Test 1 Review
Test 3 Review Information: This test will cover sections 6.5 – 6.9, 7.3 – 7.5, 8.1 – 8.5, 8.8, 9.1 – 9.6, and 10.1 – 10.6. The test will not be multiple choice. There will be a question with a lot of blank space for you to fill. You will be given a periodic table and you should bring your own calculator (non-graphing only). Information you need to know: From Test 1 & Test 2: • Selected prefixes to the metric system and the conversions between them, found in Table 1.5 • Conversion of temperature from Celsius to Kelvin • Names of the groups, found in Table 2.3 • Diatomic molecules • Polyatomic ions, given to you in your class guide (Periodic Table, Compounds, Ions and Naming folder) • Acids found in your class guide (Periodic Table, Compounds, Ions and Naming folder) • Avogadro’s Number actual yield • % Yield = ÷100 theoretical yield • How to use the solubility rules (Table 4.1) • Strong acids and strong bases (Table 4.2) • Acid + Base H2O + salt • Oxidation Number rules found in the text on page 132 moles of solute • Molarity= Liters of solution • M1V1 = M2V2; where M = molarity and V = volume • ∆E = q + w; where ∆E = change in energy (Joules, J), q = heat (J), and w = work (J) • How to assign signs to ∆E, q, w, ∆H • ∆H of a reverse reaction has the opposite sign than the forward reaction • If multiply a reaction by a value (such as 2), multiply ∆H by the same value • q = mc∆T, where q = heat (J), c = specific heat capacity (J/g*°C), ∆T = change in temperature = Tfinal – Tinitial (ºC) • ∆Hfº of an element = 0, where ∆Hfº = enthalpy of formation of a substance • ∆Hrxnº =Σn ∆Hfº products – Σn ∆Hfº reactants; where ∆Hrxnº = enthalpy change of a reaction under standard conditions; n = number of moles, and ∆Hfº = enthalpy of formation of a substance New for Test 3: • Quantum numbers (n, l, ml, and ms) and how to assign acceptable values for them
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Periodic trends for size, ionization energy, electron affinity, and electronegativity, including isoelectronic species How to draw Lewis symbols for atoms and Lewis Structures for molecules Formal charge = # valence electrons – bonds – non-bonding electrons How to rank strengths of ionic and covalent bonds and lengths of covalent bonds Bond and molecule polarity How to assign electron domain geometry, molecular geometry, and bond angles Determine hybrid orbitals, and sigma and pi bonds STP = Standard Temperature (273.15 K) and Pressure (1.0000 atm) PV = nRT, where P = pressure (atm); V = volume (L); n = moles (mol); R = constant L*atm that will be given to you ÷ ; T = temperature (K) mol*K P1V1 P2 V2 = , where P = Pressure (atm); V = volume (L); n = moles (mol); T = n1T1 n 2 T2 temperature (K) dRT g , where μ = molar mass ÷; d = density P mol L*atm that will be given to you ÷ ; T = temperature mol*K Volume (L) μ=
g mass = ÷ ; R = constant volume L (K); P = Pressure (atm); V =
μP g g , where d = density ÷; μ = molar mass ÷; P = Pressure (atm); R = RT L mol L*atm constant that will be given to you ÷ ; T = temperature (K) mol*K d=
•
PT = P1 + P2 + P3 + … + Px, where P = pressure (atm)
•
PA =χ A PT , where P = pressure (atm); and χ A =
# moles gas A (unitless) total # moles gas
Problems: 1. What are the correct 4 quantum numbers for an outermost electron in Ca? n = 4, l = 0, ml = 0, ms = + ½ n = 4 since the outermost shell of Ca is in period 4 l = 0 since the outermost electrons are in the s area of the periodic table ml = 0 since there is only one orientation of the s orbital ms = + ½; this could also equal – ½, both answers would be correct
2. Which of the following sets of quantum numbers cannot exist, and why? a) n = 4, l = 4, ml = -3, ms = + ½ b) n = 5, l = 2, ml = +2, ms = + ½ c) n = 2, l = 1, ml = 0, ms = + ½ d) n = 1, l = 0, ml = 0, ms = + ½ Answer A cannot exist because the l value is too high. The l value should be a whole number somewhere between 0 and n-1. The l value does not meet this criterion. 3. State which of the following corresponds to a shell, subshell, orientation, or electron spin. a) l = 2, b) ms = + ½, c) n = 2, d) ml = -2 a) b) c) d)
corresponds to a subshell corresponds to an electron spin corresponds to a shell corresponds to the orientation of an orbital
4. Which of the following atoms is largest and why? a) Ca b) K c) Ar d) Cl e) S Both Ca and K have an n level = 4, while Ar, Cl, and S have an n level = 3. Since Ca and K have the higher n levels, one of those two must be the largest. Next, we look at number of protons. Since Ca has a higher number of protons (20 versus 19 for K), the protons will hold the electrons tighter, and thus Ca will be smaller than K. So K is the largest. 5. Which of the following atoms/ions is largest and why? a) O2b) F- c) Na+ d) Mg2+ This is an isoelectronic series (they all have the same number of electrons). Since the series is isoelectronic, positive ions are smaller than neutral atoms, and neutral atoms are smaller than negative ions. So, O2- is the biggest, and Mg2+ is the smallest. 6. Define ionization energy and state which of the following has the highest ionization energy. a) B b) C c) N d) O e) F Ionization energy is the energy required to remove an electron from an atom. The smaller an atom, the harder it is to remove an electron because the attraction between the positive nucleus and the negative electrons is strong. Since F is the smallest of the atoms, it should have the highest ionization energy – it will be most difficult to remove an electron from F.
