Q.2.19 P - Chemistry: Molecular Nature Of Matter And Change

Question

State the mass law(s) demonstrated by the following experimental results, and explain your reasoning: 

Experiment 1: A student heats 1.27 g of copper and 3.50 g of iodine to produce 3.81 g of a white compound; 0.96 g of iodine remains.

Experiment 2: A second student heats 2.55 g of copper and 3.50 g of iodine to form 5.25 g of a white compound, and 0.80 g of copper remains.

 

Step-by-Step Solution

Answer

Individually both experiment 1 and experiment 2 demonstrate the “Law of Conservation of Mass” while collectively they demonstrate the “Law of Definite Composition.”

Step 1: Explanation of Experiment 1

Experiment 1 is given as: 

 $$\begin{gathered} \mathit{C}\mathit{o}\mathit{o}\mathit{p}\mathit{e}\mathit{r}\mathbf{+}\mathit{I}\mathit{o}\mathit{d}\mathit{i}\mathit{n}\mathit{e}\mathbf{=}\mathit{W}\mathit{h}\mathit{i}\mathit{t}\mathit{e}\mathbf{ }\mathit{c}\mathit{o}\mathit{m}\mathit{p}\mathit{o}\mathit{u}\mathit{n}\mathit{d}\mathbf{ }\mathbf{+}\mathit{I}\mathit{o}\mathit{d}\mathit{i}\mathit{i}\mathit{n}\mathit{e}\mathbf{ }\mathit{r}\mathit{e}\mathit{m}\mathit{a}\mathit{i}\mathit{n}\mathit{s} \\ \mathbf{1}\mathbf{.}\mathbf{27}\mathit{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{3}\mathbf{.}\mathbf{50}\mathit{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{3}\mathbf{.}\mathbf{81}\mathit{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{96}\mathit{g} \end{gathered}$$

 From the equation, we can say that,

 $\mathit{M}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{R}\mathit{e}\mathit{a}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\mathbf{ }\left(4.77g\right)\mathbf{=}\mathit{M}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{p}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t}\mathit{s}\mathbf{ }\left(4.77 g\right)$

 As remains constant before and after the reaction hence it demonstrates the “law of Conservation of Mass.”

Step 2: Explanation of Experiment 2

Experiment 2 is given as: 

 $$\begin{gathered} \mathit{C}\mathit{o}\mathit{p}\mathit{p}\mathit{e}\mathit{r}\mathbf{ }\mathbf{+}\mathit{I}\mathit{o}\mathit{d}\mathit{i}\mathit{n}\mathit{e}\mathbf{=}\mathit{w}\mathit{h}\mathit{i}\mathit{t}\mathit{e}\mathbf{ }\mathit{C}\mathit{o}\mathit{m}\mathit{p}\mathit{o}\mathit{u}\mathit{n}\mathit{d}\mathbf{+}\mathit{C}\mathit{o}\mathit{p}\mathit{p}\mathit{e}\mathit{r}\mathbf{ }\mathit{r}\mathit{e}\mathit{m}\mathit{a}\mathit{i}\mathit{n}\mathit{s} \\ \mathbf{ }\mathbf{ }\mathbf{2}\mathbf{.}\mathbf{55}\mathit{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{3}\mathbf{.}\mathbf{50}\mathit{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{5}\mathbf{.}\mathbf{25}\mathit{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{80}\mathit{g} \end{gathered}$$

 

 From the equation, we can say that,

$\mathit{M}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{r}\mathit{e}\mathit{a}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\mathbf{ }\left(6.05g\right)\mathbf{=}\mathit{m}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{p}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t}\mathit{s}\left(5.25g+0.80g=6.05g\right)$ 

As remains constant before and after the reaction hence it demonstrates the “law of Conservation of Mass.”

Step 3: Both Experiments

Both experiments collectively demonstrate the Law of Definite Composition. The ratio of the reactants reacted the same in both experiments.