Question

Experimental analysis of a compound containing the elements x,y,z on analysis gave the following data. x = 32 %, y = 24 %, z = 44 %. The relative number of atoms of x, y and z are 2, 1 and 0.5, respectively. (Molecular mass of the compound is 400 g) Find out.
i) The atomic masses of the element x,y,z.
ii) Empirical formula of the compound and
iii) Molecular formula of the compound.​

Answer 1

Answer:

i) The atomic masses of the element x is 32 g/mol,y is 48 g/mol, and z is 176 g/mol.

ii) Empirical formula of the compound $x_4y_2z_1$.

iii) Molecular formula of the compound $x_4y_2z_1$.

Explanation:

Molecular mass of the compound = 400 g/mol

Percentage of of element x = 32%

The relative number of atoms of x, y and z are 2, 1 and 0.5, respectively.

Empirical formula of the compound :

$x_2y_1z_{0.5} =x_2y_1z_{\frac{1}{2}}=x_4y_2z_1$

Molecular formula of the compound =$x_4y_2z_1$

Percentage of of element x = 32%

$32\%=\frac{4\times x g/mol}{400 g/mol}=32 g/mol$

Percentage of of element y = 24%

$24\%=\frac{2\times y g/mol}{400 g/mol}=48 g/mol$

Percentage of of element z = 44 %

$44\%=\frac{1\times z g/mol}{400 g/mol}=176 g/mol$

Empirical mass of the compound :$32 g/mol\times 4+48 g/mol/times 2+1\times 176 g/mol=400 g/mol$

Empirical mass = molecular mass = 400 g/mol

So, the molecular formula will also empirical formula $x_4y_2z_1$.

Answer 2

Explanation:

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