Respuesta :
Answer : The empirical formula of a compound is, [tex]C_6H_{10}S_2O_1[/tex] and the molecular of the compound is, [tex]C_{18}H_{30}S_6O_3[/tex]
Solution : Given,
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given.
Mass of C = 44.4 g
Mass of H = 6.21 g
Mass of S = 39.5 g
Mass of O = 9.86 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of S = 32 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of C = [tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{44.4g}{12g/mole}=3.7moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.21g}{1g/mole}=6.21moles[/tex]
Moles of S = [tex]\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{39.5g}{32g/mole}=1.23moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{9.86g}{16g/mole}=0.62moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{3.7}{0.62}=5.96\approx 6[/tex]
For H = [tex]\frac{6.21}{0.62}=10.01\approx 10[/tex]
For S = [tex]\frac{1.23}{0.62}=1.98\approx 2[/tex]
For O = [tex]\frac{0.62}{0.62}=1[/tex]
The ratio of C : H : S : O = 6 : 10 : 2 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_6H_{10}S_2O_1[/tex]
The empirical formula weight = 6(12) + 10(1) + 2(32) + 1(16) = 162 gram/eq
Now we have to calculate the molecular formula of the compound.
Formula used :
[tex]n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}=\frac{486}{162}=3[/tex]
Molecular formula = [tex](C_6H_{10}S_2O_1)_n=(C_6H_{10}S_2O_1)_3=C_{18}H_{30}S_6O_3[/tex]
Therefore, the molecular of the compound is, [tex]C_{18}H_{30}S_6O_3[/tex]
