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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; and Molar ; STP; Direct and Inverse Relationships; Empirical/Molecular Formulas;

Units: P = = 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)

GAS LAWS V V P V P V 1 1 2 (V T) 1 1 2 2 P1V1 = P2V2 ( P ∝ ) = µ = V T1 T2 T1 T2 Boyle's Law: PV = constant Charles' Law: V/T = constant Combined Gas Law: PV/T = constant PV = nRT gRT (P)(molar mass) molar mass = D = PV RT € Law € derived from € derived from Ideal Gas Law STP = Standard & V V 1 = 2 (V µ n) Pressure; P = 1.0atm; T = 0.0˚C n n (273K); at STP 1 mol gas occupies 1 2 Equal volumes have equal mol 22.4L € € STP Avogadro's Law: V/n = constant V = (1000ml = 1L); T = temperature - always must be in K (273.15 + ˚C = K); n = #moles of gas; L atm € R = = 0.0821 ; g = grams of gas; D = density of gas (must be in g/L) mol K 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?

P1V1 P2V2 P1V1T2 (1.75atm)(2.00L)(273.15+ 75.0K) Answer 1: 2.04atm { = ; rearrange to solve for P2: P2 = ; P2 = = 2.04atm } T1 T2 V2T1 (2.00L)(273.15+ 25.0K)

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? gRT Answer 2: 44.0g/mol { molar mass = ; T = 27.5 + 273.15 = 300.65K; P = 740torr x (1atm/760torr) = 0.974atm; PV (0.1137g)(0.0821Latm / molK)(300.65K) V = 65.5ml x (1L/1000ml) = 0.0655L; molar mass = = 44.00g / mol} (0.974atm)(0.0655L)

Example 3: How many are in 83.0ml of gas at 32.5˚C and 0.945atm? 21 PV Answer 3: 1.88 x 10 molecules {PV = nRT; n = ; V = (83.0ml) x (1L/1000ml) = 0.0830L; T = 32.5 + 273.15 = 305.65K; RT 23 PV (0.945atm)(0.0830L) ⎛ 6.022 x 10 molecules⎞ 21 n = = = 0.00313mol; 0.00313mol⎜ ⎟ = 1.88 x 10 molecules } RT (0.0821Latm / molK)(305.65K) ⎝ 1mol ⎠

Relationships between Gas Variables Using PV = nRT and DRT = P(molar mass): if variables are on the same side of the equation they have an inverse relationship; if they are on opposite sides of the equation they have a direct relationship

(Linear Graph: Direct relationship; Curve Graph: Inverse relationship) P and V P and n P and T V and T V (L) Pressure (atm) Pressure (atm) Pressure (atm)

Volume (L) moles (n) T (K) T (K) V and n D and T D and P D and Molar Mass Density (g/L) Density (g/L) Density (g/L) Volume (L)

moles (n) Temperature (K) Pressure (atm) Molar Mass (g/mol)

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 ; divide by smallest number of moles 1 1 2 1 3 5. Fractions: 0.5 = /2 ® x2; 0.33 or 0.66 = /3 or /3 ® x3; 0.25 or 0.75 = /4 or /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 molar mass 1. Determine ratio = molecular formula molar massempirical formula 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. (4.202g)(0.0821Latm / molK)(273.15+ 35.0K) molar massmolecular formula 92.03 molar mass = = 92.03g / mol ; ratio = = = 2; (585torr)(1atm / 760torr)(1.50L) molar massempiricalformula 46.01 EF x ratio = MF; N(1 x 2)O(2 x 2) = N2O4}

STOICHIOMETRY

molarity A molarity B or atoms or molecules A molecules B M = molA/L M = molB/L 1 mol = 1 mol = 6.022 x 1023 Chemical Reaction or 6.022 x 1023 Chemical Formula moles A moles B

molar molar mass 22.4 L = 1 mol 22.4 L = 1 mol mass grams A at STP or at STP or grams B PV = nRT PV = nRT

gas A: gas B: P, V, T P, V, T

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

N + 3H 2 NH 1. 2 2 3 ⎛ 1molA⎞ ⎛ molB⎞ ⎛ L B ⎞ 2. gA = L B ; do in 2 parts: ⎝⎜ gA ⎠⎟ ⎝⎜ molA⎠⎟ ⎝⎜ molB⎠⎟ ⎛ 1molA⎞ ⎛ molB⎞ Step 1. gA = molB and then Step 2. Ideal Gas Law: PV = nRT ⎝⎜ gA ⎠⎟ ⎝⎜ molA⎠⎟

⎛ 1molN2 ⎞ ⎛ 2molNH3 ⎞ 3. Step 1: 84.0gN2 ⎜ ⎟ ⎜ ⎟ = 6.00molNH3 ⎝ 28.0gN2 ⎠ ⎝ 1molN2 ⎠ nRT 4. Step 2: PV = nRT; rearrange for volume: V = P (6.00mol)(0.0821Latm / molK)(55.0 + 273.15K) V = = 134.7L 1.20atm

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?

