Determining the Element Through Chemical Stoichiometry: A Step-by-Step Guide

In chemistry, understanding the relationships between elements and their compounds is fundamental. One common application is to determine an unknown element based on its reaction with a known substance. This article provides a detailed guide on how to determine the element (M) through stoichiometric calculations, using the reaction of (M) with fluorine to form the hexafluoride (MF_6).

Introduction to Chemical Stoichiometry

Chemical stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. When elements react with each other, the masses of the reactants and products are related by the atomic masses of the elements involved. This relationship can be described by the chemical equation and the molar ratios of the reactants and products.

Problem Presentation: Determining the Element (M)

The problem involves determining the element (M) that reacts with excess fluorine (F) to form the hexafluoride (MF_6). Here are the steps involved in solving this problem:

Step 1: Determine the Mass of Fluorine Used

The mass of the hexafluoride (MF_6) produced is 0.547 g, and the mass of the element (M) is 0.250 g. The mass of fluorine (F) used can be calculated by subtracting the mass of (M) from the total mass of (MF_6).

[text{Mass of } F text{Mass of } MF_6 - text{Mass of } M]

[text{Mass of } F 0.547 text{ g} - 0.250 text{ g} 0.297 text{ g}]

Step 2: Calculate the Moles of the Element (M)

The molar mass of the element (M) can be represented as (M_M). The number of moles of (M) is calculated as the mass of (M) divided by its molar mass.

[text{Moles of } M frac{text{Mass of } M}{M_M} frac{0.250 text{ g}}{M_M}]

Step 3: Calculate the Moles of Fluorine

The molar mass of fluorine (F) is approximately 19.00 g/mol. The number of moles of fluorine (F2) used is calculated as the mass of (F_2) divided by its molar mass.

[text{Moles of } F frac{text{Mass of } F}{M_F} frac{0.297 text{ g}}{19.00 text{ g/mol}} approx 0.0156 text{ mol}]

Step 4: Determine the Stoichiometry of the Reaction

The reaction between (M) and (F) can be represented as:

[text{M} 6text{F}_2 rightarrow text{MF}_6]

From this equation, we can see that 1 mole of (M) reacts with 6 moles of (F_2). Therefore, the moles of (M) can be related to the moles of (F_2) as follows:

[text{Moles of } M frac{1}{6} times text{Moles of } F_2]

[text{Moles of } M frac{1}{6} times 0.0156 text{ mol} approx 0.00260 text{ mol}]

Step 5: Calculate the Molar Mass of (M)

Now, we can find the molar mass of (M) by dividing the mass of (M) by the moles of (M).

[text{Molar mass of } M frac{text{Mass of } M}{text{Moles of } M} frac{0.250 text{ g}}{0.00260 text{ mol}} approx 96.15 text{ g/mol}]

Step 6: Identify the Element

The molar mass of approximately 96.15 g/mol corresponds to the element Zirconium (Zr), which has a molar mass of about 91.22 g/mol. Considering the formation of hexafluoride and slight variations in measurements, it is reasonable to identify (M) as Zirconium (Zr).

Additional Example: Molybdenum

In another scenario, the required chemical equation is:

[text{M} 3text{F}_2 rightarrow text{MF}_6]

The mass of (M) consumed is 0.250g and the mass of the product (MF_6) formed is 0.547g. Therefore, the mass of (F_2) consumed is 0.297g.

The molecular weight of (F_2) is 38 g. Using stoichiometric equivalence, the atomic mass of (M) can be determined as:

[text{m}_M frac{0.250 times 3 times 38}{0.297} 95.959 text{ g/mol}]

This closely corresponds to the atomic mass of Molybdenum (Mo), which is 95.94 g/mol. Therefore, the element can be identified as Molybdenum.

Conclusion

Understanding chemical stoichiometry is crucial for formulating accurate predictions and solving problems in chemistry. This article has provided a step-by-step guide on how to determine the element (M) through chemical reactions and stoichiometric calculations. Whether it is Zirconium or Molybdenum, the process is analogous, demonstrating the consistency and reliability of stoichiometric calculations in chemistry.