Molecular and Formula Masses

The molecular mass of a substance is the sum of the average masses of the atoms in one molecule of a substance. It is calculated by adding together the atomic masses of the elements in the substance, each multiplied by its subscript (written or implied) in the molecular formula. Because the units of atomic mass are atomic mass units, the units of molecular mass are also atomic mass units. The procedure for calculating molecular masses is illustrated in Example (PageIndex{1}).

Example (PageIndex{1}): Molecular Mass of Ethanol

This number (Avogadro's number) is 6.023 X 10 23. It is the number of molecules of any gas present in a volume of 22.41 L and is the same for the lightest gas (hydrogen) as for a heavy gas such as carbon dioxide or bromine. Avogadro's number is one of the fundamental constants of chemistry.

Calculate the molecular mass of ethanol, whose condensed structural formula is (ce{CH3CH2OH}). Among its many uses, ethanol is a fuel for internal combustion engines.

Given: molecule

Avogadro’s number is defined as the number of elementary particles (molecules, atoms, compounds, etc.) per mole of a substance. It is equal to 6.022×10 23 mol-1 and is expressed as the symbol N A. Avogadro’s number is a similar concept to that of a dozen or a gross. A dozen molecules is 12 molecules. A gross of molecules is 144 molecules. Avogadro's number is named after an Italian Scientist, Amedeo Avogadro in the 19 th Century. As per his theory, at certain pressure and temperature conditions, the volume of gas is proportional to the number of molecules in the gas irrespective of the nature of gas.

Asked for: molecular mass

Strategy:

1. Determine the number of atoms of each element in the molecule.
2. Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.
3. Add together the masses to give the molecular mass.

Solution:

A The molecular formula of ethanol may be written in three different ways: (ce{CH3CH2OH}) (which illustrates the presence of an ethyl group, CH₃CH₂₋, and an −OH group), (ce{C2CH5OH}), and (ce{C2H6O}); all show that ethanol has two carbon atoms, six hydrogen atoms, and one oxygen atom.

B Taking the atomic masses from the periodic table, we obtain

[ begin{align*} 2 times text { atomic mass of carbon} &= 2 , atoms left ( {12.011 , amu over atoms } right ) [4pt] &= 24.022 ,amu end{align*}]

[ begin{align*} 6 times text { atomic mass of hydrogen} &= 2 , atoms left ( {1.0079 , amu over atoms } right ) [4pt] &= 6.0474 ,amu end{align*}]

[ begin{align*} 1 times text { atomic mass of oxygen} &= 1 , atoms left ( {15.9994 , amu over atoms } right ) [4pt] &= 15.994 ,amu end{align*}]

C Adding together the masses gives the molecular mass:

[ 24.022 ,amu + 6.0474 ,amu + 15.9994 ,amu = 46.069 ,amu nonumber]

Exercise (PageIndex{1}): Molecular Mass of Freon

Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is (CCl_3F). Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows:

137.368 amu

Unlike molecules, which form covalent bonds, ionic compounds do not have a readily identifiable molecular unit. Therefore, for ionic compounds, the formula mass (also called the empirical formula mass) of the compound is used instead of the molecular mass. The formula mass is the sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript (written or implied). It is directly analogous to the molecular mass of a covalent compound. The units are atomic mass units.

Example (PageIndex{2}): Formula Mass of Calcium Phosphate

Calculate the formula mass of Ca₃(PO₄)₂, commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk.

Given: ionic compound

Asked for: formula mass

Strategy:

1. Determine the number of atoms of each element in the empirical formula.
2. Obtain the atomic masses of each element from the periodic table and multiply the atomic mass of each element by the number of atoms of that element.
3. Add together the masses to give the formula mass.

Solution:

A The empirical formula—Ca₃(PO₄)₂—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca²⁺ ions and two PO₄³⁻ ions. The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms.

