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Avogadro's number has sometimes been described as a conversion factor between amu and grams. Use the fluorine atom \((19.00\) amu) as an example to show the relationship between the atomic mass unit and the gram.

Short Answer

Expert verified
The atomic mass in amu equals the mass of one mole in grams (19.00 amu = 19.00 grams for fluorine).

Step by step solution

01

Understanding Avogadro's Number

Avogadro's number, approximately \(6.022 imes 10^{23}\), is the number of atoms or molecules in one mole of a substance. This constant allows us to relate atomic mass units (amu) to grams.
02

Atomic Mass of Fluorine

The atomic mass of a fluorine atom is given as 19.00 amu. This means each fluorine atom weighs 19.00 amu.
03

Converting AMU to Grams using Avogadro's Number

To find the mass of one mole of fluorine atoms in grams, we use Avogadro's number. Since Avogadro's number is the number of atoms in one mole, and each atom weighs 19.00 amu, one mole weighs 19.00 grams, as \(19.00 \, \text{amu}/\text{atom} imes 6.022 imes 10^{23} \, \text{atoms/mole} = 19.00 \, \text{grams/mole}\).
04

Conclusion

This calculation shows that the atomic mass in amu is numerically equivalent to the mass of one mole of atoms in grams. Hence, Avogadro's number acts as a conversion factor between amu and grams.

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Key Concepts

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

Atomic Mass Unit (amu)
The atomic mass unit, often abbreviated as amu, is a crucial concept in chemistry and physics. It serves as a unit of mass that helps scientists measure atomic and molecular masses on a very small scale. The atomic mass unit is defined as one twelfth of the mass of a carbon-12 atom, which is approximately equal to 1.66 x 10^-24 grams. This definition is universally accepted in scientific communities as it provides a standardized way to express the weight of atoms and molecules.

The use of amu is essential because the mass of atoms and subatomic particles is extremely small and cumbersome to express in grams directly. By using amu, scientists can more easily communicate and calculate the mass of different atoms in chemical elements.

For instance, the atomic mass of fluorine given as 19.00 amu indicates that one fluorine atom is 19 times heavier than 1 amu. By understanding how atomic mass units work, you can easily relate them to larger measurements like grams.
Mole Concept
The mole concept is a fundamental chemical principle that describes a way to count atoms, molecules, or particles in a given substance. One mole is defined as exactly 6.022 x 10^23 particles, which is known as Avogadro's number. This number is astonishingly large yet very useful for converting between atomic scale and laboratory scale measurements.

This concept plays a pivotal role in chemistry because it allows for a simple way to relate the mass of a substance to the amount of molecules it contains. By using the mole, you can convert atomic mass unit measurements into grams rather effortlessly.

For example, when you have one mole of fluorine atoms—it means you have 6.022 x 10^23 fluorine atoms. Since each fluorine atom has an atomic mass of 19 amu, one mole of fluorine would weigh 19 grams. Therefore, the mole concept is a convenient bridge between the microscopic world of atoms and the macroscopic world of grams and kilograms.
Conversion Factor
A conversion factor is a mathematical tool that allows you to adjust and convert one unit of measure into another while maintaining accuracy. In chemistry, Avogadro's number acts as a crucial conversion factor between atomic mass units (amu) and grams.

Understanding how to utilize conversion factors is vital when engaging in stoichiometric calculations or when you are trying to relate atomic-scale masses to laboratory-scale masses.

In the context of the fluorine atom, using 19.00 amu and Avogadro's number, you can convert this atomic mass into a gram mass. The calculation makes it clear: one mole of fluorine atoms, each 19.00 amu, corresponds to 19.00 grams. This direct conversion demonstrates not only how these units interrelate but also underscores why Avogadro's number is indispensable in chemical calculations.

Ultimately, the conversion factor enables you to move seamlessly between different measurement scales without losing meaning or precision.

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Most popular questions from this chapter

A sample of a compound of \(\mathrm{Cl}\) and \(\mathrm{O}\) reacts with an excess of \(\mathrm{H}_{2}\) to give \(0.233 \mathrm{~g}\) of \(\mathrm{HCl}\) and \(0.403 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) Determine the empirical formula of the compound.

Lysine, an essential amino acid in the human body, contains \(\mathrm{C}, \mathrm{H}, \mathrm{O},\) and \(\mathrm{N}\). In one experiment, the complete combustion of \(2.175 \mathrm{~g}\) of lysine gave \(3.94 \mathrm{~g}\) \(\mathrm{CO}_{2}\) and \(1.89 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In a separate experiment, \(1.873 \mathrm{~g}\) of lysine gave \(0.436 \mathrm{~g} \mathrm{NH}_{3}\). (a) Calculate the empirical formula of lysine. (b) The approximate molar mass of lysine is \(150 \mathrm{~g}\). What is the molecular formula of the compound?

Silicon tetrachloride \(\left(\mathrm{SiCl}_{4}\right)\) can be prepared by heating Si in chlorine gas: $$ \mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(l) $$ In one reaction, \(0.507 \mathrm{~mol}\) of \(\mathrm{SiCl}_{4}\) is produced. How many moles of molecular chlorine were used in the reaction?

Limestone \(\left(\mathrm{CaCO}_{3}\right)\) is decomposed by heating to quicklime \((\mathrm{CaO})\) and carbon dioxide. Calculate how many grams of quicklime can be produced from \(1.0 \mathrm{~kg}\) of limestone.

Heating \(2.40 \mathrm{~g}\) of the oxide of metal \(\mathrm{X}\) (molar mass of \(\mathrm{X}=55.9 \mathrm{~g} / \mathrm{mol}\) ) in carbon monoxide (CO) yields the pure metal and carbon dioxide. The mass of the metal product is \(1.68 \mathrm{~g}\). From the data given, show that the simplest formula of the oxide is \(\mathrm{X}_{2} \mathrm{O}_{3}\) and write a balanced equation for the reaction.

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