5. , 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 graph and label the axes 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 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.

12. Which samples of gas below have the same number molecules if the samples are all at the same temperature? i. 2.00L CO2(g) at 2.25atm ii. 4.00L F2(g) at 1.00atm iii. 3.00L Ar(g) at 1.50atm iv. 2.50L CO(g) at 3.00atm a. i, ii b. i, iii, iv c. ii, iii d. i, iii e. ii, iv

13. The solid rocket boosters for the space shuttle employed a mixture of aluminum and 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 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

14. 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

15. The complete reaction between 5.0L Br2(g) and 25.0L F2(g) yields 10.0L of a new gaseous compound. If all the gases are at 1.0atm and 0.0˚C (i.e., STP) what is the empirical formula of the new compound?

16. In lab, a student added an unknown liquid to a 275.68 ml flask (called a Dumas flask). The flask’s mass when empty was measured to be 59.0905g was then submerged into a bath of hot water at 98.76˚C and the added liquid vaporized pushing out all the air. The flask was removed, cooled, and re-weighed with a new mass of 59.7028g being measured. The laboratory pressure was 734.5mmHg. The volatile liquid was analyzed and found to contain 85.63% C and 14.37% H. What is the molecular formula of the unknown chemical? a. C3H6 b. C5H10 c. C7H14 d. C8H16 e. C8H18

ANSWERS 1. 3.8 atm {P1V1 = P2V2; (1.5)(5.0) = (x)(2.0); x = 3.75atm}

P(molar mass) DRT 2. 45.1 g/mol { D = ; molar mass = ; RT P (1.434g / L)(0.0821Latm / molK)(90.0 + 273.15K) molar mass = = 45.13g / mol} (720.0mmHg)(1atm / 760mmHg)

P1V1 P2V2 P1V1T2 (0.125atm)(2.0L)(57 + 273.15K) 3. 0.072 atm { = ; P2 = ; P2 = = 0.07245atm } T1 T2 V2T1 (4.5L)(−20.+ 273.15K)

gRT nRT 4. 5.00 atm {can use either molar mass = or PV = nRT; I’ll use PV = nRT: P = ; PV V ⎛ 1mol Ar ⎞ (2.020mol)(0.0821Latm / molK)(21.0 + 273.15K) 80.7g Ar = 2.020mol Ar ; P = = 4.998atm} ⎝⎜ 39.95g Ar ⎠⎟ 9.76L

P(molar mass) (1.00atm)(58.12g / mol) 5. 2.59 g/L { D = ; molar mass = 58.12g/mol; D = = 2.592g / L} RT (0.0821Latm / molK)(0.00 + 273.15K)

DRT (0.981g / L)(0.0821Latm / molK)(25+ 273.15K) 6. b { molar mass = ; molar mass = = 16.01g / mol; P 1.5atm molar mass matches ® CH4}

P

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

gRT (6.22 x 10−4g)(0.0821Latm / molK)(1000.+ 273.15K) 8. c { molar mass = ; molar mass = = 22.98g / mol ® Na} PV (10.0torr)(1atm / 760torr)(0.215L)

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 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 {P µ 1/V; from PV = nRT since P and V are on the same side of the equation ® inverse relationship} ii. F {P µ T; from PV = nRT since P and T are on opposite sides of the equation ® direct relationship} iii. T {V µ T; from PV = nRT since V and T are on opposite sides of the equation ® direct relationship} iv. F {n µ V; from PV = nRT since n and V are on opposite sides of the equation ® direct relationship} v. F {n µ V; from PV = nRT since n and P are on opposite sides of the equation ® direct relationship}

P1V1 P1V1 1 11. c {P1V1 = P2V2; P2 = = = P1; pressure has dropped to 1/3 of its original value} V2 3V1 3

