PM + Interview - Kidkit

Chapter 3 Example Problems
3.1
A sample of copper is vaporized and injected into a spectrometer, the results are shown below.
Use the data obtained to calculate the average mass of copper. ( The mass for Cu-63 and Cu-62
are 62.94 amu and 64.94 amu.)
63 amu --- 69.10 atoms
65 amu --- 30.90 atoms
(69.10 atoms)(62.94 amu/atoms) + (30.90 atoms)(63.94 amu/atom) = 6355
6355amu/100 atoms = 63.55 amu/atom
Solution
Of every 100 atoms of copper 69.10 are Cu-63 and 30.90 are Cu-65. The mass of 100 atoms of
copper is 63.55 amu/atom.
3.2
Calculate the mass in grams of a sample of Lead containing 10 atoms.
10 atoms x 207.2 amu/atom = 2072 amu
-- 6.022x10^23 = 1 g
-- 1 g/ 6.022x10^23
The mass of 10 atoms of lead is
2072 x
1g
= 3.441 x 10^-21 g
6.022 x 10^23 amu
3.3
Calculate both the number of moles of atoms and the number of atoms in a 15.0 g sample of
Chlorine.
15.0 g Cl x 1 mol Cl = 0.282 mol Cl atoms
35.45 g
The number of atoms in a 15.0 g of Cl is 1.70 x 10^23 atoms
0.282 mol Cl x 6.022 x 10^23 atoms = 1.70 x 10^23 toms
1 mol Cl
3.4
A sample of Phosphorus has a mass of 7.00 mg. How many phosphorus atoms are present in the
sample?
7.00 mg P x 1 g P
= 7.00 x 10 ^-3 g P
1000 mg p
7.00 x 10^-3 g P x 1 mol P
= 2.26 x 106-4 mol P
30.97 g P
2.26 x 10^-4 mol P x 6.022 x 10^23 atoms = 1.36 x 10^20 atoms
1 mol P
3.6
The formula for methane is CH4 (the 4 is a subscript)
a. Calculate the molar mass of methane.
b. How many moles of methane does a 2.00 x 10^-2 g sample of methane contain.
a. 1 C: 1 x 12.01 g = 12.01 g
4 H:
4 x 1.01 g = 4.04 g
Mass of 1 mole CH4 = 16.05 g
The molar mass of methane is 16.05 grams.
b. 2.00 x 10^-2 g methane x 1 mol methane = 1.25 x 10^-3 mol methane
16.05 g
3.7
a. Calculate the mass of sodium carbonate (Na2CO3) (the 2 and 3 are subscripts)
b. A sample of sodium carbonate contains 5.54 moles. What is the mass in grams of the
sample? What is the mass of the CO3 ions present?
a. 2 Na: 2 x 22.99 g = 45.98 g
1 C : 1 x 12.01 g = 12.01 g
3 O:
3 x 16.00 g = 48.00 g
Mass 1 mol Na2CO3 = 105.99 g
The molar mass of 1 mol Na2CO3 is 105.99 g
b. 5.54 mol Na2CO3 x 105.99 g = 587 g Na2CO3
1 mol Na2CO3
1 C: 1 x 12.01 g = 12.01 g
3 O: 3 x 16.00 g = 48.00 g
Mass of 1 mol CO3 = 60.01 g
5.54 mol CO3 x 60.01 g CO3 = 332 g CO3
1 mol CO3
3.8
How many molecules are contained in 2 x 10^-5 g of proponal ( C3H7OH) ( the 3 an d7 are
subscripts). How many atoms of hydrogen are present?
3 mol C x 12.01 g 1 mol = 36.03 g C
8 mol H x 1.01 g 1 mol = 8.08 g H
1 mol O x 16.00 g 1 mol = 16.00 g O
60.01 g
1 mole of propanol has a mass of 60.01 g
2 x 10^-5 g C3H7OH x 1 mol C3H7OH = 3 x 10^-7 mol C3H7OH
60.01 g C3H7OH
3 x 10^-7 mol C3H7OH x 6.022 x 10^23 molecules = 2 x 10^17 molecules
1 mol C3H7OH
2 x 10^17 molecules x 8 H atoms = 2 x 10^14 hydrogen atoms
molecule
3.9
Calculate the mass percent of each element in ethanol ( C2H5OH) ( the 2 and the 5 are
subscripts).
