Chapter 3
Practice Exercises
BU
3.1
⎛ 1 mol Al ⎞
mol Al = 3.47 g Al ⎜
⎟ = 0.129 mol Al
⎝ 26.98 g Al ⎠
3.2
⎛ 1 mol Si ⎞
–5
Uncertainty in moles = ±0.002 g ⎜
⎟ = ±7.12 × 10 mol Si
28.0855
g
Si
⎝
⎠
3.3
Find the mass of 5.64 × 1018 molecules of Ca(NO3)2 (MW = 164.09 g/mo)
⎛
⎞ ⎛ 164.09 g Ca ( NO )
1 mol Ca ( NO3 )2
3 2
⎟⎜
18 ⎜
g = 5.64 × 10 ⎜
23
⎟ ⎜ 1 mol Ca ( NO3 )
2
⎝ 6.022 × 10 molecules Ca ( NO3 )2 ⎠ ⎝
⎞
⎟ = 1.54 × 10–3 g
⎟
⎠
g = 1.54 × 10–3 g = 0.00154 g
Many laboratory balances can measure 1 mg (0.001 g); therefore, it is possible to weigh 5.64 × 1018
molecules of Ca(NO3)2.
3.4
Formula mass of sucrose = (12 C)(12.011 g/mol) + (22 H)(1.0079 g/mol) +
(11 O)(15.9994 g/mol) = 342.299 g/mol
0.002 g of uncertainty = ? mol of sucrose
⎛ 1 mol sucrose ⎞
–6
mol of sucrose = 0.002 g ⎜
⎟ = 5.8 × 10 mol sucrose
342.299
g
⎝
⎠
⎛ 6.022 × 1023 molecules sucrose ⎞
molecules of sucrose = 5.8 × 10–6 mol sucrose ⎜
⎟ =
⎜
⎟
1 mol sucrose
⎝
⎠
3.5× 1018 molecules of sucrose
3.5
Aluminum sulfate: Al2(SO4)3, the aluminum is Al3+
⎛ 2 mol Al3+ ⎞
⎟
3+
3+
2– ⎜
mole Al = 0.0774 mol SO4 ⎜
2 − ⎟ = 0.0516 mol Al
3
mol
SO
4
⎝
⎠
3.6
⎛ 2 mol N ⎞
mol N = ( 8.60 mol O ) ⎜
⎟ = 3.44 mol N atoms
⎝ 5 mol O ⎠
3.7
⎛ 1 mol O ⎞ ⎛ 2 mol Fe ⎞ ⎛ 55.85 g Fe ⎞
g Fe = ( 25.6 g O ) ⎜
⎟⎜
⎟⎜
⎟ = 59.6 g Fe
⎝ 16.00 g O ⎠ ⎝ 3 mol O ⎠ ⎝ 1 mol Fe ⎠
3.8
⎛ 1 mol Fe2 O3 ⎞ ⎛ 2 mol Fe ⎞ ⎛ 55.85 g Fe ⎞
g Fe = (15.0 g Fe2 O3 ) ⎜
⎟⎜
⎟⎜
⎟ = 10.5 g Fe
⎝ 159.7 g Fe 2 O3 ⎠ ⎝ 1 mol Fe 2 O3 ⎠ ⎝ 1 mol Fe ⎠
3.9
⎛ 1 mol O ⎞ ⎛ 1 mol Fe2 O3 ⎞ ⎛ 2 mol Fe ⎞ ⎛ 55.85 g Fe ⎞
g Fe = (12.0 g O) ⎜
⎟⎜
⎟⎜
⎟⎜
⎟ = 27.9 g Fe
⎝ 16.00 g O ⎠ ⎝ 3 mol O ⎠ ⎝ 1 mol Fe2 O3 ⎠ ⎝ 1 mol Fe ⎠
31
Chapter 3
3.10
⎛ 0.0870 g H ⎞
⎛ mass H ⎞
% H =⎜
⎟ × 100% = ⎜ 0.6672 g total ⎟ × 100% = 13.04%
total
mass
⎝
⎠
⎝
⎠
⎛ 0.3481 g C ⎞
⎛ mass C ⎞
% C =⎜
⎟ × 100% = ⎜ 0.6672 g total ⎟ × 100% = 52.17%
total
mass
⎝
⎠
⎝
⎠
It is likely that the compound contains another element since the percentages do not add up to 100%.
