CHEMISTRY 103 – Help Sheet #7 Gases-Part I

Do the topics appropriate for your course Prepared by Dr. Tony Jacob

https://clc.chem.wisc.edu (Resources page)

Nuggets: Gas Laws and Units; Density and Molar Mass; Empirical/Molecular Formulas; Stoichiometry; STP Units: P = pressure = F/A where F = force and A = area Unit Conversions: 1atm = 760torr = 760mmHg; 1atmosphere (atm) = 1.013 x 105 Pa; 1mmHg = 133.3Pa;

1atm = 14.70 pounds per square inch (psi) Kinetic-Molecular Theory

1. The gas is composed of molecules that are small compared to the distances between them. 2. The gas molecules are in constant random motion at varying speeds. 3. The forces between molecules are minimal; collisions with the container creates pressure. 4. Molecular collisions and collisions with the container walls occur without loss of E (elastic collisions) and

the average kinetic energy stays constant. 5. The average kinetic energy is proportional to the temperature in K.

GAS LAWS

P1V1 = P2V2 ( ) (V µ T)

Boyle's Law: PV = constant Charles' Law: V/T = constant Combined Gas Law: PV/T = constant PV = nRT

Ideal Gas Law derived from Ideal Gas Law derived from Ideal Gas Law

(V µ n)

Equal volumes have equal mol

STP = Standard Temperature & Pressure; P = 1.0atm; T = 0.0˚C (273K); at STP 1 mol gas occupies 22.4L

Avogadro's Law: V/n = constant STP V = volume (1000ml = 1L); T = temperature - always must be in K (273.15 + ˚C = K); n = #moles of gas;

R = gas constant = 0.0821L atmmol K ; g = grams of gas; D = density of gas (must be in g/L)

If R is part of the equation ® P must be in atm and V must be in L If R is not part of the equation ® P1 and P2 can be in any units but must be the same units; V1 and V2 can be in any units but must be the same units

Example 1: A 2.00-L sample of a gas at 25.0˚C and 1.75atm is heat to 75.0˚C while the volume is maintained. What is the new pressure of the gas?

Answer 1: 2.04atm { ; rearrange to solve for P2: ; }

Example 2: 0.1137g of an unknown gas at 27.5˚C and 740. torr occupies 65.5ml. What is the molar mass of the unknown gas?

Answer 2: 44.0g/mol { ; T = 27.5 + 273.15 = 300.65K; P = 740torr x (1atm/760torr) = 0.974atm;

V = 65.5ml x (1L/1000ml) = 0.0655L; }

Example 3: How many molecules are in 83.0ml of gas at 32.5˚C and 0.945atm?

Answer 3: 1.88 x 1021 molecules {PV = nRT; ; V = (83.0ml) x (1L/1000ml) = 0.0830L; T = 32.5 + 273.15 = 305.65K;

; }

€

P ∝ 1V

€

V1T1

=V2T2

€

P1V1T1

=P2V2T2

€

molar mass =gRTPV

€

D =(P)(molar mass)

RT

€

V1n1

=V2n2

P1V1T1

=P2V2T2

P2 =P1V1T2V2T1

P2 =(1.75atm)(2.00L)(273.15+ 75.0K)

(2.00L)(273.15+ 25.0K)= 2.04atm

molar mass = gRTPV

molar mass = (0.1137g)(0.0821Latm / molK)(300.65K)(0.974atm)(0.0655L)

= 44.00g / mol

n = PVRT

n = PVRT

= (0.945atm)(0.0830L)(0.0821Latm / molK)(305.65K)

= 0.00313mol 0.00313mol 6.022 x1023molecules1mol

⎛

⎝⎜

⎞

⎠⎟ = 1.88 x1021 molecules

Relationships between Gas Variables (Linear Graph: Direct relationship; Curve Graph: Inverse relationship)

Empirical formula (review): simplest ratio between elements in a formula; Determine empirical formula from mass%:

1. Assume 100g 2. % ® g 3. g ® mol (if g originally given and not mass% ® start at step 3!) 4. Write chemical formula; divide by smallest number of moles 5. Fractions: 0.5 = 1/2 ® x2; 0.33 or 0.66 = 1/3 or 2/3 ® x3; 0.25 or 0.75 = 1/4 or 3/4 ® x4

Molecular formula (review): the exact formula of a compound For N2O4, the molecular formula is N2O4 and the empirical formula is NO2

Determine Molecular Formula from the Empirical Formula using Molar Mass

1. Determine

2. Take empirical formula and multiply each subscript by the number from above ratio Example 4: The empirical formula of a nitrogen-oxygen compound is NO2. When 4.202grams of the compound was placed in a 1.50L contain at

35.0˚C, the pressure was 585torr. What is the molecular formula of the compound? Answer 4: N2O4 {From the data determine the molar mass of the unknown compound.