B Taking atomic masses from the periodic table, we obtain

[ 3 times text {atomic mass of calcium} = 3 , atoms left ( {40.078 , amu over atom } right ) = 120.234 , amu nonumber ]

[ 2 times text {atomic mass of phosphorus} = 2 , atoms left ( {30.973761 , amu over atom } right ) = 61.947522 , amu nonumber]

[ 8 times text {atomic mass of oxygen} = 8 , atoms left ( {15.9994 , amu over atom } right ) = 127.9952 , amu nonumber]

C Adding together the masses gives the formula mass of Ca₃(PO₄)₂:

[120.234 ,amu + 61.947522 , amu + 127.9952 , amu = 310.177 , amu nonumber]

Exercise (PageIndex{2}): Formula Mass of Silicon Nitride

Calculate the formula mass of (ce{Si3N4}), commonly called silicon nitride. It is an extremely hard and inert material that is used to make cutting tools for machining hard metal alloys.

(140.29 ,amu)

The Mole

Dalton’s theory that each chemical compound has a particular combination of atoms and that the ratios of the numbers of atoms of the elements present are usually small whole numbers. It also describes the law of multiple proportions, which states that the ratios of the masses of elements that form a series of compounds are small whole numbers. The problem for Dalton and other early chemists was to discover the quantitative relationship between the number of atoms in a chemical substance and its mass. Because the masses of individual atoms are so minuscule (on the order of 10⁻²³ g/atom), chemists do not measure the mass of individual atoms or molecules. In the laboratory, for example, the masses of compounds and elements used by chemists typically range from milligrams to grams, while in industry, chemicals are bought and sold in kilograms and tons. To analyze the transformations that occur between individual atoms or molecules in a chemical reaction, it is therefore essential for chemists to know how many atoms or molecules are contained in a measurable quantity in the laboratory—a given mass of sample. The unit that provides this link is the mole (mol), from the Latin moles, meaning “pile” or “heap.”

Many familiar items are sold in numerical quantities with distinct names. For example, cans of soda come in a six-pack, eggs are sold by the dozen (12), and pencils often come in a gross (12 dozen, or 144). Sheets of printer paper are packaged in reams of 500, a seemingly large number. Atoms are so small, however, that even 500 atoms are too small to see or measure by most common techniques. Any readily measurable mass of an element or compound contains an extraordinarily large number of atoms, molecules, or ions, so an extremely large numerical unit is needed to count them. The mole is used for this purpose.

A mole is defined as the amount of a substance that contains the number of carbon atoms in exactly 12 g of isotopically pure carbon-12. According to the most recent experimental measurements, this mass of carbon-12 contains 6.022142 × 10²³ atoms, but for most purposes 6.022 × 10²³ provides an adequate number of significant figures. Just as 1 mole of atoms contains 6.022 × 10²³ atoms, 1 mole of eggs contains 6.022 × 10²³ eggs. This number is called Avogadro’s number, after the 19th-century Italian scientist who first proposed a relationship between the volumes of gases and the numbers of particles they contain.

It is not obvious why eggs come in dozens rather than 10s or 14s, or why a ream of paper contains 500 sheets rather than 400 or 600. The definition of a mole—that is, the decision to base it on 12 g of carbon-12—is also arbitrary. The important point is that 1 mole of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.022 × 10²³.

To appreciate the magnitude of Avogadro’s number, consider a mole of pennies. Stacked vertically, a mole of pennies would be 4.5 × 10¹⁷ mi high, or almost six times the diameter of the Milky Way galaxy. If a mole of pennies were distributed equally among the entire population on Earth, each person would have more than one trillion dollars. The mole is so large that it is useful only for measuring very small objects, such as atoms.

