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Chapter 3 Homework Solutions
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- a) C 2 H 5 NO 2
b) MW = (2 atomsC) C
C atom
- 0107 amu
H atom
- 00794 amu
(1 atomN) N
N atom
- 0067 amu
O atom
- 9994 amu = 75.0666 amu
c) %N = ( 100 %)
- 0666 amu
atom
- 0067 amu ( 1 atom ) N
N N = 18.66 %
- a) 6, 1, 2 c) 2, 2, 1, 4 e) 3, 2, 1, 6 g) 4, 9, 4, 10, 2 b) 1, 3, 2 d) 1, 6, 3, 2 f) 2, 1, 1, 2
- a) SO3 (g) + H 2 O() H 2 SO 4 (aq)
b) B 2 S 3 (s) + 6 H 2 O() 2 H 3 BO 3 (aq) + 3 H 2 S(g) c) Pb(NO 3 ) 2 (aq) + 2 NaI PbI2 (s) + 2 NaNO 3 (aq) d) 2 Hg(NO 3 ) 2 (s) 2 HgO(s) + 4 NO 2 (g) + O 2 (g) e) Cu(s) + 2 H 2 SO 4 (aq) CuSO 4 (aq) + SO 2 (g) + 2 H 2 O()
- a) 2 Ca(s) + O 2 (g) 2 CaO(s)
b) C 3 H 6 O() + 4 O 2 (g) 3 CO 2 (g) + 3 H 2 O()
- a) 4 Al(s) + 3 O 2 (g) 2 Al 2 O 3 (s)
b) Cu(OH) 2 (s) CuO(s) + H 2 O(g) c) C 7 H16 () + 11 O 2 (g) 7 CO 2 (g) + 8 H 2 O() d) 2 C 5 H 12 O() + 15 O 2 (g) 10 CO 2 (g) + 12 H 2 O()
- a) 2, 9, 6, 6; combustion d) 1, 3, 2; combination b) 1, 1, 2; decomposition e) 1, 1, 2; combination c) 1, 6, 5, 3; combustion
- a) MWN2O = (2 atomsN) N
N atom
- 0067 amu
O atom
- 9994 amu = 44.0128 amu
b) MWBA = (7 atomsC) C
C atom
- 0107 amu
1 00794. amu atom
H H
(2 atomsO)
15 9994. amu atom
O O
= 122.1213 amu
c) MWMH = (1 atomsMg) Mg
Mg atom
2 4.3050amu
15 9994. amu atom
O O
(2 atomsH)
1 00794. amu atom
H H
= 58.3197 amu
d) MWurea = (1 atomO O
O atom
1 5.9994amu
C atom
1 2.0107amu
(2 atomsN) 14 0067. amu atom
N N
- (4 atomsH) 1 00794. amu atom
H H
= 60.0553 amu
e) MWIA = (7 atomsC) C
C atom
- 0107 amu
- (14 atomsH) 1 00794. amu atom
H H
(2 atomsO) 15 9994. amu atom
O O
= 130.1849 amu
- To save space on this problem, I am omitting the calculation of the molecular weights.
a) %C = acetylene
C
c C
26.0373 g
mol
- 0107 g 2 mol x 100% = 92.26%C [remember only report to the hundredths
place]
b) %H = ascorbicacid
H
H H
176.1241 g
mol
- 00794 g 8 mol x 100% = 4.58%H
c) %H = amm.sulf.
H
H H
132.1395 g
mol
- 00794 g 8 mol x 100% = 6.10%H
d) %Pt = cis-platin
Pt
Pt Pt
300.045 g
mol
- 078 g 1 mol x 100% = 65.02%Pt
e) %O = estradiol
O
O O
272.3820 g
mol
- 9994 g 2 mol x 100% = 11.75%O
f) %C = capsaiscin
C
C C
305.4119 g
mol
- 0107 g 18 mol x 100% = 70.79%C
Then set up the ratio: EF
cadaverine 102.1781 g/mol
102.2g/mol
EW
MW = 1 molC5H14N2 per mol(C5H14N2)n
Thus the molecular formula is C 5 H 14 N 2. c) empirical formula = C 9 H 13 O 3 N
EWepinephrine = (9 molC) C
C mol
- 0107 g
H mol
- 00794 g
O mol
1 5.9994g
N mol
- 0067 g = 183.2044 g/mol
Then set up the ratio: EF
epinephrine 183.2044 g/mol
180 g/mol
EW
MW
= 1 molC9H13O3N per mol(C9H13O3N)n
Thus the molecular formula is C 9 H 13 O 3 N.
- This problem requires several steps to complete. First the masses of C, H, and O must be determined. This must be done indirectly.
a) massC = (0.00632 gCO2) C
C CO
C CO
CO 1 mol
- 0107 g 1 mol
1 mol
- 0095 g
1 mol = 0.00172 gC
massH = (0.00258 gH2O) H
H H2O
H H2O
H2O 1 mol
1 .00794g 1 mol
2 mol 1 8.0153g
1 mol = 0.000284 gH
massO = 0.00278 g - 0.00172 g - 0.000284 g = 0.00078 gO Then the moles of each substance must be calculated. For C & H you can use the masses you just calculated, but it is better to do it as follows:
molC = (0.00632 gCO2) CO
C CO
CO 1 mol
1 mol
- 0095 g
1 mol = 1.44 x 10-^4 molC
molH = (0.00258 gH2O)
1 18.0153 g
2 1
mol mol mol
H2O H2O
H H2O
= 2.86 x 10-^4 molH
molO = (0.00078 gO)
1 mol 15.9994 g
O O
= 4.8 x 10-^5 molO
Now set up the molar ratios.
