3.1 The Mole and Molar Masses
Learning Objective
 To calculate the molecular mass of a covalent compound and the formula mass of an ionic compound and to calculate the number of atoms, molecules, or formula units in a sample of a substance.
As you learned in Chapter 1 “Introduction to Chemistry”, the mass number is the sum of the numbers of protons and neutrons present in the nucleus of an atom. The mass number is an integer that is approximately equal to the numerical value of the atomic mass. Although the mass number is unitless, it is assigned units called atomic mass units (amu). Because a molecule or a polyatomic ion is an assembly of atoms whose identities are given in its molecular or ionic formula, we can calculate the average atomic mass of any molecule or polyatomic ion from its composition by adding together the masses of the constituent atoms. The average mass of a monatomic ion is the same as the average mass of an atom of the element because the mass of electrons is so small that it is insignificant in most calculations.
Molecular and Formula Masses
The molecular massThe sum of the average masses of the atoms in one molecule of a substance, each multiplied by its subscript. 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 1.
Example 1
Calculate the molecular mass of ethanol, whose condensed structural formula is CH_{3}CH_{2}OH. Among its many uses, ethanol is a fuel for internal combustion engines.
Given: molecule
Asked for: molecular mass
Strategy:
A Determine the number of atoms of each element in the molecule.
B 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.
C Add together the masses to give the molecular mass.
Solution:
A The molecular formula of ethanol may be written in three different ways: CH_{3}CH_{2}OH (which illustrates the presence of an ethyl group, CH_{3}CH_{2}−, and an −OH group), C_{2}H_{5}OH, and C_{2}H_{6}O; 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
$$\begin{array}{l}\text{2}\times \text{atomicmassofcarbon}=\text{2}\overline{)\text{atoms}}\left(\frac{\text{12}\text{.011amu}}{\overline{)\text{atom}}}\right)=24.022\text{amu}\\ \text{6}\times \text{atomicmassofhydrogen}=\text{6}\overline{)\text{atoms}}\left(\frac{\text{1}\text{.0079amu}}{\overline{)\text{atom}}}\right)=6.0474\text{amu}\\ \text{1}\times \text{atomicmassofoxygen}=\text{1}\overline{)\text{atom}}\left(\frac{\text{15}\text{.9994amu}}{\overline{)\text{atom}}}\right)=15.9994\text{amu}\end{array}$$C Adding together the masses gives the molecular mass:
24.022 amu + 6.0474 amu + 15.9994 amu = 46.069 amuAlternatively, we could have used unit conversions to reach the result in one step, as described in Essential Skills 2 (Chapter 3, Section 7 “Essential Skills 2”):
$$\left[\text{2}\overline{)\text{atomsC}}\left(\frac{12.011\text{amu}}{\text{1}\overline{)\text{atomC}}}\right)\right]+\left[\text{6}\overline{)\text{atomsH}}\left(\frac{1.0079\text{amu}}{\text{1}\overline{)\text{atomH}}}\right)\right]+\left[\text{1}\overline{)\text{atomsO}}\left(\frac{15.9994\text{amu}}{\text{1}\overline{)\text{atomO}}}\right)\right]=46.069\text{amu}$$The same calculation can also be done in a tabular format, which is especially helpful for more complex molecules:
$$\begin{array}{rrrr}\hfill \text{2C}& \hfill \text{(2atoms)(12}\text{.011amu/atom)}& \hfill =& \hfill \text{24}\text{.022amu}\\ \hfill \text{6H}& \hfill \text{(6atoms)(1}\text{.0079amu/atom)}& \hfill =& \hfill \text{6}\text{.0474amu}\\ \hfill +\text{1O}& \hfill \text{(1atom)(15}\text{.9994amu/atom)}& \hfill =& \hfill \text{15}\text{.9994amu}\\ \hfill {\text{C}}_{\text{2}}{\text{H}}_{\text{6}}\text{O}& \hfill \text{molecularmassofethanol}& \hfill =& \hfill \text{46}\text{.069amu}\end{array}$$Exercise