7. Define electron affinity and state which of the following has the largest electron affinity. a) Al b) Si c) P d) S e) Cl Electron affinity is the ability of an atom to attract an electron. This is also based on size. The smaller an atom, the easier it is to attract an electron because there is a high attraction between the positive nucleus and the negative electron. Cl is the smallest of these atoms, and it should have the highest electron affinity – it will be easiest to give an electron to Cl. 8. Which of the following would have the highest second ionization energy? a) Na b) Mg c) Al d) Si e) P The second ionization energy is the amount of energy required to remove another electron from a positive ion. These energies are usually quite high since the attraction between positive and negative is strong in cations. However, if removing a second electron will ruin a noble gas configuration, the energy for this is very high, since full orbitals (and half-full orbitals) are particularly stable. So Na will have the highest second ionization energy since removing a second electron will destroy a full orbital. 9. Draw the Lewis symbol for P. P has 5 valence electrons.
10. Draw the Lewis symbol for S2-. S2- has 8 valence electrons. S 11. Draw the Lewis structure for CCl4. Valence electrons = 1(C) + 4 (Cl) = 1(4) + 4(7) = 32 valence electrons. Cl Cl
C Cl
Cl
12. Draw the Lewis structure for C2Cl2. Valence electrons = 2(C) + 2(Cl) = 2(4) + 2(7) = 22 valence electrons. Cl
C
C
Cl
13. State the molecular geometry (shape) and the bond angles around each of the atoms with an arrow in the following structure. H H
C
tetrahedral H 109.5o
C
N
linear, 180o
14. State the electron domain geometry (geometry) and bond angles around each of the atoms with an arrow in the following structure.
15. State the hybridization of each atom with an arrow in the following structure. sp3 sp2
O H
C
N H
sp2
H
16. Calculate the formal charge on each atom in the following structure.
H H
H
H
O
N
C
C
H
C
H
O
H
Formal charge = # valence electrons – bonds – non-bonding electrons All H’s = 1 – 1 – 0 = 0 N = 5 – 4 – 0 = +1 All C’s = 4 – 4 – 0 = 0 Double-bonded O = 6 – 2 – 4 = 0 Single-bonded O = 6 – 1 – 6 = -1 17. Which of the following has the highest lattice energy? a) MgO b) Na2O c) Li2O d) CaO e) K2O First, we look at the charges on the elements. Those ions with the highest charges will have the highest lattice energy. So, MgO and CaO both have +2 cations and -2 anions, while Na2O, Li2O, and K2O all have +1 cations and -2 anions. So, MgO and CaO will have the highest lattice energy out of the set of 5. To determine which of these two has the highest lattice energy, we look at size. Smaller things get closer together and thus have higher lattice energies. O is the same size in both species, so we will compare Mg and Ca. Mg is smaller than Ca, so MgO will have the highest lattice energy. 18. Which of the following has the strongest C to N bond? Which has the shortest C to N bond? H H
a)
C
N
H
H
H
b)
H
H
H
C
N
c) H
C
N
The compound in A will have the longest and weakest C to N bond because it is a single bond. The compound in C will have the strongest and shortest C to N bond because it is a triple bond.
