Respuesta :
Answer : The empirical formula of a compound is, [tex]Na_3BO_3[/tex]
Solution : Given,
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 = 53.976 g
Mass of B = 8.461 g
Mass of O = [100 - (53.976 + 8.461)] = 37.563 g
Molar mass of Na = 23 g/mole
Molar mass of B = 11 g/mole
Molar mass of O = 16 g/mole
Step 1 : convert given masses into moles.
Moles of Na = [tex]\frac{\text{ given mass of Na}}{\text{ molar mass of Na}}= \frac{53.976g}{23g/mole}=2.347moles[/tex]
Moles of B = [tex]\frac{\text{ given mass of B}}{\text{ molar mass of B}}= \frac{8.461g}{11g/mole}=0.769moles[/tex]
Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{37.563g}{16g/mole}=2.347moles[/tex]
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For Na = [tex]\frac{2.347}{0.769}=3.05\approx 3[/tex]
For B = [tex]\frac{0.769}{0.769}=1[/tex]
For O = [tex]\frac{2.347}{0.769}=3.05\approx 3[/tex]
The ratio of Na : B : O = 3 : 1 : 3
The mole ratio of the element is represented by subscripts in empirical formula.
The Empirical formula = [tex]Na_3B_1O_3[/tex] = [tex]Na_3BO_3[/tex]