Calculate the molecular mass of trichlorofluoromethane, also known as Freon11, whose condensed structural formula is CCl_{3}F. Until recently, it was used as a refrigerant. The structure of a molecule of Freon11 is as follows:
Answer: 137.368 amu
Unlike molecules, which have covalent bonds, ionic compounds do not have a readily identifiable molecular unit. So for ionic compounds we use the formula mass (also called the empirical formula massAnother name for formula mass.) of the compound rather than the molecular mass. The formula massThe sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript. 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. Once again, the units are atomic mass units.
Note the Pattern
Atomic mass, molecular mass, and formula mass all have the same units: atomic mass units.
Example 2
Calculate the formula mass of Ca_{3}(PO_{4})_{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:
A Determine the number of atoms of each element in the empirical formula.
B 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.
C Add together the masses to give the formula mass.
Solution:
A The empirical formula—Ca_{3}(PO_{4})_{2}—indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca^{2+} ions and two PO_{4}^{3−} 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
$$\begin{array}{l}\text{3}\times \text{atomicmassofcalcium}=\text{3}\overline{)\text{atoms}}\left(\frac{\text{40}\text{.078amu}}{\overline{)\text{atom}}}\right)=120.234\text{amu}\\ \text{2}\times \text{atomicmassofphosphorus}=\text{2}\overline{)\text{atoms}}\left(\frac{\text{30}\text{.973761amu}}{\overline{)\text{atom}}}\right)=61.947522\text{amu}\\ \text{8}\times \text{atomicmassofoxygen}=\text{8}\overline{)\text{atoms}}\left(\frac{\text{15}\text{.9994amu}}{\overline{)\text{atom}}}\right)=127.9952\text{amu}\end{array}$$C Adding together the masses gives the formula mass of Ca_{3}(PO_{4})_{2}:
120.234 amu + 61.947522 amu + 127.9952 amu = 310.177 amuWe could also find the formula mass of Ca_{3}(PO_{4})_{2} in one step by using unit conversions or a tabular format:
$$\left[\text{3}\overline{)\text{atomsCa}}\left(\frac{\text{40}\text{.078amu}}{\text{1}\overline{)\text{atomCa}}}\right)\right]+\left[\text{2}\overline{)\text{atomsP}}\left(\frac{\text{30}\text{.973761amu}}{\text{1}\overline{)\text{atomP}}}\right)\right]+\left[\text{8}\overline{)\text{atomsO}}\left(\frac{\text{15}\text{.9994amu}}{\text{1}\overline{)\text{atomO}}}\right)\right]=310.177\text{amu}$$ $$\begin{array}{rrrr}\hfill \text{3Ca}& \hfill \text{(3atoms)(40}\text{.078amu/atom)}& \hfill =& \hfill \text{120}\text{.234amu}\\ \hfill \text{2P}& \hfill \text{(2atoms)(30}\text{.973761amu/atom)}& \hfill =& \hfill \text{61}\text{.947522amu}\\ \hfill +\text{8O}& \hfill \text{(8atoms)(15}\text{.9994amu/atom)}& \hfill =& \hfill \text{127}\text{.9952amu}\\ \hfill {\text{Ca}}_{3}{\text{P}}_{2}{\text{O}}_{8}& \hfill {\text{formulamassofCa}}_{3}{{\text{(PO}}_{4})}_{2}& \hfill =& \hfill \text{310}\text{.177amu}\end{array}$$Exercise
Calculate the formula mass of Si_{3}N_{4}, commonly called silicon nitride. It is an extremely hard and inert material that is used to make cutting tools for machining hard metal alloys.
Answer: 140.29 amu
The Mole
