# A compound is found be 36.5% Na, 25.3% S, and 38.0% O. What is its empirical formula?

May 19, 2016

$N {a}_{2} S {O}_{3}$

#### Explanation:

To solve for an empirical formula, when given the percent compositions, first, divide each percent composition by the atomic mass of THAT element.

1. (36.5% Na)/(22.99 g/(mol) Na (25.3% S)/(32.07 g/(mol) S (38.0% O)/(16 g/(mol) O
(considering that this is not oxygen gas). For these atomic masses, I rounded to the nearest hundredth.

You get: $1.587$ $0.788$ $2.375$ (in corresponding order to the above equations- these values don't have units).

1. Then, you divide each of these values by the SMALLEST of these values.
$\frac{1.587}{0.788}$ $\frac{0.788}{0.788}$ $\frac{2.375}{0.788}$
Then, still corresponding to Na, S, and O respectively, we get:
$2.013$ mol Na= $2$ mol Na
$1$ mol S
$3.013$ mol O= $3$ mol O
(after rounding the answers)

Therefore, the empirical formula is $N {a}_{2} S {O}_{3}$