Question

There is 45% oxygen, 33% sodium and 22% sulfur. What is the empirical formula?

Answer 1

Hello

To solve this question, let's change the percentages into grams.

`\begin{gathered} O=45g \\ Na=33g \\ S=22g \end{gathered}`

Step 2

convert the grams into moles

but the formula of mole is given as

`n=\frac{\text{mass}}{\text{molar mass}}`

Let's do this

`\begin{gathered} O=(45)/(16)=2.8125moles \\ Na=(33)/(23)=1.434moles \\ S=(22)/(32)=0.6875\text{moles} \end{gathered}`

Lastly, we take the ratio of the atoms

`\begin{gathered} (O)/(S)=(2.8125)/(0.6875)=4.09 \\ (Na)/(S)=(1.434)/(0.6875)=2.085 \end{gathered}`

From the calculations above, using the ratio of the atoms, the emperical formula of the compound is Na₂SO₄