# Molecular Composition of Gases. Volume-Mass Relationships of Gases.

## Presentation on theme: "Molecular Composition of Gases. Volume-Mass Relationships of Gases."— Presentation transcript:

Molecular Composition of Gases

Volume-Mass Relationships of Gases

Measuring and Comparing the Volumes of Reacting Gases  Early 1800s, Gay-Lussac studied gas volume relationships with chemical reaction between H and O  Observed 2 L H can react with 1 L O to form 2 L of water vapor at constant temperature and pressure Hydrogen gas + oxygen gas  water vapor 2 L 1 L 2 L 2 volumes1 volume 2 volumes

 Reaction shows 2:1:2 relationship between volumes of reactants and product  Ratio applies to any proportions (mL, L, cm 3 )  Gay-Lussac also noticed ratios by volume between other reactions of gases Hydrogen gas + chlorine gas  hydrogen chloride gas 1 L 1 L 2 L

Law of Combining Volumes of Gases  1808 Gay-Lussac summarized results in Gay-Lussac’s law of combining volumes of gases  at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers

Avogadro’s Law  Important point of Dalton’s atomic theory: atoms are indivisible  Dalton also thought particles of gaseous elements exist in form of single atoms  Believed one atom of one element always combines with one atom of another element to form single particle of product

 Gay-Lussac’s results presented problem for Dalton’s theory  Ex. Reactions like formation of water Hydrogen gas + oxygen gas  water vapor 2 L 1 L 2 L  Seems that oxygen involved would have to divide into two parts

 1811 Avogadro found way to explain Gay- Lussac’s simple ratios of combining volumes without violating Dalton’s idea of indivisible atoms  Rejected Dalton’s idea that reactant elements are always in monatomic form when they combine to form products  Reasoned these molecules could contain more than 1 atom

 Avogadro’s law  equal volumes of gases at the same temperature and pressure contain equal numbers of molecules  At the same temp and pressure, volume of any given gas varies directly with the number of molecules

1 mol CO2 at STP = 22.4 L 1 mol O2 at STP = 22.4 L 1 mol H2 at STP = 22.4 L

 Consider reaction of H and Cl to produce HCl  According to Avogadro’s law, equal volumes of H and Cl contain same number of molecules  b/c he rejected Dalton’s theory that elements are always monatomic, he concluded H and Cl components must each consist of 2 or more atoms joined together

 Simplest assumption was that H and Cl molecules had 2 atoms each  Leads to following balanced equation: H 2 (g) + Cl 2 (g) → 2HCl(g) 1 volume 1 volume 2 volumes 1 molecule 1 molecule 2 molecules  If the simplest formula for hydrogen chloride, HCl indicates molecule contains 1 H and 1 Cl  Then the simplest formulas for hydrogen and chlorine must be H 2 and Cl 2

 Avogadro’s law also indicates that gas volume is directly proportional to the amount of gas, at given temp and pressure V = kn  k = constant  n = amount of gas in moles

Molar Volume of Gases  Remember 1 mol of substance contains 6.022 x 10 23  According to Avogadro’s law, 1 mol of any gas occupies same volume as 1 mol of any other gas at same temperature and pressure, even though masses are different  Standard molar volume of gas  volume occupied by 1 mol of gas at STP  = 22.4 L

 Knowing volume of gas, you can use 1mol/22.4 L as conversion factor  Can find number of moles  Can find mass  Can also use molar volume to find volume if you have number of moles or mass

Sample Problem  A chemical reaction produces 0.0680 mol of oxygen gas. What volume in liters is occupied by this gas sample at STP?

1. Analyze  Given: moles of O2 = 0.0680 mol  Unknown: volume of O2 in liters at STP

2. Plan  moles of O2 → liters of O2 at STP  The standard molar volume can be used to find the volume of a known molar amount of a gas at STP

3. Compute

Practice Problems  At STP, what is the volume of 7.08 mol of nitrogen gas?  159 L N 2  A sample of hydrogen gas occupies 14.1 L at STP. How many moles of the gas are present?  0.629 mol H 2  At STP, a sample of neon gas occupies 550. cm 3. How many moles of neon gas does this represent?  0.0246 mol Ne

Sample Problem  A chemical reaction produced 98.0 mL of sulfur dioxide gas, SO 2, at STP. What was the mass (in grams) of the gas produced?

1. Analyze  Given: volume of SO 2 at STP = 98.0 mL  Unknown: mass of SO 2 in grams

2. Plan  liters of SO 2 at STP→moles of SO 2 →grams of SO 2

3. Compute  = 0.280 g SO 2

Practice Problems  What is the mass of 1.33 × 10 4 mL of oxygen gas at STP?  19.0 g O 2  What is the volume of 77.0 g of nitrogen dioxide gas at STP?  37.5 L NO 2  At STP, 3 L of chlorine is produced during a chemical reaction. What is the mass of this gas?  9 g Cl 2

The Ideal Gas Law

 A gas sample can be characterized by 4 quantities 1. Pressure 2. Volume 3. Temperature 4. Number of moles

 Number of moles present always affects at least one of the other 3 quantities  Collision rate per unit area of container wall depends on number of moles  Increase moles, increase collision rate, increase pressure

 Pressure, volume, temperature, moles are all interrelated  A mathematical relationship exists to describe behavior of gas for any combination of these conditions  Ideal gas law  mathematical relationship among pressure, volume, temperature, and number of moles of a gas

Derivation of Ideal Gas Law  Derived by combining the other gas laws  Boyle’s law: at constant temp, the volume of a given mass of gas is inversely proportional to the pressure.