19. State whether the following molecules are polar or non-polar. If polar, draw an arrow showing the direction of the dipole. a) HF b) F2 c) NH3 d) Cl2 e) HCl
H
F
HF is polar:
F2 is non-polar:
H
N
H
F
H
H
NH3 is polar:
F
Cl2 is non-polar:
Cl
Cl
Cl
HCl is polar: 20. A balloon is filled with helium at sea level, where the pressure is 743 torr and the temperature is 28 °C. The volume of the balloon is 1.43 L. What will be the volume of the balloon if a child accidentally releases it, and it reaches a point in the atmosphere where the temperature is 16 °C and the pressure is 487 torr? Useful Info: 760 torr = 1 atm. P1 =743 torr
1 atm =0.978 atm 760 torr
T1 = 28 + 273.15 = 301 K
P2 =487 torr
1 atm =0.641 atm 760 torr
T2 = 16 + 273.15 = 289 K
P1V1 P2 V2 = T1 T2
V2 =
P1V1T2 ( 0.978 atm ) ( 1.43 L ) ( 289 K ) = =2.09 L P2T1 ( 0.641 atm ) ( 301 K )
21. A 54.3 g sample of UF6 occupies 18.2 L at 97.3 °C. What is the pressure exerted by the gas? Useful Info: R = 0.0821(L*atm)/(mol*K) mol UF6 =54.3 g UF6
1 mol UF6 =0.154 mol UF6 352.03 g UF6
T = 97.3 + 273.15 = 370.5 K PV = nRT
P=
nRT = V
( 0.154 mol ) 0.0821
L*atm ÷( 370.5 K ) mol*K =0.257 atm 18.2 L
22. a) How many moles of gaseous arsine, AsH3, will occupy 0.0400 L at STP? b) What is the density of gaseous arsine? Useful Info: R = 0.0821 (L*atm)/(mol*K)
a) PV = nRT
n=
( 1.000 atm ) ( 0.0400 L ) =0.00178 mol PV = L*atm RT 0.0821 ÷( 273.15 K ) mol*K
g 77.94 ÷( 1.000 atm ) μP g mol = =3.46 b) d= L*atm RT L 0.0821 ÷( 273.15 K ) mol*K 23. How many grams of C5H12 are required to react with 12.5 L of O2 at 25 °C and 13.5 atm? Useful Info: R = 0.0821 (L*atm)/(mol*K) C5H12 (l) + 8 O2 (g) 5 CO2 (g) + 6 H2O (g) T = 25 °C + 273.15 = 298 K
PV = nRT
n=
( 13.5 atm ) ( 12.5 L ) =6.90 mol O PV = 2 L*atm RT 0.0821 298 K ( ) ÷ mol*K
mass (g) C5 H12 = 6.90 mol O 2
1 mol C5 H12 72.17 g C5 H12 = 62.2 g C5 H12 8 mol O 2 1 mol C5 H12
24. How many liters of O2 will be produced at 25 °C and 0.980 atm if 25.6 g of C 5H12 react? Useful Info: R = 0.0821 (L*atm)/(mol*K) C5H12 (l) + 8 O2 (g) 5 CO2 (g) + 6 H2O (g) T = 25 °C + 273.15 = 298 K mol O 2 = 25.6 g C5 H12
PV = nRT
V=
1 mol C5 H12 8 mol O 2 = 2.84 mol O 2 72.17 g C5 H12 1 mol C5 H12
nRT = P
( 2.84 mol ) 0.0821
L*atm ÷( 298 K ) mol*K =70.9 L 0.980 atm
25. A sample of 32.5 g of a noble gas occupies 301 mL at 27 °C and 66.5 atm. What is its identity? Useful Info: R = 0.0821 (L*atm)/(mol*K) T = 27 °C + 273.15 = 300. K
μ=
dRT = P
( 32.5 g ) 0.0821
V=301 mL
1L = 0.301 L 1000 mL
L*atm ÷( 300. K ) g mol*K = 39.9 mol ( 0.301 L ) ( 66.5 atm )
The noble gas is argon (Ar).