Mass of C in 1 mol = 2 mol x 12.01 g/mol = 24.02 g
Mass of H in 1 mol = 6 mol x 1.01 g/mol = 6.06 g
Mass of O in 1 mol = 1 mol x 16.00 g/mol = 16.00 g
Mass of 1 mol of C2H5OH
Mass percent of C = 24.02 g C
= 46.08 g
x 100% = 52.13% C
46.08 g C2H5OH
Mass percent of H = 6.06 g H
x 100% = 13.15% H
46.08 g C2H5OH
Mass percent of O = 16.00 g O
x 100% = 34.72 % O
46.08 g C2H5OH
3.10
Calculate the mass percent of each element in dichloroethane CL2C2H4 ( both 2’s and the 4 are
subscripts)
C: 2 mol x 12.01 g/mol = 24.02 g
H: 4 mol x 1.01 g/ mol = 4.04 g
Cl: 2 mol x 35.445 g/mol = 70.90 g
Mass 1 mol Cl2C2H4 = 98.96 g
Mass percent of C = 24.02 g x 100% = 24.27 %
98.96 g
Mass percent of H = 4.04 g x 100% = 4.08%
98.96 g
Mass percent of Cl = 70.90 g x 100% = 71.65%
98096 g
3.11
Determine the empirical and molecular formulas for penicillin F that gives the following:
53.81% C 6.45% H 8.97% N 10.27% S 20.49% O
The molar mass is 312.4 g/mol.
53.81 g C x 1 mol C = 4.480 mol C
12.01 g
6.45 g H x 1 mol H = 6.39 mol H
1.01 g
8.97 g N x 1 mol N = 0.640 mol N
14.01 g
10.27 g S x 1 mol S = 0.320 mol S
32.00 g
20.49 g O x 1 mol O = 1.281 mol O
16.00 g
= 312.4 g/mol
molar mass
=2
empirical formula mass 172.24 g/mol
Molecular formula (C14H20N2SO4)= C28H40N4SO4
3.12
A substance was analyzed and found to consists of 71.65%Cl, 24.27%C, and 4.07%H
The compound has a molar mass or 98.96 g/mol.
Find the empirical and molecular formula of the compound.
71.65 g Cl x 1 mol Cl = 2.021 mol Cl
1
35.45 g
24.27 g C x 1 mol C = 2.021 mol C
1
12.01 g C
4.07 g H x 1 mol H = 4.040 mol H
1
1.01 g
98.96 g/mol = (ClCH2) = Cl2C2H4
49.48 g/mol
3.13
A compound contains 51% carbon, 5.3% hydrogen, 28.9% nitrogen, and 14.8% oxygen by
mass. It has a molar mass of 194.2 g/mol. Determine the molecular formula for the compound.
51.0 g C x 194.2 g = 99.0 g C
100.0 g
1 mol
mol compound
5.30 g H x 194.2 g = 10.3 g H
100.0 g
1 mol
mol compound
28.9 g N x 194.2 g = 56.1 g N
100.0 g
1 mol
mol compound
14.8 g O x 194.2 g = 28.7 g O
100.0 g
1 mol
mol compound
Now convert to moles:
Carbon: 9.90 g C x 1 mol = 8.24 mol O
1 mol
12.01 g
Hydrogen: 10.3 g H x 1 mol = 10.2 mol H
1 mol
1.01 g
Nitrogen: 56.1 g N x 1 mol = 4.00 mol N
1 mol
14.01 g
Oxygen: 28.7 g O x 1 mol = 1.79 mol O
1 mol
16.00 g
C8H10N4O2 ( the 8, 10, and 2 are subscripts)
3.14
When hydrogen sulfide is bubbled through a solution of silver nitrate it produces solid silver
sulfide and nitric acid. Write the balanced equation.