 Charles’s law: At constant pressure, volume of given mass of gas is directly proportional to Kelvin temperature V α T  Avogadro’s law: at constant temp and pressure, volume of given mass of gas is directly proportional to the number of moles V α n

 Volume is proportional to pressure, temp and moles in each equation  Combine the 3: Can change proportion to equality by adding constant, this time R

 This equation says the volume of a gas varies directly with the number of moles of gas and its Kelvin temperature  Volume also varies inversely with pressure

 Ideal gas law combines Boyle’s, Charles’s, Gay-Lussac’s, and Avogadro’s laws  Ex: PV = nRT  n and T are constant, nRT is constant b/c R is also constant  This makes PV = constant which is Boyle’s law

The Ideal Gas Constant  Ideal gas constant  R  Value depends on units for volume, pressure, temp Unit of RValue of R Unit of PUnit of VUnit of TUnit of n 62.4mmHgLKMol 0.0821AtmLKMol 8.314Pam3m3 KMol 8.314kPaLKMol

Sample Problem  What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K?

1. Analyze  Given:  V of N2 = 10.0 L  n of N2 = 0.500 mol  T of N2 = 298 K  Unknown:  P of N2 in atm

2. Plan  n,V,T → P  The gas sample undergoes no change in conditions  Therefore, the ideal gas law can be rearranged and used to find the pressure as follows

3. Compute = 1.22 atm

Practice Problems  What pressure, in atmospheres, is exerted by 0.325 mol of hydrogen gas in a 4.08 L container at 35°C?  2.01 atm  A gas sample occupies 8.77 L at 20°C.What is the pressure, in atmospheres, given that there are 1.45 mol of gas in the sample?  3.98 atm

 What is the volume, in liters, of 0.250 mol of oxygen gas at 20.0°C and 0.974 atm pressure?  6.17 L O 2  A sample that contains 4.38 mol of a gas at 250 K has a pressure of 0.857 atm. What is the volume?  105 L  How many liters are occupied by 0.909 mol of nitrogen at 125°C and 0.901 atm pressure?  33.0 L N 2

 What mass of chlorine gas, Cl 2, in grams, is contained in a 10.0 L tank at 27°C and 3.50 atm of pressure?  101 g Cl 2  How many grams of carbon dioxide gas are there in a 45.1 L container at 34°C and 1.04 atm?  81.9 g CO 2  What is the mass, in grams, of oxygen gas in a 12.5 L container at 45°C and 7.22 atm?  111 g O 2

 A sample of carbon dioxide with a mass of 0.30 g was placed in a 250 mL container at 400. K. What is the pressure exerted by the gas?  0.90 atm

Finding Molar Mass or Density from Ideal Gas Law  If P, V, T and mass are known you can calculate number of moles (n) in sample  Can calculate molar mass (g/mol)  Equation shows relationship between density, P, T, molar mass  Mass divided by molar mass gives moles  Substitute m/M for n in equation PV=nRT

 Density (D) = mass (m) per unit volume (V)  D = m/V

Sample Problem  At 28°C and 0.974 atm, 1.00 L of gas has a mass of 5.16 g. What is the molar mass of this gas?

1. Analyze  Given:  P of gas = 0.974 atm  V of gas = 1.00 L  T of gas = 28°C + 273 = 301 K  m of gas = 5.16 g  Unknown:  M of gas in g/mol

2. Plan  P, V, T, m → M  You can use the rearranged ideal gas law provided earlier to find the answer

3. Compute  = 131 g/mol

Practice Problems  What is the molar mass of a gas if 0.427 g of the gas occupies a volume of 125 mL at 20.0°C and 0.980 atm?  83.8 g/mol  What is the density of a sample of ammonia gas, NH 3, if the pressure is 0.928 atm and the temperature is 63.0°C?  0.572 g/L NH 3

 The density of a gas was found to be 2.0 g/L at 1.50 atm and 27°C. What is the molar mass of the gas?  33 g/mol  What is the density of argon gas,Ar, at a pressure of 551 torr and a temperature of 25°C?  1.18 g/L Ar

Stoichiometry of Gases

 You can apply gas laws to calculate stoichiometry of reactions involving gases  Coefficients in balanced equations represent mole AND volume ratios

Volume-Volume Calculations  Propane, C 3 H 8, is a gas that is sometimes used as a fuel for cooking and heating. The complete combustion of propane occurs according to the following equation. C 3 H 8 (g) + 5O 2 (g) → 3CO 2 (g) + 4H 2 O(g)  (a) What will be the volume, in liters, of oxygen required for the complete combustion of 0.350 L of propane? (b) What will be the volume of carbon dioxide produced in the reaction? Assume that all volume measurements are made at the same temperature and pressure.