26. A sample of 35.2 g O2, 14.8 g Ar, and 15.3 g CO2 has a pressure of 45.8 atm. What is the partial pressure of each gas? mol O 2 = 35.2 g O 2
1 mol O 2 =1.10 mol O 2 32.00 g O 2
mol Ar = 14.8 g Ar
1 mol Ar = 0.370 mol Ar 39.95 g Ar
mol CO2 = 15.3 g CO 2
1 mol CO 2 = 0.348 mol CO 2 44.01 g CO 2
MolTotal = mol O2+mol Ar + mol CO2 = 1.10 mol + 0.370 mol + 0.348 mol = 1.82 mol 1.10 mol PO2 =χ O2 PT = ÷45.8 atm = 0.604* ( 45.8 atm ) = 27.7 atm 1.82 mol 0.370 mol PAr =χ Ar PT = ÷45.8 atm = 0.204* ( 45.8 atm ) = 9.31 atm 1.82 mol 0.348 mol PCO2 =χ CO2 PT = ÷45.8 atm = 0.191* ( 45.8 atm ) =8.76 atm 1.82 mol We can double-check our calculations by using the other version of Dalton’s Law: PT =PO2 +PAr +PCO2 =27.7 atm + 9.31 atm + 8.76 atm = 45.8 atm Since this number matches our total pressure, we have done the calculations correctly!
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Periodic trends for size, ionization energy, electron affinity, and electronegativity, including isoelectronic species How to draw Lewis symbols for atoms and Lewis Structures for molecules Formal charge = # valence electrons – bonds – non-bonding electrons How to rank strengths of ionic and covalent bonds and lengths of covalent bonds Bond and molecule polarity How to assign electron domain geometry, molecular geometry, and bond angles Determine hybrid orbitals, and sigma and pi bonds STP = Standard Temperature (273.15 K) and Pressure (1.0000 atm) PV = nRT, where P = pressure (atm); V = volume (L); n = moles (mol); R = constant L*atm that will be given to you ÷ ; T = temperature (K) mol*K P1V1 P2 V2 = , where P = Pressure (atm); V = volume (L); n = moles (mol); T = n1T1 n 2 T2 temperature (K) dRT g , where μ = molar mass ÷; d = density P mol L*atm that will be given to you ÷ ; T = temperature mol*K Volume (L) μ=
g mass = ÷ ; R = constant volume L (K); P = Pressure (atm); V =
μP g g , where d = density ÷; μ = molar mass ÷; P = Pressure (atm); R = RT L mol L*atm constant that will be given to you ÷ ; T = temperature (K) mol*K d=
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PT = P1 + P2 + P3 + … + Px, where P = pressure (atm)
•
PA =χ A PT , where P = pressure (atm); and χ A =
# moles gas A (unitless) total # moles gas
Problems: 1. What are the correct 4 quantum numbers for an outermost electron in Ca? n = 4, l = 0, ml = 0, ms = + ½ n = 4 since the outermost shell of Ca is in period 4 l = 0 since the outermost electrons are in the s area of the periodic table ml = 0 since there is only one orientation of the s orbital ms = + ½; this could also equal – ½, both answers would be correct
2. Which of the following sets of quantum numbers cannot exist, and why? a) n = 4, l = 4, ml = -3, ms = + ½ b) n = 5, l = 2, ml = +2, ms = + ½ c) n = 2, l = 1, ml = 0, ms = + ½ d) n = 1, l = 0, ml = 0, ms = + ½ Answer A cannot exist because the l value is too high. The l value should be a whole number somewhere between 0 and n-1. The l value does not meet this criterion. 3. State which of the following corresponds to a shell, subshell, orientation, or electron spin. a) l = 2, b) ms = + ½, c) n = 2, d) ml = -2 a) b) c) d)
corresponds to a subshell corresponds to an electron spin corresponds to a shell corresponds to the orientation of an orbital
4. Which of the following atoms is largest and why? a) Ca b) K c) Ar d) Cl e) S Both Ca and K have an n level = 4, while Ar, Cl, and S have an n level = 3. Since Ca and K have the higher n levels, one of those two must be the largest. Next, we look at number of protons. Since Ca has a higher number of protons (20 versus 19 for K), the protons will hold the electrons tighter, and thus Ca will be smaller than K. So K is the largest. 5. Which of the following atoms/ions is largest and why? a) O2b) F- c) Na+ d) Mg2+ This is an isoelectronic series (they all have the same number of electrons). Since the series is isoelectronic, positive ions are smaller than neutral atoms, and neutral atoms are smaller than negative ions. So, O2- is the biggest, and Mg2+ is the smallest. 6. Define ionization energy and state which of the following has the highest ionization energy. a) B b) C c) N d) O e) F Ionization energy is the energy required to remove an electron from an atom. The smaller an atom, the harder it is to remove an electron because the attraction between the positive nucleus and the negative electrons is strong. Since F is the smallest of the atoms, it should have the highest ionization energy – it will be most difficult to remove an electron from F.
