Question

A compound is found to contain 26.73 % phosphorus, 12.09 % nitrogen, and 61.18 % chlorine by mass. What is the empirical formula for this compound

Answer 1

`PNCl_2`

Step-by-step explanation:

Hello!

In this case, when determining empirical formulas by knowing the by-mass percent, we first must assume the percentages as masses so we can compute the moles of each element:

`n_P=(26.73g)/(30.97g/mol)=0.863mol\\\\n_N=(12.09g)/(14.01g/mol)=0.863mol\\\\n_C_l=(61.18g)/(35.45g/mol)=1.726mol`

Now, for the determination of the subscript of each element in the empirical formula, we divide the moles by the fewest moles (P or N):

`P=(0.863mol)/(0.863mol)=1\\\\N= (0.863mol)/(0.863mol)=1\\\\Cl=(1.726mol)/(0.863mol)2`

Thus, the empirical formula is:

`PNCl_2`

Regards!