If the empirical formula is C_2H_5, and the molecular mass is 58*g*mol^-1, what is the molecular formula?

Sep 28, 2015

If I am reading your question correctly, it is ${H}_{3} C - C {H}_{2} C {H}_{2} C {H}_{3}$ or ${C}_{4} {H}_{10}$.

Explanation:

Butane is a ${C}_{4}$ alkane. The isomer depicted is the straight chain, n -butane. It has 1 other isomer.

Sep 29, 2015

${\text{C"_4"H}}_{10}$

Explanation:

I assume that you have to determine the molecular formula of butane starting from its empirical formula and its molar mass, which is $\cong \text{58 g}$.

If you started from butane's percent composition of carbon and hydrogen, and provided that your calculations were correct, then the empirical formula should have come out as ${\text{C"_2"H}}_{5}$.

Now, you know that the empirical formula tells you the simplest ratio of atoms between the elements that make up a compound.

In your case, you know that a butane molecule contains 5 atoms of hydrogen for every 2 atoms of carbon.

Since you know what the compound's molar mass is, you can say that

(2 xx M_"M carbon" + 5 xx M_"M hydrogen") * color(blue)(n) = "58 g"

You know that the molecule contains at least two atoms of carbon and five atoms of hydrogen, but you don't exactly how many of each you have $\to$ this is what $\textcolor{b l u e}{n}$ represents.

So, use the molar masses of carbon and hydrogen to get the value of $\textcolor{b l u e}{n}$

(2 xx "12.01 g/mol" + 5 xx "1.01 g/mol") * color(blue)(n) = "58 g"

color(blue)(n) = (58color(red)(cancel(color(black)("g"))))/(29.07color(red)(cancel(color(black)("g")))) = "1.995 ~= 2

This means tha tthe molecular formula of butane is

$\left({\text{C"_2"H"_5)_2 = color(green)("C"_4"H}}_{10}\right)$