Answer: The empirical formula is [tex]NaBO_2[/tex]
Explanation:
If percentage are given then we are taking total mass is 100 grams.
So, the mass of each element is equal to the percentage given:
Mass of Na= 34.5 g
Mass of B= 16.4 g
Mass of O = 48.6 g
Step 1 : convert given masses into moles.
Moles of Na =[tex]\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= \frac{34.5g}{23g/mole}=1.5moles[/tex]
Moles of B =[tex]\frac{\text{ given mass of B}}{\text{ molar mass of B}}= \frac{16.4g}{11g/mole}=1.5moles[/tex]
Moles of O =[tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{48.6g}{16g/mole}=3moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Na = [tex]\frac{1.5}{1.5}=1[/tex]
For B = [tex]\frac{1.5}{1.5}=1[/tex]
For O =[tex]\frac{3}{1.5}=2[/tex]
The ratio of Na: B: O= 1: 1: 2
Hence the empirical formula is [tex]NaBO_2[/tex]
