10.2: Mass, Volume, and the Mole
Lesson Objectives
 Use molar mass to make conversions between mass and moles of a substance.
 Explain Avogadro’s hypothesis and how it relates to the volume of a gas at standard temperature and pressure.
 Convert between moles and volume of a gas at STP.
 Calculate the density of gases at STP.
 Use the mole road map to make twostep conversions between mass, number of particles, and gas volume.
Lesson Vocabulary
 Avogadro’s hypothesis
 molar volume
 standard temperature and pressure (STP)
Check Your Understanding
Recalling Prior Knowledge
 What is a mole? How is the mole related to the number of particles in a sample of matter?
 What is the molar mass of a substance and how is it calculated?
Chemistry is about the study of chemical reactions. Quantitatively, chemicals react with one another in very specific ratios based upon the number of reacting particles. A more in depth study of chemical reactions requires the ability to make conversions between mass, volume and moles.
MoleMass Relationship
You now know that the molar mass of any substance is the mass in grams of one mole of representative particles of that substance. The representative particles can be atoms, molecules, or formula units of ionic compounds. This relationship is frequently used to make calculations in the laboratory. Suppose that you need 3.00 moles of calcium chloride (CaCl_{2}) for a certain experiment. Since calcium chloride is a solid (Figure below), it would be convenient to measure out this amount by using a balance, but to do so, you would need to know what mass of CaCl_{2} is equivalent to 3.00 moles. The molar mass of CaCl_{2} is 110.98 g/mol, so a conversion factor can be constructed based on the fact that 1 mol CaCl_{2} = 110.98 g CaCl_{2}. Dimensional analysis will then allow you to calculate the mass of CaCl_{2} that you should use.



\begin{align*}3.00 \ \text{mol CaCl}_2 \times \frac{110.98 \ \text{g CaCl}_2}{1 \ \text{mol CaCl}_2} = 333 \ \text{g CaCl}_2\end{align*}
3.00 mol CaCl2×110.98 g CaCl21 mol CaCl2=333 g CaCl2

\begin{align*}3.00 \ \text{mol CaCl}_2 \times \frac{110.98 \ \text{g CaCl}_2}{1 \ \text{mol CaCl}_2} = 333 \ \text{g CaCl}_2\end{align*}

When you measure out 333 g of CaCl_{2}, the resulting sample will contain 3.00 moles of CaCl_{2}.
Calcium chloride is used as a drying agent and as a road deicer.
Sample Problem 10.4: Converting Moles to Mass
Chromium metal is used for decorative electroplating of car bumpers and other surfaces. Find the mass of 0.560 moles of chromium.
Step 1: List the known quantities and plan the problem.
Known
 0.560 mol Cr
 molar mass of Cr = 52.00 g/mol
Unknown
 0.560 mol Cr = ? g
The molar mass of chromium will allow us to convert from moles of Cr to grams.
Step 2: Calculate



\begin{align*}0.560 \ \text{mol Cr} \times \frac{52.00 \ \text{g Cr}}{1 \ \text{mol Cr}} = 29.1 \ \text{g Cr}\end{align*}
0.560 mol Cr×52.00 g Cr1 mol Cr=29.1 g Cr

\begin{align*}0.560 \ \text{mol Cr} \times \frac{52.00 \ \text{g Cr}}{1 \ \text{mol Cr}} = 29.1 \ \text{g Cr}\end{align*}

