Question (a)calculate the empirical formula of a compound that has the following composition by mass. 29.0% na, 40.5% s, and the remainder being oxygen.

question b:
if the molecular mass of the compound in question(a) is found to be 158 g/mol, what is the molecular formula of the compound?

for question a i got naso2 is that right?

Empirical formula is  Na₂S₂O₃.

Molecular formula = Na₂S₂O₃

Explanation:

Given data:

Percentage of sodium = 29.0%

Percentage of sulfur = 40.5%

Molecular formula = ?

Empirical formula = ?

Solution:

Number of gram atoms of Na = 29.0/23 =1.261

Number of gram atoms of S = 40.5 / 32 = 1.266

Number of gram atoms of O = 30.5 / 16 = 1.906

Atomic ratio:

Na               :            S                  :    O

1.261/1.261 :       1.266/1.261     :   1.906/1.261

      1          :            1                    :         1.5

Na : S : O = 2  ( 1: 1 : 1.5)

Na : S : O = 2 : 2 : 3

Empirical formula is  Na₂S₂O₃.

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 158 / 158

n = 1

Molecular formula = n (empirical formula)

Molecular formula = 1 (Na₂S₂O₃)

Molecular formula = Na₂S₂O₃

what does the table look like