Question

A compound which contains one atom of X and two atoms of Y for every three atoms of Z is made of mixing \[{\rm{5}}\,\,{\rm{gm}}\]of X, \[{\rm{1}}{\rm{.15}} \times {\rm{1}}{{\rm{0}}^{23}}\]atoms of Y and \[0.03\]moles of Z atoms. Given that only \[{\rm{4}}{\rm{.40}}\,\,{\rm{gm}}\]of compound results, calculate the atomic weight of Y if the atomic weight of X and Z are \[{\rm{60}}\]and \[{\rm{80}}\]respectively.

Answer 1

Hint: The atomic mass or atomic weight of an element is the average relative mass of an atom of the element as compared to the mass of \[{}^{{\rm{12}}}{\rm{C}}\]atom taken as \[12\]units.

Complete Step by Step Solution:
A mole is the amount of substance that contains as many particles (atoms, molecules, ions, etc.) as there are atoms exactly in \[{\rm{0}}{\rm{.012}}\,{\rm{kg}}\]\[{\rm{(i}}{\rm{.e}}{\rm{. 12 g)}}\]of carbon \[({}^{{\rm{12}}}{\rm{C)}}\]. It must be remembered that mass of one mole of atom is equal to the gram atomic mass of the element, mass of one mole of molecule is equal to gram molecular mass of the substance and mass of one mole of formula units in case of an ionic compound is equal to gram formula mass of the ionic compound.

The relationship between moles (n), mass (m), and molar mass (M) is as shown below:
\[{\rm{moles}}\,{\rm{(n) = }}\dfrac{{{\rm{mass(m)}}}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}}}\]

Empirical formula of a compound gives the simple whole number ratio of the atoms of various elements present in one molecule of the compound.
As per given data,
The compound consists of one atom of X and two atoms of Y for every three atoms of Z. Therefore, the empirical formula of the compound will be \[{\rm{X}}{{\rm{Y}}_2}{{\rm{Z}}_3}\].
Total mass of the product \[{\rm{ = }}\,\,{\rm{4}}{\rm{.4}}\,\,{\rm{g}}\]
Mass of X \[{\rm{ = }}\,\,{\rm{5}}\,\,{\rm{g}}\]
So, moles of X \[ = \,\,\dfrac{5}{{60}}\, = \,\,0.083\]
Atoms of Y \[{\rm{ = }}\,{\rm{ 1}}{\rm{.15}} \times {\rm{1}}{{\rm{0}}^{23}}\]
So, moles of Y \[{\rm{ = }}\,{\rm{ }}\dfrac{{{\rm{1}}{\rm{.15}} \times {\rm{1}}{{\rm{0}}^{23}}}}{{6.022 \times {\rm{1}}{{\rm{0}}^{23}}}}\, = \,0.19\]
moles of Z is given as \[0.03\]moles
Z is the limiting reagent here.

If \[0.03\]moles of Z are present, then from the empirical formula it can be said that:
moles of Y \[ = \,\,0.02\]
moles of X \[ = \,\,0.01\]

Find the mass of X, Y and Z as shown below:
mass of X \[{\rm{ = }}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{X}}\,\,{\rm{ \times }}\,\,{\rm{its}}\,\,{\rm{atomic}}\,\,{\rm{mass}}\,\,{\rm{ = }}\,\,{\rm{0}}{\rm{.01}} \times {\rm{60}}\,\,{\rm{ = }}\,{\rm{0}}{\rm{.6}}\,\,{\rm{g}}\]
mass of Z \[{\rm{ = }}\,\,{\rm{moles}}\,\,{\rm{of}}\,\,{\rm{Z}}\,{\rm{ \times }}\,\,{\rm{its}}\,\,{\rm{atomic}}\,\,{\rm{mass}}\,\,{\rm{ = }}\,\,{\rm{0}}{\rm{.03}} \times 8{\rm{0}}\,\,{\rm{ = }}\,2.4\,\,{\rm{g}}\]
mass of Y \[{\rm{ = }}\,\,{\rm{total}}\,\,{\rm{mass}}\,\, - {\rm{(mass}}\,\,{\rm{of}}\,\,{\rm{X}}\,\,{\rm{ + }}\,\,{\rm{mass}}\,\,{\rm{of}}\,\,{\rm{Z)}}\,\,{\rm{ = }}\,\,{\rm{4}}{\rm{.4}}\,\, - \,\,{\rm{(0}}{\rm{.6 + 2}}{\rm{.4)}}\,\,{\rm{g}}\,{\rm{ = }}\,\,{\rm{1}}{\rm{.4}}\,\,{\rm{g}}\]
Find the molar mass of Y as:
\[{l}{\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{\rm{Y = }}\dfrac{{{\rm{mass(m)}}\,\,{\rm{of}}\,\,{\rm{Y}}}}{{{\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,\,{\rm{of}}\,\,{\rm{Y}}}}\\ \Rightarrow {\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,\,{\rm{of}}\,\,{\rm{Y}} = \dfrac{{{\rm{mass(m)}}\,\,{\rm{of}}\,\,{\rm{Y}}}}{{{\rm{moles}}\,{\rm{(n)}}\,\,{\rm{of}}\,\,{\rm{Y}}}}\\ \Rightarrow {\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,\,{\rm{of}}\,\,{\rm{Y}} = \dfrac{{{\rm{1}}{\rm{.4}}\,\,{\rm{g}}}}{{{\rm{0}}{\rm{.02}}\,\,{\rm{mol}}}}\\ \Rightarrow {\rm{molar}}\,\,{\rm{mass}}\,\,{\rm{(M)}}\,\,\,{\rm{of}}\,\,{\rm{Y}} = \,\,{\rm{70}}\dfrac{{\rm{g}}}{{{\rm{mol}}}}\]

Hence, atomic mass of Y is \[{\rm{70}} \ {\rm{amu}}\] (atomic mass unit)
Therefore, the atomic mass of Y is \[{\rm{70}}\,\,{\rm{amu}}\]

Note: The reactant which is completely used and determines the amount of product formed is known as limiting reagent or reactant. The theoretical yield of a product is the maximum yield obtainable as calculated on the basis of the amount of limiting reactant used. Atomic mass has no units. However, it is expressed in a.m.u. which only signifies that it is taken on atomic mass unit scale.