Calculating Grams Of Magnesium Hydroxide From Formula Units

In chemistry, accurately converting between different units of measurement is a fundamental skill. A common task involves converting the number of formula units of a compound to its mass in grams. This article will guide you through the process of calculating the mass of magnesium hydroxide ($Mg(OH)_2$) in grams, given a specific number of formula units. We will break down the steps, explain the underlying concepts, and provide a clear, step-by-step solution. Understanding these conversions is essential for various applications, from laboratory experiments to industrial processes. Whether you are a student learning stoichiometry or a professional needing a refresher, this guide will help you master this crucial skill. Let's delve into the specifics of converting formula units to grams using $Mg(OH)_2$ as our example.

Understanding Formula Units and Molar Mass

Before we dive into the calculation, it's crucial to grasp the concepts of formula units and molar mass. A formula unit represents the smallest electrically neutral collection of ions in an ionic compound. In the case of $Mg(OH)_2$, one formula unit consists of one magnesium ion ($Mg^{2+}$) and two hydroxide ions ($OH^-$). This is a discrete unit representing the compound's composition at the atomic level. Molar mass, on the other hand, is the mass of one mole of a substance, expressed in grams per mole (g/mol). One mole is defined as $6.022 imes 10^{23}$ entities (atoms, molecules, ions, etc.), a number known as Avogadro's number. The molar mass of a compound is numerically equivalent to the sum of the atomic masses of all the atoms in its formula, which can be found on the periodic table. To calculate the molar mass of $Mg(OH)_2$, we add the atomic mass of magnesium (Mg), two times the atomic mass of oxygen (O), and two times the atomic mass of hydrogen (H). This calculation bridges the gap between the microscopic world of atoms and molecules and the macroscopic world of grams that we can measure in the laboratory. Understanding molar mass allows us to convert between the number of moles, the number of formula units, and the mass of a substance, which is fundamental to stoichiometric calculations.

Calculating the Molar Mass of $Mg(OH)_2$

The first step in converting formula units to grams is to calculate the molar mass of $Mg(OH)_2$. To do this, we need to look up the atomic masses of each element in the compound from the periodic table. The atomic mass of magnesium (Mg) is approximately 24.31 g/mol, the atomic mass of oxygen (O) is approximately 16.00 g/mol, and the atomic mass of hydrogen (H) is approximately 1.01 g/mol. Now, we calculate the molar mass of $Mg(OH)_2$ by adding the atomic masses of each element, considering the number of atoms of each element in the formula:

Molar mass of $Mg(OH)_2$ = (1 × atomic mass of Mg) + (2 × atomic mass of O) + (2 × atomic mass of H)

Molar mass of $Mg(OH)_2$ = (1 × 24.31 g/mol) + (2 × 16.00 g/mol) + (2 × 1.01 g/mol)

Molar mass of $Mg(OH)_2$ = 24.31 g/mol + 32.00 g/mol + 2.02 g/mol

Molar mass of $Mg(OH)_2$ = 58.33 g/mol

So, the molar mass of $Mg(OH)_2$ is approximately 58.33 grams per mole. This value is crucial because it provides the conversion factor between moles and grams. This means that one mole of $Mg(OH)_2$ weighs 58.33 grams. With the molar mass calculated, we are now equipped to convert formula units to grams using the appropriate conversion factors. This molar mass serves as a bridge, linking the number of moles to the mass in grams, a critical step in many chemical calculations. Knowing the molar mass of $Mg(OH)_2$ allows us to accurately determine the mass corresponding to a given number of formula units, making it an indispensable value in stoichiometry.

Converting Formula Units to Moles

Now that we have the molar mass of $Mg(OH)_2$, the next step is to convert the given number of formula units into moles. We are given $7.5 imes 10^{23}$ formula units of $Mg(OH)_2$. To convert formula units to moles, we use Avogadro's number ($6.022 imes 10^{23}$ entities/mol) as a conversion factor. Avogadro's number represents the number of formula units (or molecules, atoms, etc.) in one mole of a substance. The conversion factor is set up so that the units of formula units cancel out, leaving us with moles. This is achieved by dividing the given number of formula units by Avogadro's number. The setup looks like this:

Moles of $Mg(OH)_2$ = (Number of formula units) / (Avogadro's number)

Moles of $Mg(OH)_2$ = $(7.5 imes 10^{23} ext{ f.un.}) / (6.022 imes 10^{23} ext{ f.un./mol})$

Performing this calculation gives us the number of moles of $Mg(OH)_2$. This step is vital because it translates the discrete number of formula units into a more manageable chemical quantity – moles. Moles are central to stoichiometric calculations, allowing us to relate the amounts of different substances in a chemical reaction. Converting to moles is a crucial intermediate step that allows us to use the molar mass to find the mass in grams. This conversion demonstrates the fundamental principle of using Avogadro's number to move between the count of individual particles and the collective unit of the mole, bridging the gap between the microscopic and macroscopic scales in chemistry.

