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 = $\frac{30.46}{100}\times 100g=30.46g$

Similarly, mass of oxygen = $\frac{69.54}{100}\times 100g=69.54$

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 = $\frac{30.46g}{14g/mol}=2.18moles$

Moles of oxygen = $\frac{69.54g}{16g/mol}=4.35mol$

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

Mole ratio of nitrogen = $\frac{2.18}{2.18}=1$

Mole ratio of oxyegn = $\frac{4.35}{2.18}=1.99\approx 2$

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 $NO_2=(14\times 1)+(16\times 2)g/mol$

Empirical molar mass = 46 g/mol

Molar mass of the compound = 92 g/mol

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

$n=\frac{92g/mol}{46g/mol}=2$

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

Molecular formula = $N_2O_4$