O
C
4.8x 10 mol
1.44x 10 mol = 3.0 molC per molO O
H
4.8x 10 mol
2.86x 10 mol = 5.96 molH per molO
This yields an empirical formula of C 3 H 6 O. b) By a similar series of steps one finds for nicotine: massC = 3.8868 mg massH = 0.4569 mg massN =0.906 mg molC = 0.3236 mol molH = 0.4533 mol molN = 0.0647 mol 5.00 molC/molN 7.00 molH/molN Empirical formula = C 5 H 7 N. The molecular weight provided is incorrect. It should be 160 ± 5, yielding a molecular
formula of C 10 H 14 N 2.
- This problem is solved in a manner that is similar to 46, 50, and 52 except that we can treat the magnesium sulfate and water as units without subdividing them into atoms. masswater = 5.061 g – 2.472 g = 2.589 g
molMgSO4 = (2.472 gMgSO4) MgSO
MgSO
- 3676 g
1 mol = 0.02054 molMgSO
molH2O = (2.589 gH2O) H2O
H2O
- 0153 g
1 mol = 0.1437 molH2O
MgSO
H2O
02054 g
1437 mol = 6.997 molH2O per molMgSO4 the formula is MgSO 4 •7H 2 O
a) molCO2 = (0.400 molC6H12O6) C6H12O
CO 1 mol
2 mol = 0.800 molCO
b) massC6H12O6 = (7.50 gC2H5OH) C6H12O
C6H12O C2H5OH
C6H12O C2H5OH
C2H5OH mol
- 1559 g 2 mol
1 mol 46.0684g
1 mol
= 14.7 gC6H12O
c) massCO2 = (7.50 gC2H5OH) CO
CO C2H5OH
CO C2H5OH
C2H5OH mol
4 4.0095g 2 mol
2 mol 46.0684g
1 mol = 7.16 gCO
- a) Fe 2 O 3 (s) + 3 CO(g) 2 Fe(s) + 3 CO 2 (g)
b) massCO = (0.150 kgFe2O3) CO
CO Fe2O
CO Fe2O
Fe2O Fe2O
Fe2O 1 mol
28.0101g 1 mol
3 mol
- 688 g
1 mol 1 kg
1000 g
= 78.9 gCO c) As in (b), massFe = 105 gFe; massCO2 = 124 gCO d) massreactants = 150 g + 78.9 g = 229 g massproducts = 105 g + 124 g = 229 g massreactants = massproducts
- a) To work this problem, assume you use up all of one of the reagents, then calculate how much of the other would be required to accomplish this.
molAl(OH)3 (required) = (0.500 molH2SO4) H2SO
Al(OH) 3 mol
2 mol = 0.333 molAl(OH)
This means that if all of the sulfuric acid were consumed, then 0.333 mol of aluminum hydroxide would be consumed, but you have 0.500 moles of aluminum hydroxide, which is more than this. Thus, if all of the sulfuric acid were used up, you’d have some aluminum hydroxide left over, hence the sulfuric acid is the limiting reagent.
b) molAl2(SO4)3 = (0.500 molH2SO4) H2SO
Al2(SO4) 3 mol
1 mol = 0.167 molAl2(SO4)
massHC2H3O2 (formed) = (7.50 gPb(OAc)2) HC2H3O
HC2H3O Pb(C2H3O2) 2
HC2H3O Pb(C2H3O2) 2
Pb(C2H3O2) 2 1 mol
6 0.0520g 1 mol
2 mol 325.3g
1 mol
= 2.77 gHC2H3O
massPbSO4 (formed) = (7.50 gPb(OAc)2) PbSO
PbSO Pb(C2H3O2) 2
PbSO Pb(C2H3O2) 2
Pb(C2H3O2) 2 1 mol
3 03.3g 1 mol
1 mol 325.3g
1 mol
= 6.99 gPbSO
- If you can’t do this one, it’s ok, but this is an interesting problem. The key is to figure out how to begin. You are given the percent bromine in the sample. If you assume 100 g of compound, this gives you the mass of bromine (52.92 g). From this you can determine the number of moles of bromine. The formula tells you the molar ratio of potassium to bromine (1:1). Knowing the number of moles of potassium you can calculate the mass of potassium. Knowing the masses of bromine & potassium allows you to calculate the mass of oxygen. At this point, the problem is solved just like all of the other empirical formula problems. The actual math follows below.
molBr = (52.92 gBr) Br
Br
- 904 g
1 mol = 0.6623 molBr
molK = molBr = 0.6623 molK
massK = (0.6623 molK) K
K 1 mol
- 0983 g = 25.89 gK
massO = 100.00 g – 52.92 g – 25.89 g = 21.19 gO
molO = (21.19 gO) O
O 15.9994 g
1 mol = 1.324 molO
molar ratio: Br
O
6623 mol
324 mol = 2.0 molO per molBr formula is KBrO 2
First begin with a balanced equation: 2 C 8 H 18 () + 25 O 2 (g) 18 H 2 O() + 16 CO 2 (g)
Now calculate the volume of gasoline needed: Vgas = (225 mi) 20.5mi
1 gal = 11.0 galgas
From this you can calculate the mass of the gasoline and from that the mass of CO 2 produced.
massCO2 = (11.0 galgas) CO
CO C8H
CO C8H
C8H 1 mol
4401 g 2 mol
16 mol 1142 g
1 mol mL
069 g L
1000 mL 1.057qt
1 L gal
4 qt. .
.
= 88,500 gCO2 (or 88.5 kg)
