Question

question 1
what is the empirical formula of a compound that is 17% sodium, 47.4% sulfur and 35.6% oxygen?

Explanation:

Step1: Assume 100g of the compound

Assume we have 100g of the compound. Then the masses are: \(m_{Na}=17g\), \(m_{S}=47.4g\), \(m_{O}=35.6g\)

Step2: Calculate the number of moles

The molar mass of \(Na\) is \(M_{Na} = 22.99g/mol\), of \(S\) is \(M_{S}=32.07g/mol\), and of \(O\) is \(M_{O} = 16.00g/mol\).
The number of moles:
\(n_{Na}=\frac{m_{Na}}{M_{Na}}=\frac{17g}{22.99g/mol}\approx0.74mol\)
\(n_{S}=\frac{m_{S}}{M_{S}}=\frac{47.4g}{32.07g/mol}\approx1.48mol\)
\(n_{O}=\frac{m_{O}}{M_{O}}=\frac{35.6g}{16.00g/mol}\approx2.23mol\)

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (\(n_{Na}\approx0.74mol\))
\(r_{Na}=\frac{n_{Na}}{n_{Na}} = 1\)
\(r_{S}=\frac{n_{S}}{n_{Na}}=\frac{1.48mol}{0.74mol}=2\)
\(r_{O}=\frac{n_{O}}{n_{Na}}=\frac{2.23mol}{0.74mol}\approx3\)

Answer:

\(NaS_{2}O_{3}\)