Answer to Question #229424 in General Chemistry for Joel uduak dick

Question

A compound has the following percent composition 28.03%Mg, 21.60%Si, 1.16%H and 49.21%O with a molar mass of 520.8g/mol. Determine the empirical and molecular formula of the compound (Si:28, Mg:24, H:1, O:16)

Answer 1

Assuming that there are 100 grams of the compound, the percentage of each element will be numerically equal to its mass in grams. Therefore, the molar ratio of elements can be found by dividing each mass by the atomic mass of the respective element:

$n(Mg)=\frac{28.03g}{24.31g/mol}=1.153mol$

$n(Si)=\frac{21.60g}{28.09g/mol}=0.7690 mol$

$n(H)=\frac{1.16g}{1.008g/mol}=1.15mol$

$n(O)=\frac{49.21g}{16.00g/mol}=3.076mol$

Dividing each quantity by the smallest one (which is 0.769 mol), we get the following molar ratios:

Mg : Si : H : O = 1.5 : 1 : 1.5 : 4. The lowest whole number ratio, which is 3 : 2 : 3 : 8, will determine the empirical formula of the compound - Mg₃Si₂H₃O₈. To calculate the molecular formula, the molar mass of the empirical formula unit should be calculated and compared to the molar mass of the actual compound, which is known.

$M(Mg_3Si_2H_3O_8)=3\times24.31+2\times28.09+3\times1.008+8\times16.00=260.1g/mol$

Now it is obvious that the molar mass of the actual compound is twice the molar mass of the empirical formula unit: $\frac{520.8g/mol}{260.1g/mol}=2$ .

Therefore, the subscripts of the empirical formula should be multiplied by 2 to get the molecular formula: Mg₆Si₄H₆O₁₆.

Answer: Mg₃Si₂H₃O₈ (empirical); Mg₆Si₄H₆O₁₆ (molecular)