Hello
To solve this question, let's change the percentages into grams.
[tex]\begin{gathered} O=45g \\ Na=33g \\ S=22g \end{gathered}[/tex]Step 2
convert the grams into moles
but the formula of mole is given as
[tex]n=\frac{\text{mass}}{\text{molar mass}}[/tex]Let's do this
[tex]\begin{gathered} O=\frac{45}{16}=2.8125moles \\ Na=\frac{33}{23}=1.434moles \\ S=\frac{22}{32}=0.6875\text{moles} \end{gathered}[/tex]Lastly, we take the ratio of the atoms
[tex]\begin{gathered} \frac{O}{S}=\frac{2.8125}{0.6875}=4.09 \\ \frac{Na}{S}=\frac{1.434}{0.6875}=2.085 \end{gathered}[/tex]From the calculations above, using the ratio of the atoms, the emperical formula of the compound is Na₂SO₄
