Respuesta :
Answer : The molecular formula of a compound is, [tex]C_{19}H_{38}O[/tex]
Solution :
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 = 80.78 g
Mass of H = 13.56 g
Mass of O = 5.66 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 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{80.78g}{12g/mole}=6.732moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{13.56g}{1g/mole}=13.56moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{5.66g}{16g/mole}=0.3538moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{6.732}{0.3538}=19.0\approx 19[/tex]
For H = [tex]\frac{13.56}{0.3538}=38.3\approx 38[/tex]
For O = [tex]\frac{0.3538}{0.3538}=1[/tex]
The ratio of C : H : O = 19 : 38 : 1
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_{19}H_{38}O_1=C_{19}H_{38}O[/tex]
The empirical formula weight = 19(12) + 38(1) + 1(16) = 282 gram/eq
Now we have to calculate the molar mass of pheromone.
Formula used :
[tex]\Delta T_f=i\times K_f\times m\\\\T^o-T_s=i\times K_f\times\frac{\text{Mass of pheromone}}{\text{Molar mass of pheromone}\times \text{Mass of benzene in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point
[tex]\Delta T_s[/tex] = freezing point of solution = [tex]3.37^oC[/tex]
[tex]\Delta T^o[/tex] = freezing point of benzene = [tex]5.50^oC[/tex]
i = Van't Hoff factor = 1 (for non-electrolyte)
[tex]K_f[/tex] = freezing point constant for benzene = [tex]5.12^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex](5.50-3.37)^oC=1\times (5.12^oC/m)\times \frac{1.00g}{\text{Mass of pheromone}\times 0.00850kg}[/tex]
[tex]\text{Mass of pheromone}=282.8g/mol[/tex]
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{282.8}{282}=1[/tex]
Molecular formula = [tex](C_{19}H_{38}O)_n=(C_{19}H_{38}O)_1=C_{19}H_{38}O[/tex]
Therefore, the molecular of the compound is, [tex]C_{19}H_{38}O[/tex]