3.11
% N = 0.2012/0.5462 × 100% = 36.84% N
% O = 0.3450/0.5462 × 100% = 63.16% O
Since these two values constitute 100%, there are no other elements present.
3.12
We first determine the number of grams of each element that are present in one mol of sample:
2 mol N × 14.01 g/mol = 28.02 g N
4 mol O × 16.00 g/mol = 64.00 g O
The percentages by mass are then obtained using the formula mass of the compound (92.02 g):
% N = (28.02/92.02) × 100% = 30.45% N
% O = (64.00/92.02) × 100% = 69.55% O
3.13
N2O:
NO:
NO2:
N2O3:
N2O4:
N2O5:
Formula mass = 44.02 g/mol
2 mol N × 14.01 g/mol = 28.02 g N
1 mol O × 16.00 g/mol = 16.00 g O
Formula mass = 30.01 g/mol
1 mol N × 14.01 g/mol = 14.01 g N
1 mol O × 16.00 g/mol = 16.00 g O
Formula mass = 46.01 g/mol
1 mol N × 14.01 g/mol = 14.01 g N
2 mol O × 16.00 g/mol = 32.00 g O
Formula mass = 76.02 g/mol
2 mol N × 14.01 g/mol = 28.02 g N
3 mol O × 16.00 g/mol = 48.00 g O
Formula mass = 92.02 g/mol
2 mol N × 14.01 g/mol = 28.02 g N
4 mol O × 16.00 g/mol = 64.00 g O
Formula mass = 108.02 g/mol
2 mol N × 14.01 g/mol = 28.02 g N
5 mol O × 16.00 g/mol = 80.00 g O
% N = (28.02/44.02) × 100% = 63.65% N
% O = (16.00/44.02) × 100% = 36.34% O
% N = (14.01/30.01) × 100% = 46.68% N
% O = (16.00/30.01) × 100% = 53.32% O
% N = (14.01/46.01) × 100% = 30.45% N
% O = (32.00/46.01) × 100% = 69.55% O
% N = (28.02/76.02) × 100% = 36.86% N
% O = (48.00/76.02) × 100% = 63.14% O
% N = (28.02/92.02) × 100% = 30.45% N
% O = (64.00/92.02) × 100% = 69.55% O
% N = (28.02/108.02) × 100% = 25.94% N
% O = (80.00/108.02) × 100% = 74.06% O
The compound N2O3 corresponds to the data in Practice Exercise 3.11.
3.14
We first determine the number of mol of each element as follows:
⎛ 1 mol N ⎞
mol N = ( 0.712 g N ) ⎜
⎟ = 0.0508 mol N
⎝ 14.01 g N ⎠
We need to know the number of grams of O. Since there is a total of 1.525 g of compound and the only
other element present is N, the mass of O = 1.525 g – 0.712 g = 0.813 g O.
⎛ 1 mol O ⎞
mol O = ( 0.813 g O ) ⎜
⎟ = 0.0508 mol O
⎝ 16.00 g O ⎠
Since these two mole amounts are the same, the empirical formula is NO.
3.15
First, find the number of moles of each element, then determine the empirical formula by comparing the
ratio of the number of moles of each element.
Start with the number of moles of S:
⎛ 1 mol S ⎞
mol S = 0.7625 g S ⎜
⎟ = 0.02378 mol S
⎝ 32.066 g S ⎠
32
Chapter 3
Then find the number of moles of O: since there are only two elements in the compound, S and O, the
remaining mass is O
g O = 1.525 g compound – 0.7625 g S = 0.7625 g O
⎛ 1 mol O ⎞
mol O = 0.7625 g O ⎜
⎟ = 0.04766 mol O
⎝ 15.9994 g O ⎠
The empirical formula is
S0.02378O0.4766
The empirical formula must be in whole numbers, so divide by the smaller subscript:
S 0.02378 O 0.04766 which becomes SO2
0.02378
3.16
0.02378
⎛ 2000 lb Al ⎞ ⎛ 454 g Al ⎞ ⎛ 1 mol Al ⎞
5
mol Al = 5.68 tons Al ⎜
⎟ = 1.91 × 10 mol Al
⎟⎜
⎟⎜
⎝ 1 ton Al ⎠ ⎝ 1 lb Al ⎠ ⎝ 26.98 g Al ⎠
⎛ 2000 lb O ⎞ ⎛ 454 g O ⎞ ⎛ 1 mol O ⎞
5
mol O = 5.04 tons O ⎜
⎟ ⎜ 1 lb O ⎟ ⎜ 16.00 g O ⎟ = 2.86 × 10 mol O
1
ton
O
⎝
⎠⎝
⎠⎝
⎠
Empirical Formula: Al1.91×105 O 2.86×105
In whole numbers: Al1.91×105 O 2.86×105 which becomes AlO1.5 and multiply the subscripts by 2: Al2O3
1.91×105
3.17
1.91×105
We first determine the number of mol of each element as follows:
⎛ 1 mol N ⎞
mol N = ( 0.522 g N ) ⎜
⎟ = 0.0373 mol N
⎝ 14.01 g N ⎠
We need to know the number of grams of O. Since there is a total of 2.012 g of compound and the only
other element present is N, the mass of O = 2.012 g – 0.522 g = 1.490 g O.
