Question

a compound is found to contain 6.360 % silicon, 36.18 % bromine, and 57.46 % iodine by mass. question 1: the empirical formula for this compound is. question 2: the molar mass for this compound is 441.7 g/mol. the molecular formula for this compound is

Explanation:

Step1: Assume 100 g of the compound

Since we have mass - percentages, assuming 100 g of the compound means we have 6.360 g of Si, 36.18 g of Br, and 57.46 g of I.

Step2: Calculate the moles of each element

The molar mass of Si is \(M_{Si}=28.09\ g/mol\), the molar mass of Br is \(M_{Br} = 79.90\ g/mol\), and the molar mass of I is \(M_{I}=126.90\ g/mol\).
The moles of Si, \(n_{Si}=\frac{6.360\ g}{28.09\ g/mol}\approx0.226\ mol\).
The moles of Br, \(n_{Br}=\frac{36.18\ g}{79.90\ g/mol}\approx0.453\ mol\).
The moles of I, \(n_{I}=\frac{57.46\ g}{126.90\ g/mol}\approx0.453\ mol\).

Step3: Find the mole - ratio

Divide each number of moles by the smallest number of moles (\(n_{Si} = 0.226\ mol\)).
For Si: \(\frac{0.226\ mol}{0.226\ mol}=1\).
For Br: \(\frac{0.453\ mol}{0.226\ mol}\approx2\).
For I: \(\frac{0.453\ mol}{0.226\ mol}\approx2\).
So the empirical formula is \(SiBr_{2}I_{2}\).

Step4: Calculate the empirical - formula mass

The empirical - formula mass of \(SiBr_{2}I_{2}\) is \(M = 28.09+2\times79.90 + 2\times126.90=28.09 + 159.8+253.8 = 441.69\ g/mol\).

Step5: Determine the molecular formula

Since the molar mass of the compound is \(441.7\ g/mol\) and the empirical - formula mass is approximately \(441.69\ g/mol\), the ratio \(n=\frac{\text{Molar mass}}{\text{Empirical - formula mass}}=\frac{441.7\ g/mol}{441.69\ g/mol}\approx1\).
So the molecular formula is the same as the empirical formula, \(SiBr_{2}I_{2}\).

Answer:

Question 1: \(SiBr_{2}I_{2}\)
Question 2: \(SiBr_{2}I_{2}\)