In Chapter 1 “Introduction to Chemistry”, we described 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. We also described 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 reactionA process in which a substance is converted to one or more other substances with different compositions and properties., it is therefore absolutely 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)The quantity of a substance that contains the same number of units (e.g., atoms or molecules) as the number of carbon atoms in exactly 12 g of isotopically pure carbon12., from the Latin moles, meaning “pile” or “heap” (not from the small subterranean animal!).
Many familiar items are sold in numerical quantities that have unusual names. For example, cans of soda come in a sixpack, 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 extraordinarily 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 carbon12. According to the most recent experimental measurements, this mass of carbon12 contains 6.022142 × 10^{23} atoms, but for most purposes 6.022 × 10^{23} provides an adequate number of significant figures. Just as 1 mol of atoms contains 6.022 × 10^{23} atoms, 1 mol of eggs contains 6.022 × 10^{23} eggs. The number in a mole is called Avogadro’s numberThe number of units (e.g., atoms, molecules, or formula units) in 1 mol: $6.022\times {10}^{23}$., after the 19thcentury 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 carbon12—is also arbitrary. The important point is that 1 mol of carbon—or of anything else, whether atoms, compact discs, or houses—always has the same number of objects: 6.022 × 10^{23}.
Note the Pattern
One mole always has the same number of objects: 6.022 × 10^{23}.
To appreciate the magnitude of Avogadro’s number, consider a mole of pennies. Stacked vertically, a mole of pennies would be 4.5 × 10^{17} 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 get more than one trillion dollars. Clearly, the mole is so large that it is useful only for measuring very small objects, such as atoms.
The concept of the mole allows us to count a specific number of individual atoms and molecules by weighing measurable quantities of elements and compounds. To obtain 1 mol of carbon12 atoms, we would weigh out 12 g of isotopically pure carbon12. 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 × 10^{23}). 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 in which 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 carbon12 atom (12 amu), the mass of 1 mol of magnesium atoms (24.305 g) is slightly more than twice that of 1 mol of carbon12 (12 g). Similarly, the mass of 1 mol of helium (atomic mass = 4.002602 amu) is 4.002602 g, which is about onethird that of 1 mol of carbon12. Using the concept of the mole, we can now restate Dalton’s theory: 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 (H_{2}O) has 2 mol of hydrogen atoms and 1 mol of oxygen atoms.
Molar Mass
The molar massThe mass in grams of 1 mol of a substance. of a substance is defined as the mass in grams of 1 mol of that substance. One mole of isotopically pure carbon12 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 × 10^{23} 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.
Note the Pattern
The molar mass of any substance is its atomic mass, molecular mass, or formula mass in grams per mole.
The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 10^{23} carbon atoms—is therefore 12.011 g/mol:
Substance (formula)  Atomic, Molecular, or Formula Mass (amu)  Molar Mass (g/mol) 