1. Analyze  Given:  balanced chemical equation  V of propane = 0.350 L  Unknown:  a. V of O 2 in L;  b. V of CO 2 in L

2. Plan  a. V of C 3 H 8 → V of O 2 ;  b. V of C 3 H 8 → V of CO 2  All volumes are to be compared at the same temperature and pressure  Therefore, volume ratios can be used like mole ratios to find the unknowns.

3. Compute  = 0.175 L O 2  = 1.05 L CO 2

Practice Problem  Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen gas is needed to react completely with 4.55 L of oxygen gas to produce water vapor?  9.10 L H 2

 What volume of oxygen gas is needed to react completely with 0.626 L of carbon monoxide gas, CO, to form gaseous carbon dioxide? Assume all volume measurements are made at the same temperature and pressure.  0.313 L O 2

Volume-Mass and Mass-Volume Calculations  gas volume A → moles A → moles B → mass B  or  mass A → moles A → moles B → gas volume B  You must know the conditions under which both the known and unknown gas volumes have been measured  The ideal gas law is useful for calculating values at standard and nonstandard conditions

SAMPLE PROBLEM  Calcium carbonate, CaCO 3, also known as limestone, can be heated to produce calcium oxide (lime), an industrial chemical with a wide variety of uses. The balanced equation for the reaction follows. Δ CaCO 3 (s) → CaO(s) + CO 2 (g)  How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide gas at STP?

1. Analyze  Given:  balanced chemical equation  desired volume of CO 2 produced at STP = 5.00 L  Unknown:  mass of CaCO 3 in grams

2. Plan  The known volume is given at STP  This tells us the pressure and temperature  The ideal gas law can be used to find the moles of CO 2  The mole ratios from the balanced equation can then be used to calculate the moles of CaCO 3 needed  (Note that volume ratios do not apply here because calcium carbonate is a solid)

3. Compute  = 0.223 mol CO2

Practice Problem  What mass of sulfur must be used to produce 12.61 L of gaseous sulfur dioxide at STP according to the following equation? S 8 (s) + 8O 2 (g) → 8SO 2 (g)  18.0 g S 8

 How many grams of water can be produced from the complete reaction of 3.44 L of oxygen gas, at STP, with hydrogen gas?  5.53 g H 2 O

 Tungsten, W, a metal used in light-bulb filaments, is produced industrially by the reaction of tungsten oxide with hydrogen. WO 3 (s) + 3H 2 (g) →W(s) + 3H 2 O(l)  How many liters of hydrogen gas at 35°C and 0.980 atm are needed to react completely with 875 g of tungsten oxide?  292 L H 2

 What volume of chlorine gas at 38°C and 1.63 atm is needed to react completely with 10.4 g of sodium to form NaCl?  3.54 L Cl 2  How many liters of gaseous carbon monoxide at 27°C and 0.247 atm can be produced from the burning of 65.5 g of carbon according to the following equation? 2C(s) + O 2 (g) → 2CO(g)  544 L CO

Effusion and Diffusion

Graham’s Law of Effusion  Rates of effusion and diffusion depend on relative velocities of gas molecules  Velocity varies inversely with mass (lighter molecules move faster)

 Average kinetic energy ½ mv 2  For two gases, A and B, at same temp: ½ M A v A 2 = ½ M B v B 2  M A and M B = molar masses of A and B  Multiply by 2 M A v A 2 = M B v B 2

 Suppose you wanted to compare the velocities of the two gases  You would first rearrange the equation above to give the velocities as a ratio

 Take square root of each side This equation shows that velocities of two gases are inversely proportional to the square roots of their molar masses

 b/c rates of effusion are directly proportional to molecular velocities, can write

Graham’s Law of Effusion  The rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses

Application of Graham’s Law  Graham’s experiments dealt with densities of gases  Density varies directly with molar mass  So…square roots of molar masses from equation can be replaced with square roots of densities

Sample Problem  Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.

1. Analyze  Given:  identities of two gases, H 2 and O 2  Unknown:  relative rates of effusion

2. Plan  molar mass ratio → ratio of rates of effusion  The ratio of the rates of effusion of two gases at the same temperature and pressure can be found from Graham’s law

3. Compute  Hydrogen effuses 3.98 times faster than oxygen.

Practice Problems  A sample of hydrogen effuses through a porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas.  160 g/mol  Compare the rate of effusion of carbon dioxide with that of hydrogen chloride at the same temperature and pressure.  CO 2 will effuse about 0.9 times as fast as HCl

 If a molecule of neon gas travels at an average of 400 m/s at a given temperature, estimate the average speed of a molecule of butane gas, C 4 H 10, at the same temperature.  about 235 m/s
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8905228972434998, "perplexity": 3390.5938239944494}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125936981.24/warc/CC-MAIN-20180419150012-20180419170012-00562.warc.gz"}