Question

A compound of nitrogen and oxygen is 30.46 percent by mass of N and 69.54 percent by mass O. The molar mass if the compound was determined to be 92g/mol. what is the emperical formula and molecular formular

Answer 1

Answer: Empirical formula is
`NO_2` and molecular formula is
`N_2O_4`

Explanation: To find the empirical and molecular formula, we follow few steps:

Step 1: Converting mass percent into mass

We are given the percentage of elements by mass. So, the total mass take will be 100 grams.

Therefore, mass of nitrogen =
`(30.46)/(100)* 100g=30.46g`

Similarly, mass of oxygen =
`(69.54)/(100)* 100g=69.54`

Step 2: Converting the masses into their respective moles

We use the formula:

`Moles=\frac{\text{Given mass}}{\text{Molar mass}}`

Molar mass of Nitrogen = 14 g/mol

Molar mass of oxygen = 16 g/mol

Moles of nitrogen =
`(30.46g)/(14g/mol)=2.18moles`

Moles of oxygen =
`(69.54g)/(16g/mol)=4.35mol`

Step 3: Getting the mole ratio of nitrogen and oxygen by dividing the calculated moles by the lowest mole value.

Mole ratio of nitrogen =
`(2.18)/(2.18)=1`

Mole ratio of oxyegn =
`(4.35)/(2.18)=1.99\approx 2`

Step 4: The mole ratio of elements are represented as the subscripts in a empirical formula, we get

Empirical formula =
`N_1O_2=NO_2`

Step 5: For molecular formula, we divide the molar mass of the compound by the empirical molar mass.

Empirical molar mass of
`NO_2=(14* 1)+(16* 2)g/mol`

Empirical molar mass = 46 g/mol

Molar mass of the compound = 92 g/mol

`n=\frac{\text{Molar mass}}{\text{Empirical molar mass}}`

`n=(92g/mol)/(46g/mol)=2`

Now, multiplying each of the subscript of empirical formula by 'n', we get

Molecular formula =
`N_2O_4`