Q19P - 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: 

$$\begin{gathered} \mathbf{Copper}\mathbf{+}\mathbf{Iodine}\mathbf{=}\mathbf{white}\mathbf{ }\mathbf{compound}\mathbf{+}\mathbf{Iodine}\mathbf{ }\mathbf{remains} \\ \mathbf{1}\mathbf{.}\mathbf{27}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{3}\mathbf{.}\mathbf{81}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{96}\mathbf{ }\mathbf{g} \end{gathered}$$ 

 From the equation, we can say that,

 $\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{reactant}\mathbf{ }\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{77}\mathbf{ }\mathbf{g}\mathbf{)}\mathbf{=}\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{products}\mathbf{ }\mathbf{(}\mathbf{4}\mathbf{.}\mathbf{77}\mathbf{ }\mathbf{g}\mathbf{)}$

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: 

 $$\begin{gathered} \mathbf{Copper}\mathbf{+}\mathbf{Iodine}\mathbf{=}\mathbf{white}\mathbf{ }\mathbf{compound}\mathbf{+}\mathbf{Iodine}\mathbf{ }\mathbf{remains} \\ \mathbf{2}\mathbf{.}\mathbf{55}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{5}\mathbf{.}\mathbf{25}\mathbf{ }\mathbf{g}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{ }\mathbf{g} \end{gathered}$$

 From the equation, we can say that,

 $\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{reactant}\mathbf{ }\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{05}\mathbf{ }\mathbf{g}\mathbf{)}\mathbf{=}\mathbf{Mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{products}\mathbf{ }\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{25}\mathbf{ }\mathit{g}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{ }\mathit{g}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{05}\mathbf{ }\mathit{g}\mathbf{)}$

 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.

 $$\begin{gathered} \frac{\mathbf{g}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cu}\mathbf{ }\mathbf{reacted}}{\mathbf{g}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{I}\mathbf{ }\mathbf{reacted}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{27}}{\mathbf{2}\mathbf{.}\mathbf{54}} \\ \frac{\mathbf{g}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Cu}\mathbf{ }\mathbf{reacted}}{\mathbf{g}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{I}\mathbf{ }\mathbf{reacted}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{75}}{\mathbf{3}\mathbf{.}\mathbf{50}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5} \end{gathered}$$