3.1: Fórmula Masa y el Concepto Mole

Objetivos de aprendizaje

• Calcular masas de fórmula para compuestos covalentes e iónicos
• Definir la cantidad unidad mol y la cantidad relacionada con el número de Avogadro
• Explicar la relación entre la masa, los moles y el número de átomos o moléculas, y realizar cálculos que deriven estas cantidades entre sí

Podemos argumentar que la ciencia química moderna comenzó cuando los científicos comenzaron a explorar los aspectos cuantitativos y cualitativos de la química. Por ejemplo, la teoría atómica de Dalton fue un intento de explicar los resultados de mediciones que le permitieron calcular las masas relativas de elementos combinados en diversos compuestos. Comprender la relación entre las masas de átomos y las fórmulas químicas de los compuestos permite describir cuantitativamente la composición de las sustancias.

Masa de Fórmula

En un capítulo anterior describimos el desarrollo de la unidad de masa atómica, el concepto de masas atómicas promedio y el uso de fórmulas químicas para representar la composición elemental de sustancias. Estas ideas pueden extenderse para calcular la masa de fórmula de una sustancia sumando las masas atómicas promedio de todos los átomos representados en la fórmula de la sustancia.

Masa de Fórmula para Sustancias Covalentes

Para las sustancias covalentes, la fórmula representa los números y tipos de átomos que componen una sola molécula de la sustancia; por lo tanto, la masa de fórmula puede denominarse correctamente masa molecular. Considere el cloroformo (CHCl ₃), un compuesto covalente que alguna vez se usó como anestésico quirúrgico y que ahora se usó principalmente en la producción de tetrafluoroetileno, el bloque de construcción del polímero “antiadherente”, el teflón. La fórmula molecular del cloroformo indica que una sola molécula contiene un átomo de carbono, un átomo de hidrógeno y tres átomos de cloro. La masa molecular promedio de una molécula de cloroformo es, por lo tanto, igual a la suma de las masas atómicas promedio de estos átomos. La figura\(\PageIndex{1}\) resume los cálculos utilizados para derivar la masa molecular del cloroformo, que es de 119.37 amu.

Figura\(\PageIndex{1}\): La masa promedio de una molécula de cloroformo, CHCl ₃, es 119.37 amu, que es la suma de las masas atómicas promedio de cada uno de sus átomos constituyentes. El modelo muestra la estructura molecular del cloroformo.

Se muestra una tabla y un diagrama. El cuadro está conformado por seis columnas y cinco filas. La fila del encabezado dice: “Elemento”, “Cantidad”, un espacio en blanco, “Masa atómica promedio (a m u)”, un espacio en blanco y “Subtotal (a m u)”. La primera columna contiene los símbolos “C”, “H”, “C l” y una celda fusionada en blanco que recorre el ancho de las primeras cinco columnas. La segunda columna contiene los números “1”, “1" y “3", así como la celda fusionada. La tercera columna contiene el símbolo de multiplicación en cada celda excepto la última celda fusionada. La cuarta columna contiene los números “12.01”, “1.008” y “35.45” así como la celda fusionada. La quinta columna contiene el símbolo “=” en cada celda excepto en la última celda fusionada. La sexta columna contiene los valores “12.01”, “1.008”, “106.35” y “119.37”. Hay una gruesa línea negra por debajo del número 106.35. La celda fusionada bajo las primeras cinco columnas dice “Masa molecular”. A la izquierda de la tabla hay un diagrama de una molécula. Tres esferas verdes están unidas a una esfera negra ligeramente más pequeña, que también está unida a una esfera blanca más pequeña. Las esferas verdes se encuentran debajo y a los lados de la esfera negra, mientras que la esfera blanca se ubica directamente desde la esfera negra.