Converting Moles to Grams

After converting formula units to moles, the final step is to convert moles of $Mg(OH)_2$ to grams. We use the molar mass of $Mg(OH)_2$ (58.33 g/mol) as a conversion factor. To convert moles to grams, we multiply the number of moles by the molar mass. This conversion factor is set up so that the units of moles cancel out, leaving us with grams. Let's first calculate the moles of $Mg(OH)_2$ using the formula from the previous section:

Moles of $Mg(OH)_2$ = $(7.5 imes 10^{23} ext{ f.un.}) / (6.022 imes 10^{23} ext{ f.un./mol})$

Moles of $Mg(OH)_2$ ≈ 1.245 moles

Now, we use the molar mass to convert moles to grams:

Grams of $Mg(OH)_2$ = (Moles of $Mg(OH)_2$) × (Molar mass of $Mg(OH)_2$)

Grams of $Mg(OH)_2$ = (1.245 mol) × (58.33 g/mol)

Grams of $Mg(OH)_2$ ≈ 72.59 grams

Therefore, there are approximately 72.59 grams of $Mg(OH)_2$ in $7.5 imes 10^{23}$ formula units. This final conversion provides the answer in a unit that is easily measurable in a laboratory setting. Converting moles to grams is a routine task in chemistry, essential for preparing solutions, performing reactions, and analyzing products. This calculation highlights the practical application of molar mass as a bridge between the number of moles and the mass of a substance, making it a cornerstone of quantitative chemistry.

Identifying the Correct Conversion Factor

In the conversion process, it's crucial to use the correct conversion factors and place them in the appropriate positions. In the step where we convert formula units to moles, the conversion factor is Avogadro's number ($6.022 imes 10^{23}$ formula units/mol). To ensure the units cancel out correctly, Avogadro's number should be placed in the denominator, and 1 mole should be in the numerator. Thus, the conversion factor that goes in the green box is:

$(1 ext{ mol}) / (6.022 imes 10^{23} ext{ f.un.})$

This placement ensures that the formula units in the numerator of the given quantity cancel out with the formula units in the denominator of the conversion factor, leaving the answer in moles. Correctly identifying and positioning the conversion factor is essential for accurate calculations in chemistry. The conversion factor acts as a bridge, transforming the units from one form to another while preserving the underlying quantity. In this case, Avogadro's number serves as the bridge between the count of individual formula units and the collective unit of the mole. Understanding how to correctly apply conversion factors is a fundamental skill in stoichiometry, crucial for solving a wide range of chemical problems.

Step-by-Step Solution

To summarize, here is the step-by-step solution to the problem:

1. Calculate the molar mass of $Mg(OH)_2$:
   • Mg: 1 × 24.31 g/mol = 24.31 g/mol
   • O: 2 × 16.00 g/mol = 32.00 g/mol
   • H: 2 × 1.01 g/mol = 2.02 g/mol
   • Molar mass of $Mg(OH)_2$ = 24.31 + 32.00 + 2.02 = 58.33 g/mol
2. Convert formula units to moles:
   • Moles of $Mg(OH)_2$ = $(7.5 imes 10^{23} ext{ f.un.}) / (6.022 imes 10^{23} ext{ f.un./mol})$ ≈ 1.245 moles
3. Convert moles to grams:
   • Grams of $Mg(OH)_2$ = (1.245 mol) × (58.33 g/mol) ≈ 72.59 grams
4. Identify the correct conversion factor:
   • The conversion factor that goes in the green box is: $(1 ext{ mol}) / (6.022 imes 10^{23} ext{ f.un.})$

This structured approach allows for a clear and accurate conversion from formula units to grams. By breaking down the problem into manageable steps, the process becomes less daunting and easier to understand. Each step builds upon the previous one, leading to the final answer. This method underscores the importance of a systematic approach in solving quantitative chemistry problems. By following these steps, you can confidently convert between different units of measurement and tackle more complex stoichiometric calculations.

Conclusion

Converting formula units to grams is a fundamental skill in chemistry, essential for various calculations and applications. In this article, we demonstrated how to convert $7.5 imes 10^{23}$ formula units of $Mg(OH)_2$ to grams. We first calculated the molar mass of $Mg(OH)_2$, then converted formula units to moles using Avogadro's number, and finally converted moles to grams using the molar mass. We also identified the correct conversion factor for converting formula units to moles. The process involves understanding the definitions of formula units and molar mass, applying conversion factors correctly, and performing the necessary calculations. Mastering these conversions is crucial for success in chemistry, allowing for accurate quantification and manipulation of chemical substances. The ability to convert between different units of measurement is a cornerstone of quantitative chemistry, enabling chemists to make precise measurements, predict reaction outcomes, and synthesize new compounds. By practicing these conversions, you build a solid foundation for more advanced topics in chemistry and related fields. This skill is not only valuable in academic settings but also essential in industrial and research laboratories where accurate measurements and conversions are paramount.