⎛ 1 mol O ⎞
mol O = (1.490 g O ) ⎜
⎟ = 0.0931 mol O
⎝ 16.00 g O ⎠
Since these two mole amounts are the same, the empirical formula is N0.0373O0.0931; to have the empirical
formula in whole numbers, first divide by the smaller number of moles: N 0.0373 O 0.0931 which is NO2.5,
0.0373
0.0373
now to have whole numbers, multiply the subscripts by 2: N2O5.
3.18
It is convenient to assume that we have 100 g of the sample, so that the % by mass values may be taken
directly to represent masses. Thus there is 32.4 g of Na, 22.6 g of S and (100.00 – 32.4 – 22.6) = 45.0 g of
O. Now, convert these masses to a number of mol:
⎛ 1 mol Na ⎞
mol Na = ( 32.4 g Na ) ⎜
⎟ = 1.40 mol Naθ
⎝ 23.00 g Na ⎠
⎛ 1 mol S ⎞
mol S = ( 22.6 g S ) ⎜
⎟ = 0.705 mol S
⎝ 32.06 g S ⎠
⎛ 1 mol O ⎞
mol O = ( 45.0 g O ) ⎜
⎟ = 2.81 mol O
⎝ 16.00 g O ⎠
Next, we divide each of these mol amounts by the smallest in order to deduce the simplest whole number
ratio:
For Na: 1.40 mol/0.705 mol = 1.99
For S: 0.705 mol/0.705 mol = 1.00
For O: 2.81 mol/0.705 mol = 3.99
The empirical formula is Na2SO4.
33
Chapter 3
3.19
It is convenient to assume that we have 100 g of the sample, so that the % by mass values may be taken
directly to represent masses. Thus there is 81.79 g of C, 6.10 g of H and (100.00 – 81.79 – 6.10) = 12.11 g
of O. Now, convert these masses to a number of mol:
⎛ 1 mol C ⎞
mol C = ( 81.79 g C ) ⎜
⎟ = 6.81 mol C
⎝ 12.01 g C ⎠
⎛ 1 mol H ⎞
mol H = ( 6.10 g H ) ⎜
⎟ = 6.05 mol H
⎝ 1.008 g H ⎠
⎛ 1 mol O ⎞
mol O = (12.11 g O ) ⎜
⎟ = 0.757 mol O
⎝ 16.00 g O ⎠
Next, we divide each of these mol amounts by the smallest in order to deduce the simplest whole number
ratio:
For C: 6.81 mol/0.757 mol = 9.00
For H: 6.05 mol/0.757 mol = 7.99
For O: 0.757 mol/0.757 mol = 1.00
The empirical formula is C9H8O.
3.20
Find the moles of S and C using the stoichiometric ratios, and then find the empirical formula from the
ratio of moles of S and C.
FM SO2 = 64.06 g mol–1 FM CO2 = 44.01 g mol–1
⎛ 1 mol SO2 ⎞ ⎛ 1 mol S
mol S = 0.640 g SO2 ⎜
⎟⎜
⎝ 64.06 g SO 2 ⎠ ⎝ 1 mol SO 2
⎞
–3
⎟ = 9.99 × 10
⎠
⎛ 1 mol CO 2 ⎞ ⎛ 1 mol C ⎞
–3
mol C = 0.220 g CO2 ⎜
⎟⎜
⎟ = 5.00 × 10
44.01
g
CO
1
mol
CO
2 ⎠⎝
2 ⎠
⎝
Empirical Formula C5.00×10−3 S9.99×10−3 divide both subscripts by 5.00 × 10–3 to get CS2.
