Determine the average number of the moles for both magnesium and oxygen and calculate the empirical formula based on the class average

Lab group #
1
2
Mole of Oxygen
Mole of Magnesium
0.012
0.011
0.018
0.014
3 (our data)
4
5
6
0.018
0.017
0.018
0.017
0.018
0.012
0.018
0.017
a. Determine the average number of the moles for both magnesium and oxygen and calculate the empirical formula based on the class average.
b. How does your data compare to the class average.
c. Determine the standard deviation for the moles of oxygen and magnesium.
D. Based on the standard deviation briefly discuss of the class’s overall results

The Correct Answer and Explanation is :

a. Determining the Average Moles and Empirical Formula

To determine the empirical formula of magnesium oxide (MgO) based on the class data, we first calculate the average moles of magnesium and oxygen:

• Average Moles of Oxygen:
  calculatorturn0calculator0
• Average Moles of Magnesium:
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The empirical formula represents the simplest whole-number ratio of elements in a compound. To find this ratio:

1. Calculate the Ratio of Moles:

• Divide the average moles of magnesium by the average moles of oxygen:

1. Interpret the Ratio:

• A ratio close to 1:1 indicates a 1:1 ratio of magnesium to oxygen atoms, suggesting the empirical formula is MgO.

b. Comparison with Class Average

Comparing your group’s data to the class average involves assessing how your measurements align with the mean values. If your group’s mole ratios are close to 1:1, your results are consistent with the class average, supporting the empirical formula MgO. Significant deviations might indicate experimental errors or variations in procedure.

c. Determining the Standard Deviation

The standard deviation measures the spread of data points from the mean, indicating the precision of measurements. To calculate the standard deviation for the moles of oxygen and magnesium:

1. Calculate the Mean (Average):

• calculatorturn0calculator0
• calculatorturn0calculator1

1. Calculate Each Deviation from the Mean:

• Subtract the mean from each data point.

1. Square Each Deviation:

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• calculatorturn0calculator1

1. Calculate the Variance:

• calculatorturn0calculator0
• calculatorturn0calculator1

1. Calculate the Standard Deviation:

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• calculatorturn0calculator1

d. Discussion Based on Standard Deviation

A low standard deviation indicates that the data points are close to the mean, suggesting high precision in measurements. Conversely, a high standard deviation implies greater variability, which may result from experimental errors, inconsistencies in procedure, or equipment limitations. In this experiment, a standard deviation close to zero would indicate that the class’s results are consistent and reliable, supporting the empirical formula MgO. Significant deviations would warrant a review of experimental methods to identify and correct potential sources of error.

For a visual explanation of calculating the empirical formula of magnesium oxide, you might find the following video helpful:

videoEmpirical Formula of Magnesium Oxide Calculationturn0search0