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Chapter 11: Problem 65

\(11-65 .\) (a) What is the difference between an empirical formula and a molecular formula? (b) Is it possible for the empirical and molecular formulas of a compound to be the same?

Short Answer

Expert verified
The empirical formula is the simplest ratio, while the molecular formula is the actual number of atoms; they can be the same.

Step by step solution

01

Understanding Empirical Formula

An empirical formula is the simplest whole-number ratio of elements in a compound. It does not necessarily represent the actual number of atoms present in a compound but shows the smallest ratio between them.
02

Understanding Molecular Formula

A molecular formula gives the actual number of atoms of each element in a molecule of the compound. It represents the true structure of a molecule, thus providing more detailed information.
03

Comparing Empirical and Molecular Formulas

While the empirical formula provides the simplest ratio, the molecular formula shows the total count of atoms. The molecular formula is often a multiple of the empirical formula if more than the basic ratio of atoms is present in the molecule.
04

Analyzing Possibility of Matching Formulas

Yes, it is possible for the empirical and molecular formulas to be the same if the actual molecule already is in its simplest form, meaning the simplest ratio also represents the total number of atoms in the molecule.

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Key Concepts

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

Empirical Formula
An empirical formula is like a basic recipe in chemistry. It shows the simplest whole-number ratio of the different atoms within a compound. Think of it as the foundational guideline to various compounds, without focusing on the complex details. It gives us an idea of how the elements relate to each other in the simplest terms.

For example:
  • A compound with an empirical formula of CH could be made up of any number of carbon and hydrogen atoms, as long as they maintain a 1:1 ratio.
  • In fact, the compound could be C extsubscript{2}H extsubscript{2}, C extsubscript{3}H extsubscript{3}, etc., all of which adhere to this simplest ratio.
The empirical formula is useful for providing a straightforward snapshot of the elemental makeup of a compound. However, it won't give you the full picture of the exact number of atoms, as it simply represents the lowest possible ratio.
Molecular Formula
Now, if the empirical formula is the basic recipe, think of the molecular formula as the precise dish. It indicates the total number of each type of atom in a single molecule of the compound, offering a detailed portrayal of its structure. In essence, it walks you through the actual blueprint of the compound.

For instance:
  • Consider glucose which has a molecular formula of C extsubscript{6}H extsubscript{12}O extsubscript{6}. It means each molecule contains exactly 6 carbon, 12 hydrogen, and 6 oxygen atoms.
  • This molecular formula can be simplified into an empirical formula by reducing it to CH extsubscript{2}O, representing the simplest ratio of the atoms.
The molecular formula is crucial when you need to know the exact details and structure of a compound. It shows how the compound is constructed at the atomic level, which can be particularly important in chemical reactions and molecular biology.
Atomic Ratio
Atomic ratio is a term that ties closely with both empirical and molecular formulas. It provides the ratio of different elements present within a compound, reflecting how these atoms compare within the structure. The atomic ratio helps reveal key insights into the properties and behavioral tendencies of that compound in reactions.

Here's a simple breakdown:
  • The atomic ratio is the fundamental aspect of an empirical formula, as it is showcased in its simplest form.
  • If we revisit the CH example, the atomic ratio of 1:1 means for every carbon atom, there鈥檚 one hydrogen atom.
  • The same concept applies to molecular formulas, as well, albeit with potentially more complex ratios that need further breakdown.
Understanding the atomic ratio is essential. It helps predict how compounds will interact in chemical reactions, as well as their intrinsic properties. Whether analyzing empirical or molecular formulas, the atomic ratio offers valuable foundational information about a compound's composition.

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Most popular questions from this chapter

CALCULATIONS INVOLVING CHEMICAL REACTIONS The combustion of propane may be described by the chemical equation $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ How many grams of oxygen are required to burn completely \(10.0\) grams of propane?

A 12.42-gram sample of a mixture of \(\mathrm{NaCl}(s)\) and \(\mathrm{CaCl}_{2}(s)\) is dissolved in water and all the chloride is precipitated as \(\mathrm{AgCl}(s) .\) If the \(\mathrm{AgCl}(s)\) precipitate has a mass of \(31.70\) grams, calculate the mass percentage of each component in the original mixture.

Determine the number of grams of (a) carbon in \(58.5\) grams of calcium carbonate (b) lead in \(10.0\) grams of lead(IV) oxide (c) oxygen in \(5.0\) grams of copper(II) sulfate pentahydrate (d) hydrogen in \(100.0\) grams of hydrogen sulfide

II-83. A scientist is analyzing a white crystalline compound with a pungent odor similar to mothballs. Combustion analysis of \(5.28\) milligrams of the sample yields \(18.13\) milligrams of \(\mathrm{CO}_{2}(g)\) and \(2.97\) milligrams of \(\mathrm{H}_{2} \mathrm{O}(l) .\) Analysis of the compound in a mass spectrometer finds it to have a formula mass of 128 . What is the molecular formula of the compound?

When left exposed to air and water, iron metal will rust, forming an oxide. If a bar of pure iron is allowed to rust and then both the rust and remaining iron are weighed, will the total mass be greater than, less than, or equal to the mass of the original iron bar? If all the rust is scraped off, will the mass of the bar now be greater than, less than, or equal to the original mass of the bar? Justify your answers.
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