3.21
Since the entire amount of carbon that was present in the original sample appears among the products only
as CO2, we calculate the amount of carbon in the sample as follows:
⎛ 1 mol CO 2 ⎞ ⎛ 1 mol C
g C = ( 7.406 g CO 2 ) ⎜
⎟⎜
⎝ 44.01 g CO 2 ⎠ ⎝ 1 mol CO 2
⎞ ⎛ 12.01 g C ⎞
⎟⎜
⎟ = 2.021 g C
⎠ ⎝ 1 mol C ⎠
Similarly, the entire mass of hydrogen that was present in the original sample appears among the products
only as H2O. Thus the mass of hydrogen in the sample is:
⎛ 1 mol H 2 O ⎞ ⎛ 2 mol H ⎞ ⎛ 1.008 g H ⎞
g H = ( 3.027 g H 2 O ) ⎜
⎟⎜
⎟⎜
⎟ = 0.3386 g H
⎝ 18.02 g H 2 O ⎠ ⎝ 1 mol H 2 O ⎠ ⎝ 1 mol H ⎠
The mass of oxygen in the original sample is determined by difference:
5.048 g – 2.021 g – 0.3386 g = 2.688 g O
Next, these mass amounts are converted to the corresponding mol amounts:
⎛ 1 mol C ⎞
mol C = ( 2.021 g C ) ⎜
⎟ = 0.1683 mol C
⎝ 12.01 g C ⎠
34
Chapter 3
⎛ 1 mol H ⎞
mol H = ( 0.3386 g H ) ⎜
⎟ = 0.3359 mol H
⎝ 1.008 g H ⎠
⎛ 1 mol O ⎞
mol O = ( 2.688 g O ) ⎜
⎟ = 0.1680 mol O
⎝ 16.00 g O ⎠
The simplest formula is obtained by dividing each of these mol amounts by the smallest:
For C: 0.1683 mol/0.1680 mol= 1.002
for H: 0.3359 mol/0.1680 mol= 1.999
For O: 0.1680 mol/0.1680 mol = 1.000
These values give us the simplest formula directly, namely CH2O.
3.22
To find the molecular formula, divide the molecular mass by the formula mass of the empirical formula,
then multiply the subscripts of the empirical formula by that value.
Formula mass of CH2Cl: 49.48 g mol–1
Formula mass of CHCl: 48.47 g mol–1
100
289
= 2.02 and
= 5.84
For CH2Cl
49.48
49.48
100
289
For CHCl:
= 2.06 and
= 5.96
48.47
48.47
The CH2Cl rounds better using the molecular mass of 100, therefore multiply the subscripts by 2 and the
formula is C2H4Cl2.
For CHCl, the molecular mass of 289 gives a multiple of 6, therefore the formula is C6H6Cl6.
3.23
The formula mass of the empirical unit is 1 N + 2 H = 16.03. Since this is half of the molecular mass, the
molecular formula is N2H4.
3.24
AlCl3(aq) + Na3PO4(aq) J AlPO4(s) + 3NaCl(aq)
3.25
3CaCl2(aq) + 2K3PO4(aq) J Ca3(PO4)2(s) + 6KCl(aq)
3.26
⎛ 1 mol O2 ⎞
mol O2 = ( 6.76 mol SO3 ) ⎜
⎟ = 3.38 mol O 2
⎝ 2 mol SO3 ⎠
3.27
⎛ 1 mol H 2SO 4 ⎞
mol H 2SO 4 = ( 0.366 mol NaOH ) ⎜
⎟ = 0.183 mol H 2SO4
⎝ 2 mol NaOH ⎠
3.28
3.29
⎞
⎛ 1 mol Fe ⎞ ⎛ 1 mol Al2 O3 ⎞ ⎛
102.0 g Al2 O3
g Al2 O3 = ( 86.0 g Fe ) ⎜
⎟
⎟⎜
⎟⎜
⎝ 55.85 g Fe ⎠ ⎝ 2 mol Fe ⎠ ⎝ 1 mol Al2 O3 mol Al2 O3 ⎠
= 78.5 g Al2O3
