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10.2: Mass, Volume, and the Mole

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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 two-step conversions between mass, number of particles, and gas volume.

Lesson Vocabulary

• molar volume
• standard temperature and pressure (STP)

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.

Mole-Mass 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 in the laboratory. Suppose that for a certain experiment you need 3.00 moles of calcium chloride (CaCl2). Since calcium chloride is a solid (Figure below), it would be convenient to use a balance to measure the mass that is needed. The molar mass of CaCl2 is 110.98 g/mol. The conversion factor that can be used is then based on the equality that 1 mol = 110.98 g CaCl2. Dimensional analysis will allow you to calculate the mass of CaCl2 that you should measure.

$3.00 \ \text{mol CaCl}_2 \times \frac{110.98 \ \text{g CaCl}_2}{1 \ \text{mol CaCl}_2} = 333 \ \text{g CaCl}_2$

When you measure the mass of 333 g of CaCl2, you are measuring 3.00 moles of CaCl2.

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

• molar mass of Cr = 52.00 g/mol
• 0.560 mol Cr

Unknown

• 0.560 mol Cr = ? g

One conversion factor will allow us to convert from the moles of Cr to mass.

Step 2: Calculate

$0.560 \ \text{mol Cr} \times \frac{52.00 \ \text{g Cr}}{1 \ \text{mol Cr}} = 29.1 \ \text{g Cr}$

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 of the 0.560 mol.

Practice Problem
1. Find the mass of the following amounts.
1. 2.15 mol hydrogen sulfide, H2S
2. 3.95 × 10−3 mol lead(II) iodide, PbI2

A similar conversion factor utilizing molar mass can be used to convert from the mass of an substance to moles. In a laboratory situation, you may perform a reaction and produce a certain amount of a product which can be massed. It will often then be necessary to determine the number of moles of the product that was formed. 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 = 2.81 g

Unknown

• mol Cu(OH)2

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.

$2.81 \ \text{g Cu(OH)}_2 \times \frac{1 \ \text{mol Cu(OH)}_2}{97.57 \ \text{g Cu(OH)}_2} = 0.0288 \ \text{mol Cu(OH)}_2$

The relatively small mass of product formed results in a small number of moles.

Practice Problem
1. Calculate the number of moles represented by the following masses.
1. 2.00 × 102 g of silver
2. 37.1 g of silicon dioxide, SiO2

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 mass of a substance in grams. We can combine the two types of problems into one. Figure below illustrates that mass and number of particles are both related to grams. In order to convert from mass to number of particles or vice-versa, it will first require a conversion to moles.

Conversion from number of particles to mass or from mass to number of particles requires two steps.

Sample Problem 10.6: Converting Mass to Particles

How many molecules is 20.0 g of chlorine gas, Cl2?

Step 1: List the known quantities and plan the problem.

Known

• molar mass Cl2 = 70.90 g/mol
• 20.0 g Cl2

Unknown

• number of molecules of Cl2

Use two conversion factors. The first converts grams of Cl2 to moles. The second converts moles of Cl2 to the number of molecules.

Step 2: Calculate

$20.0 \ \text{g Cl}_2 \times \frac{1 \ \text{mol Cl}_2}{70.90 \ \text{g Cl}_2} \times \frac{6.02 \times 10^{23} \ \text{molecules Cl}_2}{1 \ \text{mol Cl}_2} = 1.70 \times 10^{23} \ \text{molecules Cl}_2$

The problem is done using two consecutive conversion factors. There is no need to explicitly calculate the moles of Cl2.

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.

Practice Problems
1. How many formula units are contained in 270.2 g of zinc nitrate, Zn(NO3)2?
2. What is the mass of 5.84 × 1021 atoms of xenon?

Mole-Volume Relationship

Volume is a third way to measure the amount of matter, after item count and mass. With liquids and solids, volume varies 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 the particles. 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 explained that the volumes of all gases 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 made up primarily 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, H2, contains the same number of particles as the same volume of a heavy gas with large particles such as sulfur hexafluoride, SF6.

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, the comparison of gas volumes must be done at one standard temperature and pressure. Standard temperature and pressure (STP) is defined as 0°C (273.15 K) and 1 atm pressure. The molar volume of a gas is the volume of one mole of a gas at STP. At STP, one mole (6.02 × 1023 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. Samples of helium (He), nitrogen (N2), and methane (CH4) are at STP. Each contains 1 mole or 6.02 × 1023 particles. However, the mass of each gas is different and corresponds to the molar mass of that gas: 4.00 g/mol for He, 28.0 g/mol for N2, and 16.0 g/mol for CH4.

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/Chapter7-Gas_Laws/Chapter7-Labs/Mg-HCl_Lab_web_01-02.doc.