Step 3: Think about your result.
Since the desired amount was slightly more than one half of a mole, the mass should be slightly more than one half of the molar mass. The answer has three significant figures because the given value (0.560 mol) also has three significant figures.
 Find the masses of the following amounts.
 2.15 mol of hydrogen sulfide, H_{2}S
 3.95 × 10^{−3} mol of lead(II) iodide, PbI_{2}
A similar conversion factor based on molar mass can be used to convert the mass of a known substance to moles. In a laboratory situation, you might perform a reaction and produce a certain amount of a product. It will often be necessary to then determine the number of moles of the product that was formed, but this cannot be measured directly. However, you can use a balance to measure the mass of the product, and the number of moles can be easily calculated. The next Sample Problem illustrates this situation.
Sample Problem 10.5: Converting Mass to Moles
A certain reaction produces 2.81 g of copper(II) hydroxide, Cu(OH)_{2}. Determine the number of moles produced in the reaction.
Step 1: List the known quantities and plan the problem.
Known
 mass of Cu(OH)_{2} produced = 2.81 g
Unknown
 amount of Cu(OH)_{2} produced in moles
One conversion factor will allow us to convert from mass to moles.
Step 2: Calculate
First, it is necessary to calculate the molar mass of Cu(OH)_{2} from the molar masses of Cu, O, and H. The molar mass is 97.57 g/mol.



\begin{align*}2.81 \ \text{g Cu(OH)}_2 \times \frac{1 \ \text{mol Cu(OH)}_2}{97.57 \ \text{g Cu(OH)}_2} = 0.0288 \ \text{mol Cu(OH)}_2\end{align*}
2.81 g Cu(OH)2×1 mol Cu(OH)297.57 g Cu(OH)2=0.0288 mol Cu(OH)2

\begin{align*}2.81 \ \text{g Cu(OH)}_2 \times \frac{1 \ \text{mol Cu(OH)}_2}{97.57 \ \text{g Cu(OH)}_2} = 0.0288 \ \text{mol Cu(OH)}_2\end{align*}