⎛ 1 mol CaO ⎞ ⎛ 1 mol CO 2 ⎞ ⎛ 44.01 g CO2 ⎞
2
g CO2 = (1.50 × 102 g CaO) ⎜
⎟ = 1.18 × 10 g CO2
⎟⎜
⎟⎜
⎝ 56.08 g CaO ⎠ ⎝ 1 mol CaO ⎠ ⎝ 1 mol CO 2 ⎠
35
Chapter 3
3.30
First determine the number of grams of CaCO3 that would be required to react completely with the given
amount of HCl:
⎛ 1 mol HCl ⎞ ⎛ 1 mol CaCO3 ⎞ ⎛ 100.088 g CaCO3 ⎞
g CaCO3 = (125 g HCl) ⎜
⎟ = 171.57 g CaCO3
⎟⎜
⎟⎜
⎝ 36.461 g HCl ⎠ ⎝ 2 mol HCl ⎠ ⎝ 1 mol CaO3 ⎠
Since this is more than the amount that is available, we conclude that CaCO3 is the limiting reactant. The
rest of the calculation is therefore based on the available amount of CaCO3:
⎛ 1 mol CaCO3 ⎞ ⎛ 1 mol CO 2 ⎞ ⎛ 44.01 g CO 2 ⎞
g CO2 = (125 g CaCO3) ⎜
⎟⎜
⎟⎜
⎟
⎝ 100.088 g CaCO3 ⎠ ⎝ 1 mol CaCO3 ⎠ ⎝ 1 mol CO 2 ⎠
= 55.0 g CO2
For the number of grams of left over HCl, the excess reagent, find the amount of HCl used and then
subtract that from the amount of HCl started with, 125 g.
⎛ 1 mol CaCO3 ⎞⎛ 2 mol HCl ⎞ ⎛ 36.461 g HCl ⎞
g HCl used = (125 g CaCO3) ⎜
⎟⎜
⎟⎜
⎟
⎝ 100.088 g CaCO3 ⎠⎝ 1 mol CaCO3 ⎠ ⎝ 1 mol HCl ⎠
= 91.1 g HCl
g HCl remaining = 125 g – 91.1 g = 34 g HCl remaining
3.31
First determine the number of grams of O2 that would be required to react completely with the given
amount of ammonia:
⎛ 1 mol NH3 ⎞ ⎛ 5 mol O2 ⎞ ⎛ 32.00 g O 2 ⎞
g O 2 = ( 30.00 g NH3 ) ⎜
⎟⎜
⎟⎜
⎟
⎝ 17.03 g NH3 ⎠ ⎝ 4 mol NH3 ⎠ ⎝ 1 mol O 2 ⎠
= 70.46 g O2
Since this is more than the amount that is available, we conclude that oxygen is the limiting reactant. The
rest of the calculation is therefore based on the available amount of oxygen:
⎛ 1 mol O 2 ⎞ ⎛ 4 mol NO ⎞ ⎛ 30.01 g NO ⎞
g NO = ( 40.00 g O 2 ) ⎜
⎟⎜
⎟⎜
⎟
⎝ 32.00 g O 2 ⎠ ⎝ 5 mol O 2 ⎠ ⎝ 1 mol NO ⎠
= 30.01 g NO
3.32
First determine the number of grams of salicylic acid, HOOCC6H4OH that would be required to react
completely with the given amount of acetic anhydride, C4H6O3:
g HOOCC6H4OH = (15.6 g C4H6O3) ×
⎛ 1 mol C4 H 6 O3 ⎞ ⎛ 2 mol HOOCC6 H 4 OH ⎞⎛ 138.12 g HOOCC6 H 4 OH ⎞
⎜
⎟⎜
⎟⎜
⎟
1 mol C4 H 6 O3
⎝ 102.09 g C4 H 6 O3 ⎠ ⎝
⎠⎝ 1 mol HOOCC6 H 4 OH ⎠
= 42.2 g HOOCC6H4OH
Since more salicylic acid is required than is available, it is the limiting reagent. Once 28.2 g of salicylic
acid is reacted the reaction will stop, even though there are 15.6 g of acetic anhydride present. Therefore
the salicylic acid is the limiting reactant. The theoretical yield of aspirin HOOCC6H4O2C2H3 is therefore
based on the amount of salicylic acid added. This is calculated below:
g HOOCC6H4O2C2H3 = (28.2 g HOOCC6H4OH) ×
⎛ 1 mol HOOCC6 H 4 OH ⎞ ⎛ 2 mol HOOCC6 H 4 O2 C2 H3 ⎞ ⎛ 180.16 g HOOCC6 H 4 O 2 C2 H3 ⎞
⎜
⎟⎜
⎟⎜
⎟
⎝ 138.12 g HOOCC6 H 4 OH ⎠ ⎝ 2 mol HOOCC6 H 4 OH ⎠ ⎝ 1 mol HOOCC6 H 4 O 2 C2 H3 ⎠
= 36.78 g HOOCC6H4O2C2O3
36
Chapter 3
Now the percentage yield can be calculated from the amount of acetyl salicylic acid actually produced, 30.7 g:
⎛ 30.7 g HOOCC6 H 4 O 2 C2 H3 ⎞
⎛ actual yield ⎞
percent yield = ⎜
⎟ × 100%
⎟ × 100% = ⎜
⎝ theoretical yield ⎠
⎝ 36.78 g HOOCC6 H 4 O2 C2 H3 ⎠
= 83.5%
3.33
First determine the number of grams of C2H5OH that would be required to react completely with the given
amount of sodium dichromate:
⎛ 1 mol Na 2 Cr2 O7 ⎞ ⎛ 3 mol C2 H5 OH ⎞⎛ 46.08 g C2 H5 OH ⎞
g C2 H5 OH = ( 90.0 g Na 2 Cr2 O7 ) ⎜
⎟⎜
⎟⎜
⎟
⎝ 262.0 g Na 2 Cr2 O7 ⎠ ⎝ 2 mol Na 2 Cr2 O7 ⎠⎝ 1 mol C2 H5OH ⎠
= 23.7 g C2 H5 OH
Once this amount of C2 H5 OH is reacted the reaction will stop, even though there are 24.0 g C2H5OH
present, because the Na 2 Cr2 O7 will be used up. Therefore Na 2 Cr2 O7 is the limiting reactant. The
theoretical yield of acetic acid (HC2H3O2) is therefore based on the amount of Na 2 Cr2 O7 added. This is
calculated below:
⎛ 1 mol Na 2 Cr2 O7 ⎞ ⎛ 3 mol HC2 H3O 2 ⎞ ⎛ 60.06 g HC2 H3O 2 ⎞
g HC2 H3O 2 = ( 90.0 g Na 2 Cr2 O7 ) ⎜
⎟⎜
⎟⎜
⎟
⎝ 262.0 g Na 2 Cr2 O 7 ⎠ ⎝ 2 mol Na 2 Cr2 O7 ⎠ ⎝ 1 mol HC2 H3O2 ⎠
= 30.9 g HC2 H3O2
Now the percentage yield can be calculated from the amount of acetic acid actually produced, 26.6 g:
⎛ 26.6 g HC2 H3O 2 ⎞
⎛ actual yield ⎞
percent yield = ⎜
⎟ × 100 = 86.1%
⎟ × 100 = ⎜
⎝ theoretical yield ⎠
⎝ 30.9 g HC2 H3O 2 ⎠
Review Questions
U
3.1
The mole is the SI unit for the amount of a substance. A mole is equal in quantity to Avogadro’s number
(6.022 × 1023) of particles, or the formula mass in grams of a substance.
3.2
To estimate the number of atoms in a gram of iron, using atomic mass units, u, convert g to kg, then use the
relationship, 1.661 × 10–27 kg = 1 u, finally using the atomic mass of Fe (55.85 u) to find the number of
atoms:
⎞ ⎛ 1 molecule ⎞
⎛ 1 kg ⎞ ⎛
1u
22
1 g Fe ⎜
⎟⎜
⎟ ⎜⎜
⎟ = 1.08 × 10 atoms Fe
−27
⎟
1000
g
55.85
u
⎠
kg ⎠ ⎝
⎝
⎠ ⎝ 1.661 × 10
3.3
Moles are used for calculations instead of atomic mass units because they have the right units for
converting from grams to moles and vice versa.
3.4
There are the same number of molecules in 2.5 moles of H2O and 2.5 moles of H2.