The concept of the mole allows scientists to count a specific number of individual atoms and molecules by weighing measurable quantities of elements and compounds. To obtain 1 mol of carbon-12 atoms, one weighs out 12 g of isotopically pure carbon-12. Because each element has a different atomic mass, however, a mole of each element has a different mass, even though it contains the same number of atoms (6.022 × 10²³). This is analogous to the fact that a dozen extra large eggs weighs more than a dozen small eggs, or that the total weight of 50 adult humans is greater than the total weight of 50 children. Because of the way the mole is defined, for every element the number of grams in a mole is the same as the number of atomic mass units in the atomic mass of the element. For example, the mass of 1 mol of magnesium (atomic mass = 24.305 amu) is 24.305 g. Because the atomic mass of magnesium (24.305 amu) is slightly more than twice that of a carbon-12 atom (12 amu), the mass of 1 mol of magnesium atoms (24.305 g) is slightly more than twice that of 1 mol of carbon-12 (12 g). Similarly, the mass of 1 mol of helium (atomic mass = 4.002602 amu) is 4.002602 g, which is about one-third that of 1 mol of carbon-12. Using the concept of the mole, Dalton’s theory can be restated: 1 mol of a compound is formed by combining elements in amounts whose mole ratios are small whole numbers. For example, 1 mol of water (H₂O) has 2 mol of hydrogen atoms and 1 mol of oxygen atoms.

Molar Mass

The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 10²³ atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively.

The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 10²³ carbon atoms—is therefore 12.011 g/mol:

The molar mass of naturally-occurring carbon is different from that of carbon-12, and is not an integer because carbon occurs as a mixture of carbon-12, carbon-13, and carbon-14. One mole of carbon still has 6.022 × 10²³ carbon atoms, but 98.89% of those atoms are carbon-12, 1.11% are carbon-13, and a trace (about 1 atom in 1012) are carbon-14. (For more information, see Section 1.6 'Isotopes and Atomic Masses'.) Similarly, the molar mass of uranium is 238.03 g/mol, and the molar mass of iodine is 126.90 g/mol. When dealing with elements such as iodine and sulfur, which occur as a diatomic molecule (I₂) and a polyatomic molecule (S₈), respectively, molar mass usually refers to the mass of 1 mol of atoms of the element—in this case I and S, not to the mass of 1 mol of molecules of the element (I₂ and S₈).

The molar mass of ethanol is the mass of ethanol (C₂H₅OH) that contains 6.022 × 10²³ ethanol molecules. As in Example (PageIndex{1}), the molecular mass of ethanol is 46.069 amu. Because 1 mol of ethanol contains 2 mol of carbon atoms (2 × 12.011 g), 6 mol of hydrogen atoms (6 × 1.0079 g), and 1 mol of oxygen atoms (1 × 15.9994 g), its molar mass is 46.069 g/mol. Similarly, the formula mass of calcium phosphate [Ca₃(PO₄)₂] is 310.177 amu, so its molar mass is 310.177 g/mol. This is the mass of calcium phosphate that contains 6.022 × 10²³ formula units.

The mole is the basis of quantitative chemistry. It provides chemists with a way to convert easily between the mass of a substance and the number of individual atoms, molecules, or formula units of that substance. Conversely, it enables chemists to calculate the mass of a substance needed to obtain a desired number of atoms, molecules, or formula units. For example, to convert moles of a substance to mass, the following relationship is used:

or, more specifically,

[ moles left ( {grams over mole } right ) = grams ]

Conversely, to convert the mass of a substance to moles:

The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules involved in the reaction and as the relative number of moles. For example, in the balanced equation:

the production of two moles of water would require the consumption of 2 moles of (H_2) and one mole of (O_2). Therefore, when considering this particular reaction

• 2 moles of H₂
• 1 mole of O₂ and
• 2 moles of H₂O

would be considered to be stoichiometrically equivalent quantitites.

These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of (H_2O) would be produced from 1.57 moles of (O_2)?

[ (1.57; mol; O_2) left( dfrac{2; mol;H_2O}{1;mol;O_2} right) = 3.14; mol; H_2O]

The ratio ( left( dfrac{2; mol; H_2O}{1;mol;O_2} right)) is the stoichiometric relationship between (H_2O) and (O_2) from the balanced equation for this reaction.

Example (PageIndex{3}): Combustion of Butane

For the combustion of butane ((C_4H_{10})) the balanced equation is:

[ce{2C4H10(l) + 13O2(g) rightarrow 8CO2(g) + 10H2O(l)}]

Calculate the mass of (CO_2) that is produced in burning 1.00 gram of (C_4H_{10}).