Step 3: Think about your result.
The molar mass is approximately 100 g/mol, so a quick estimate can be obtained by dividing the original value by 100 (moving the decimal point two places to the left). The relatively small mass of product formed results in a small number of moles.
 Calculate the number of moles represented by the following masses.
 2.00 × 10^{2} g of silver
 37.1 g of silicon dioxide, SiO_{2}
Conversions Between Mass and Number of Particles
In the last lesson, “The Mole Concept,” you learned how to convert back and forth between moles and the number of representative particles. Now you have seen how to convert back and forth between moles and the mass of a substance in grams. We can also combine these two types of problems. Figure below illustrates that mass, number of particles, and moles are all interrelated. In order to convert between mass and number of particles, a conversion to moles is required first.
A conversion from number of particles to mass or from mass to number of particles requires two conversion factors.
Sample Problem 10.6: Converting Mass to Particles
How many molecules are present in a 20.0 g sample of chlorine gas, Cl_{2}?
Step 1: List the known quantities and plan the problem.
Known
 sample mass = 20.0 g Cl_{2}
 molar mass of Cl_{2} = 70.90 g/mol
Unknown
 number of molecules of Cl_{2}
Use two conversion factors. The first converts grams of Cl_{2} to moles. The second converts moles of Cl_{2} to the number of molecules.
Step 2: Calculate
\begin{align*}20.0 \ \text{g Cl}_2 \times \frac{1 \ \text{mol Cl}_2}{70.90 \ \text{g Cl}_2} \times \frac{6.02 \times 10^{23} \ \text{molecules Cl}_2}{1 \ \text{mol Cl}_2} = 1.70 \times 10^{23} \ \text{molecules Cl}_2\end{align*}
The problem is done using two consecutive conversion factors. There is no need to explicitly calculate the moles of Cl_{2}.
Step 3: Think about your result.
Since the given mass is less than half of the molar mass of chlorine, the resulting number of molecules is less than half of Avogadro’s number.
 How many formula units are contained in 270.2 g of zinc nitrate, Zn(NO_{3})_{2}?
 What is the mass of 5.84 × 10^{21} atoms of xenon?
MoleVolume Relationship
Avogadro’s Hypothesis and Molar Volume
In addition to number of particles and total mass, volume offers a third way to measure the amount of matter in a sample. With liquids and solids, the volume of a given number of particles can vary greatly depending on the density of the substance. This is because solid and liquid particles are packed close together with very little space in between. However, gases are largely composed of empty space between the actual gas particles (Figure below).
Gas particles are very small compared to the large amounts of empty space between them.
In 1811, Amedeo Avogadro suggested that the amount of gas in a given volume can be easily determined. Avogadro’s hypothesis states that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles. Since the total volume that a gas occupies is primarily composed of the empty space between the particles, the actual size of the particles themselves is nearly negligible. A given volume of a gas with small light particles, such as hydrogen (H_{2}), contains the same number of particles as the same volume of a heavy gas with larger particles, such as sulfur hexafluoride (SF_{6}).
Gases are compressible, meaning that when put under high pressure, the particles are forced closer to one another. This decreases the amount of empty space and reduces the volume of the gas. Gas volume is also affected by temperature. When a gas is heated, its molecules move faster and the gas expands. Because of the variation in gas volume due to pressure and temperature changes, gas volumes must be compared at the same temperature and pressure. Standard temperature and pressure (STP) is defined as 0°C (273.15 K) and 1 atm of pressure. The molar volume of a gas is the volume of one mole of the gas at a given temperature and pressure. At STP, one mole (6.02 × 10^{23} representative particles) of any gas occupies a volume of 22.4 L (Figure below).
Any gas occupies 22.4 L at standard temperature and pressure (0°C and 1 atm).
Figure below illustrates how molar volume can be seen when comparing different gases. The given samples of helium (He), nitrogen (N_{2}), and methane (CH_{4}) are at STP. Each bulb contains 1 mole, or 6.02 × 10^{23} particles. However, because the gases have different molar masses (4.00 g/mol for He, 28.0 g/mol for N_{2}, and 16.0 g/mol for CH_{4}), the mass of each sample is different, even though each has the same number of particles.
Avogadro’s hypothesis states that equal volumes of any gas at the same temperature and pressure contain the same number of particles. At standard temperature and pressure, 1 mole of any gas occupies 22.4 L.
You can watch a video experiment in which the molar volume of hydrogen gas at STP is determined at http://www.youtube.com/watch?v=6dmtLj2dLi0.
The lab document for this video can be found at http://www.dlt.ncssm.edu/core/Chapter7Gas_Laws/Chapter7Labs/MgHCl_Lab_web_0102.doc.
Conversions Between Moles and Gas Volume
Molar volume at STP can be used to convert from moles to volume and from volume to moles for gaseous samples. The fact that 1 mole = 22.4 L is the basis for the conversion factor. This equality is only true at STP, so be sure that those are the specified conditions before using this as a conversion factor.
Sample Problem 10.7: Converting Gas Volume to Moles
Many metals react with acids to produce hydrogen gas. A certain reaction produces 86.5 L of hydrogen gas at STP. How many moles of H_{2} were produced?
Step 1: List the known quantities and plan the problem.
Known
 volume of product = 86.5 L H_{2}
 1 mol = 22.4 L
Unknown
 moles of H_{2}
Use the molar volume to convert from liters to moles.
Step 2: Calculate



\begin{align*}86.5 \ \text{L H}_2 \times \frac{1 \ \text{mol H}_2}{22.4 \ \text{L H}_2} = 3.86 \ \text{mol H}_2\end{align*}
86.5 L H2×1 mol H222.4 L H2=3.86 mol H2

\begin{align*}86.5 \ \text{L H}_2 \times \frac{1 \ \text{mol H}_2}{22.4 \ \text{L H}_2} = 3.86 \ \text{mol H}_2\end{align*}

