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

3 Answers
May 15, 2018

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. .

May 15, 2018

#"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)#

May 15, 2018

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"#