For a compound NaAc, the percent composition is that: 29.28% C, 3.68% H, 39.01% O, and 28.03% Na. What is the empirical formula of this compound?

Apr 5, 2017

${\text{NaC"_2"H"_3"O}}_{2}$

Explanation:

Convert the percentages to grams. If there were 100 g, each 1 % would equal 1 g.

So,

$\text{29.28 % C = 29.28g C}$
$\text{3.68 % H = 3.68 g H}$
$\text{39.01% O" = "39.01 g O}$
$\text{28.03% Na" = "28.03 g Na}$

Then convert the grams to moles

 ("29.28 g")/("12.01 g/mol") = "2.438 mol C"

("3.68 g")/ ("1.008 g/mol") = "3.65 mol H"

("39.01 g")/("16.00 g/mol") = "2.438 mol O"

 ("28.03 g")/("22.99 g/mol")= "1.219 mol Na"

Convert these numbers of moles to mole ratios by dividing all of them by the smallest mole number.

$\frac{2.438}{1.219} = \frac{2.000}{1} = \text{2 O: 1 Na}$

$\frac{2.438}{1.219} = \frac{2.000}{1} = \text{2 C : 1 Na}$

$\frac{3.65}{1.219} = \frac{2.99}{1} = \text{3 H : 1 Na}$

So, the molar ratio is $\text{1 Na : 2 C : 2 O : 3 H}$,

and the empirical formula is

${\text{NaC"_2"H"_3"O}}_{2}$