Respuesta :
This is an incomplete question, here is a complete question.
A compound is 42.9% C, 2.4% H, 16.7% N and 38.1% O by mass. Addition of 6.45 g of this compound to 50.0 mL benzene, C₆H₆ (d= 0.879 g/mL; Kf= 5.12 degrees Celsius/m), lowers the freezing point from 5.53 to 1.37 degrees Celsius. What is the molecular formula of this compound?
Answer : The molecular of the compound is, [tex]C_6H_4N_2O_4[/tex]
Explanation :
First we have to calculate the mass of benzene.
[tex]\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}[/tex]
[tex]\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g[/tex]
Now we have to calculate the molar mass of unknown compound.
Given:
Mass of unknown compound (solute) = 6.45 g
Mass of benzene (solvent) = 43.95 g = 0.04395 kg
Formula used :
[tex]\Delta T_f=K_f\times m\\\\\Delta T_f=K_f\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of benzene in Kg}}[/tex]
where,
[tex]\Delta T_f[/tex] = change in freezing point = [tex]5.53-1.37=4.16^oC[/tex]
[tex]\Delta T_s[/tex] = freezing point of solution
[tex]\Delta T^o[/tex] = freezing point of benzene
Molal-freezing-point-depression constant [tex](K_f)[/tex] for benzene = [tex]5.12^oC/m[/tex]
m = molality
Now put all the given values in this formula, we get
[tex]4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}[/tex]
[tex]\text{Molar mass of unknown compound}=180.6g/mol[/tex]
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 = 42.9 g
Mass of H = 2.4 g
Mass of N = 16.7 g
Mass of O = 38.1 g
Molar mass of C = 12 g/mole
Molar mass of H = 1 g/mole
Molar mass of N = 14 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{42.9g}{12g/mole}=3.575moles[/tex]
Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles[/tex]
Moles of N = [tex]\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = [tex]\frac{3.575}{1.193}=2.99\approx 3[/tex]
For H = [tex]\frac{2.4}{1.193}=2.01\approx 2[/tex]
For N = [tex]\frac{1.193}{1.193}=1[/tex]
For O = [tex]\frac{2.381}{1.193}=1.99\approx 2[/tex]
The ratio of C : H : N : O = 3 : 2 : 1 : 2
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]C_3H_2N_1O_2[/tex]
The empirical formula weight = 3(12) + 2(1) + 1(14) + 2(16) = 84 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{180.6}{84}=2[/tex]
Molecular formula = [tex](C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4[/tex]
Therefore, the molecular of the compound is, [tex]C_6H_4N_2O_4[/tex]
