Question

Gallium oxide, Ga.Oy, forms when gallium is combined with oxygen. A 1.25 g of Ga is allowed to react with excess oxygen and 1.68 g of the gallium oxide formed. What is the chemical formula of the gallium oxide if the molar mass is 187.44 g/mol?

Answer 1

Answer : The chemical formula of a compound is, $Ga_2O_3$

Solution :  Given,

Mass of gallium = 1.25 g

Mass of gallium oxide = 1.68 g

Mass of oxygen = Mass of gallium oxide - Mass of gallium

Mass of oxygen = 1.68 - 1.25 = 0.43 g

Molar mass of Ga = 69.72 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of Ga = $\frac{\text{ given mass of Ga}}{\text{ molar mass of Ga}}= \frac{1.25g}{69.72g/mole}=0.0179moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.43g}{16g/mole}=0.027moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Ga = $\frac{0.0179}{0.0179}=1$

For O = $\frac{0.027}{0.0179}=1.5$

The ratio of Ga : O = 1 : 1.5

To make in whole number we multiple ratio by 2, we get:

The ratio of Ga : O = 2 : 3

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $Ga_2O_3$

The empirical formula weight = 2(69.72) + 3(16) = 187.44 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{187.44}{187.44}=1$

Molecular formula = $(Ga_2O_3)_n=(Ga_2O_3)_1=Ga_2O_3$

Therefore, the chemical of the compound is, $Ga_2O_3$