7. Define electron affinity and state which of the following has the largest electron affinity. a) Al b) Si c) P d) S e) Cl Electron affinity is the ability of an atom to attract an electron. This is also based on size. The smaller an atom, the easier it is to attract an electron because there is a high attraction between the positive nucleus and the negative electron. Cl is the smallest of these atoms, and it should have the highest electron affinity – it will be easiest to give an electron to Cl. 8. Which of the following would have the highest second ionization energy? a) Na b) Mg c) Al d) Si e) P The second ionization energy is the amount of energy required to remove another electron from a positive ion. These energies are usually quite high since the attraction between positive and negative is strong in cations. However, if removing a second electron will ruin a noble gas configuration, the energy for this is very high, since full orbitals (and half-full orbitals) are particularly stable. So Na will have the highest second ionization energy since removing a second electron will destroy a full orbital. 9. Draw the Lewis symbol for P. P has 5 valence electrons.
10. Draw the Lewis symbol for S2-. S2- has 8 valence electrons. S 11. Draw the Lewis structure for CCl4. Valence electrons = 1(C) + 4 (Cl) = 1(4) + 4(7) = 32 valence electrons. Cl Cl
C Cl
Cl
12. Draw the Lewis structure for C2Cl2. Valence electrons = 2(C) + 2(Cl) = 2(4) + 2(7) = 22 valence electrons. Cl
C
C
Cl
13. State the molecular geometry (shape) and the bond angles around each of the atoms with an arrow in the following structure. H H
C
tetrahedral H 109.5o
C
N
linear, 180o
14. State the electron domain geometry (geometry) and bond angles around each of the atoms with an arrow in the following structure.
15. State the hybridization of each atom with an arrow in the following structure. sp3 sp2
O H
C
N H
sp2
H
16. Calculate the formal charge on each atom in the following structure.
H H
H
H
O
N
C
C
H
C
H
O
H
Formal charge = # valence electrons – bonds – non-bonding electrons All H’s = 1 – 1 – 0 = 0 N = 5 – 4 – 0 = +1 All C’s = 4 – 4 – 0 = 0 Double-bonded O = 6 – 2 – 4 = 0 Single-bonded O = 6 – 1 – 6 = -1 17. Which of the following has the highest lattice energy? a) MgO b) Na2O c) Li2O d) CaO e) K2O First, we look at the charges on the elements. Those ions with the highest charges will have the highest lattice energy. So, MgO and CaO both have +2 cations and -2 anions, while Na2O, Li2O, and K2O all have +1 cations and -2 anions. So, MgO and CaO will have the highest lattice energy out of the set of 5. To determine which of these two has the highest lattice energy, we look at size. Smaller things get closer together and thus have higher lattice energies. O is the same size in both species, so we will compare Mg and Ca. Mg is smaller than Ca, so MgO will have the highest lattice energy. 18. Which of the following has the strongest C to N bond? Which has the shortest C to N bond? H H
a)
C
N
H
H
H
b)
H
H
H
C
N
c) H
C
N
The compound in A will have the longest and weakest C to N bond because it is a single bond. The compound in C will have the strongest and shortest C to N bond because it is a triple bond.
