What is the molecular formula of a compound that has a molecular mass of 54 and the empirical formula C2H3?

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What is the molecular formula of a compound that has a molecular mass of 54 and the empirical formula C2H3?

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Answer 1 · 2017-05-23T02:02:55

The molecular formula is $C_{4} H_{6}$ $"C"_4"H"_6"$ .

Refer to the explanation for the process.

Explanation:

You have been given the empirical formula for a hydrocarbon. You need to determine the empirical formula mass. Then divide the molecular mass by the empirical mass. Then multiply the subscripts of the empirical formula by the result.

${Empirical Mass of C}_{2} H_{3}$ $color(blue)("Empirical Mass of C"_2"H"_3"$
Multiply the subscript of each element by its atomic mass and add.

$C_{2} H_{3} :$ $"C"_2"H"_3:$ $(2 \times 12.011 u C) + (3 \times 1.008 u H) = {27.046 u C}_{2} H_{3}$ $(2xx12.011"u C") + (3xx1.008 "u H")="27.046 u C"_2"H"_3"$

$Determine Molecular Formula$ $color(blue)("Determine Molecular Formula"$
Divide the molecular mass by the empirical mass.

$\frac{54}{27.046} = 2.0$ $54/(27.046)=2.0$ rounded to two sig figs.

Multiply the subscripts in the empirical formula by 2 to get the molecular formula.

$C_{2 \times 2} H_{3 \times 2} {=C}_{4} H_{6}$ $""C_(2xx2)""H_(3xx2)"=C"_4"H"_6"$

The molecular formula is $C_{4} H_{6}$ $"C"_4"H"_6"$ .

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What is the molecular formula of a compound that has a molecular mass of 54 and the empirical formula C2H3?

1 Answer

Explanation:

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