Solution

Thus, the overall sequence of steps to solve this problem is:

First of all we need to calculate how many moles of butane we have in a 1.00 gram sample:

[ (1.00; g; C_4H_{10}) left(dfrac{1; mol; C_4H_{10}}{58.0;g; C_4H_{10}}right) = 1.72 times 10^{-2} ; mol; C_4H_{10}]

Now, the stoichiometric relationship between (C_4H_{10}) and (CO_2) is:

[left( dfrac{8; mol; CO_2}{2; mol; C_4H_{10}}right)]

Therefore:

[ left(dfrac{8; mol; CO_2}{2; mol; C_4H_{10}} right) times 1.72 times 10^{-2} ; mol; C_4H_{10} = 6.88 times 10^{-2} ; mol; CO_2]

The question called for the determination of the mass of (CO_2) produced, thus we have to convert moles of (CO_2) into grams (by using the molecular weight of (CO_2)):

[ 6.88 times 10^{-2} ; mol; CO_2 left( dfrac{44.0; g; CO_2}{1; mol; CO_2} right) = 3.03;g ; CO_2]

Be sure to pay attention to the units when converting between mass and moles. Figure (PageIndex{1}) is a flowchart for converting between mass; the number of moles; and the number of atoms, molecules, or formula units. The use of these conversions is illustrated in Examples (PageIndex{3}) and (PageIndex{4}).

Figure (PageIndex{1}): A Flowchart for Converting between Mass; the Number of Moles; and the Number of Atoms, Molecules, or Formula Units

Example (PageIndex{4}): Ethylene Glycol

For 35.00 g of ethylene glycol (HOCH₂CH₂OH), which is used in inks for ballpoint pens, calculate the number of

1. moles.
2. molecules.

Given: mass and molecular formula

Asked for: number of moles and number of molecules

Strategy:

1. Use the molecular formula of the compound to calculate its molecular mass in grams per mole.
2. Convert from mass to moles by dividing the mass given by the compound’s molar mass.
3. Convert from moles to molecules by multiplying the number of moles by Avogadro’s number.

Solution:

A The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example (PageIndex{1}): The molar mass of ethylene glycol is 62.068 g/mol.

B The number of moles of ethylene glycol present in 35.00 g can be calculated by dividing the mass (in grams) by the molar mass (in grams per mole):

[ { text {mass of ethylene glycol (g)} over text {molar mass (g/mol)} } = text {moles ethylene glycol (mol) }]

So

It is always a good idea to estimate the answer before you do the actual calculation. In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1 mol, so we have probably not made a major error in the calculations.

C To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number:

[ text {molecules of ethylene glycol} = 0.5639 , mol left ( {6.022 times 10^{23} , molecules over 1 , mol } right ) ]

[ = 3.396 times 10^{23} , molecules ]

Because we are dealing with slightly more than 0.5 mol of ethylene glycol, we expect the number of molecules present to be slightly more than one-half of Avogadro’s number, or slightly more than 3 × 10²³ molecules, which is indeed the case.

Exercise (PageIndex{4}): Freon-11

For 75.0 g of CCl₃F (Freon-11), calculate the number of

1. moles.
2. molecules.

0.546 mol

3.29 × 10²³ molecules

Example (PageIndex{5})

Calculate the mass of 1.75 mol of each compound.

1. (ce{S2Cl2}) (common name: sulfur monochloride; systematic name: disulfur dichloride)
2. (ce{Ca(ClO)2}) (calcium hypochlorite)

Given: number of moles and molecular or empirical formula

Asked for: mass

Strategy:

A Calculate the molecular mass of the compound in grams from its molecular formula (if covalent) or empirical formula (if ionic).

B Convert from moles to mass by multiplying the moles of the compound given by its molar mass.

Solution:

We begin by calculating the molecular mass of (ce{S2Cl2}) and the formula mass of (ce{Ca(ClO)2}).

A The molar mass of (ce{S2Cl2}) is 135.036 g/mol.