Step 3: Think about your result.
The volume of gas produced is nearly four times larger than the molar volume. The fact that the gas is hydrogen plays no role in the calculation.
 How many moles of gas are present in 57.20 L of argon at a pressure of 1 atm and a temperature of 0°C?
 At STP, what is the volume in milliliters of 0.0395 mol of fluorine gas, F_{2}?
Gas Density
As you know, density is defined as the mass per unit volume of a substance. Since gases all occupy the same volume on a per mole basis, the density of a particular gas at a given temperature and pressure is dependent only on its molar mass. A gas with a small molar mass will have a lower density than a gas with a large molar mass (Figure below). Recall that gas densities are typically reported in g/L. Gas density can be calculated by combining molar mass and molar volume.
Balloons filled with helium gas float in air because the density of helium is less than the density of air.
Sample Problem 10.8: Gas Density
What is the density of nitrogen gas at STP?
Step 1: List the known quantities and plan the problem.
Known
 molar mass of N_{2} = 28.02 g/mol
 1 mol = 22.4 L
Unknown
 density of N_{2} = ? g/L
Molar mass divided by molar volume yields the gas density at STP.
Step 2: Calculate



\begin{align*}\frac{28.02 \ \text{g}}{1 \ \text{mol}} \times \frac{1 \ \text{mol}}{22.4 \ \text{L}} = 1.25 \ \text{g/L}\end{align*}
28.02 g1 mol×1 mol22.4 L=1.25 g/L

\begin{align*}\frac{28.02 \ \text{g}}{1 \ \text{mol}} \times \frac{1 \ \text{mol}}{22.4 \ \text{L}} = 1.25 \ \text{g/L}\end{align*}

When these two ratios are multiplied in this way, the mol unit cancels, leaving g/L as the units for the answer.
Step 3: Think about your result.
The molar mass of nitrogen is slightly larger than molar volume, so its density is slightly greater than 1 g/L.
Alternatively, the molar mass of a gas can be determined if the density of the gas at STP is known.
Sample Problem 10.9: Molar Mass from Gas Density
What is the molar mass of a gas whose density is 0.761 g/L at STP?
Step 1: List the known quantities and plan the problem.
Known
 density = 0.761 g/L
 1 mol = 22.4 L
Unknown
 molar mass = ? g/mol
Molar mass is equal to density (in g/L) multiplied by molar volume.
Step 2: Calculate



\begin{align*}\frac{0.761 \ \text{g}}{1 \ \text{L}} \times \frac{22.4 \ \text{L}}{1 \ \text{mol}} = 17.0 \ \text{g/mol}\end{align*}
0.761 g1 L×22.4 L1 mol=17.0 g/mol

\begin{align*}\frac{0.761 \ \text{g}}{1 \ \text{L}} \times \frac{22.4 \ \text{L}}{1 \ \text{mol}} = 17.0 \ \text{g/mol}\end{align*}

Step 3: Think about your result.
Because the density of the gas is less than 1 g/L, the molar mass is less than 22.4 g/mol.
 What is the density of sulfur dioxide, SO_{2}, at STP?
 The density of an unknown noble gas is measured to be 1.78 g/L at STP. Calculate the molar mass and identify the noble gas.
Mole Road Map
Previously, we saw how the conversions between mass and number of particles required two steps, with moles as the intermediate. This concept can now be extended to also include gas volume at STP. The diagram in Figure below is referred to as a mole road map.
The mole road map shows the conversion factors needed to interconvert between mass, number of particles, and volume of a gas at STP.
The mole is the at the center of any calculation involving the amount of a substance. Sample Problem 10.10 is one of many different problems that can be solved using the mole road map.
Sample Problem 10.10: Mole Road Map
What is the volume of 79.3 g of neon gas at STP?
Step 1: List the known quantities and plan the problem.
Known
 molar mass of Ne = 20.18 g/mol
 1 mol = 22.4 L
Unknown
 sample volume = ? L
This problem can be solved by converting grams → moles → gas volume.
Step 2: Calculate



\begin{align*}79.3 \ \text{g Ne} \times \frac{1 \ \text{mol Ne}}{20.18 \ \text{g Ne}} \times \frac{22.4 \ \text{L Ne}}{1 \ \text{mol Ne}} = 88.0 \ \text{L Ne}\end{align*}
79.3 g Ne×1 mol Ne20.18 g Ne×22.4 L Ne1 mol Ne=88.0 L Ne

\begin{align*}79.3 \ \text{g Ne} \times \frac{1 \ \text{mol Ne}}{20.18 \ \text{g Ne}} \times \frac{22.4 \ \text{L Ne}}{1 \ \text{mol Ne}} = 88.0 \ \text{L Ne}\end{align*}