PV 12. d {PV = nRT; n = ; since R and T are constant, then n is proportional to PV ( n ∝ PV); also, the #molecules is proportional RT to n ( # molecules ∝ n ); therefore, the #molecules is proportional to PV ( # molecules ∝ PV ); calculate PV for each option; “a”: PV = (2.00L)(2.25atm) = 4.50; “b”: PV = (4.00L)(1.0atm) = 4.00; “c”: PV = (3.00L)(1.50atm) = 4.50; “d”: PV = (2.50L)(3.00atm) = 7.50; note that “i” and “iii” have the same PV values, and will therefore have the same #moles and the same #molecules; to verify this pick a random temperature and see how the numbers work out, for example, T = 298K; PV (2.25atm)(2.00L) PV (1.50atm)(3.00L) “i”: n = = = 0.1839mol; “iii”: n = = = 0.1839mol; note how the RT (0.0821Latm / molK)(298K) RT (0.0821Latm / molK)(298K) denominator is the same (same T) and how the numerator is determined by PV}

PV (1.05atm)(534L) 13. b {find mol H2O: n = = = 6.334mol H O ; RT (0.0821Latm / molK)(805+ 273.15K) 2

⎛ 3mol NH4ClO4 ⎞ ⎛ 117.5g NH4ClO4 ⎞ 6.334mol H2O⎜ ⎟ ⎜ ⎟ = 372.1g NH4ClO4} ⎝ 6mol H2O ⎠ ⎝ 1mol NH4ClO4 ⎠

⎛ 1mol I2 ⎞ ⎛ 5mol F2 ⎞ 14. b { 350.g I2 ⎜ ⎟ ⎜ ⎟ = 6.895mol F2; PV = nRT; ⎝ 253.8 I2 ⎠ ⎝ 1mol I2 ⎠ nRT (6.895mol)(0.0821Latm / molK)(37.0 + 273.15K) V = = = 189.3L } P (705torr)(1atm / 760torr)

15. BrF5 {start by writing a reaction: xBr2(g) + yF2(g) ® zBraFb(g); since all the gases are at the same T and P, from Avogadro’s Law, the volumes are proportional to moles; hence, change the reaction to the following: 5Br2(g) + 25F2(g) ® 10BraFb(g); since ⎛ 2L Br ⎞ there were initially 10L of Br ( ), there must be 10L at the end; hence, the subscript “a” must be 1; 5.0L Br2 ⎜ ⎟ = 10L Br ⎝ 1L Br2 ⎠ ⎛ 1L Br ⎞ 10Br F with “a” = 1 means 10Br F would yield 10L of Br ( ); repeating for F, since there were a b 1 b 10.0L BrFb ⎜ ⎟ = 10L Br ⎝ 1L BrFb ⎠ ⎛ 2L F ⎞ initially 50L of F ( ), there must be 50L at the end; hence, the subscript “b” must be 5; 25.0L F2 ⎜ ⎟ = 50L F ⎝ 1L F2 ⎠ ⎛ 5L F ⎞ 10Br F with “b” = 5 means 10Br F would yield 50L of Br ( ); the empirical formula: BrF } a b 1 5 10.0L BrF5 ⎜ ⎟ = 50L F 5 ⎝ 1L BrF5 ⎠

gRT 16. b {use the information to determine the molar mass; molar mass = ; mass = 59.7028g – 59.0905 = 0.6123g; PV T = 98.76 + 273.15 = 371.91K; P = 734.5mmHg(1atm/760mmHg) = 0.96645atm; V = 275.68ml(1L/1000ml) = 0.27568L; gRT (0.6123g)(0.0821L atm / mol K)(371.91) molar mass = = = 70.17g / mol ; from mass%C and mass%H ® empirical formula: PV (0.96645atm)(0.27568L) 1. assume 100g 2. % ® grams: C: 85.63% x 100g = 85.63g C; H: 14.37% x 100g = 14.37g H ⎛ 1mol C ⎞ ⎛ 1mol H ⎞ 3. g ® mol: 85.63g C = 7.1299mol C; 14.37g H = 14.2560mol H ⎝⎜ 12.01g C⎠⎟ ⎝⎜ 1.008g H⎠⎟ 4. Write EF: C7.1299H14.2560 and divide by smallest number of mol: 7.1299: C7.1299H14.2560 ⇒ C1H1.999 = CH2 7.1299 7.1299 5. No fractions Calculate molecular formula: molar massmolecular formula 70.17 1. = = 5.003 molar massempiricalformula 14.026 2. Multiply empirical formula by ratio: C(1 x 5)H(2 x 5) = C5H10 = molecular formula}