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

May 15, 2018

Would it not be ${C}_{4} {H}_{8}$...?

#### Explanation:

The molecular formula is always a whole number multiple of the empirical formula....we were quoted an empirical formula of $C {H}_{2}$...and so, since $\text{{empirical formula}"xxn=underbrace(56*"amu")_"quoted molecular mass}$

$\left(12.01 + 2 \times 1.01\right)$ $\text{amu"xxn=56*"amu}$...

CLEARLY $n = 4$...and molecular formula is ${C}_{4} {H}_{8}$. This is a RING or an olefin to account for its ONE degree of unsaturation. .

May 15, 2018

${\text{C"_4"H}}_{8}$

#### Explanation:

First, find the mass of the empirical formula:

${\text{CH}}_{2}$

$12.01 + 1.00 \left(2\right)$
$12.01 + 2.00$
$14.01$

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

$\left(56 \text{ amu")/(14.01 " amu}\right) = 3.99$

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

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

$\textcolor{red}{{\text{C"_4"H}}_{8}}$

May 15, 2018

The answer is ${C}_{\text{4"H_"8.}}$

#### Explanation:

Here the empirical formula is $C {H}_{\text{2}}$ 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:

$\left(C {H}_{\text{2}}\right) n = 56 a \mu$

Mass of $C {H}_{\text{2}} =$ Mass of C + 2 * Mass of H
or, Mass $= 12 + 2 \cdot 1 = 14$ grams

So,
$\left(14\right) n = 56$
$\mathmr{and} , n = \frac{56}{14}$
$\mathmr{and} , n = 4$

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