Question

What is the percent composition of Sb2F3.What is the percent composition of F in this molecule? Round the answer to five significant figures.

Answer 1

Answer:

$\%F=18.967\%$

Explanation:

Hello,

In this case, a percent composition is computed by considering the atomic mass of each element in the compound as well as the number of atoms and the molar mass of the compound. Thus, Sb2F3 has a molar mass of 300.52 g/mol, antimony has an atomic mass of 121.76 g/mol and fluorine 19.0 g/mol, therefore, the percent composition of fluorine which has three atoms with five significant figures is:

$\% F=\frac{3*19.0g/mol}{300.52g/mol}*100\%\\ \\\%F=18.967\%$

Best regards.