Question

Titanium oxide ($Ti{O_2}$) is heated in a stream of hydrogen gas to give water and a new oxide $T{i_x}{O_y}$ . If $1.6g$ of $Ti{O_2}$ produces $1.44g$ of $T{i_x}{O_y}$ , the formula of new oxide is:
[Atomic weight of $Ti = 48$ ]
(A) $TiO$
(B) $T{i_2}{O_3}$
(C) $T{i_2}{O_5}$
(D) None of these

Answer 1

We need to understand the concept of empirical formula and how can it be calculated. As we are asked to find the new oxide $T{i_x}{O_y}$, we basically need to find the empirical formula by finding the values of x and y. The empirical formula tells us the relative ratios of the atoms of a compound. Hence by calculating the percentage composition of both the Titanium and Oxygen, we can find the formula of the new oxide.

Complete step by step answer:
Empirical formula is calculated using the following rules:
The mass of each of the constituent elements is converted to moles using the respective atomic weights.
Each mole value is divided by the smallest number of moles calculated.
The mole ratio of the elements is represented as the subscripts as the empirical formula.
Given that $Ti{O_2}$ is heated in a stream of hydrogen gas to give water and a new oxide $T{i_x}{O_y}$ . Let us first understand this statement by writing the reaction.
$Ti{O_2} + {H_2}\left( g \right) \to {H_2}O + T{i_x}{O_y}$
The atomic weight of $Ti = 48g/mol$
And Atomic weight of oxygen $= 16g/mol$
Let us first calculate the mass of $Ti$ in $Ti{O_2}$.
% of Ti in $Ti{O_2}$= $\left( {\dfrac{{48}}{{48 + 32}}} \right) \times 100 = 60\%$
Therefore mass of Ti in $Ti{O_2} = 60\%$of $1.6g = 0.96g$
Now, let us calculate the % of Ti in $T{i_x}{O_y}$ since we know the amount (in grams) if Ti used.
% of Ti in $T{i_x}{O_y}$=$\dfrac{{0.96}}{{1.44}} \times 100 = 66.66\%$
It is therefore clear that $T{i_x}{O_y}$ contains $66.66\%$ of Ti and $33.34\%$ of O. In other words, $100$ grams of $T{i_x}{O_y}$ contains $66.66g$ of Ti and $33.34g$ of O.
We now calculate the number of moles of Ti and O in $T{i_x}{O_y}$ .
$66.66g$ of Ti in $T{i_x}{O_y}$=$\dfrac{{1mol}}{{48g/mol}} \times 66.66 = 1.39mol$
$33.34g$ of O in $T{i_x}{O_y}$=$\dfrac{{1mol}}{{16g/mol}} \times 33.34 = 2.08mol$
Each mole value is divided by the smallest number of moles calculated which is $1.39mol$ For Ti: $\dfrac{{1.39}}{{1.39}} = 1$ and for O: $\dfrac{{2.08}}{{1.39}} = 1.5$
Therefore, the formula for the new oxide is $T{i_1}{O_{1.5}}$ . Converting it into whole numbers, we have the final formula as $T{i_2}{O_3}$.

Hence option B is correct.

Note: We must be noted that the percentage composition calculation of each of the constituent elements of a compound is the most necessary step to calculate the empirical formula of a compound. Also, empirical formula and molecular formula need not be the same. Empirical formula shows the simplest whole-number ratio of atoms in a compound whereas molecular formula shows the number of each type of atom in a molecule.