Q42P - Chemistry: Molecular Nature Of Matter And Change

Question

A 0.370 mol sample of a metal oxide $\left(M_2{0}_3\right)$ weighs 55.4 g.
 (a) How many moles of O are in the sample?
 (b) How many grams of M are in the sample?
 (c) What element is represented by the symbol M?

Step-by-Step Solution

Answer

(a) The moles of O is 1.11 mol.

(b) The grams of M is 37.64 g.

(c) The element is represented as Antimony (Sb).

Step 1: Calculation of the moles of O

The moles of sample is 0.370 mol. Since the chemical formula of the compound is ${\mathbf{M}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{3}}$ 

This states that 3 mol of O are present in 1 mol of ${\mathbf{M}}_{\mathbf{2}}{\mathbf{0}}_{\mathbf{3}}$  

Thus, the moles of O is:

  $$\begin{gathered} \mathbf{n}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{\times }\mathbf{\left(}{\mathbf{M}}_{\mathbf{2}}{\mathbf{0}}_{\mathbf{3}}\mathbf{\right)} \\ \mathbf{n}\mathbf{\left(}\mathbf{o}\mathbf{\right)}\mathbf{=}\mathbf{3}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{370}\mathbf{ }\mathbf{mol} \\ \mathbf{n}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{11}\mathbf{mol} \end{gathered}$$

Step 2: Determine the mass of M

The mass of M is calculated as:

 $$\begin{gathered} \mathbf{m}\mathbf{\left(}\mathbf{M}\mathbf{\right)}\mathbf{=}\mathbf{m}\left(M_2{0}_3\right)\mathbf{-}\mathbf{m}\left(O\right) \\ \mathbf{m}\mathbf{\left(}\mathbf{M}\mathbf{\right)}\mathbf{=}\mathbf{55}\mathbf{.}\mathbf{4}\mathbf{ }\mathbf{g}\mathbf{-}\left(1.11\mathrm{mol}\times 16.00 g/\mathrm{mol}\right) \\ \mathbf{m}\mathbf{\left(}\mathbf{M}\mathbf{\right)}\mathbf{=}\mathbf{46}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{g}\mathbf{-}\mathbf{17}\mathbf{.}\mathbf{76}\mathbf{ }\mathbf{g} \\ \mathbf{m}\left(M\right)\mathbf{=}\mathbf{37}\mathbf{.}\mathbf{64}\mathbf{ }\mathbf{g} \end{gathered}$$

Step 3: Determine the representation of element having symbol M

The molar mass of M is calculated as:

 $$\begin{gathered} \mathbf{Molar}\mathbf{ }\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{M}\mathbf{=}\frac{\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{M}}{\mathbf{moles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{M}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{37}\mathbf{.}\mathbf{64}\mathbf{ }\mathbf{g}}{\mathbf{0}\mathbf{.}\mathbf{74}\mathbf{ }\mathbf{mol}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{50}\mathbf{.}\mathbf{86}\mathbf{g}\mathbf{/}\mathbf{mol} \end{gathered}$$

Thus, the element is Antimony.