

The atomic mass of elements given as:

C = 12 g

H = 1 g

N = 14 g

Calculating number of moles as :

$N = \frac{Mass}{Molecular-Mass}$

 $C = \frac{54.48}{12} = 4.54$

  $H= \frac{13.75}{1} = 13.75$

  $N = \frac{31.77}{14} = 2.26$

Now, we will divide the least number of moles from number of moles of  C, H and N respectively, we will round off the values to an integer value:

For C:     $\frac{4.54}{2.26} = 2$

For H:      $\frac{13.75}{2.26} = 6$

For N:      $\frac{2.26}{2.26} = 1$

Hence, C, H and N are in the ratio of 2:6:1, so empirical formula will be for given compound that is 54.48% carbon, 13.75% hydrogen and 31.77 % nitrogen is $C_2H_6N$.