Question

2.38g of uranium was heated strongly in a current of air . the resulting oxide weighed 2.806g .determine the empirical formula of the oxide at mass U=238 O=16

Answer 1

Weight of oxygen :

2.806 - 2.38 = 0.426

Moles of Oxygen :

0.426 / 16 = 0.02663

Moles of uranium :

2.38 / 238 = 0.01 moles

Divide both by 0.01 (the smallest value of the two)

0.01 / 0.01 = 1

0.02663 / 0.01 = 2.665

To nearest whole number is 3

EMPERICAL FORMULA FOR THE OXIDE.

UO₃

Answer 2

Answer : The empirical formula of uranium oxide is, $U_5O_{13}$.

Solution :  Given,

Mass of U = 2.38 g

Mass of resulting oxide = 2.806 g

Mass of O = 2.806 - 2.38 = 0.426 g

Molar mass of U =  238 g/mole

Molar mass of O = 16 g/mole

Step 1 : convert given masses into moles.

Moles of U = $\frac{\text{ given mass of U}}{\text{ molar mass of U}}= \frac{2.38g}{238g/mole}=0.01moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{0.426g}{16g/mole}=0.026moles$

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

For U = 0.01/0.01 = 1

For O = 0.026/0.01 = 2.6

To make as a whole number multiply by 5.

U : O = 5 : 13

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

Therefore, the Empirical formula = $U_5O_{13}$