carbon (C)  12.011 (atomic mass)  12.011 
ethanol (C_{2}H_{5}OH)  46.069 (molecular mass)  46.069 
calcium phosphate [Ca_{3}(PO_{4})_{2}]  310.177 (formula mass)  310.177 
The molar mass of naturally occurring carbon is different from that of carbon12 and is not an integer because carbon occurs as a mixture of carbon12, carbon13, and carbon14. One mole of carbon still has 6.022 × 10^{23} carbon atoms, but 98.89% of those atoms are carbon12, 1.11% are carbon13, and a trace (about 1 atom in 10^{12}) are carbon14. (For more information, see Chapter 1, Section 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 we deal with elements such as iodine and sulfur, which occur as a diatomic molecule (I_{2}) and a polyatomic molecule (S_{8}), 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 (I_{2} and S_{8}).
The molar mass of ethanol is the mass of ethanol (C_{2}H_{5}OH) that contains 6.022 × 10^{23} ethanol molecules. As you calculated in Example 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 [Ca_{3}(PO_{4})_{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 × 10^{23} formula units. Figure 3.1 “ Samples of 1 Mol of Some Common Substances” shows samples that contain precisely one molar mass of several common substances.
Figure 3.1 Samples of 1 Mol of Some Common Substances
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, we use the relationship
Equation 3.1
(moles)(molar mass) → massor, more specifically,
$$\overline{)\text{moles}}\left(\frac{\text{grams}}{\overline{)\text{mole}}}\right)=\text{grams}$$Conversely, to convert the mass of a substance to moles, we use
Equation 3.2
$$\begin{array}{l}\left(\frac{\text{mass}}{\text{molarmass}}\right)\to \text{moles}\\ \left(\frac{\text{grams}}{\text{grams/mole}}\right)=\overline{)\text{grams}}\left(\frac{\text{mole}}{\overline{)\text{grams}}}\right)=\text{moles}\end{array}$$Be sure to pay attention to the units when converting between mass and moles.
Figure 3.2 “A Flowchart for Converting between Mass; the Number of Moles; and the Number of Atoms, Molecules, or Formula Units” 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 Example 3 and Example 4.
Figure 3.2 A Flowchart for Converting between Mass; the Number of Moles; and the Number of Atoms, Molecules, or Formula Units
Example 3
For 35.00 g of ethylene glycol (HOCH_{2}CH_{2}OH), which is used in inks for ballpoint pens, calculate the number of
 moles.
 molecules.
Given: mass and molecular formula
Asked for: number of moles and number of molecules
Strategy:
A Use the molecular formula of the compound to calculate its molecular mass in grams per mole.
B Convert from mass to moles by dividing the mass given by the compound’s molar mass.
C 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 1:
$$\begin{array}{rrrr}\hfill \text{2C}& \hfill \text{(2atoms)(12}\text{.011amu/atom)}& \hfill =& \hfill \text{24}\text{.022amu}\\ \hfill \text{6H}& \hfill \text{(6atoms)(1}\text{.0079amu/atom)}& \hfill =& \hfill \text{6}\text{.0474amu}\\ \hfill +\text{2O}& \hfill \text{(2atoms)(15}\text{.9994amu/atom)}& \hfill =& \hfill \text{31}\text{.9988amu}\\ \hfill {\text{C}}_{2}{\text{H}}_{6}{\text{O}}_{2}& \hfill \text{molecularmassofethyleneglycol}& \hfill =& \hfill \text{62}\text{.068amu}\end{array}$$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):
$$\frac{\text{massofethyleneglycol(g)}}{\text{molarmass(g/mol)}}=\text{molesethyleneglycol(mol)}$$So
$$\text{35}\text{.00}\overline{)\text{gethyleneglycol}}\left(\frac{\text{1molethyleneglycol}}{\text{62}\text{.068}\overline{)\text{gethyleneglycol}}}\right)=0.5639\text{molethyleneglycol}$$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:
$$\begin{array}{ll}\text{moleculesofethyleneglycol}\hfill & =\text{0}\text{.5639}\overline{)\text{mol}}\left(\frac{\text{6}\text{.022}\times {\text{10}}^{\text{23}}\text{molecules}}{\text{1}\overline{)\text{mol}}}\right)\hfill \\ \hfill & =\text{3}\text{.396}\times {\text{10}}^{\text{23}}\text{molecules}\hfill \end{array}$$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 onehalf of Avogadro’s number, or slightly more than 3 × 10^{23} molecules, which is indeed the case.
Exercise
For 75.0 g of CCl_{3}F (Freon11), calculate the number of
 moles.
 molecules.
Answer:
 0.546 mol
 3.29 × 10^{23} molecules
Example 4
Calculate the mass of 1.75 mol of each compound.
 S_{2}Cl_{2} (common name: sulfur monochloride; systematic name: disulfur dichloride)
 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 S_{2}Cl_{2} and the formula mass of Ca(ClO)_{2}.

