Experiment Data Table This table is used to record and analyze the results of an experiment involving the measurement of mass and moles in different test tubes. Here’s a detailed breakdown of each column and corresponding values: | | Tube A | Tube B | Tube C | |---|--------|--------|--------| | 1. Mass of tube + NaHCO₃ | 104.63 g | 105.04 g | 103.92 g | | 2. Mass of tube + boiling chips | 100.25 g | 100.25 g | 100.35 g | | 3. Mass of NaHCO₃ | | | | | 4. Moles of NaHCO₃ | | | | | 5. Mass of tube + NaCl | 103.28 g | 103.23 g | 103.81 g | | 6. Mass of tube (same as Line 2) | 100.25 g | 100.15 g | 100.35 g | | 7. Mass of NaCl | | | | | 8. Moles of NaCl | | | | | 9. Ratio: moles of NaCl/moles of NaHCO₃ | | | | Instructions: - Row 3 and 7 data should be recorded to 2 decimal places. - Row 4, 8 should reflect values to 3 significant figures. - Ensure to include the correct units for all values. Additional Notes: - The table is utilized to compute the masses of substances based on differences and physical states, with further calculations to derive moles and ratios essential for stoichiometric analysis. - Average Ratio for the three test tubes can be calculated from the ratios found in Row 9. This template supports experiments focusing on chemical reactions, specifically those involving sodium bicarbonate (NaHCO₃) and sodium chloride (NaCl).

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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Question

**Experiment Data Table**

This table is used to record and analyze the results of an experiment involving the measurement of mass and moles in different test tubes. Here’s a detailed breakdown of each column and corresponding values:

| | Tube A | Tube B | Tube C |
|---|--------|--------|--------|
| **1. Mass of tube + NaHCO₃** | 104.63 g | 105.04 g | 103.92 g |
| **2. Mass of tube + boiling chips** | 100.25 g | 100.25 g | 100.35 g |
| **3. Mass of NaHCO₃** | | | |
| **4. Moles of NaHCO₃** | | | |
| **5. Mass of tube + NaCl** | 103.28 g | 103.23 g | 103.81 g |
| **6. Mass of tube (same as Line 2)** | 100.25 g | 100.15 g | 100.35 g |
| **7. Mass of NaCl** | | | |
| **8. Moles of NaCl** | | | |
| **9. Ratio: moles of NaCl/moles of NaHCO₃** | | | |

**Instructions:**
- **Row 3 and 7** data should be recorded to 2 decimal places.
- **Row 4, 8** should reflect values to 3 significant figures.
- Ensure to include the correct units for all values.

**Additional Notes:**
- The table is utilized to compute the masses of substances based on differences and physical states, with further calculations to derive moles and ratios essential for stoichiometric analysis.
- Average Ratio for the three test tubes can be calculated from the ratios found in Row 9.

This template supports experiments focusing on chemical reactions, specifically those involving sodium bicarbonate (NaHCO₃) and sodium chloride (NaCl).

Expert Solution

Step 1

$M a s s o f N a H C O_{3} = {M a s s o f t u b e + N a H C O_{3}} - {M a s s o f t u b e} M o l a r m a s s o f N a H C O_{3} = 84 g / m o l N u m b e r o f m o l e s o f N a H C O_{3} = \frac{g i v e n m a s s}{m o l a r m a s s} T u b e A : M a s s o f t u b e + N a H C O_{3} = 104.63 g M a s s o f t u b e + b o i l i n g c h i p s = 100.25 g M a s s o f N a H C O_{3} = 104.63 g - 100.25 g = 4.38 g N u m b e r o f m o l e s o f N a H C O_{3} = \frac{4.38 g}{84 g / m o l} = 0.0521 m o l S i m i l a r l y, T u b e B : M a s s o f N a H C O_{3} = 105.04 g - 100.15 g = 4.89 g N u m b e r o f m o l e s o f N a H C O_{3} = \frac{4.89 g}{84 g / m o l} = 0.0582 m o l S i m i l a r l y, T u b e C : M a s s o f N a H C O_{3} = 103.92 g - 100.35 g = 3.57 g N u m b e r o f m o l e s o f N a H C O_{3} = \frac{3.57 g}{84 g / m o l} = 0.0425 m o l$

Step 2

$M a s s o f N a O H = {M a s s o f t u b e + N a O H} - {M a s s o f t u b e} M o l a r m a s s o f N a O H = 40 g / m o l N u m b e r o f m o l e s o f N a O H = \frac{g i v e n m a s s}{m o l a r m a s s} T u b e A : M a s s o f t u b e + N a O H = 103.89 g M a s s o f t u b e + b o i l i n g c h i p s = 100.25 g M a s s o f N a O H = 103.89 g - 100.25 g = 3.64 g N u m b e r o f m o l e s o f N a O H = \frac{3.64 g}{40 g / m o l} = 0.0910 m o l S i m i l a r l y, T u b e B : M a s s o f N a O H = 103.23 g - 100.15 g = 3.08 g N u m b e r o f m o l e s o f N a O H = \frac{3.08 g}{40 g / m o l} = 0.0770 m o l S i m i l a r l y, T u b e C : M a s s o f N a O H = 103.81 g - 100.35 g = 3.46 g N u m b e r o f m o l e s o f N a O H = \frac{3.46 g}{40 g / m o l} = 0.0865 m o l$

Step 3

$R a t i o = \frac{M o l e s o f N a C l}{M o l e s o f N a H C O_{3}} T u b e A : R a t i o = \frac{0.0910 m o l}{0.0521 m o l} = 1.75 T u b e B : R a t i o = \frac{0.0770 m o l}{0.0582 m o l} = 1.31 T u b e C : R a t i o = \frac{0.0865 m o l}{0.0425 m o l} = 2.04$