Question

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?

Answer 1

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.