Asimismo, la masa molecular de una molécula de aspirina, C ₉ H ₈ O ₄, es la suma de las masas atómicas de nueve átomos de carbono, ocho átomos de hidrógeno y cuatro átomos de oxígeno, que asciende a 180.15 amu (Figura\(\PageIndex{2}\)).

Figura\(\PageIndex{2}\): The average mass of an aspirin molecule is 180.15 amu. The model shows the molecular structure of aspirin, C₉H₈O₄.

A table and diagram are shown. The table is made up of six columns and five rows. The header row reads: “Element,” “Quantity,” a blank space, “Average atomic mass (a m u),” a blank space, and “Subtotal (a m u).” The first column contains the symbols “C,” “H,” “O,” and a merged cell. The merged cell runs the length of the first five columns. The second column contains the numbers “9,” “8,” and “4” as well as the merged, cell. The third column contains the multiplication symbol in each cell except for the last, merged cell. The fourth column contains the numbers “12.01,” “1.008,” and “16.00” as well as the merged cell. The fifth column contains the symbol “=” in each cell except for the last, merged cell. The sixth column contains the values: “108.09,” “8.064,” “64.00,” and “180.15.” There is a thick black line below the number 64.00. The merged cell under the first five columns reads “Molecular mass.” To the left of the table is a diagram of a molecule. Six black spheres are located in a six-sided ring and connected by alternating double and single black bonds. Attached to each of the four black spheres is one smaller white sphere. Attached to the farthest right black sphere is a red sphere, connected to two more black spheres, all in a row. Attached to the last black sphere of that row are two more white spheres. Attached to the first black sphere of that row is another red sphere. A black sphere, attached to two red spheres and a white sphere is attached to the black sphere on the top right of the six-sided ring.

Example \(\PageIndex{1}\): Computing Molecular Mass for a Covalent Compound

Ibuprofen, C₁₃H₁₈O₂, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Advil and Motrin. What is the molecular mass (amu) for this compound?

Solution

Molecules of this compound are comprised of 13 carbon atoms, 18 hydrogen atoms, and 2 oxygen atoms. Following the approach described above, the average molecular mass for this compound is therefore:

Exercise \(\PageIndex{1}\)

Acetaminophen, C₈H₉NO₂, is a covalent compound and the active ingredient in several popular nonprescription pain medications, such as Tylenol. What is the molecular mass (amu) for this compound?

Answer

151.16 amu

Formula Mass for Ionic Compounds

Ionic compounds are composed of discrete cations and anions combined in ratios to yield electrically neutral bulk matter. The formula mass for an ionic compound is calculated in the same way as the formula mass for covalent compounds: by summing the average atomic masses of all the atoms in the compound’s formula. Keep in mind, however, that the formula for an ionic compound does not represent the composition of a discrete molecule, so it may not correctly be referred to as the “molecular mass.”

As an example, consider sodium chloride, NaCl, the chemical name for common table salt. Sodium chloride is an ionic compound composed of sodium cations, Na⁺, and chloride anions, Cl⁻, combined in a 1:1 ratio. The formula mass for this compound is computed as 58.44 amu (Figure \(\PageIndex{3}\)).

Figure \(\PageIndex{3}\): Table salt, NaCl, contains an array of sodium and chloride ions combined in a 1:1 ratio. Its formula mass is 58.44 amu.

A table and diagram are shown. The table is made up of six columns and four rows. The header row reads: “Element,” “Quantity,” a blank space, “Average atomic mass (a m u),” a blank space and “Subtotal (a m u).” The first column contains the symbols “N a”, “C l,” and a merged cell. The merged cell runs the length of the first five columns. The second column contains the numbers “1” and “1” as well as the merged cell. The third column contains the multiplication symbol in each cell except for the last, merged cell. The fourth column contains the numbers “22.99” and “35.45” as well as the merged cell. The fifth column contains the symbol “=” in each cell except for the last, merged cell. The sixth column contains the values “22.99,” “35.45,” and “58.44.” There is a thick black line below the number “35.45.” The merged cell under the first five columns reads “Formula mass.” To the left of the table is a diagram of a chemical structure. The diagram shows green and purple spheres placed in an alternating pattern, making up the corners of eight stacked cubes to form one larger cube. The green spheres are slightly smaller than the purple spheres.

