question 2\n1 pts\na compound is 31.1% s, the rest is chlorine. the compound has a molecular mass of 205.94…

Answer:

(S_2Cl_2)

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)