Question

3.5 g of nitrogen combines with 2 g of oxygen to form oxide of nitrogen. What is the
empirical formula of this oxide.
step by step answers please

Answer 1

Answer:

N2O

Explanation:

no. of moles in N2 = 3.5/14 =1/4 = 0.25 --------eq1

no of moles in O2 = 2/16= 1/8 = 0.125 ----------eq2

Divide smallest no. (1/8) in both equation

0.25 / 0.125 = 2 -------eq3

&

0.125/0.125 = 1 -----------eq4

Eq3 represents N-atoms

Eq4 represents O-atoms

Therefore, empirical formula is 'N2O'

2 is written under N

Answer 2

Given: 3.5 g of nitrogen combines with 2 g of oxygen to form oxide of nitrogen.

To find: We have to find the empirical formula of this oxide.

Solution:

The molecular weight of nitrogen is 28.

The molecular weight of oxygen is 32.

Number of moles of nitrogen will be-

$\frac{3.5}{28} = \frac{1}{8}$

Number of moles of oxygen will be-

$\frac{2}{32} = \frac{1}{16}$

Smallest number of mole is 1/16.

So, we have divide number of moles by 1/16.

Number of nitrogen atom is

$\frac{1}{8} \div \frac{1}{16} = 2$

So, 2 nitrogen atoms will present.

Number of oxygen atom is-

$\frac{1}{16} \div \frac{1}{16} = 1$

So, 1 oxygen atom will present.

Thus emperical formula of the oxide is $N_2O$.