Note that the average masses of neutral sodium and chlorine atoms were used in this computation, rather than the masses for sodium cations and chlorine anions. This approach is perfectly acceptable when computing the formula mass of an ionic compound. Even though a sodium cation has a slightly smaller mass than a sodium atom (since it is missing an electron), this difference will be offset by the fact that a chloride anion is slightly more massive than a chloride atom (due to the extra electron). Moreover, the mass of an electron is negligibly small with respect to the mass of a typical atom. Even when calculating the mass of an isolated ion, the missing or additional electrons can generally be ignored, since their contribution to the overall mass is negligible, reflected only in the nonsignificant digits that will be lost when the computed mass is properly rounded. The few exceptions to this guideline are very light ions derived from elements with precisely known atomic masses.

Example \(\PageIndex{2}\): Computing Formula Mass for an Ionic Compound

Aluminum sulfate, Al₂(SO₄)₃, is an ionic compound that is used in the manufacture of paper and in various water purification processes. What is the formula mass (amu) of this compound?

Solution

The formula for this compound indicates it contains Al³⁺ and SO₄²⁻ ions combined in a 2:3 ratio. For purposes of computing a formula mass, it is helpful to rewrite the formula in the simpler format, Al₂S₃O₁₂. Following the approach outlined above, the formula mass for this compound is calculated as follows:

Exercise \(\PageIndex{2}\)

Calcium phosphate, \(\ce{Ca3(PO4)2}\), is an ionic compound and a common anti-caking agent added to food products. What is the formula mass (amu) of calcium phosphate?

Answer

310.18 amu

 

The Mole

The identity of a substance is defined not only by the types of atoms or ions it contains, but by the quantity of each type of atom or ion. For example, water, \(\ce{H2O}\), and hydrogen peroxide, \(\ce{H2O2}\), are alike in that their respective molecules are composed of hydrogen and oxygen atoms. However, because a hydrogen peroxide molecule contains two oxygen atoms, as opposed to the water molecule, which has only one, the two substances exhibit very different properties. Today, we possess sophisticated instruments that allow the direct measurement of these defining microscopic traits; however, the same traits were originally derived from the measurement of macroscopic properties (the masses and volumes of bulk quantities of matter) using relatively simple tools (balances and volumetric glassware). This experimental approach required the introduction of a new unit for amount of substances, the mole, which remains indispensable in modern chemical science.

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter. A mole is defined as the amount of substance containing the same number of discrete entities (such as atoms, molecules, and ions) as the number of atoms in a sample of pure ¹²C weighing exactly 12 g. One Latin connotation for the word “mole” is “large mass” or “bulk,” which is consistent with its use as the name for this unit. The mole provides a link between an easily measured macroscopic property, bulk mass, and an extremely important fundamental property, number of atoms, molecules, and so forth.

The number of entities composing a mole has been experimentally determined to be \(6.02214179 \times 10^{23}\), a fundamental constant named Avogadro’s number (\(N_A\)) or the Avogadro constant in honor of Italian scientist Amedeo Avogadro. This constant is properly reported with an explicit unit of “per mole,” a conveniently rounded version being \(6.022 \times 10^{23}/\ce{mol}\).

Consistent with its definition as an amount unit, 1 mole of any element contains the same number of atoms as 1 mole of any other element. The masses of 1 mole of different elements, however, are different, since the masses of the individual atoms are drastically different. The molar mass of an element (or compound) is the mass in grams of 1 mole of that substance, a property expressed in units of grams per mole (g/mol) (Figure \(\PageIndex{4}\)).