19. State whether the following molecules are polar or non-polar. If polar, draw an arrow showing the direction of the dipole. a) HF b) F2 c) NH3 d) Cl2 e) HCl
H
F
HF is polar:
F2 is non-polar:
H
N
H
F
H
H
NH3 is polar:
F
Cl2 is non-polar:
Cl
Cl
Cl
HCl is polar: 20. A balloon is filled with helium at sea level, where the pressure is 743 torr and the temperature is 28 °C. The volume of the balloon is 1.43 L. What will be the volume of the balloon if a child accidentally releases it, and it reaches a point in the atmosphere where the temperature is 16 °C and the pressure is 487 torr? Useful Info: 760 torr = 1 atm. P1 =743 torr
1 atm =0.978 atm 760 torr
T1 = 28 + 273.15 = 301 K
P2 =487 torr
1 atm =0.641 atm 760 torr
T2 = 16 + 273.15 = 289 K
P1V1 P2 V2 = T1 T2
V2 =
P1V1T2 ( 0.978 atm ) ( 1.43 L ) ( 289 K ) = =2.09 L P2T1 ( 0.641 atm ) ( 301 K )
21. A 54.3 g sample of UF6 occupies 18.2 L at 97.3 °C. What is the pressure exerted by the gas? Useful Info: R = 0.0821(L*atm)/(mol*K) mol UF6 =54.3 g UF6
1 mol UF6 =0.154 mol UF6 352.03 g UF6
T = 97.3 + 273.15 = 370.5 K PV = nRT
P=
nRT = V
( 0.154 mol ) 0.0821
L*atm ÷( 370.5 K ) mol*K =0.257 atm 18.2 L
22. a) How many moles of gaseous arsine, AsH3, will occupy 0.0400 L at STP? b) What is the density of gaseous arsine? Useful Info: R = 0.0821 (L*atm)/(mol*K)
a) PV = nRT
n=
( 1.000 atm ) ( 0.0400 L ) =0.00178 mol PV = L*atm RT 0.0821 ÷( 273.15 K ) mol*K
g 77.94 ÷( 1.000 atm ) μP g mol = =3.46 b) d= L*atm RT L 0.0821 ÷( 273.15 K ) mol*K 23. How many grams of C5H12 are required to react with 12.5 L of O2 at 25 °C and 13.5 atm? Useful Info: R = 0.0821 (L*atm)/(mol*K) C5H12 (l) + 8 O2 (g) 5 CO2 (g) + 6 H2O (g) T = 25 °C + 273.15 = 298 K
PV = nRT
n=
( 13.5 atm ) ( 12.5 L ) =6.90 mol O PV = 2 L*atm RT 0.0821 298 K ( ) ÷ mol*K
mass (g) C5 H12 = 6.90 mol O 2
1 mol C5 H12 72.17 g C5 H12 = 62.2 g C5 H12 8 mol O 2 1 mol C5 H12
24. How many liters of O2 will be produced at 25 °C and 0.980 atm if 25.6 g of C 5H12 react? Useful Info: R = 0.0821 (L*atm)/(mol*K) C5H12 (l) + 8 O2 (g) 5 CO2 (g) + 6 H2O (g) T = 25 °C + 273.15 = 298 K mol O 2 = 25.6 g C5 H12
PV = nRT
V=
1 mol C5 H12 8 mol O 2 = 2.84 mol O 2 72.17 g C5 H12 1 mol C5 H12
nRT = P
( 2.84 mol ) 0.0821
L*atm ÷( 298 K ) mol*K =70.9 L 0.980 atm
25. A sample of 32.5 g of a noble gas occupies 301 mL at 27 °C and 66.5 atm. What is its identity? Useful Info: R = 0.0821 (L*atm)/(mol*K) T = 27 °C + 273.15 = 300. K
μ=
dRT = P
( 32.5 g ) 0.0821
V=301 mL
1L = 0.301 L 1000 mL
L*atm ÷( 300. K ) g mol*K = 39.9 mol ( 0.301 L ) ( 66.5 atm )
The noble gas is argon (Ar).
26. A sample of 35.2 g O2, 14.8 g Ar, and 15.3 g CO2 has a pressure of 45.8 atm. What is the partial pressure of each gas? mol O 2 = 35.2 g O 2
1 mol O 2 =1.10 mol O 2 32.00 g O 2
mol Ar = 14.8 g Ar
1 mol Ar = 0.370 mol Ar 39.95 g Ar
mol CO2 = 15.3 g CO 2
1 mol CO 2 = 0.348 mol CO 2 44.01 g CO 2
MolTotal = mol O2+mol Ar + mol CO2 = 1.10 mol + 0.370 mol + 0.348 mol = 1.82 mol 1.10 mol PO2 =χ O2 PT = ÷45.8 atm = 0.604* ( 45.8 atm ) = 27.7 atm 1.82 mol 0.370 mol PAr =χ Ar PT = ÷45.8 atm = 0.204* ( 45.8 atm ) = 9.31 atm 1.82 mol 0.348 mol PCO2 =χ CO2 PT = ÷45.8 atm = 0.191* ( 45.8 atm ) =8.76 atm 1.82 mol We can double-check our calculations by using the other version of Dalton’s Law: PT =PO2 +PAr +PCO2 =27.7 atm + 9.31 atm + 8.76 atm = 45.8 atm Since this number matches our total pressure, we have done the calculations correctly!
