QUESTION 24Mass of evaporating dish and hydrated copper chloride sample (g)Mass of evaporating dish (g)Mass of hydrated copper chloride (g)Mass of evaporating dish and dehydrated copper chloride sample (g) 40.90Mass of dehydrated copper chloride (g)41.1140.11Mass of water (g)Mass of copper (g)0.37Mass of chlorine (g)i) Does the predicted formula based on the charges match the empirical formula from the given data? Explain.For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac).IUSParagraphArial14px国田田日IIEx!! 田田

Mass of copper and water can be calculated as follows. Mass of evaporating dish and hydrated copper…

Answered: QUESTION 24 Mass of evaporating dish…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Step 1

Mass of copper and water can be calculated as follows.

$\begin{array}{rcl} Mass of evaporating dish and hydrated copper chloride & = & 41.11 g \\ Mass of evaporating dish & = & 40.11 g \\ Mass of sample & = & (41.11 - 40.11) g \\ = & 1 g \\ Mass of evaporating dish and dehydrated copper chloride & = & 40.90 g \\ Mass of water & = & (41.11 - 40.90) g \\ = & 0.21 g \end{array}$

Mass of copper is given as $0.37 g$ .

Mass of chlorine in the sample is calculated by substracting the mass of water and copper from mass of sample.

$\begin{array}{rcl} Mass of chlorine & = & 1 g - (0.21 + 0.37) g \\ = & 0.42 g \end{array}$

The common on chlorine atom is $- 1$ and common charge on copper is $+ 2$ .

Hence the predicted formula for dehydrated copper chloride is ${CuCl}_{2}$ .

Step 2

Now let's find the no of moles of copper, chlorine and water in the given sample.

$\begin{array}{rcl} Molar mass of Cu & = & 63.55 g / mol \\ Given mass & = & 0.37 g \\ no of moles & = & \frac{mass}{molar mass} \\ = & \frac{0.37 g}{63.55 g / mol} \\ = & 5.82 \times 10^{- 3} mol \\ Molar mass of Cl & = & 35.45 g / mol \\ Given mass & = & 0.42 g \\ no of moles & = & \frac{mass}{molar mass} \\ = & \frac{0.42 g}{35.45 g / mol} \\ = & 0.01185 mol \\ Molar mass of H_{2} O & = & 18 g / mol \\ Given mass & = & 0.21 g \\ no of moles & = & \frac{mass}{molar mass} \\ = & \frac{0.21 g}{18 g / mol} \\ = & 0.01167 mol \end{array}$

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