B The mass of 1.75 mol of (ce{S2Cl2}) is calculated as follows:

[moles S_2Cl_2 left [text {molar mass}left ({ g over mol} right )right ] rightarrow mass of S_2Cl_2 , (g) ]

[ 1.75 , mol S_2Cl_2left ({135.036 , g S_2Cl_2 over 1 , mol S_2Cl_2 } right ) = 236 , g S_2Cl_2 ]

A The molar mass of Ca(ClO)₂ 142.983 g/mol.

B The mass of 1.75 mol of Ca(ClO)₂ is calculated as follows:

Because 1.75 mol is less than 2 mol, the final quantity in grams in both cases should be less than twice the molar mass, which it is.

Exercise (PageIndex{5})

Calculate the mass of 0.0122 mol of each compound.

1. Si₃N₄ (silicon nitride), used as bearings and rollers
2. (CH₃)₃N (trimethylamine), a corrosion inhibitor

1.71 g

0.721 g

The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules involved in the reaction and as the relative number of moles. For example, in the balanced equation:

the production of two moles of water would require the consumption of 2 moles of (H_2) and one mole of (O_2). Therefore, when considering this particular reaction

• 2 moles of H₂
• 1 mole of O₂ and
• 2 moles of H₂O

would be considered to be stoichiometrically equivalent quantitites.

These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of (H_2O) would be produced from 1.57 moles of (O_2)?

[ (1.57; mol; O_2) left( dfrac{2; mol ;H_2O}{1;mol;O_2} right) = 3.14; mol; H_2O nonumber]

The ratio ( left( dfrac{2; mol;H_2O}{1;mol;O_2} right)) is the stoichiometric relationship between (H_2O) and (O_2) from the balanced equation for this reaction.

Example (PageIndex{6})

For the combustion of butane ((C_4H_{10})) the balanced equation is:

[ ce{2C4H_{10} (l) + 13O2(g) rightarrow 8CO2(g) + 10H2O(l)} nonumber]

Calculate the mass of (CO_2) that is produced in burning 1.00 gram of (C_4H_{10}).

Solution

First of all we need to calculate how many moles of butane we have in a 1.00 gram sample:

[ (1.00; g; C_4H_{10}) left(dfrac{1; mol; C_4H_{10}}{58.0;g; C_4H_{10}}right) = 1.72 times 10^{-2} ; mol; C_4H_{10} nonumber]

Now, the stoichiometric relationship between (C_4H_{10}) and (CO_2) is:

[left( dfrac{8; mol; CO_2}{2; mol; C_4H_{10}}right) nonumber]

Therefore:

[ left(dfrac{8; mol; CO_2}{2; mol; C_4H_{10}} right) times 1.72 times 10^{-2} ; mol; C_4H_{10} = 6.88 times 10^{-2} ; mol; CO_2 nonumber]

The question called for the determination of the mass of (CO_2) produced, thus we have to convert moles of (CO_2) into grams (by using the molecular weight of (CO_2)):

[ 6.88 times 10^{-2} ; mol; CO_2 left( dfrac{44.0; g; CO_2}{1; mol; CO_2} right) = 3.03;g ; CO_2 nonumber]

Thus, the overall sequence of steps to solve this problem were:

In a similar way we could determine the mass of water produced, or oxygen consumed, etc.

Interpreting Chemical Equations (Part 2)

Previously, we discussed that chemical equations give the relative amounts of reactants and products consumed or produced in a reaction. This was discussed in terms of the the number of atoms, molecules, or formula units of a reactant or a product. A chemical equation can also be interpreted interms of moles of a reactant or product. As illustrated below, the coefficients allow interpreted in any of the following ways:

• Two NH₄⁺ ions and one Cr₂O₇²⁻ ion yield 1 formula unit of Cr₂O₃, 1 N₂ molecule, and 4 H₂O molecules.
• One mole of (NH₄)₂Cr₂O₇ yields 1 mol of Cr₂O₃, 1 mol of N₂, and 4 mol of H₂O.
• A mass of 252 g of (NH₄)₂Cr₂O₇ yields 152 g of Cr₂O₃, 28 g of N₂, and 72 g of H₂O.
• A total of 6.022 × 10²³ formula units of (NH₄)₂Cr₂O₇ yields 6.022 × 10²³ formula units of Cr₂O₃, 6.022 × 10²³ molecules of N₂, and 24.09 × 10²³ molecules of H₂O.