A The molar mass of S_{2}Cl_{2} is obtained from its molecular mass as follows:
$$\begin{array}{rrrr}\hfill \text{2S}& \hfill \text{(2atoms)(32}\text{.065amu/atom)}& \hfill =& \hfill \text{64}\text{.130amu}\\ \hfill +\text{2Cl}& \hfill \text{(2atoms)(35}\text{.453amu/atom)}& \hfill =& \hfill \text{70}\text{.906amu}\\ \hfill {\text{S}}_{\text{2}}{\text{Cl}}_{\text{2}}& \hfill {\text{molecularmassofS}}_{2}{\text{Cl}}_{2}& \hfill =& \hfill \text{1}35.036\text{amu}\end{array}$$The molar mass of S_{2}Cl_{2} is 135.036 g/mol.
B The mass of 1.75 mol of S_{2}Cl_{2} is calculated as follows:
$$\begin{array}{l}{\text{molesS}}_{\text{2}}{\text{Cl}}_{\text{2}}\left[\text{molarmass}\left(\frac{\text{g}}{\text{mol}}\right)\right]\to {\text{massofS}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{(g)}\\ \text{1}\text{.75}\overline{){\text{molS}}_{\text{2}}{\text{Cl}}_{\text{2}}}\left(\frac{\text{135}{\text{.036gS}}_{\text{2}}{\text{Cl}}_{\text{2}}}{\text{1}\overline{){\text{molS}}_{\text{2}}{\text{Cl}}_{\text{2}}}}\right)={\text{236gS}}_{\text{2}}{\text{Cl}}_{\text{2}}\end{array}$$ 
A The formula mass of Ca(ClO)_{2} is obtained as follows:
$$\begin{array}{rrrr}\hfill \text{1Ca}& \hfill \text{(1atom)(40}\text{.078amu/atom)}& \hfill =& \hfill \text{40}\text{.078amu}\\ \hfill \text{2Cl}& \hfill \text{(2atoms)(35}\text{.453amu/atom)}& \hfill =& \hfill \text{70}\text{.906amu}\\ \hfill +\text{2O}& \hfill \text{(2atoms)(15}\text{.9994amu/atom)}& \hfill =& \hfill \text{31}\text{.9988amu}\\ \hfill {\text{Ca(ClO)}}_{2}& \hfill {\text{formulamassofCa(ClO)}}_{2}& \hfill =& \hfill \text{1}42.983\text{amu}\end{array}$$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:
$$\begin{array}{l}{\text{molesCa(ClO)}}_{\text{2}}\left[\frac{{\text{molarmassCa(ClO)}}_{\text{2}}}{{\text{1molCa(ClO)}}_{\text{2}}}\right]={\text{massCa(ClO)}}_{\text{2}}\\ \text{1}\text{.75}\overline{){\text{molCa(ClO)}}_{\text{2}}}\left[\frac{\text{142}{\text{.983gCa(ClO)}}_{\text{2}}}{\text{1}\overline{){\text{molCa(ClO)}}_{\text{2}}}}\right]={\text{250gCa(ClO)}}_{\text{2}}\end{array}$$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
Calculate the mass of 0.0122 mol of each compound.
 Si_{3}N_{4} (silicon nitride), used as bearings and rollers
 (CH_{3})_{3}N (trimethylamine), a corrosion inhibitor
Answer:
 1.71 g
 0.721 g
Summary
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 carbon12 and consists of Avogadro’s number (6.022 × 10^{23}) of atoms of carbon12. 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 × 10^{23} atoms, molecules, or formula units of that substance.
Key Takeaway
 To analyze chemical transformations, it is essential to use a standardized unit of measure called the mole.
Conceptual Problems

