Answer
$H_{2}SO_{3}$
H = 2.44%
S = 39.06%
O = 58.5%
Work Step by Step
We can tell from the acid name with $-ous$ suffix that the compound is a sulfite ion with its charge canceled out by $H^{+}$ ions. Since sulfite($SO_{3}^{-2}$) has a $-2$ charge we need 2 $H^{+}$ to balance the charge. Therefore the formula is $H_{2}SO_{3}$. Adding up the molar masses of the constituent elements we get $82.07\frac{g}{mol}$. So to find the percent composition we just multiply the molar mass of each element (found on the periodic table) by how many times it occurs, divide by the molar mass of the whole molecule and multiply by 100%. Here is Sulfur as an example:
$\frac{(32.06\frac{g}{mol})}{82.07\frac{g}{mol}}\times100\%= 39.1\%$
