Question

A 12.95 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 13.82 g. Add subscripts below to correctly identify the empirical formula of the new oxide.

Answer 1

(18.75 g Mo2O3) x [(1 mole Mo2O3)/(239.88 g)] x [(2 moles Mo)/(1 mole Mo2O3)] = 0.1563 moles

We know that Mox = 0.1563 moles
Mo = 15.00 g Mo, which in turn means that there must be (22.50 - 15.00) grams of O or 7.50 g O, which converts to: 

(7.50 g O) x [(1 mole O) / (15.999 g O)] = 0.4688 moles O 

Taking the ratio of Mo to O results in: 

O/Mo = [(0.4688)/(0.1563)] = 3 

Answer 2

Answer : The empirical formula of the new oxide will be $Mo_2O_5$.

Explanation :

First we have to calculate the moles of $Mo_2O_3$.

Molar mass of $Mo_2O_3$ = 240 g/mole

$\text{ Moles of }Mo_2O_3=\frac{\text{ Mass of }Mo_2O_3}{\text{ Molar mass of }Mo_2O_3}=\frac{12.95g}{240g/mole}=0.0539moles$

Now we have to calculate the moles of 'Mo'.

As, 1 mole of $Mo_2O_3$ contains 2 mole 'Mo'

So, 0.0539 mole of $Mo_2O_3$ contains $0.0539\times 2=0.1078$ mole 'Mo'

Now we have to calculate the mass of 'Mo'.

$\text{Mass of }Mo=\text{Moles of }Mo\times \text{Molar mass of }Mo$

Molar mass of Mo = 95.96 g/mole

$\text{Mass of }Mo=0.1078mole\times 95.96g/mole=10.34g$

Now we have to calculate the mass of oxygen.

Mass of oxygen = 13.82 - 10.34 = 3.48 g

Now we have to calculate the moles of oxygen.

$\text{ Moles of oxygen}=\frac{\text{ Mass of oxygen}}{\text{ Molar mass of oxygen}}=\frac{3.48g}{16g/mole}=0.2175moles$

Now we have to calculate the ratio of 'O' and 'Mo'.

$\frac{O}{Mo}=\frac{0.2175}{0.1078}=2.02\approx 2$

Now 3 oxygen are added in $Mo_2O_3$, we get the new formula of  oxide.

The new formula of oxide = $Mo_2O_5$

Hence, the empirical formula of the new oxide will be $Mo_2O_5$.