Question

question 2
1 pts
a compound is 31.1% s, the rest is chlorine. the compound has a molecular mass of 205.94 amu. what is its molecular formula?

Explanation:

Step1: Assume 100g of the compound

If we assume 100g of the compound, then the mass of \(S\) is \(m(S)=31.1g\), and the mass of \(Cl\) is \(m(Cl)=(100 - 31.1)g=68.9g\)

Step2: Calculate the number of moles of each element

The molar mass of \(S\) is \(M(S)=32.07g/mol\), and the molar mass of \(Cl\) is \(M(Cl)=35.45g/mol\)
The number of moles of \(S\), \(n(S)=\frac{m(S)}{M(S)}=\frac{31.1g}{32.07g/mol}\approx0.97mol\)
The number of moles of \(Cl\), \(n(Cl)=\frac{m(Cl)}{M(Cl)}=\frac{68.9g}{35.45g/mol}\approx1.94mol\)

Step3: Find the empirical formula

Divide the number of moles by the smaller number of moles (\(n = 0.97\))
For \(S\): \(\frac{n(S)}{n(S)} = 1\)
For \(Cl\): \(\frac{n(Cl)}{n(S)}=\frac{1.94}{0.97}\approx2\)
The empirical formula is \(SCl\)

Step4: Calculate the empirical formula mass

The empirical formula mass \(M_{empirical}=(32.07 + 35.45)g/mol=67.52g/mol\)

Step5: Find the ratio \(n\)

\(n=\frac{M_{molecular}}{M_{empirical}}=\frac{205.94g/mol}{67.52g/mol}\approx3\) (more accurately, \(205.94\div67.52 = 3.05\approx3\) when considering significant - figures and experimental errors. But if we recalculate more precisely:
\(n=\frac{205.94}{32.07 + 2\times35.45}=\frac{205.94}{32.07+70.9}=\frac{205.94}{102.97}\approx2\))
Multiply the sub - scripts in the empirical formula by \(n\). The molecular formula is \(S_2Cl_2\)

Answer:

\(S_2Cl_2\)