3.5
(a)
⎛ 1 mol S ⎞
⎜ 2 mol O ⎟
⎝
⎠
(b)
⎛ 2 mol As ⎞
⎜ 3 mol O ⎟
⎝
⎠
⎛ 1 mol S ⎞ ⎛ 2 mol O ⎞
⎜
⎟ ⎜
⎟
⎝ 1 mol SO2 ⎠ ⎝ 1 mol SO 2 ⎠
⎛ 3 mol O ⎞ ⎛ 2 mol As ⎞ ⎛ 3 mol O ⎞
⎜ 2 mol As ⎟ ⎜ 1 mol As O ⎟ ⎜ 1 mol As O ⎟
⎝
⎠ ⎝
2 3⎠
2 3⎠
⎝
⎛ 2 mol O ⎞
⎜ 1 mol S ⎟
⎝
⎠
37
Chapter 3
(c)
(d)
⎛ 2 mol K ⎞ ⎛ 1 mol S ⎞ ⎛ 4 mol O ⎞
⎜
⎟ ⎜
⎟ ⎜
⎟
⎝ 1 mol K 2SO 4 ⎠ ⎝ 1 mol K 2SO 4 ⎠ ⎝ 1 mol K 2SO 4 ⎠
⎛ 1 mol S ⎞ ⎛ 1 mol S ⎞ ⎛ 4 mol O ⎞ ⎛ 4 mol O ⎞
⎜ 2 mol K ⎟ ⎜ 4 mol O ⎟ ⎜ 2 mol K ⎟ ⎜ 1 mol S ⎟
⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠
⎛ 2 mol K ⎞
⎜ 1 mol S ⎟
⎝
⎠
⎛ 2 mol K ⎞
⎜ 4 mol O ⎟
⎝
⎠
⎛
⎞ ⎛
⎞ ⎛
⎞
⎛
⎞
2 mol Na
1 mol H
1 mol P
4 mol O
⎜
⎟ ⎜
⎟ ⎜
⎟
⎜
⎟
⎝ 1 mol Na 2 HPO 4 ⎠ ⎝ 1 mol Na 2 HPO 4 ⎠ ⎝ 1 mol Na 2 HPO 4 ⎠
⎝ 1 mol Na 2 HPO 4 ⎠
⎛ 2 mol Na ⎞ ⎛ 2 mol Na ⎞
⎛ 2 mol Na ⎞ ⎛ 1 mol H ⎞ ⎛ 1 mol H ⎞ ⎛ 1 mol H ⎞
⎜ 1 mol H ⎟ ⎜ 1 mol P ⎟
⎜ 4 mol O ⎟ ⎜ 2 mol Na ⎟ ⎜ 1 mol P ⎟ ⎜ 4 mol O ⎟
⎝
⎠ ⎝
⎠
⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠
⎛ 1 mol P ⎞
⎜ 2 mol Na ⎟
⎝
⎠
⎛ 1 mol P ⎞
⎜ 1 mol H ⎟
⎝
⎠
⎛ 1 mol P ⎞
⎜ 4 mol O ⎟
⎝
⎠
⎛ 4 mol O ⎞
⎜ 2 mol Na ⎟
⎝
⎠
⎛ 4 mol O ⎞
⎜ 1 mol H ⎟
⎝
⎠
⎛ 4 mol O ⎞
⎜ 1 mol P ⎟
⎝
⎠
3.6
The molecular mass is required to convert grams of a substance to moles of that same substance.
3.7
There are 2 moles of iron atoms in 1 mole of Fe2O3. The stoichiometric equivalent between Fe and Fe2O is
2 mol Fe ≡ 1 mol Fe2O3.
For the number of iron atoms in 1 mole of Fe2O3:
23
⎞
⎛
⎞⎛
1 mol Fe2O3 ⎜ 2 mol Fe ⎟ ⎜ 6.022 × 10 Fe atoms ⎟ = 1.204 × 1024 atoms Fe
⎟
1 mol Fe
⎝ 1 mole Fe2 O3 ⎠ ⎜⎝
⎠
3.8
(a)
⎛ 3 mol Mn ⎞
⎜ 4 mol O ⎟
⎝
⎠
⎛ 4 mol O ⎞
⎜ 3 mol Mn ⎟
⎝
⎠
(b)
⎛ 2 mol Sb ⎞
⎜ 5 mol S ⎟
⎝
⎠
⎛ 5 mol S ⎞
⎜ 2 mol Sb ⎟
⎝
⎠
(c)
(d)
3.9
⎛
⎞
2 mol N
⎜
⎟
⎜ 1 mol ( NH 4 ) SO4 ⎟
2
⎝
⎠
⎛
⎞
4 mol O
⎜
⎟
⎜ 1 mol ( NH 4 ) SO4 ⎟
2
⎝
⎠
⎛ 8 mol H ⎞ ⎛ 1 mol S ⎞
⎜ 4 mol O ⎟ ⎜ 2 mol N ⎟
⎝
⎠ ⎝
⎠
⎛ 2 mol Hg ⎞
⎜ 2 mol Cl ⎟
⎝
⎠
⎛ 3 mol Mn
⎜
⎝ 1 mol Mn 3O4
⎛ 2 mol Sb ⎞
⎜
⎟
⎝ 1 mol Sb 2S5 ⎠
⎞
⎟
⎠
⎛
8 mol H
⎜
⎜ 1 mol ( NH 4 ) SO4
2
⎝
⎞
⎟
⎟
⎠
⎛ 2 mol N ⎞
⎜ 8 mol H ⎟
⎝
⎠
⎛ 2 mol Cl ⎞
⎜
⎟
⎝ 2 mol Hg ⎠
⎛ 4 mol O ⎞
⎜
⎟
⎝ 1 mol Mn 3O4 ⎠
⎛ 5 mol S ⎞
⎜
⎟
⎝ 1 mol Sb 2S5 ⎠
⎛
1 mol S
⎜
⎜ 1 mol ( NH 4 ) SO4
2
⎝
⎛ 2 mol N ⎞
⎜ 1 mol S ⎟
⎝
⎠
⎛ 1 mol S ⎞
⎜ 8 mol H ⎟
⎝
⎠
⎛ 2 mol N ⎞