Conversions Between Moles and Gas Volume

Molar volume at STP can be used to convert from moles to gas volume and from gas volume to moles. The equality of 1 mole = 22.4 L is the basis for the 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 hydrogen were produced?

Step 1: List the known quantities and plan the problem.

Known

• 86.5 L H2
• 1 mol = 22.4 L

Unknown

• moles of H2

Apply a conversion factor to convert from liters to moles.

Step 2: Calculate

$86.5 \ \text{L H}_2 \times \frac{1 \ \text{mol H}_2}{22.4 \ \text{L H}_2} = 3.86 \ \text{mol H}_2$

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.

Practice Problems
1. How many moles of gas is 57.20 L of argon?
2. What is the volume in milliliters of 0.0395 mol of fluorine gas, F2?

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 is dependent 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 from 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

• N2 = 28.02 g/mol
• 1 mol = 22.4 L

Unknown

• density = ? g/L

Molar mass divided by molar volume yields the gas density at STP.

Step 2: Calculate

$\frac{28.02 \ \text{g}}{1 \ \text{mol}} \times \frac{1 \ \text{mol}}{22.4 \ \text{L}} = 1.25 \ \text{g/L}$

When set up with a conversion factor, the mol unit cancels, leaving g/L as the unit in the result.

The molar mass of nitrogen is slightly larger than molar volume, so the 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 multiplied by molar volume.

Step 2: Calculate

$\frac{0.761 \ \text{g}}{1 \ \text{L}} \times \frac{22.4 \ \text{L}}{1 \ \text{mol}} = 17.0 \ \text{g/mol}$

Because the density of the gas is less than 1 g/L, the molar mass is less than 22.4.

Practice Problems
1. What is the density of sulfur dioxide, SO2, at STP?
2. 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.

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 resulting diagram is referred to as a mole road map (Figure below).

The mole road map shows the conversion factors needed to interconvert between mass, number of particles, and volume of a gas.

The mole is the at the center of any calculation involving 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

• Ne = 20.18 g/mol
• 1 mol = 22.4 L

Unknown

• volume = ? L

The conversion factors will be grams → moles → gas volume.

Step 2: Calculate

$79.3 \ \text{g Ne} \times \frac{1 \ \text{mol Ne}}{20.18 \ \text{g Ne}} \times \frac{22.4 \ \text{L Ne}}{1 \ \text{mol Ne}} = 88.0 \ \text{L Ne}$

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 vice-versa by using a two-step 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 at standard temperature and pressure is called its molar volume and is equal to 22.4 L. 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 all possible conversions between mass, number of representative particles, and gas volume.

Lesson Review Questions

Reviewing Concepts

1. What do you need to know in order to calculate the number of moles of a substance from its mass?
2. 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.
3. Why does the temperature and pressure need to be specified when working with the molar volume of a gas?
4. How would you expect the molar volume of a gas to change (increase or decrease) with the following changes in conditions?
1. The temperature is increased.
2. The pressure of the gas is increased.

Problems

1. Given the following two quantities: 0.50 mol of CH4 and 1.0 mol of HCl
1. Which has more atoms?
2. Which has more molecules?
3. Which has the greater mass?
4. Both are gases. Which has the greater volume at the same temperature and pressure?
2. How many moles is each of the following?
1. 61.3 g of HBr
2. 19.8 g of BeF2
3. 265 g of AgNO3
4. 0.412 kg of O2
5. 513 L of CO2 gas at STP
6. 1300. ml of He gas at STP
3. What is the mass in grams of each of the following?
1. 3.20 mol of magnesium
2. 6.55 × 10-3 mol of (NH4)3PO4
3. 12.20 mol of SnSO4
4. 4.05 × 1023 atoms of mercury
5. 6.13 × 1024 molecules of I2
6. 15.4 L of N2O gas at STP
7. 247 L of C3H8 gas at STP
4. Determine the volume of the following gas quantities at STP.
1. 2.78 mol of He
2. 0.315 mol of CH4
3. 212 g of N2
4. 8.91 g of OF2
5. 3.36 × 1021 molecules of NH3
6. 7.81 × 1023 atoms of Kr
5. Make the following conversions.
1. 612 g of Zn to atoms
2. 18.77 L of CO gas to molecules
3. 2.10 g of Ba(NO3)2 to formula units
4. 25.0 ml of Ne gas to atoms
6. What is the density in g/L of each of the following gases at STP?
1. Cl2
2. He
7. 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?
8. Determine the number of C, H, and O atoms in 50.0 g of sucrose, C12H22O11.
9. The density of aluminum metal is 2.70 g/cm3. How many aluminum atoms are present in a cube of aluminum that measures 1.50 cm on each side?

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?

Aug 02, 2012

Aug 21, 2014