Describe the relationship between an atomic mass unit and a gram.

Is it correct to say that ethanol has a formula mass of 46? Why or why not?

If 2 mol of sodium react completely with 1 mol of chlorine to produce sodium chloride, does this mean that 2 g of sodium reacts completely with 1 g of chlorine to give the same product? Explain your answer.

Construct a flowchart to show how you would calculate the number of moles of silicon in a 37.0 g sample of orthoclase (KAlSi_{3}O_{8}), a mineral used in the manufacture of porcelain.

Construct a flowchart to show how you would calculate the number of moles of nitrogen in a 22.4 g sample of nitroglycerin that contains 18.5% nitrogen by mass.
Please be sure you are familiar with the topics discussed in Essential Skills 2 (Chapter 3, Section 7 “Essential Skills 2”) before proceeding to the Conceptual Problems.
Answer
Numerical Problems

Derive an expression that relates the number of molecules in a sample of a substance to its mass and molecular mass.

Calculate the molecular mass or formula mass of each compound.
 KCl (potassium chloride)
 NaCN (sodium cyanide)
 H_{2}S (hydrogen sulfide)
 NaN_{3} (sodium azide)
 H_{2}CO_{3} (carbonic acid)
 K_{2}O (potassium oxide)
 Al(NO_{3})_{3} (aluminum nitrate)
 Cu(ClO_{4})_{2} [copper(II) perchlorate]

Calculate the molecular mass or formula mass of each compound.
 V_{2}O_{4} (vanadium(IV) oxide)
 CaSiO_{3} (calcium silicate)
 BiOCl (bismuth oxychloride)
 CH_{3}COOH (acetic acid)
 Ag_{2}SO_{4} (silver sulfate)
 Na_{2}CO_{3} (sodium carbonate)
 (CH_{3})_{2}CHOH (isopropyl alcohol)

Calculate the molar mass of each compound.


Calculate the molar mass of each compound.


For each compound, write the condensed formula, name the compound, and give its molar mass.


For each compound, write the condensed formula, name the compound, and give its molar mass.


Calculate the number of moles in 5.00 × 10^{2} g of each substance. How many molecules or formula units are present in each sample?
 CaO (lime)
 CaCO_{3} (chalk)
 C_{12}H_{22}O_{11} [sucrose (cane sugar)]
 NaOCl (bleach)
 CO_{2} (dry ice)

Calculate the mass in grams of each sample.
 0.520 mol of N_{2}O_{4}
 1.63 mol of C_{6}H_{4}Br_{2}
 4.62 mol of (NH_{4})_{2}SO_{3}

Give the number of molecules or formula units in each sample.
 1.30 × 10^{−2} mol of SCl_{2}
 1.03 mol of N_{2}O_{5}
 0.265 mol of Ag_{2}Cr_{2}O_{7}

Give the number of moles in each sample.
 9.58 × 10^{26} molecules of Cl_{2}
 3.62 × 10^{27} formula units of KCl
 6.94 × 10^{28} formula units of Fe(OH)_{2}

Solutions of iodine are used as antiseptics and disinfectants. How many iodine atoms correspond to 11.0 g of molecular iodine (I_{2})?

What is the total number of atoms in each sample?
 0.431 mol of Li
 2.783 mol of methanol (CH_{3}OH)
 0.0361 mol of CoCO_{3}
 1.002 mol of SeBr_{2}O

What is the total number of atoms in each sample?
 0.980 mol of Na
 2.35 mol of O_{2}
 1.83 mol of Ag_{2}S
 1.23 mol of propane (C_{3}H_{8})

What is the total number of atoms in each sample?
 2.48 g of HBr
 4.77 g of CS_{2}
 1.89 g of NaOH
 1.46 g of SrC_{2}O_{4}

Decide whether each statement is true or false and explain your reasoning.
 There are more molecules in 0.5 mol of Cl_{2} than in 0.5 mol of H_{2}.
 One mole of H_{2} has 6.022 × 10^{23} hydrogen atoms.
 The molecular mass of H_{2}O is 18.0 amu.
 The formula mass of benzene is 78 amu.

Complete the following table.
Substance Mass (g) Number of Moles Number of Molecules or Formula Units Number of Atoms or Ions MgCl_{2} 37.62 AgNO_{3} 2.84 BH_{4}Cl 8.93 × 10^{25} K_{2}S 7.69 × 10^{26} H_{2}SO_{4} 1.29 C_{6}H_{14} 11.84 HClO_{3} 2.45 × 10^{26} 
Give the formula mass or the molecular mass of each substance.
 PbClF
 Cu_{2}P_{2}O_{7}
 BiONO_{3}
 Tl_{2}SeO_{4}

Give the formula mass or the molecular mass of each substance.
 MoCl_{5}
 B_{2}O_{3}
 UO_{2}CO_{3}
 NH_{4}UO_{2}AsO_{4}
Please be sure you are familiar with the topics discussed in Essential Skills 2 (Chapter 3, Section 7 “Essential Skills 2”) before proceeding to the Numerical Problems.