Figure \(\PageIndex{4}\): Each sample contains \(6.022 \times 10^{23}\) atoms —1.00 mol of atoms. From left to right (top row): 65.4 g zinc, 12.0 g carbon, 24.3 g magnesium, and 63.5 g copper. From left to right (bottom row): 32.1 g sulfur, 28.1 g silicon, 207 g lead, and 118.7 g tin. (credit: modification of work by Mark Ott).

This figure contains eight different substances displayed on white circles. The amount of each substance is visibly different.

Because the definitions of both the mole and the atomic mass unit are based on the same reference substance, ¹²C, the molar mass of any substance is numerically equivalent to its atomic or formula weight in amu. Per the amu definition, a single ¹²C atom weighs 12 amu (its atomic mass is 12 amu). According to the definition of the mole, 12 g of ¹²C contains 1 mole of ¹²C atoms (its molar mass is 12 g/mol). This relationship holds for all elements, since their atomic masses are measured relative to that of the amu-reference substance, ¹²C. Extending this principle, the molar mass of a compound in grams is likewise numerically equivalent to its formula mass in amu (Figure \(\PageIndex{5}\)).

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Figure \(\PageIndex{5}\): Each sample contains \(6.022 \times 10^{23}\) molecules or formula units—1.00 mol of the compound or element. Clock-wise from the upper left: 130.2 g of C₈H₁₇OH (1-octanol, formula mass 130.2 amu), 454.4 g of HgI₂ (mercury(II) iodide, formula mass 454.4 amu), 32.0 g of CH₃OH (methanol, formula mass 32.0 amu) and 256.5 g of S₈ (sulfur, formula mass 256.5 amu). (credit: Sahar Atwa).

Table \(\PageIndex{1}\): Mass of one mole of elements

Element	Average Atomic Mass (amu)	Molar Mass (g/mol)	Atoms/Mole
C	12.01	12.01	\(6.022 \times 10^{23}\)
H	1.008	1.008	\(6.022 \times 10^{23}\)
O	16.00	16.00	\(6.022 \times 10^{23}\)
Na	22.99	22.99	\(6.022 \times 10^{23}\)
Cl	33.45	35.45	\(6.022 \times 10^{23}\)

While atomic mass and molar mass are numerically equivalent, keep in mind that they are vastly different in terms of scale, as represented by the vast difference in the magnitudes of their respective units (amu versus g). To appreciate the enormity of the mole, consider a small drop of water after a rainfall. Although this represents just a tiny fraction of 1 mole of water (~18 g), it contains more water molecules than can be clearly imagined. If the molecules were distributed equally among the roughly seven billion people on earth, each person would receive more than 100 billion molecules.

Video \(\PageIndex{1}\): The mole is used in chemistry to represent \(6.022 \times 10^{23}\) of something, but it can be difficult to conceptualize such a large number. Watch this video and then complete the “Think” questions that follow. Explore more about the mole by reviewing the information under “Dig Deeper.”

The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.

Example \(\PageIndex{3}\): Deriving Moles from Grams for an Element

According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?

Solution

The mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu, and so its molar mass is 39.10 g/mol. The given mass of K (4.7 g) is a bit more than one-tenth the molar mass (39.10 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol.

The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):

The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”

\[ \mathrm{4.7\; \cancel{g} K \left ( \dfrac{mol\; K}{39.10\;\cancel{g}}\right)=0.12\;mol\; K} \nonumber \]

The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.

Exercise \(\PageIndex{3}\): Beryllium

Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g?

Answer

0.360 mol

Example \(\PageIndex{4}\): Deriving Grams from Moles for an Element

A liter of air contains \(9.2 \times 10^{−4}\) mol argon. What is the mass of Ar in a liter of air?

Solution

The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10⁻³) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):

In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):

\[\mathrm{9.2 \times10^{-4}\; \cancel{mol} \; Ar \left( \dfrac{39.95\;g}{\cancel{mol}\;Ar} \right)=0.037\;g\; Ar} \nonumber \]

The result is in agreement with our expectations, around 0.04 g Ar.

