Question

Calculate the empirical formula of a molecule with the following percent compositions:

55.0% gallium (Ga) and 45.0% fluorine (F). Complete the table below to help with the calculations. Round off all calculations to the nearest tenth.
Element Mass Percent Mass of 100 g of the Compound Number of Moles in 100 g
Ga
a) GaF2
b) Ga2F3
c) Ga3F2
d) GaF3

Answer 1

The empirical formula of the compound with 55.0% gallium and 45.0% fluorine is GaF3. We calculate this by assuming a 100 g sample, converting the given percentages into grams, determining moles using atomic masses of Ga and F, and then finding the simplest whole-number ratio of these elements.

Step-by-step explanation:

To calculate the empirical formula for a molecule with 55.0% gallium (Ga) and 45.0% fluorine (F), we start by assuming we have a 100 g sample. This simplifies the math, allowing us to use the mass percent directly as the mass of each element in grams.

1. Mass of Ga: 55.0 g; Mass of F: 45.0 g.
2. Convert these masses to moles using atomic weights: Ga = 69.7 g/mol, F = 19.0 g/mol.
   
   • Number of moles of Ga = 55.0 g ÷ 69.7 g/mol = 0.789 moles.
   • Number of moles of F = 45.0 g ÷ 19.0 g/mol = 2.368 moles.
3. Divide each by the smallest number of moles to find the ratio:
   
   • Mole ratio of Ga = 0.789 ÷ 0.789 = 1.
   • Mole ratio of F = 2.368 ÷ 0.789 = 3.
4. The empirical formula is GaF3.