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

2 Answers
Sep 28, 2015

Answer:

If I am reading your question correctly, it is #H_3C-CH_2CH_2CH_3# or #C_4H_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

Answer:

#"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 #~= "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 #"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 #-># this is what #color(blue)(n)# represents.

So, use the molar masses of carbon and hydrogen to get the value of #color(blue)(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

#("C"_2"H"_5)_2 = color(green)("C"_4"H"_10)#