# Figure 3.2.6: Average atomic mass (2023)

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There are 21 elements that have only one isotope, so all their atoms have the same mass. All other elements have two or more isotopes, so their atoms have at least two different masses. But all elements obey certain laws of proportion when combined with other elements, so they arepretendThey only have one type of atom with a certain mass. To solve this conundrum, we define atomic mass as the weighted average mass of all naturally occurring (sometimes radioactive) isotopes of an element.

The weighted average is defined as

$\text{atomic mass} = \left(\dfrac{\%\text{isotope-abundance 1}}{100}\right)\times \left(\text{isotope-mass 1}\right) + \left ( \ dfrac{\%\text{Isotope overview 2}}{100}\right)\times \left(\text{Isotope mass 2}\right)~ ~ ~ + ~ ~ ... \number$

Similar conditions will be added for all isotopes. The calculation is similar to what most colleges use to calculate grade point averages:

$\text{GPA} = \left(\tfrac{\text{hours in course 1}}{\text{total hours}}\right)\times \left(\text{grade in course 1}\ right ) + \left(\tfrac{\text{credit course 2}}{\text{total credits}}\right)\times \left(\text{credit course 2}\right)~ + ~ ... \ None number$

## any agreement

deadline"Average Atomic Weight"or just"atomic weight"Usually used to refer to the actual"relative atomic mass"Atomic weights are technically dimensionless because they cannot be specified as absolute values. Historically, they were calculated based on mass ratios (early chemists might say that a magnesium atom is 24.305/15.999 times heavier than an oxygen atom, as this is the mass ratio of magnesium to oxygen in MgO). Atomic weights are now calculated from the positions of peaks in the mass spectrum.

While peak positions can be labeled in amu, this is only possible if the mass spectrometer is calibrated with a standard whose mass is known only relative to another and which is also technically dimensionless. To solve this problem, we havedefinitionamu is 1/12 $${}_{\text{6}}^{\text{12}}\text{C}$$ of the atomic mass. $${}_{\text{6}}^{\text{12}}\text{C}$$ can then be used to calibrate the mass spectrometer. For convenience, we often use the "token" dimension.amu/middle atomfor atomic weight org/molis of molar massa.

Atomic weight calculations include IUPAC isotope values ​​and "naturally occurring isotopes" as defined by the Commission on Atomic Weights (IUPAC/CIAAW), including half-lives greater than 1 x 1010Year. Thus, the atomic weights of thorium, protactinium, and uranium are 232.0, 231.0, and 238.0, respectively, but no isotopes of other radioactive elements are long enough to get atomic weights.

##### Example $$\PageIndex{1}$$: Isotope

Naturally occurring lead has been found to consist of four isotopes:

• 1.40% $${}_{\text{82}}^{\text{204}}\text{Pb}$$ has an isotopic mass of 203.973.
• 24.10% $${}_{\text{82}}^{\text{206}}\text{Pb}$$ has an isotopic mass of 205.974.
• 22.10% $${}_{\text{82}}^{\text{207}}\text{Pb}$$ has an isotopic mass of 206,976.
• 52.40% $${}_{\text{82}}^{\text{208}}\text{Pb}$$ has an isotope mass of 207,977.

Calculate the atomic mass of the average naturally occurring lead sample.

###### solution

Suppose you have 1 mole of lead. This is 1.40% ($$\tfrac{1.40}{100}$$ × 1 mol) $${}_{\text{82}}^{\text{204}}\text{Pb}$$, The molar mass is 203.973 g mol–1.time20482pipeshall

\begin{align*} \text{m}_{\text{204}} &=n_{\text{204}}\times \text{ }M_{\text{204}} \\[4pt] &=\left( \frac{\text{1}\text{.40}}{\text{100}}\times \text{ 1 mol} \right)\text{ (203}\text{.973 g mol}^{\text{-1}}\text{)} \\[4pt] &=\text{2}\text{0,86 g} \end{align*} \liczba

Comparable to other isotopes

\begin{align*}\text{m}_{\text{206}}&=n_{\text{206}}\times \text{ }M_{\text{206}}\\[4pt] &=\left( \frac{\text{24}\text{.10}}{\text{100}}\times \text{ 1 mol} \right)\text{ (205}\text{.974 g mol}^{\text{-1}}\text{)}\\[4pt] &=\text{49}\text{0.64 g} \\[6pt]\text{m}_{\text{207 }}&=n_{\text{207}}\times \text{ }M_{\text{207}}\\[4pt] &=\left( \frac{\text{22}\text{.10} }{\text{100}}\times \text{ 1 mol} \right)\text{ (206}\text{0,976 g mol}^{\text{-1}}\text{)}\\[ 4pt ] &=\text{45}\text{0,74 g} \\[6pt] \text{m}_{\text{208}}&=n_{\text{208}}\times \text{ } M_ {\text{208}}\\[4pt] &=\left( \frac{\text{52}\text{.40}}{\text{100}}\times \text{ 1 mol} \right ) \text{ (207}\text{0,977 g mol}^{\text{-1}}\text{)}\\[4pt] &=\text{108}\text{0,98 g} \end {对齐 *} \brak numeru

When all four results are added together, we get the mass of 1 mole of the isotope mixture

$2,86\, g + 49,64\, g + 45,74\, g + 108,98\, g = 207,22\, g\liczba$

Therefore, as mentioned in the previous discussion, lead has an atomic mass of 207.2 g/mol.