Exercise \(\PageIndex{4}\)

What is the mass of 2.561 mol of gold?

Answer

504.4 g

Example \(\PageIndex{6}\): Deriving Number of Atoms from Mass for an Element

Copper is commonly used to fabricate electrical wire (Figure \(\PageIndex{6}\)). How many copper atoms are in 5.00 g of copper wire?

Figure \(\PageIndex{6}\): Copper wire is composed of many, many atoms of Cu. (credit: Emilian Robert Vicol)

Solution

The number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu, and then using Avogadro’s number (N_A) to convert this molar amount to number of Cu atoms:

Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth N_A, or approximately 10²² Cu atoms. Carrying out the two-step computation yields:

\[\mathrm{5.00\:\cancel{g}\:Cu\left(\dfrac{\cancel{mol}\:Cu}{63.55\:\cancel{g}}\right)\left(\dfrac{6.022\times10^{23}\:atoms}{\cancel{mol}}\right)=4.74\times10^{22}\:atoms\: of\: copper} \nonumber \]

The factor-label method yields the desired cancellation of units, and the computed result is on the order of 10²² as expected.

Exercise \(\PageIndex{6}\)

A prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?

Answer

\(4.586 \times 10^{22}\; Au\) atoms

Example \(\PageIndex{7}\): Deriving Moles from Grams for a Compound

Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula C₂H₅O₂N. How many moles of glycine molecules are contained in 28.35 g of glycine?

Solution

We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example \(\PageIndex{6}\):

The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, C₂H₅O₂N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:

 

A table is shown that is made up of six columns and six rows. The header row reads: “Element,” “Quantity (mol element / mol compound,” a blank space, “Molar mass (g / mol element),” a blank space, and “Subtotal (a m u).” The first column contains the symbols “C,” “H,” “O,” “N,” and a merged cell. The merged cell runs the width of the first five columns. The second column contains the numbers “2,” “5,” “2,” and “1” as well as the merged cell. The third column contains the multiplication symbol in each cell except for the last, merged cell. The fourth column contains the numbers “12.01,” “1.008,” “16.00,” and “14.007” as well as the merged cell. The fifth column contains the symbol “=” in each cell except for the last, merged cell. The sixth column contains the values “24.02,” “5.040,” “32.00,” “14.007,” and “75.07.” There is a thick black line under the number 14.007. The merged cell under the first five columns reads “Molar mass (g / mol compound). There is a ball-and-stick drawing to the right of this table. It shows a black sphere that forms a double bond with a slightly smaller red sphere, a single bond with another red sphere, and a single bond with another black sphere. The red sphere that forms a single bond with the black sphere also forms a single bond with a smaller, white sphere. The second black sphere forms a single bond with a smaller, white sphere and a smaller blue sphere. The blue sphere forms a single bond with two smaller, white spheres each.

The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass yields:

\[\mathrm{28.35\:\cancel{g}\:glycine\left(\dfrac{mol\: glycine}{75.07\:\cancel{g}}\right)=0.378\:mol\: glycine} \nonumber \]

This result is consistent with our rough estimate.

Exercise \(\PageIndex{7}\)

How many moles of sucrose, \(C_{12}H_{22}O_{11}\), are in a 25-g sample of sucrose?

Answer

0.073 mol

Example \(\PageIndex{8}\): Deriving Grams from Moles for a Compound

Vitamin C is a covalent compound with the molecular formula C₆H₈O₆. The recommended daily dietary allowance of vitamin C for children aged 4–8 years is 1.42 × 10⁻⁴ mol. What is the mass of this allowance in grams?