Step 3: Think about your result.
The given mass of neon is equal to about 4 moles, resulting in a volume that is about 4 times larger than molar volume.
Lesson Summary
 The molar mass of a substance is used to convert grams to moles and moles to grams. Mass can be converted to the number of representative particles and viceversa by using a twostep process.
 Avogadro’s hypothesis states that equal volumes of all gases at the same temperature and pressure contain the same number of particles. The volume of 1 mole of any gas is called its molar volume and is equal to 22.4 L at standard temperature and pressure. Molar volume allows conversions to be made between moles and volume of gases at STP.
 Gas density can be calculated from the molar mass and molar volume.
 The mole road map outlines various possible conversions between mass, number of representative particles, and gas volume.
Lesson Review Questions
Reviewing Concepts
 What do you need to know in order to calculate the number of moles of a substance from its mass?
 Atoms of xenon gas are much larger than atoms of helium gas. Explain why the volume of 1 mole of xenon is the same as the volume of 1 mole of helium.
 Why does the temperature and pressure need to be specified when working with the molar volume of a gas?
 How would you expect the molar volume of a gas to change (increase or decrease) with the following changes in conditions?
 The temperature is increased.
 The external pressure exerted on the gas is increased.
Problems
 Given the following two quantities: 0.50 mol of CH_{4} and 1.0 mol of HCl,
 Which has more atoms?
 Which has more molecules?
 Which has the greater mass?
 Which has the greater volume at the same temperature and pressure (both are gases)?
 How many moles of each substance are present in the following samples?
 61.3 g of HBr
 19.8 g of BeF_{2}
 265 g of AgNO_{3}
 0.412 kg of O_{2}
 513 L of CO_{2} gas at STP
 1300. mL of He gas at STP
 What is the mass in grams of each of the following?
 3.20 mol of magnesium
 6.55 × 10^{3} mol of (NH_{4})_{3}PO_{4}
 12.20 mol of SnSO_{4}
 4.05 × 10^{23} atoms of mercury
 6.13 × 10^{24} molecules of I_{2}
 15.4 L of N_{2}O gas at STP
 247 L of C_{3}H_{8} gas at STP
 Determine the volume of the following gas quantities at STP.
 2.78 mol of He
 0.315 mol of CH_{4}
 212 g of N_{2}
 8.91 g of OF_{2}
 3.36 × 10^{21} molecules of NH_{3}
 7.81 × 10^{23} atoms of Kr
 Make the following conversions.
 612 g of Zn to atoms
 18.77 L of CO gas (at STP) to molecules
 2.10 g of Ba(NO_{3})_{2} to formula units
 25.0 mL of Ne gas (at STP) to atoms
 What is the density in g/L of each of the following gases at STP?
 Cl_{2}
 He
 A certain gas has a density of 2.054 g/L at STP. Calculate its molar mass. If the gas is known to consist of only nitrogen and oxygen, what is its formula?
 Determine the number of C, H, and O atoms in 50.0 g of sucrose, C_{12}H_{22}O_{11}.
 The density of aluminum metal is 2.70 g/cm^{3}. How many aluminum atoms are present in a cube of aluminum that measures 1.50 cm on each side?
Further Reading / Supplemental Links
 Gram/Mole/Volume Conversions, (http://www.sciencegeek.net/Chemistry/taters/Unit4GramMoleVolume.htm)
Points to Consider
The chemical formula of an ionic compound is an empirical formula, the simplest ratio between cations and anions in the crystal. The chemical formula of a molecular compound shows the number of each atom present in the compound.
 How is the composition of a compound related to its formula?
 How can mole calculations be used to analyze chemical formulas?
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