Answer: Empirical formula of this compound is [tex]Cr_2O_3[/tex]
Explanation:
Mass of Cr= 104.0 g
Mass of O = 48.0 g
Step 1 : convert given masses into moles.
Moles of Cr =[tex]\frac{\text{ given mass of Cr}}{\text{ molar mass of Cr}}= \frac{104.0g}{52g/mole}=2moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{48.0g}{16g/mole}=3moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Cr = [tex]\frac{2}{2}=1[/tex]
For O =[tex]\frac{3}{2}=1.5[/tex]
Converting into simple whole number ratios by multiplying by 2
The ratio of Cr : O= 2: 3
Hence the empirical formula is [tex]Cr_2O_3[/tex]
The most likely empirical formula of this compound is [tex]Cr_2O_3[/tex]