Solution

As for elements, the mass of a compound can be derived from its molar amount as shown:

The molar mass for this compound is computed to be 176.124 g/mol. The given number of moles is a very small fraction of a mole (~10⁻⁴ or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:

\[\mathrm{1.42\times10^{-4}\:\cancel{mol}\:vitamin\: C\left(\dfrac{176.124\:g}{\cancel{mol}\:vitamin\: C}\right)=0.0250\:g\: vitamin\: C} \nonumber \]

This is consistent with the anticipated result.

Exercise \(\PageIndex{8}\)

What is the mass of 0.443 mol of hydrazine, \(N_2H_4\)?

Answer

14.2 g

Example \(\PageIndex{9}\): Deriving the Number of Molecules from the Compound Mass

A packet of an artificial sweetener contains 40.0 mg of saccharin (C₇H₅NO₃S), which has the structural formula:

Given that saccharin has a molar mass of 183.18 g/mol, how many saccharin molecules are in a 40.0-mg (0.0400-g) sample of saccharin? How many carbon atoms are in the same sample?

Solution

The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Example \(\PageIndex{8}\), and then multiplying by Avogadro’s number:

Using the provided mass and molar mass for saccharin yields:

\[\mathrm{0.0400\:\cancel{g}\:\ce{C7H5NO3S}\left(\dfrac{\cancel{mol}\:\ce{C7H5NO3S}}{183.18\:\cancel{g}\:\ce{C7H5NO3S}}\right)\left(\dfrac{6.022\times10^{23}\:\ce{C7H5NO3S}\:molecules}{1\:\cancel{mol}\:\ce{C7H5NO3S}}\right)}\\
=\mathrm{1.31\times10^{20}\:\ce{C7H5NO3S}\:molecules} \nonumber \]

The compound’s formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided sample is:

\[\mathrm{1.31\times10^{20}\:\ce{C7H5NO3S}\: molecules\left(\dfrac{7\:C\: atoms}{1\:\ce{C7H5NO3S}\: molecule}\right)=9.20\times10^{21}\:C\: atoms} \nonumber \]

Exercise \(\PageIndex{9}\)

How many \(C_4H_{10}\) molecules are contained in 9.213 g of this compound? How many hydrogen atoms?



Answer

• \(9.545 \times 10^{22}\; \text{molecules}\; C_4H_{10}\)
• \(9.545 \times 10^{23 }\;\text{atoms}\; H\)

 

Summary

The formula mass of a substance is the sum of the average atomic masses of each atom represented in the chemical formula and is expressed in atomic mass units. The formula mass of a covalent compound is also called the molecular mass. A convenient amount unit for expressing very large numbers of atoms or molecules is the mole. Experimental measurements have determined the number of entities composing 1 mole of substance to be 6.022 × 10²³, a quantity called Avogadro’s number. The mass in grams of 1 mole of substance is its molar mass. Due to the use of the same reference substance in defining the atomic mass unit and the mole, the formula mass (amu) and molar mass (g/mol) for any substance are numerically equivalent (for example, one H₂O molecule weighs approximately18 amu and 1 mole of H₂O molecules weighs approximately 18 g).

Footnotes

1. 1 Omiatek, Donna M., Amanda J. Bressler, Ann-Sofie Cans, Anne M. Andrews, Michael L. Heien, and Andrew G. Ewing. “The Real Catecholamine Content of Secretory Vesicles in the CNS Revealed by Electrochemical Cytometry.” Scientific Report 3 (2013): 1447, accessed January 14, 2015, doi:10.1038/srep01447.

Glossary

Avogadro’s number (N_A)

experimentally determined value of the number of entities comprising 1 mole of substance, equal to 6.022 × 10²³ mol⁻¹

formula mass

sum of the average masses for all atoms represented in a chemical formula; for covalent compounds, this is also the molecular mass

mole

amount of substance containing the same number of atoms, molecules, ions, or other entities as the number of atoms in exactly 12 grams of ¹²C

molar mass

mass in grams of 1 mole of a substance