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Is \(\mathrm{MgCl}_{2}\) an empirical or a molecular formula for magnesium chloride? Explain.

Short Answer

Expert verified
\(\mathrm{MgCl}_{2}\) is an empirical formula for magnesium chloride as it represents the simplest whole-number ratio of Mg to Cl (1:2).

Step by step solution

01

- Understand the Definitions

First, it's essential to know what empirical and molecular formulas are. An empirical formula represents the simplest whole-number ratio of the elements in a compound. A molecular formula shows the exact number of atoms of each element in a molecule of the compound.
02

- Define the Formula of the Compound

The chemical formula provided is \(\mathrm{MgCl}_{2}\), which indicates that the compound contains 1 magnesium (Mg) atom and 2 chlorine (Cl) atoms.
03

- Determine the Ratio of Elements

Observe the ratio of magnesium to chlorine in \(\mathrm{MgCl}_{2}\). The ratio is 1:2. This ratio is already in its simplest whole-number form.
04

- Analyze if the Formula Can be Simplified

Check if the ratio of 1:2 can be simplified further. Since it is already in its simplest form, it confirms that \(\mathrm{MgCl}_{2}\) cannot be reduced any further.
05

- Conclude the Nature of the Formula

Given that \(\mathrm{MgCl}_{2}\) is in its simplest whole-number ratio and cannot be simplified further, it is an empirical formula.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Formulas
Chemical formulas are the shorthand way of representing chemical substances. They tell us which elements are present in a compound and how many atoms of each element are there. For example, \(\text{H}_2\text{O}\) tells us that water is composed of 2 hydrogen atoms and 1 oxygen atom. There are two main types of chemical formulas: empirical and molecular.
The empirical formula provides the simplest whole-number ratio of the elements in a compound. It doesn't give you the exact number of atoms, just their simplest ratio. The molecular formula, on the other hand, shows the actual number of atoms of each element in a molecule. This means it can be the same as the empirical formula or a multiple of it.
Understanding these formulas is crucial in chemistry as they help us comprehend the composition and structure of substances.
Simplest Whole-Number Ratio
The simplest whole-number ratio is a key concept when dealing with empirical formulas. It refers to the smallest possible ratio of the elements in a compound that still retains the same proportions. For example, the empirical formula for glucose, which has a molecular formula of \(\text{C}_6\text{H}_{12}\text{O}_6\), is \(\text{CH}_2\text{O}\). Here, the ratio of carbon to hydrogen to oxygen is 6:12:6, which simplifies to 1:2:1.
In practical terms, to find the empirical formula, you would look at the number of atoms of each element, determine their ratios, and then simplify them to the smallest whole numbers. Using \(\text{MgCl}_2\) as an example, we see it contains 1 magnesium atom and 2 chlorine atoms. The ratio 1:2 cannot be simplified any further, indicating that it is already in its simplest whole-number form.
Understanding this concept helps in recognizing the empirical formulas of compounds quickly and accurately.
Molecular Formulas
Molecular formulas provide more detailed information than empirical formulas. They show the exact number of each type of atom in a single molecule of a substance. For instance, benzene has a molecular formula of \(\text{C}_6\text{H}_6\), which indicates that six carbon atoms and six hydrogen atoms are present in one molecule of benzene.
Sometimes, the molecular formula and the empirical formula of a compound can be the same, especially when the ratio of the atoms cannot be simplified further. However, in many cases, the molecular formula is a simple multiple of the empirical one. As in the case of hydrogen peroxide: its molecular formula is \(\text{H}_2\text{O}_2\), and its empirical formula is \(\text{HO}\). The molecular formula indicates the full composition, while the empirical formula shows the simplest ratio.
This distinction is essential because knowing a compound's molecular formula allows us to better understand its molecular structure and properties.

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Most popular questions from this chapter

Two successive reactions, \(\mathrm{D} \longrightarrow \mathrm{E}\) and \(\mathrm{E} \longrightarrow \mathrm{F},\) have yields of \(48 \%\) and \(73 \%\), respectively. What is the overall percent yield for conversion of \(\mathrm{D}\) to \(\mathrm{F} ?\)

Oxygen is required for the metabolic combustion of foods. Calculate the number of atoms in \(38.0 \mathrm{~g}\) of oxygen gas, the amount absorbed from the lungs in about 15 min when a person is at rest.

The first sulfur-nitrogen compound was prepared in 1835 and has been used to synthesize many others. In the early \(1980 \mathrm{~s}\) researchers made another such compound that conducts electricity like a metal. Mass spectrometry of the compound shows a molar mass of \(184.27 \mathrm{~g} / \mathrm{mol}\), and analysis shows it to contain 2.288 g of \(\mathrm{S}\) for every \(1.000 \mathrm{~g}\) of \(\mathrm{N}\). What is its molecular formula?

Solid iodine trichloride is prepared in two steps: first, a reaction between solid iodine and gaseous chlorine to form solid iodine monochloride; second, treatment of the solid with more chlorine gas. (a) Write a balanced equation for each step. (b) Write a balanced equation for the overall reaction. (c) How many grams of iodine are needed to prepare \(2.45 \mathrm{~kg}\) of final product?

What is the molecular formula of each compound? (a) Empirical formula \(\mathrm{CH}(\mathscr{M}=78.11 \mathrm{~g} / \mathrm{mol})\) (b) Empirical formula \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{2}(\mathscr{M}=74.08 \mathrm{~g} / \mathrm{mol})\) (c) Empirical formula \(\mathrm{HgCl}(\mathscr{M}=472.1 \mathrm{~g} / \mathrm{mol})\) (d) Empirical formula \(\mathrm{C}_{7} \mathrm{H}_{4} \mathrm{O}_{2}(\mathscr{A}=240.20 \mathrm{~g} / \mathrm{mol})\)

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