What is the molecular formula of a compound that has a molecular mass of 54 and the empirical formula C_2H_3C2H3?

1 Answer
anor277
May 14, 2017

C_4H_6C4H6.............

Explanation:

The "molecular formula"molecular formula is always a whole number multiple of the "empirical formula"empirical formula.

i.e. "molecular formula"=nxx("empirical formula")molecular formula=n×(empirical formula)

54*g*mol^-1=nxx(2xx12.011+3xx1.00794)*g*mol^-154⋅g⋅mol−1=n×(2×12.011+3×1.00794)⋅g⋅mol−1

=nxx(27.04)*g*mol^-1=n×(27.04)⋅g⋅mol−1

Clearly, n=(54.0*g*mol^-1)/(27.04*g*mol^-1)=2n=54.0⋅g⋅mol−127.04⋅g⋅mol−1=2

"Molecular formula"=(C_2H_3)xx2=C_4H_6Molecular formula=(C2H3)×2=C4H6 as required.

This compound has two degrees of unsaturation, so it is a butadiene, or a cylocbutene, or an allene derivative, or an alkyne.........

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