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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

Sagot :

Sebassandinidn

Answer:

[tex]PNCl_2[/tex]

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:

[tex]n_P=\frac{26.73g}{30.97g/mol}=0.863mol\\\\n_N=\frac{12.09g}{14.01g/mol}=0.863mol\\\\n_C_l=\frac{61.18g}{35.45g/mol}=1.726mol[/tex]

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):

[tex]P=\frac{0.863mol}{0.863mol}=1\\\\N= \frac{0.863mol}{0.863mol}=1\\\\Cl=\frac{1.726mol}{0.863mol}2[/tex]

Thus, the empirical formula is:

[tex]PNCl_2[/tex]

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