3. a compound was analyzed and found to contain 9.8 g of nitrogen, 0.70 g of
hydrogen, and 33.6 g of oxygen. what is the empirical formula of the compound?

HNO₃

Explanation:

Data given

Nitrogen = 9.8 g

Hydrogen =  0.70 g

Oxygen = 33.6 g

Empirical formula = ?

Solution:

Convert the masses to moles

For Nitrogen

Molar mass of N = 14 g/mol

                              no. of mole = mass in g / molar mass

Put value in above formula

                          no. of mole = 9.8 g/ 14 g/mol

                          no. of mole = 0.7

                           mole of N = 0.7 mol

For Hydrogen

Molar mass of H = 1 g/mol

                     no. of mole = mass in g / molar mass

Put value in above formula

                     no. of mole = 0.70 g/ 1 g/mol

                      no. of mole = 0.7

mole of H = 0.7 mol

For Oxygen

Molar mass of O = 16 g/mol

                       no. of mole = mass in g / molar mass

Put value in above formula

                       no. of mole = 33.6 g / 16 g/mol

                       no. of mole = 2.1

mole of O = 2.1 mol

Now we have values in moles as below

N = 0.7

H = 0.7

O = 2.1

Divide the all values on the smallest values to get whole number ratio

N = 0.7 / 0.7 = 1

H = 0.7 / 0.7 = 1

O = 2.1 / 0.7 = 3

So all have following values

N = 1

H = 1

O = 3

So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.

increased stability in the completion of fluorine's outer shell.