Ag2S + 2 HNO3
H2S + 2 AgNO3
Check:
2H
2H
1S
1S
2 Ag
2 Ag
2 NO3
2 NO3
3.15
When Lithium metal is burned in air it yields lithium oxide.
4 Li + O2
2 LiO2
Check:
4 Li
2O
4 Li
2O
3.16
What mass of gaseous carbon dioxide can be absorbed by 3.00 kg of lithium hydroxide? Use the
following balanced equation:
2 LiOH + CO2
Li2CO3 + H2)
3.00 kg LiOH x 10^3 g x 1 mol LiOH = 125 mol LiOH
1
1 kg
23.95 g
125 mol LiOH x 1 mol CO2 = 62.5 mol CO2
1
2 mol LiOH
62.5 mol LiOH x 44.0 g CO2 = 2.75 x 10^3 g CO2
1
1 mol CO2
3.17
Which is a better antacid, NaHCO3 or Mg(OH)2?
Given the equations
NaHCO3 + HCl
Mg(OH)2 + 2HCl
NaCl + H2O + CO2
2 H2O + MgCl2
1.00 g NaHCO3 x 1 mol NaHCO3 = 1.19 x 10^-2 mol NaHCO3
1
84.01 g
1.19 x 10^-2 mol NaHCO3 x 1mol HCl = 1.19 x 10^-2 mol HCl
1 mol NaHCO3
1.00 g Mg(OH)2 x 1 mol Mg(OH)2 = 1.71 x 10^-2 mol Mg(OH02
58.32 g Mg(OH)2
1.71 x 10^-2 mol Mg(OH)2 x 2 mol HCl = 3.42 x 10^-2 mol HCl
1 mol Mg(OH)2
1.00 g of Mg(OH)2 will neutralize 3.42 x 10^-2 mol HCl, so it is the better antacid.
3.18
If a sample containing 20.0 g NH3 is reacted with 92.0 g of CuO, which is the limiting reactant?
How many grams of N2 will be formed?
Use the following equation:
2 NH3 + 3 CuO
N2 + 3 CuO + 3 H2O
20.0 g NH3 x 1 mol NH3 = 1.17 mol NH3
17.03 g
92.0 g CuO x 1 mol CuO = 1.16 mol CuO
79.55 g
1.16 mol NH3 x 3mol CuO = 1.74 mol CuO
2 mol NH3
mol CuO = 3 = 1.5
mol NH3
2
mol CuO = 1.16 = 0.991
mol NH3
1.17
Since the actual ratio is too small ( smaller than 1.5), CuO is the limiting reactant.
3.19
Suppose 700 kg CO is reacted with 10.0 kg H2. Calculate the theoretical yield of methanol. If
4.25 x 10^4 g CH3OH is actually produced, what is the percent yield of methanol.
2 H2 + CO
CH3OH
70.01 g CO x 10^3 g x 1 mol CO = 2.50 x 10^3 mol CO
1
1 kg CO 28.02 g CO
10.0 kg H2 x 10^3 g x 1 mol H2 = 4.96 x 10^3 mol H2
1
1 kg H2 2.016 g H2
mol H2 = 2 = 2
mol CO 1
mol H2 = 4.96 x 10^3 mol H2 = 1.98
mol CO 2.50 x 10^3 mol Co
H2 is the limiting reactant.
4.96 x 10^3 mol H2 x 1 mol CH3OH = 2.48 x 10^-3 mol CH3OH
2 mol H2
2.48 x 10^3 mol CH3OH x 32.04 g CH3OH = 7.95 x 10^4 g CH3OH
1 mol CH3OH
Actual Yield x 100% = 3.57 x 10^4 g CH3OH x 100% = 44.9%
Theoretical Yield
7.95 x 10^4 g CH3OH

Download Report