# 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?

Oct 9, 2016

We calculate (i) the percentage composition, and then (ii) the emprical formula, to give an empirical formula of ${C}_{5} {H}_{10} N {O}_{5} N {a}_{2}$.

#### 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 $100 \cdot g$ of unknown compound, and dividing the individual, elemental masses by the ATOMIC mass of each component element:

$C :$ $=$ $\frac{32.1 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.67 \cdot m o l$.

$H :$ $=$ $\frac{5.39 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1}$ $=$ $5.35 \cdot m o l$.

$O :$ $=$ $\frac{42.8 \cdot g}{15.999 \cdot g \cdot m o {l}^{-} 1}$ $=$ $2.67 \cdot m o l$.

$N :$ $=$ $\frac{7.50 \cdot g}{14.01 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.535 \cdot m o l$.

$N a :$ $=$ $\frac{21.49 \cdot g}{22.99 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.935 \cdot m o l$.

Now we divide each molar quantity thru by the SMALLEST MOLAR QUANTITY, that of nitrogen, to give an empirical formula of ${C}_{5} {H}_{10} N {O}_{5} N {a}_{2}$. 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.