Question

A gas is found to contain 2.34g of Nitrogen and 5.34g of oxygen. Simplest formula of the compound is:
A.${N_2}O$
B.$NO$
C.${N_2}{O_3}$
D.$N{O_2}$

Answer 1

This problem can be solved on the basis of mole concept. We shall find the moles of the compounds given and use it to find the ratio of the moles of the elements present in the compound.

Formula used: ${\text{n = }}\dfrac{{{\text{mass}}}}{{{\text{at}}{\text{.wt}}{\text{.}}}}$

Complete step by step answer:
The mass of Nitrogen in the gas =$2.34$g, at. Wt. of nitrogen = 7
The mass of Oxygen in the gas = $5.34$g, at. Wt. of oxygen = 8.
According to the formula,
no. of moles of a substance, ${\text{n = }}\dfrac{{{\text{mass}}}}{{{\text{at}}{\text{.wt}}{\text{.}}}}$
Therefore, number of moles for nitrogen:
${{\text{n}}_{{{\text{N}}_{\text{2}}}}} = \dfrac{{2.34}}{7}$
And number of moles for oxygen:
${{\text{n}}_{{{\text{O}}_{\text{2}}}}} = \dfrac{{5.34}}{8}$
Ratio of the number of moles of nitrogen to the number of moles of oxygen is:
$\dfrac{{2.34}}{7} \times \dfrac{{5.34}}{8} = 1:2$
Therefore the formula of the compound is $N{O_2}$.

Hence, option D is correct.

Notes:
1.One mole of a substance is defined as the weight of $6.023 \times {10^{23}}$atoms/molecules/ions/electrons of that substance. This is equal to the gram-atomic weight of elements and gram molecular weight of compounds.
2.According to the Gay-Lussac’s law of combining volumes, the gases combine with each other to form products and these reactants and products bear a simple whole number ratio to each other under standard conditions of temperature and pressure.
Mathematically,
$aA + bB \to cC + dD$
Where, $a:b:c:d$ are in whole number ratios.
3.For the Gay-Lussac’s law to be valid, both the reactants as well as the products have to be in gaseous forms.
4.The law is only valid under standard conditions of temperature and pressure, i.e., $298{\text{K}}$ temperature and $1{\text{ atm}}$ pressure.