Question

A compound with the empirical formula CH2 has a mass of 56 amu. What is the molecular formula?

Answer 1

Would it not be `C_4H_8`...?

Explanation:

The molecular formula is always a whole number multiple of the empirical formula....we were quoted an empirical formula of `CH_2`...and so, since `"{empirical formula}"xxn=underbrace(56*"amu")_"quoted molecular mass"`

`(12.01+2xx1.01)` `"amu"xxn=56*"amu"`...

CLEARLY `n=4`...and molecular formula is `C_4H_8`. This is a RING or an olefin to account for its ONE degree of unsaturation. .

Answer 2

`"C"_4"H"_8`

Explanation:

First, find the mass of the empirical formula:

`"CH"_2`

`12.01 + 1.00(2)`
`12.01 + 2.00`
`14.01`

Divide the molecular formula's mass by the empirical formula's mass:

`(56 " amu")/(14.01 " amu")=3.99`

You can round `3.99` up to `4`.

Finally, multiply each term in the empirical formula by `4` to get your final answer:

`color(red)("C"_4"H"_8)`

Answer 3

The answer is `C_"4"H_"8."`

Explanation:

Here the empirical formula is `CH_"2"` and its mass of `56` amu is given:

It must be clear that the molecular formula is gonna be some factor of the empirical formula given, and let's suppose that factor `n`;

Thus:

`(CH_"2")n = 56 amu`

Mass of `CH_"2" =` Mass of C + 2 * Mass of H
or, Mass `= 12 + 2 * 1 = 14` grams

So,
`(14)n =56`
`or, n = 56/14`
`or, n = 4`

So, Molecular formula = `(CH_"2")*4 = C_"4"H_"8"`