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, ![C_6H_4N_2O_4](https://tex.z-dn.net/?f=C_6H_4N_2O_4)
Explanation :
First we have to calculate the mass of benzene.
![\text{Mass of benzene}=\text{Density of benzene}\times \text{Volume of benzene}](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20benzene%7D%3D%5Ctext%7BDensity%20of%20benzene%7D%5Ctimes%20%5Ctext%7BVolume%20of%20benzene%7D)
![\text{Mass of benzene}=0.879g/mL\times 50.0mL=43.95g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20benzene%7D%3D0.879g%2FmL%5Ctimes%2050.0mL%3D43.95g)
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 :
![\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}}](https://tex.z-dn.net/?f=%5CDelta%20T_f%3DK_f%5Ctimes%20m%5C%5C%5C%5C%5CDelta%20T_f%3DK_f%5Ctimes%5Cfrac%7B%5Ctext%7BMass%20of%20unknown%20compound%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20unknown%20compound%7D%5Ctimes%20%5Ctext%7BMass%20of%20benzene%20in%20Kg%7D%7D)
where,
= change in freezing point = ![5.53-1.37=4.16^oC](https://tex.z-dn.net/?f=5.53-1.37%3D4.16%5EoC)
= freezing point of solution
= freezing point of benzene
Molal-freezing-point-depression constant
for benzene = ![5.12^oC/m](https://tex.z-dn.net/?f=5.12%5EoC%2Fm)
m = molality
Now put all the given values in this formula, we get
![4.16^oC=(5.12^oC/m)\times \frac{6.45g}{\text{Molar mass of unknown compound}\times 0.04395kg}](https://tex.z-dn.net/?f=4.16%5EoC%3D%285.12%5EoC%2Fm%29%5Ctimes%20%5Cfrac%7B6.45g%7D%7B%5Ctext%7BMolar%20mass%20of%20unknown%20compound%7D%5Ctimes%200.04395kg%7D)
![\text{Molar mass of unknown compound}=180.6g/mol](https://tex.z-dn.net/?f=%5Ctext%7BMolar%20mass%20of%20unknown%20compound%7D%3D180.6g%2Fmol)
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 = ![\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{42.9g}{12g/mole}=3.575moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20C%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20C%7D%7D%3D%20%5Cfrac%7B42.9g%7D%7B12g%2Fmole%7D%3D3.575moles)
Moles of H = ![\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{2.4g}{1g/mole}=2.4moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20H%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20H%7D%7D%3D%20%5Cfrac%7B2.4g%7D%7B1g%2Fmole%7D%3D2.4moles)
Moles of N = ![\frac{\text{ given mass of N}}{\text{ molar mass of N}}= \frac{16.7g}{14g/mole}=1.193moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20N%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20N%7D%7D%3D%20%5Cfrac%7B16.7g%7D%7B14g%2Fmole%7D%3D1.193moles)
Moles of O = ![\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{38.1g}{16g/mole}=2.381moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7B%20given%20mass%20of%20O%7D%7D%7B%5Ctext%7B%20molar%20mass%20of%20O%7D%7D%3D%20%5Cfrac%7B38.1g%7D%7B16g%2Fmole%7D%3D2.381moles)
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C = ![\frac{3.575}{1.193}=2.99\approx 3](https://tex.z-dn.net/?f=%5Cfrac%7B3.575%7D%7B1.193%7D%3D2.99%5Capprox%203)
For H = ![\frac{2.4}{1.193}=2.01\approx 2](https://tex.z-dn.net/?f=%5Cfrac%7B2.4%7D%7B1.193%7D%3D2.01%5Capprox%202)
For N = ![\frac{1.193}{1.193}=1](https://tex.z-dn.net/?f=%5Cfrac%7B1.193%7D%7B1.193%7D%3D1)
For O = ![\frac{2.381}{1.193}=1.99\approx 2](https://tex.z-dn.net/?f=%5Cfrac%7B2.381%7D%7B1.193%7D%3D1.99%5Capprox%202)
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 = ![C_3H_2N_1O_2](https://tex.z-dn.net/?f=C_3H_2N_1O_2)
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 :
![n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B%5Ctext%7BMolecular%20formula%7D%7D%7B%5Ctext%7BEmpirical%20formula%20weight%7D%7D)
![n=\frac{180.6}{84}=2](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B180.6%7D%7B84%7D%3D2)
Molecular formula = ![(C_3H_2N_1O_2)_n=(C_3H_2N_1O_2)_2=C_6H_4N_2O_4](https://tex.z-dn.net/?f=%28C_3H_2N_1O_2%29_n%3D%28C_3H_2N_1O_2%29_2%3DC_6H_4N_2O_4)
Therefore, the molecular of the compound is, ![C_6H_4N_2O_4](https://tex.z-dn.net/?f=C_6H_4N_2O_4)