; ;

EF x ratio = MF; N(1 x 2)O(2 x 2) = N2O4}

0102030405060708090

100

0.00 5.00 10.00 15.00 20.00 25.00

Pressure(a

tm)

Volume(L)

PandV

0102030405060708090100

0.00 0.20 0.40 0.60 0.80 1.00

Pressure(a

tm)

moles(n)

Pandn

0102030405060708090

100

0 25 50 75 100 125 150 175 200

Pressure(a

tm)

T(K)

PandT

0.02.55.07.510.012.515.017.520.022.525.0

0 1000 2000 3000 4000 5000

V(L)

T(K)

TandV

0

100

200

300

400

500

600

0.00 1.00 2.00 3.00 4.00 5.00 6.00

Volume(L)

moles(n)

nandV

ratio = molar massmolecular formulamolar massempirical formula

molar mass = (4.202g)(0.0821Latm / molK)(273.15+ 35.0K)(585torr)(1atm / 760torr)(1.50L)

= 92.03g / mol ratio =molarmassmolecular formulamolarmassempiricalformula

= 92.0346.01

= 2

STOICHIOMETRY

Example 5: How many liters NH3(g) collected at 55.0˚C and 1.20atm can be prepared from 84.0g N2(g) using: N2(g) + 3H2(g) ® 2NH3(g) This is a “grams A ® gas B” calculation (see diagram); it requires 3 conversions (steps from the flow chart) Answer 5: 135L

1.

2. ; do in 2 parts:

Step 1. and then Step 2. Ideal Gas Law: PV = nRT

3. Step 1:

4. Step 2: PV = nRT; rearrange for volume:

Example 6: How many liters NO2(g) can be prepared from 10.0L of NO(g) and 12.0L of O2(g) using the reaction shown? All gases are at the same

temperature and pressure. 2NO(g) + O2(g) ® 2NO2(g) Answer 6: 10.0L NO2 can be produced; NO(g) is the limiting reagent; O2 is the excess reagent

{This is a “LA ® LB” calculation simplified with Avogadro’s Law since all the gases are at the same T and P. This means that: V µ mol. Since there are 2 reactant quantities this is also a limiting reagent question.

NO(g) (LNO ® LNO2): ; O2 (LO2 ® LNO2):

(the Liter-to-Liter ratio is allowed because the gases are at the same T and P) The smaller quantity of NO2 is the amount that can be produced; the reactant yielding this smaller amount is the limiting reagent and the reactant yielding the larger amount is the excess reagent; 10.0L NO2 can be produced; NO(g) is the limiting reagent (produced the smaller amount); O2(g) is the excess reagent (produced the larger amount)}

1. O2(g) at 1.5 atm in a 5.0-L container is compressed to 2.0L. What is the new pressure in the system? 2. The density of the vapor of a compound at 90.0 ˚C and 720.0 mmHg is 1.434 g/L. What is the compound’s molar mass? 3. A gas sample is heated from -20. ˚C to 57 ˚C and the volume is increased from 2.0L to 4.5L. If the initial pressure is 0.125 atm, what is the final pressure? 4. Argon gas, Ar, is contained in a steel cylinder with a volume of 9.76 L. The temperature is 21.0 ˚C, and the mass of the Ar is 80.7 g. What is the pressure of the Ar?

