Calculate the theoretical yield (in grams) of magneisum oxide in this experiment. **Determination of the Empirical Formula of Magnesium Oxide****Data Table 1: Empirical Formula of Magnesium Oxide**| | Trial 1 | Trial 2 ||-------------------|------------|------------|| Mass of empty crucible | 15.6800 g | 16.5808 g || Mass of crucible with Mg | 15.7857 g | 16.7053 g || Observations of reaction and product | It began to light up, a bright orange and after cooling turned into an ashy substance. || Mass of crucible and product | 15.8531 g | 16.7960 g |**Questions and Calculations: SHOW WORK on all calculations.**1. **Calculate the mass of magnesium used in this experiment.** - **Trial 1:** $ 15.7857 \, \text{g} - 15.6800 \, \text{g} = 0.1057 \, \text{g} $ - **Trial 2:** $ 16.7053 \, \text{g} - 16.5808 \, \text{g} = 0.1245 \, \text{g} $2. **Calculate the mass of product made in this experiment.** - **Trial 1:** $ 15.8531 \, \text{g} - 15.6800 \, \text{g} = 0.1731 \, \text{g} $ - **Trial 2:** $ 16.7960 \, \text{g} - 16.5808 \, \text{g} = 0.2152 \, \text{g} $***It is vital to calculate these values correctly, or the remaining calculations will all be incorrect. Feel free to check with the instructor here before continuing with the remaining calculations.***

Here the reaction of the experiment is .......... 2 Mg + O2 →2 MgO ........ (1) Here we have to…

Answered: Trial 1 15.48009 15.78579 Mass of empty…

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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Calculate the theoretical yield (in grams) of magneisum oxide in this experiment.

**Determination of the Empirical Formula of Magnesium Oxide**

**Data Table 1: Empirical Formula of Magnesium Oxide**

| | Trial 1 | Trial 2 |
|-------------------|------------|------------|
| Mass of empty crucible | 15.6800 g | 16.5808 g |
| Mass of crucible with Mg | 15.7857 g | 16.7053 g |
| Observations of reaction and product | It began to light up, a bright orange and after cooling turned into an ashy substance. |
| Mass of crucible and product | 15.8531 g | 16.7960 g |

**Questions and Calculations: SHOW WORK on all calculations.**

1. **Calculate the mass of magnesium used in this experiment.**

- **Trial 1:**
$ 15.7857 \, \text{g} - 15.6800 \, \text{g} = 0.1057 \, \text{g} $

- **Trial 2:**
$ 16.7053 \, \text{g} - 16.5808 \, \text{g} = 0.1245 \, \text{g} $

2. **Calculate the mass of product made in this experiment.**

- **Trial 1:**
$ 15.8531 \, \text{g} - 15.6800 \, \text{g} = 0.1731 \, \text{g} $

- **Trial 2:**
$ 16.7960 \, \text{g} - 16.5808 \, \text{g} = 0.2152 \, \text{g} $

***It is vital to calculate these values correctly, or the remaining calculations will all be incorrect. Feel free to check with the instructor here before continuing with the remaining calculations.***

Expert Solution

Step 1

Here the reaction of the experiment is .......... 2 Mg + O₂ $\to$ 2 MgO ........ (1)

Here we have to calculate the theoretical yield of MgO in gram for this reaction.

Theoretical yield means the mass of MgO formed if 100% reaction would occur.

Now we will calculate theoretical yield of MgO in both the trials.

In trial 1 ....

Mass of empty crucible = 15.6800 g

Mass of crucible with magnesium = 15.7857 g

So, mass of Mg = (15.7857- 15.6800) g = 0.1057 g

Mass of product with crucible = 15.8531 g

Therefore, mass of product(MgO) = (15.8531- 15.6800) g = 0.1731 g

So, no of moles of reactant (Mg) = 0.1057/ 24.305 = 4.34889 X 10^{$\land$ -3} moles $\approx$ 4.35X 10^{$\land$ -3}moles [Molar mass of Mg = 24.305 g/mole]

From (1)

From 2 moles of Mg , 2 moles of MgO is formed.

So, from 4.35X 10^{$\land$ -3}moles of Mg = 4.35X 10^{$\land$ -3}moles of MgO formed.

Therefore, Theoretical yield of MgO = 4.35X 10^{$\land$ -3}X 40.30 = 0.1753 g. [ Molar mass of MgO = 40.30g/mole]

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