Figure (PageIndex{4}): The Relationships among Moles, Masses, and Formula Units of Compounds in the Balanced Chemical Reaction for the Ammonium Dichromate Volcano

These are all chemically equivalent ways of stating the information given in the balanced chemical equation, using the concepts of the mole, molar or formula mass, and Avogadro’s number. The ratio of the number of moles of one substance to the number of moles of another is called the mole ratio. For example, the mole ratio of (H_2O) to (N_2) in Equation (ref{3.1.1}) is 4:1. The total mass of reactants equals the total mass of products, as predicted by Dalton’s law of conservation of mass:

[252 ;g ;text{of}; ce{(NH_4)_2Cr_2O_7} ]

yield

[152 + 28 + 72 = 252 ; g ; text{of products.}]

The chemical equation does not, however, show the rate of the reaction (rapidly, slowly, or not at all) or whether energy in the form of heat or light is given off. These issues are considered in more detail in later chapters.

An important chemical reaction was analyzed by Antoine Lavoisier, an 18th-century French chemist, who was interested in the chemistry of living organisms as well as simple chemical systems. In a classic series of experiments, he measured the carbon dioxide and heat produced by a guinea pig during respiration, in which organic compounds are used as fuel to produce energy, carbon dioxide, and water. Lavoisier found that the ratio of heat produced to carbon dioxide exhaled was similar to the ratio observed for the reaction of charcoal with oxygen in the air to produce carbon dioxide—a process chemists call combustion. Based on these experiments, he proposed that “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.” Lavoisier was correct, although the organic compounds consumed in respiration are substantially different from those found in charcoal. One of the most important fuels in the human body is glucose ((C_6H_{12}O_6)), which is virtually the only fuel used in the brain. Thus combustion and respiration are examples of chemical reactions.

Example (PageIndex{2}): Combustion of Glucose

The balanced chemical equation for the combustion of glucose in the laboratory (or in the brain) is as follows:

[ ce{C_6H_{12}O6(s) + 6O2(g) rightarrow 6CO2(g) + 6H2O(l)}]

Construct a table showing how to interpret the information in this equation in terms of

1. a single molecule of glucose.
2. moles of reactants and products.
3. grams of reactants and products represented by 1 mol of glucose.
4. numbers of molecules of reactants and products represented by 1 mol of glucose.

Given: balanced chemical equation

Asked for: molecule, mole, and mass relationships

Strategy:

1. Use the coefficients from the balanced chemical equation to determine both the molecular and mole ratios.
2. Use the molar masses of the reactants and products to convert from moles to grams.
3. Use Avogadro’s number to convert from moles to the number of molecules.

Solution:

This equation is balanced as written: each side has 6 carbon atoms, 18 oxygen atoms, and 12 hydrogen atoms. We can therefore use the coefficients directly to obtain the desired information.

1. One molecule of glucose reacts with 6 molecules of O₂ to yield 6 molecules of CO₂ and 6 molecules of H₂O.
2. One mole of glucose reacts with 6 mol of O₂ to yield 6 mol of CO₂ and 6 mol of H₂O.
3. To interpret the equation in terms of masses of reactants and products, we need their molar masses and the mole ratios from part b. The molar masses in grams per mole are as follows: glucose, 180.16; O₂, 31.9988; CO₂, 44.010; and H₂O, 18.015.

[ begin{align*} text{mass of reactants} &= text{mass of products} [4pt] g , glucose + g , O_2 &= g , CO_2 + g , H_2O end{align*}]

[ 372.15 , g = 372.15 , g ]

C One mole of glucose contains Avogadro’s number (6.022 × 10²³) of glucose molecules. Thus 6.022 × 10²³ glucose molecules react with (6 × 6.022 × 10²³) = 3.613 × 10²⁴ oxygen molecules to yield (6 × 6.022 × 10²³) = 3.613 × 10²⁴ molecules each of CO₂ and H₂O.