atoms or molecules A

atoms or molecules B

moles A moles B

grams A grams B

molarmass

molarmass

Chemical Reaction or Chemical Formula

molarity A

22.4 L = 1 molat STP orPV = nRT

1 mol = 6.022 x 102 3

1 mol = 6.022 x 102 3

gas A: P, V, T

22.4 L = 1 molat STP orPV = nRT

molarity B

gas B: P, V, T

M = molA/L M = molB/L

NH3N2 + H2 23

gA1molA

gA⎛⎝⎜

⎞⎠⎟

molBmolA

⎛⎝⎜

⎞⎠⎟

L BmolB

⎛⎝⎜

⎞⎠⎟= L B

gA1molAgA

⎛⎝⎜

⎞⎠⎟molBmolA

⎛⎝⎜

⎞⎠⎟= molB

84.0gN21molN228.0gN2

⎛

⎝⎜⎞

⎠⎟2molNH31molN2

⎛

⎝⎜⎞

⎠⎟= 6.00molNH3

V = nRTP

V = (6.00mol)(0.0821Latm / molK)(55.0 + 273.15K)1.20atm

= 134.7L

10.0L NO2L NO22L NO

⎛⎝⎜

⎞⎠⎟= 10.0L NO2 12.0L O2

2L NO21L O2

⎛

⎝⎜⎞

⎠⎟= 24.0L NO2

5. Butane, C4H10, is an easily liquefied gaseous fuel. Calculate the density of butane gas at STP in g/L. 6. What is the chemical formula of an unknown gas if its density is 0.981g/L at a pressure of 1.5atm at 25˚C? a. O2 b. CH4 c. N2 d. CO e. CO2 7. Draw a diagram to show how the number of moles and the pressure are related at constant volume and T. 8. At 1000. ˚C and 10.0 torr, a 6.22 x 10-4g chemical sample occupies 215ml. What is the chemical? a. Ne b. He c. Na d. Ar e. H2O 9. There are four 1-L flasks at STP. Flasks 1 through 4 contain O2, Ar, NH3, and CO, respectively. I. Which flask contains the gas with the largest number of moles? a. flask 1 b. flask 2 c. flask 3 d. flask 4 e. all have the same number II. Which flask contains the largest number of molecules? a. flask 1 b. flask 2 c. flask 3 d. flask 4 e. all have the same number III. Which flask contains the most mass? a. flask 1 b. flask 2 c. flask 3 d. flask 4 e. all have the same number IV. Which flask contains the largest number of atoms? a. flask 1 b. flask 2 c. flask 3 d. flask 4 e. all have the same number V. Which flask contains the gas with the greatest density? a. flask 1 b. flask 2 c. flask 3 d. flask 4 e. all have the same number 10. For each choice (not a multiple choice question), state whether the mathematical relationship is True (T) or False (F). Assume all other variables are constant; i.e., in question “i” assume that temperature and moles are constant. i. Pressure µ Volume ii. Pressure µ 1/Temperature iii. Volume µ Temperature iv. Moles µ 1/Volume v. Moles µ 1/Pressure 11. A molecule with an empirical formula of C2H3O occupies a volume of 412ml when 1.05g at –2.75˚C is at a pressure of 500. torr. What is the molecular formula? a. C2H3O b. C4H6O2 c. C2H6O d. C6H9O3 e. none of the above 12. A container holding gas has its volume tripled while the temperature and moles are held constant. The pressure must have a. remained unchanged. b. tripled. c. decreased by a factor of 3. d. been cut in half. e. increased by a factor of 9. 13. Small amounts of hydrogen gas can be prepared by reacting aluminum with hydrochloric acid: 2Al(s) + 6HCl(aq) ® 2AlCl3(s) + 3H2(g) How many grams of aluminum are required to prepare 2.50L of H2 gas at 765 mmHg and 22.0˚C?

14. The solid rocket boosters for the space shuttle employed a mixture of aluminum and ammonium perchlorate as fuel. The balanced reaction is 3Al(s) + 3NH4ClO4(s) ® Al2O3(s) + AlCl3(s) + 3NO(g) + 6H2O(g) How many grams of ammonium perchlorate (molar mass = 117.5 g/mol) are necessary to produce 534 liters of water vapor at a temperature of 805˚C and a pressure of 1.05 atm pressure? a. 3.17 grams b. 372 grams c. 744 grams d. 498 grams e. 997 grams 15. How many liters of O2(g) at 25˚C and 7.0 x 102 torr are needed to react with 10.g manganese to form manganese(III) oxide? 4Mn(s) + 3O2(g) ® 2Mn2O3(s) 16. IF5(g) can be prepared by the reaction: I2(s) + 5F2(g) ® 2IF5(g) What volume of F2(g) at 37.0˚C and 705 torr is needed to react completely with 350. grams I2(s)? a. 378L b. 189L c. 75.6L d. 45.3L e. 37.8L 17. Given the combustion of methane gas, CH4, below, what volume of CH4 is required to react with 1.75 L of oxygen. Assume all gases are at 0.00˚C and 1.00atm. CH4(g) + 2O2(g) ® CO2(g) + 2H2O(g) ANSWERS 1. 3.8 atm {P1V1 = P2V2; (1.5)(5.0) = (x)(2.0); x = 3.75atm}

2. 45.1 g/mol { ; ;

}

3. 0.072 atm { ; ; }

4. 5.00 atm {can use either or PV = nRT; I’ll use PV = nRT: ;