An important conclusion regarding the presence of isotopes should be emphasized here. While very accurate results are obtained, the atomic masses may vary slightly depending on where the elemental sample was collected. To this end, IUPAC CIAAW recently redefined the atomic weights of 10 elements with two or more isotopesThe percentage of different isotopes usually depends on the source of the element.

For example, the atomic weight of oxygen in Antarctic fallout is 15.99903, but oceanic N2The atomic weight of O is 15.9997."fractionering"Isotopes are produced at slightly different rates by chemical and physical processes caused by small differences in their masses. The difference can be even greater when the isotope comes from transmutation.

For example, the percentage of lead $${}_{\text{82}}^{\text{206}}\text{Pb}$$ from uranium decay is much higher than the 24.1% of the average sample. Therefore, the atomic mass of lead in uranium ore is less than 207.2 and closer to 205.974, the isotopic mass $${}_{\text{82}}^{\text{206}}\text{Pb}$$.

Thompson in 1913 after the discovery of elemental isotopes by J.J.,It has been suggested that scales of relative atomic mass (atomic mass) should use the atomic mass of a particular isotope of an element as a reference. Finally, the default $${}_{\text{6}}^{\text{12}}\text{C}$$ was chosen and an atomic mass value of exactly 12,000,000 was specified.

Therefore, the atomic mass is given byatomic scaleis the ratio of the weighted average mass (calculated by example) of the atomic masses of all isotopes of each naturally occurring element to the individual atomic mass$${}_{\text{6}}^{\text {12 }} \text { C}$$ atom. Since carbon is composed of two isotopes, 98.99% $${}_{\text{6}}^{\text{12}}\text{C}$$ isotope masses are 12,000 and 1.11% $${ }_{ \text{6}}^{\text{13}}\text{C}$$ has an isotopic mass of 13,003 and the average atomic mass of carbon is

$\frac{\text{98}\text{.89}}{\text{100}\text{.00}}\text{ }\times \text{ 12}\text{.000 + }\frac {\text{1}\text{.11}}{\text{100}\text{.00}}\text{ }\times \text{ 13}\text{.003}=\text{12}\ tekst{.011} \liczba$

For example.

## Conventional atomic weights and "range"

Deviations from the average isotopic composition are usually small, the so-calledConventional atomic mass valuesIs defined by IUPAC/CIAAW as the element with the greatest abundance variability. You can use them for almost all chemical calculations. But at the same time,atomic scaleThese elements have been redefined as reach, or"interval", for all the little differences that can make a difference in your work.The table below shows typical values.

Table $$\PageIndex{1}$$ Atomic weights
element name the symbol Conventional atomic weight atomic weight
Pol Second 10.81 [10.806; 10.821]
coal C 12.011 [12.0096; 12.0116]
Chlorine Potassium 35.45 [35.446; 35.457]
hydrogen H 1.008 [1,00784; 1.00811]
to light plum 6,94 [6.938; 6.997]
Nitrogen Nee 14.007 [14.00643; 14.00728]
oxygen Europa 15.999 [15,99903; 15.99971]
silicon I 28.085 [28.084; 28.086]
sulfur klein 32.06 [32.059; 32.076]
van time 204,38 [204.382; 204.385]

However, when studying nuclear reactions, attention should be paid to the weight of isotopes. This will be discussed in later chaptersnuclear chemistry.

## Moore's SI definition

Moore's SI definition also depends on the isotope $${}_{\text{6}}^{\text{12}}\text{C}$$, which can now be defined. A mole is the amount of matter in a system that contains as many elementary elements as there are atoms in exactly 0.012 kilograms\({}_{\text{6}}^{\text{12}}\text{ C} \ ). thisbasis organizationThey can be atoms, molecules, ions, electrons or other microscopic particles.

This definition of a mole makes the mass (in grams) of 1 mole of an element numerically equal to the average mass (in grams) of atoms. This official definition of a mole allows for a more accurate determination of Avogadro's constant than previously possible. The currently accepted value isNeeA= 6,02214179 × 1023Moore–1.This is accurate to 0.00000001% and has five more significant digits than 6.022 × 1023Moore–1, the number used to define the moleForHowever, it is very rare that more than four significant digits are needed in an Avogadro constant. The value is 6.022 × 1023Moore–1It is certainly sufficient for most of the calculations you need.

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