In tabular form:

Exercise (PageIndex{2}): Ammonium Nitrate Explosion

Ammonium nitrate is a common fertilizer, but under the wrong conditions it can be hazardous. In 1947, a ship loaded with ammonium nitrate caught fire during unloading and exploded, destroying the town of Texas City, Texas. The explosion resulted from the following reaction:

[ 2NH_4NO_{3;(s)} rightarrow 2N_{2;(g)} + 4H_2O_{(g)} + O_{2;(g)} ]

Construct a table showing how to interpret the information in the equation in terms of

1. individual molecules and ions.
2. moles of reactants and products.
3. grams of reactants and products given 2 mol of ammonium nitrate.
4. numbers of molecules or formula units of reactants and products given 2 mol of ammonium nitrate.

Answer:

Avogadro's Number Meaning Tagalog

Summary

To analyze chemical transformations, it is essential to use a standardized unit of measure called the mole. The molecular mass and the formula mass of a compound are obtained by adding together the atomic masses of the atoms present in the molecular formula or empirical formula, respectively; the units of both are atomic mass units (amu). The mole is a unit used to measure the number of atoms, molecules, or (in the case of ionic compounds) formula units in a given mass of a substance. The mole is defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon-12, Avogadro’s number (6.022 × 10²³) of atoms of carbon-12. The molar mass of a substance is defined as the mass of 1 mol of that substance, expressed in grams per mole, and is equal to the mass of 6.022 × 10²³ atoms, molecules, or formula units of that substance.

AVOGADRO'S NUMBER

A principle stated in 1811 by the Italian chemist Amadeo Avogadro (1776-1856) that equal volumes of gases at the same temperature and pressure contain the same number of molecules regardless of their chemical nature and physical properties. This number (Avogadro's number) is 6.023 X 10²³. It is the number of molecules of any gas present in a volume of 22.41 L and is the same for the lightest gas (hydrogen) as for a heavy gas such as carbon dioxide or bromine.

Avogadro's number is one of the fundamental constants of chemistry. It permits one to compare the different atoms or molecules of given substances where the same number of atoms or molecules are being compared.
It also makes possible determination of how much heavier a simple molecule of one gas is than that of another, as a result the relative molecular weights of gases can be ascertained by comparing the weights of equal volumes.

Avogadro - his contribution to chemistry

In order to understand the contribution that Avogadro made, we must consider some of the ideas
being developed at this time. Chemistry was just beginning to become an exact science. The Law of
Definite Proportions and the Law of Multiple Proportions were well accepted by 1808, at
which time John Dalton published his New System of Chemical Philosophy.

Dalton proposed that the atoms of each element had a characteristic atomic weight, and that it was
atoms that were the combining units in chemical reactions. Dalton had no method of measuring
atomic weights unambiguously, so made the incorrect assumption that in the most common
compound between two elements, there was one atom of each.

At around this time, Gay-Lussac was studying the chemical reactions of gases, and found that the
ratios of volumes of the reacting gases were small integer numbers. This provided a more logical
method of assigning atomic weights. Gay-Lussac did not carry through the full implications of his
work. However, Dalton realised that a simple integral relation between volumes of reacting gases
implied an equally simple relation between reacting particles. Dalton still equated particles with
atoms, and could not accept how one particle of oxygen could yield two particles of water. This was
a direct threat to the relatively new atomic theory, and therefore Dalton tried to discredit the work of
Gay-Lussac.

In 1811, Avogadro published an article in Journal de physique that clearly drew the distinction
between the molecule and the atom. He pointed out that Dalton had confused the concepts of atoms
and molecules. The 'atoms' of nitrogen and oxygen are in reality 'molecules' containing two atoms
each. Thus two molecules of hydrogen can combine with one molecule of oxygen to produce two
molecules of water.

Avogadro suggested that equal volumes of all gases at the same temperature and pressure contain the same number of molecules which is now known as Avogadro's Principle.

