A 175.0 g sample of a compound contains 56.15 g #C#, 9.43 g #H3, 74.81 g #O#, 13.11 g #N#, and 21.49 g #Na.# What is its empirical formula?

1 Answer
Oct 9, 2016

Answer:

We calculate (i) the percentage composition, and then (ii) the emprical formula, to give an empirical formula of #C_5H_10NO_5Na_2#.

Explanation:

We need the percentage composition of each element:

#%C:# #(56.15*g)/(175.0*g)xx100%=32.1%#

#%H:# #(9.43*g)/(175*g)xx100%=5.39%#

#%O:# #(74.81*g)/(175*g)xx100%=42.8%#

#%N:# #(13.11*g)/(175*g)xx100%=7.50%#

#%Na:# #(21.49*g)/(175*g)xx100%=12.28%#

Given this percentage makeup, we can work out an empirical formula by (i) assuming that there are #100*g# of unknown compound, and dividing the individual, elemental masses by the ATOMIC mass of each component element:

#C:# #=# #(32.1*g)/(12.011*g*mol^-1)# #=# #2.67*mol#.

#H:# #=# #(5.39*g)/(1.00794*g*mol^-1)# #=# #5.35*mol#.

#O:# #=# #(42.8*g)/(15.999*g*mol^-1)# #=# #2.67*mol#.

#N:# #=# #(7.50*g)/(14.01*g*mol^-1)# #=# #0.535*mol#.

#Na:# #=# #(21.49*g)/(22.99*g*mol^-1)# #=# #0.935*mol#.

Now we divide each molar quantity thru by the SMALLEST MOLAR QUANTITY, that of nitrogen, to give an empirical formula of #C_5H_10NO_5Na_2#. Note that the sodium percentage was a bit out, but clearly we want WHOLE numbers, so a bit of rounding up is acceptable.

Now of course, I have done this the long way round, and made a real meal out of the the problem. I could have missed the step where the percentages are calculated, and been a bit more direct. I did it the long way, because often you are quoted percentage composition by mass in empirical formula problems.