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.

(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

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Alleei Alleei · Answer 2 · 2019-09-19T02:35:31-04:00

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 \(Mo_2O_5\).