A compound is found to contain 7.962 % silicon , 20.10 % chlorine , and 71.94 % iodine by mass. a. The empirical formula for this compound is .
b. The molar mass for this compound is 352.8 g/mol. The molecular formula for this compound is:

A. Empirical formula => SiCl₂I₂

B. Molecular formula = SiCl₂I₂

Explanation:

From the question given above, the following data were:

Silicone (Si) = 7.962 %

Chlorine (Cl) = 20.10 %

Iodine (I) = 71.94 %

Molar mass of compound = 352.8 g/mol

Empirical formula =?

Molecular formula =?

A. Determination of the empirical formula of the compound.

Si = 7.962 %

Cl = 20.10 %

I = 71.94 %

Divide by their molar mass

Si = 7.962 / 28 = 0.284

Cl = 20.10 / 35.5 = 0.566

I = 71.94 / 127 = 0.566

Divide by the smallest

Si = 0.284 / 0.284 = 1

Cl = 0.566 / 0.284 = 2

I = 0.566 / 0.284 = 2

Empirical formula => SiCl₂I₂

B. Determination of the molecular formula of the compound.

Empirical formula => SiCl₂I₂

Molar mass of compound = 352.8 g/mol

Molecular formula =?

Molecular formula = [SiCl₂I₂]ₙ

[SiCl₂I₂]ₙ = 352.8

[28 + (2×35.5) + (2×127)]n = 352.8

[28 + 71 + 254]n = 352.8

353n = 352.8

Divide both side by 353

n = 352.8 / 353

n = 1

Molecular formula = [SiCl₂I₂]ₙ

Molecular formula = [SiCl₂I₂]₁

Molecular formula = SiCl₂I₂

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.