I’m sorry if I wasted your time but I think it’s alkali metals but I’m
Not sure
Answer: Empirical formula of this compound is 
Explanation:
Mass of Cr= 104.0 g
Mass of O = 48.0 g
Step 1 : convert given masses into moles.
Moles of Cr =
Moles of O =
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Cr = 
For O =
Converting into simple whole number ratios by multiplying by 2
The ratio of Cr : O= 2: 3
Hence the empirical formula is 
The most likely empirical formula of this compound is 
a. 1.505 x 10²⁴
b. 4.37 x 10²⁴
c. 2.34 x 10²⁴
d. 8.127 x 10²³
<h3>Further explanation</h3>
Given
Number of moles of elements
Required
The number of particles
Solution
1 mol = 6.02 x 10²³ particles(molecules, ions, atoms)
Can be formulated :
N = n x No
No = Avogadro's number
n = mol
a. 2.5 mo Ca x 6.02 x 10²³ = 1.505 x 10²⁴
b. 7.26 mol Al x 6.02 x 10²³ = 4.37 x 10²⁴
c. 3.89 mol S x 6.02 x 10²³ = 2.34 x 10²⁴
d. 1.35 mol Pb x 6.02 x 10²³ = 8.127 x 10²³