If the empirical formula is #C_2H_5#, and the molecular mass is #58*g*mol^-1#, what is the molecular formula?
If I am reading your question correctly, it is
Butane is a
I assume that you have to determine the molecular formula of butane starting from its empirical formula and its molar mass, which is
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
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
So, use the molar masses of carbon and hydrogen to get the value of
#(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)#