Problem 80 - Chemistry The Molecular Science

Question

Magnesium can be extracted from dolomite, a mineral that contains \(13.2 \% \mathrm{Mg}, 21.7 \% \mathrm{Ca}, 13.0 \% \mathrm{C},\) and \(52.1 \% \mathrm{O} .\) Determine the simplest formula for this ionic compound.

Step-by-Step Solution

Answer

The simplest formula for dolomite is \( \text{MgCa(CO}_3\text{)}_2 \).

Step 1: Convert Mass Percent to Grams

Assume we have 100 g of dolomite. This way, the mass percentages can be directly considered as grams of each element: 13.2 g Mg, 21.7 g Ca, 13.0 g C, and 52.1 g O.

Step 2: Convert Grams to Moles

Calculate the number of moles of each element by dividing the grams by the atomic mass:- Moles of Mg: \( \frac{13.2 \text{ g}}{24.31 \text{ g/mol}} = 0.543 \text{ moles} \) - Moles of Ca: \( \frac{21.7 \text{ g}}{40.08 \text{ g/mol}} = 0.542 \text{ moles} \) - Moles of C: \( \frac{13.0 \text{ g}}{12.01 \text{ g/mol}} = 1.08 \text{ moles} \) - Moles of O: \( \frac{52.1 \text{ g}}{16.00 \text{ g/mol}} = 3.26 \text{ moles} \)

Step 3: Determine the Simplest Ratio

Divide the number of moles of each element by the smallest number of moles calculated:- Mg: \( \frac{0.543}{0.542} \approx 1.00 \)- Ca: \( \frac{0.542}{0.542} \approx 1.00 \)- C: \( \frac{1.08}{0.542} \approx 2.00 \)- O: \( \frac{3.26}{0.542} \approx 6.00 \)

Step 4: Write the Empirical Formula

Using the ratios determined, the empirical formula of dolomite is \( \text{Mg}_{1}\text{Ca}_{1}\text{C}_{2}\text{O}_{6} \).

Key Concepts

Ionic Compounds

Ionic compounds are fascinating structures formed from positive and negative ions. These ions result from atoms either losing or gaining electrons, leading to a stable electron arrangement. A classic example includes minerals, like dolomite, which consist of charged particles:

• Positive ions, known as cations (e.g., magnesium and calcium in dolomite).
• Negative ions, known as anions (e.g., carbonate ions in dolomite).

The beauty of ionic compounds lies in their crystal lattice structure. This structure arises from the electrostatic attraction between oppositely charged ions. These compounds are typically solid with high melting points and can conduct electricity when dissolved in water. Understanding their formation helps in determining empirical formulas, as shown in the exercise.

Mole Calculations

Mole calculations form the essence of working with chemical formulas, allowing us to convert between mass and mole quantities. The mole is a fundamental unit in chemistry, akin to a dozen. It equates to Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles.

When calculating moles, it involves the mass of the substance and its molar mass (atomic or molecular weight). For example:

• To find moles of magnesium in dolomite, divide 13.2 grams by magnesium's atomic mass (24.31 g/mol).

This mole conversion is essential for deriving the empirical formula of a compound, showcasing the relative number of moles of each element present.

Mass Percent Conversion

Mass percent conversion is crucial in determining the composition of compounds from their chemical formulas. It tells us how much of each element contributes to the total mass of the compound. This conversion simplifies comparing and combining substances in a formula.

Here's a simple process:

• Assume a total mass (commonly 100 g for ease), making percentages equivalent to grams.
• Use these assumed masses to calculate moles by dividing by the atomic mass.

In the dolomite example, starting with a 100 g sample facilitates finding each element’s moles, vital for writing the empirical formula.

Chemical Formulas

Chemical formulas provide a straightforward way to represent substances using element symbols and subscripts. These symbols reveal the types and numbers of atoms in a compound. There are different types of formulas depending on the context:

• Empirical formulas give the simplest whole-number ratio of atoms, like \(\text{Mg}_1\text{Ca}_1\text{C}_2\text{O}_6\) for dolomite.
• Molecular formulas, showing actual numbers in molecules, can be multiples of the empirical formula.

Understanding chemical formulas is key to studying and balancing chemical reactions. They reveal not just the composition but also hint at the compound's empirical structure, aiding in predicting chemical behavior and properties. Thus, mastering formulas is fundamental for any chemistry student.