; }

5. 2.59 g/L { ; molar mass = 58.12g/mol; }

6. b { ; ;

molar mass matches ® CH4}

7. P/n = constant or P = kn where k is a constant

D = P(molar mass)RT

molar mass = DRTP

molar mass = (1.434g / L)(0.0821Latm / molK)(90.0+ 273.15K)(720.0mmHg)(1atm / 760mmHg)

= 45.13g / mol

P1V1T1

=P2V2T2

P2 =P1V1T2V2T1

P2 =(0.125atm)(2.0L)(57 + 273.15K)

(4.5L)(−20.+ 273.15K)= 0.07245atm

molar mass = gRTPV

P = nRTV

80.7g Ar 1mol Ar39.95g Ar

⎛⎝⎜

⎞⎠⎟= 2.020mol Ar P = (2.020mol)(0.0821Latm / molK)(21.0+ 273.15K)

9.76L= 4.998atm

D = P(molar mass)RT

D = (1.00atm)(58.12g / mol)(0.0821Latm / molK)(0.00+ 273.15K)

= 2.592g / L

molar mass = DRTP

molar mass = (0.981g / L)(0.0821Latm / molK)(25+ 273.15K)1.5atm

= 16.01g / mol

n

P

8. c { ; ® Na}

9. I. e {since all the flasks have the same P, V, and T ® same #mol}

II. e {since all have the same #mol ® same #molecules} III. b {since they have the same #mol, the one with the largest molar mass has the greatest mass ® Ar} IV. c {since they have the same #mol, the one with more atoms per molecule will have more atoms ® NH3} V. b {since D = m/V and the V is the same for all flasks, then the one with the greatest mass has the greatest D ® Ar}

10. i. F ii. F iii. T iv. F v. F

11. b {first find molar mass: ; ;

molar mass empirical formula = 43.04g/mol; ;

EF x 2: C(2 x 2)H(3 x 2)O(1 x 2) = C4H6O2}

12. c {P1V1 = P2V2; ; pressure has dropped to 1/3 of its original value}

13. 1.87 g {find mol H2: ;

}

14. b {find mol H2O: ;

}

15. 3.63L O2(g) { ;

PV = nRT; }

16. b { ; PV = nRT;

}

17. 0.875 L {Use Avogadro’s Law because the gases are all at the same P and T; }

molar mass = gRTPV

molar mass = (6.22 x 10−4g)(0.0821Latm / molK)(1000.+ 273.15K)(10.0torr)(1atm / 760torr)(0.215L)

= 22.98g / mol

molar mass = gRTPV

molar mass = (1.05g)(0.0821Latm / molK)(−2.75+ 273.15K)(500.torr)(1atm / 760torr)(0.412L)

= 86.00g / mol

ratio =molarmassmolecular formulamolarmassempiricalformula

= 86.0043.04

= 1.998 = 2

P2 =P1V1V2

=P1V13V1

= 13P1

n = PVRT

= (765mmHg)(1atm / 760mmHg)(2.50L)(0.0821Latm / molK)(22.0+ 273.15K)

= 0.1038mol H2

0.1039mol H22mol Al3mol H2

⎛

⎝⎜⎞

⎠⎟26.98g Al1mol Al

⎛⎝⎜

⎞⎠⎟= 1.869g Al

n = PVRT

= (1.05atm)(534L)(0.0821Latm / molK)(805+ 273.15K)

= 6.334mol H2O

6.334mol H2O3mol NH4ClO46mol H2O

⎛

⎝⎜⎞

⎠⎟117.5g NH4ClO41mol NH4ClO4

⎛

⎝⎜⎞

⎠⎟= 372.1g NH4ClO4

10.g Mn 1mol Mn54.94g Mn

⎛⎝⎜

⎞⎠⎟3mol O24mol Mn

⎛

⎝⎜⎞

⎠⎟= 0.1365mol O2

V = nRTP

= (0.1365mol)(0.0821Latm / molK)(25+ 273.15K)

(7.0 x 102 torr)(1atm / 760torr)= 3.63L

350.g I21mol I2253.8 I2

⎛

⎝⎜⎞

⎠⎟5mol F21mol I2

⎛

⎝⎜⎞

⎠⎟= 6.895mol F2

V = nRTP

= (6.895mol)(0.0821Latm / molK)(37.0+ 273.15K)(705torr)(1atm / 760torr)

= 189.3L

1.75L O21L CH42L O2

⎛⎝⎜

⎞⎠⎟= 0.875L CH4