Respuesta :
solution:
[tex]molar mass =146g/mol\\
amount of carbon=0.321g\\
moler of carbon=\frac{0.321}{12.01}=0.026----(1)\\
amount of H=0.044g=0.043--------(2)\\
mole of H=\frac{0.044}{1.01}\\
amount of surface=0.285\\
mole of (s)=\frac{0.285}{32.06}=0.0088--------(3)\\
dividing by the smallest unit is (1),(2),(3)\\
\frac{0.026}{0.088}=2.95\approx 3\\
\frac{0.043}{0.0088}=4.88\approx 5\\
\frac{0.0088}{0.0088}=1\\[/tex][tex]Empirical formula of the compound is c_{3}H_{5}S\\
at the let of C_{3}H_{5}S=73\\
molecular weight=146\\
molecularity=\frac{146}{73}=2\\
so the formula of the compound is c_{6}H_{10}S_{2}[/tex]
Answer : The molecular formula of a compound is, [tex]C_6H_{10}S_2[/tex]
Solution : Given,
Mass of C = 0.321 g
Mass of H = 0.044 g
Mass of S = 0.285 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of S = 32 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{0.321g}{12g/mole}=0.0267moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.044g}{1g/mole}=0.044moles[/tex]
Moles of S = [tex]\frac{\text{ given mass of S}}{\text{ molar mass of S}}= \frac{0.285g}{32g/mole}=0.0089moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{0.0267}{0.0089}=3[/tex]
For H = [tex]\frac{0.044}{0.0089}=4.94\approx 5[/tex]
For S = [tex]\frac{0.0089}{0.0089}=1[/tex]
The ratio of C : H : S = 3 : 5 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_3H_5S_1[/tex]
The empirical formula weight = 3(12) + 5(1) + 1(32) = 73 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}}[/tex]
[tex]n=\frac{146}{73}=2[/tex]
Molecular formula = [tex](C_3H_5S_1)_n=(C_3H_5S_1)_2=C_6H_{10}S_2[/tex]
Therefore, the molecular of the compound is, [tex]C_6H_{10}S_2[/tex]
