Answer: The empirical formula for the given compound is
![NO_2](https://img.qammunity.org/2021/formulas/chemistry/college/fr6ouvbu4181g8uj71kgl3lnkijqnuj6wh.png)
Explanation : Given,
Mass of O = 0.370 g
Mass of N = 0.130 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Oxygen =
![\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=(0.370g)/(16g/mole)=0.0231moles](https://img.qammunity.org/2021/formulas/chemistry/high-school/vc0h7rv53na7drrz94vc7u0zwrmkbcxwjt.png)
Moles of Nitrogen =
![\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=(0.130g)/(14g/mole)=0.00928moles](https://img.qammunity.org/2021/formulas/chemistry/high-school/q30yuzjuumt93k2iestxhx3llfwz0ef0zh.png)
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00928 moles.
For Oxygen =
![(0.0231)/(0.00928)=2.4\approx 2](https://img.qammunity.org/2021/formulas/chemistry/high-school/gv2vyoc5cwcopyl636wf3eqwlobr9zt3nu.png)
For Nitrogen =
![(0.00928)/(0.00928)=1](https://img.qammunity.org/2021/formulas/chemistry/high-school/ergd5bq7x0kw6ucg9ye0yhd3y9f5bsaqv2.png)
Step 3: Taking the mole ratio as their subscripts.
The ratio of O : N = 2 : 1
Hence, the empirical formula for the given compound is
![NO_2](https://img.qammunity.org/2021/formulas/chemistry/college/fr6ouvbu4181g8uj71kgl3lnkijqnuj6wh.png)