Question #40687

1 Answer
Fleur
Feb 12, 2018

#"C"_6"H"_15#

Explanation:

The molar mass of #"C"_2"H"_5# is approximately #(2 xx 12) + (5xx1) = "29 g/mol"#.

We know that the molecular formula is a whole-number multiple of the empirical formula.

Since the molecular mass is #"87 g/mol"#, there are #"87 g/mol"/"29 g/mol" = 3# units of #"C"_2"H"_5# in the actual compound.

Thus, the molecular formula is #("C"_2"H"_5)_3# or #"C"_6"H"_15#.

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