The work of Avogadro was almost completely neglected until it was forcefully presented by
Stanislao Cannizarro at the Karlsruhe Conference in 1860. He showed that Avogadro's Principle
could be used to determine not only molar masses, but also, indirectly, atomic masses. The reason
for the earlier neglect of Avogadro's work was probably the deeply rooted conviction that chemical
combination occurred by virtue of an affinity between unlike elements. After the electrical discoveries
of Galvani and Volta, this affinity was generally ascribed to the attraction between unlike charges.
The idea that two identical atoms of hydrogen might combine into the compound molecular hydrogen
was abhorrent to the chemical philosophy of the early nineteenth century.

Avogadro - his number

It was long after Avogadro that the idea of a mole was introduced. Since a molecular weight in
grams (mole) of any substance contains the same number of molecules, then according to
Avogadro's Principle, the molar volumes of all gases should be the same. The number of molecules
in one mole is now called Avogadro's number. It must be emphasised that Avogadro, of course,
had no knowledge of moles, or of the number that was to bear his name. Thus the number was never
actually determined by Avogadro himself.

As we all know today, Avogadro's number is very large, the presently accepted value being
6.0221367 x 10²³. The size of such a number is extremely difficult to comprehend. There are many
awe-inspiring illustrations to help visualize the enormous size of this number. For example:

An Avogadro's number of standard soft drink cans would cover the surface of the earth to a depth of over 200 miles.

If you had Avogadro's number of unpopped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles.

If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.

Determination of the number

Cannizarro, around 1860, used Avogadro's ideas to obtain a set of atomic weights, based upon
oxygen having an atomic weight of 16. In 1865, Loschmidt used a combination of liquid density,
gaseous viscosity, and the kinetic theory of gases, to establish roughly the size of molecules, and
hence the number of molecules in 1 cm3 of gas.

During the latter part of the nineteenth century, it was possible to obtain reasonable estimates for
Avogadro's number from sedimentation measurements of colloidal particles. Into the twentieth
century, then Mullikens oil drop experiment gave much better values, and was used for many years.

A more modern method is to calculate the Avogadro number from the density of a crystal, the
relative atomic mass, and the unit cell length, determined from x-ray methods. To be useful for this
purpose, the crystal must be free of defects. Very accurate values of these quantities for silicon have
been measured at the National Institute for Standards and Technology (NIST).

To use this approach, it is necessary to have accurate values of atomic weights, often obtained by
measuring the mass of atomic ions. For example, an ion trap, employing extremely uniform and
stable magnetic and electric fields should allow such measurements to be made to better than 1 part
in 1010. The relative atomic mass of silicon is particularly important, since silicon crystals are used in
the x-ray methods mentioned above.

As a continuation of this approach, one of the 1999 NIST Precision Measurement Grants has been
awarded to David Pritchard, physics professor at the Massachusetts Institute of Technology. He will
conduct cyclotron frequency measurements on ions that could achieve a 100-fold improvement in
the accuracy of atomic mass measurements. MIT has developed the world's most accurate mass
spectrometer capable of measuring the atomic mass of atoms to one part in 10 billion. Pritchard
proposes to simultaneously measure the cyclotron frequencies of two different ions in order to
improve the values of several fundamental constants, including Avogadro's number.

At the present time, information on Avogadro's number from many different experiments is pooled
with other observations on other physical constants. A most probable and self-consistent set of
physical constants that best fits all reliable data is then found by statistical methods.

Avogadro's Number Meaning Slang

The size of Avogadro's number is determined by our definition of the mole. What it does
demonstrate is how small an atom or molecule is compared to the amounts of material we are
familiar with in everyday life, since the definition of the mole involves amounts of material we are
completely familiar with.

Avogadro's Number Meaning

If you are a scholar on the life and works of Avogadro, and have information that might be usefully
included on this page, then why not mail me the details. Contributions will be acknowledged.

Avogadro's Number Meaning List

Special thanks to Chris Johnson for his insightful presentation.

Avogadro's Number Meaning Dictionary