⎜ 4 mol O ⎟
⎝
⎠
⎛ 1 mol S ⎞
⎜ 4 mol O ⎟
⎝
⎠
⎛ 2 mol Hg ⎞
⎜
⎟
⎝ 2 mol Hg 2 Cl2 ⎠
⎞
⎟
⎟
⎠
⎛ 8 mol H ⎞
⎜ 2 mol N ⎟
⎝
⎠
⎛ 4 mol O ⎞
⎜ 2 mol N ⎟
⎝
⎠
⎛ 4 mol O ⎞
⎜ 8 mol H ⎟
⎝
⎠
⎛ 8 mol H ⎞
⎜ 1 mol S ⎟
⎝
⎠
⎛ 4 mol O ⎞
⎜ 1 mol S ⎟
⎝
⎠
⎛ 2 mol Cl ⎞
⎜
⎟
⎝ 2 mol Hg 2 Cl2 ⎠
⎛ 1 mol Al ⎞
⎛ 26.98 g Al ⎞
⎜ 1 mol Al ⎟ and ⎜ 26.98 g Al ⎟
⎝
⎠
⎝
⎠
3.10
The statement, 1 mol O, does not indicate whether this is atomic oxygen, O, or molecular oxygen, O2. The
statement 64 g of oxygen is not ambiguous because the source of oxygen is not important.
3.11
At a minimum, the identity and mass of each atomic element present must be known. If the total mass of
the compound is known, then it is necessary to know all but one mass of the elements that compose the
compound.
3.12
The subscripts in a formula may not be changed unless one is determining the molecular formula from the
empirical formula.
38
Chapter 3
3.13
When balancing a chemical equation, changing the subscripts changes the identity of the substance.
3.14
There are three distinct empirical formulas represented AB2, AB3, and A3B8. There are two molecules with
the empirical formula AB3; AB3 and A2B6. There is one A3B8, and there are two with the formula AB2;
A6B12 and A3B6.
3.15
Avagadro's number would become 5 × 1023.
⎡
⎛ 1000 g ⎞ ⎤
Avagadro's number = ⎢ 2 × 10−27 kg × ⎜
⎟⎥
⎝ 1 kg ⎠ ⎦⎥
⎣⎢
(
)
−1
= 5 × 1023
3.16
To convert grams of a substance to molecules of the same substance, the molecular mass of the substance,
and Avagadro's number are needed.
3.17
Student B is correct.
Student A wrote a properly balanced equation. However, by changing the subscript for the product of the
reaction from an implied one, NaCl, to a two, NaCl2, this student has changed the identity of the product.
When balancing chemical equations, never change the values of the subscripts given in the unbalanced
equation.
3.18
Convert moles of B to moles of compound, A5B2; then using the stoichiometric ratio of moles of A to
moles of A5B2, determine the moles of A; and finally convert the moles of A to grams of A using the
molecular mass of A.
⎛ 1 mol A5 B2 ⎞ ⎛ 5 mol A ⎞ ⎛ 100.0 g A ⎞
(10 mol B) ⎜
⎟⎜
⎟⎜
⎟
⎝ 2 mol B ⎠ ⎝ 1 mol A5 B2 ⎠ ⎝ 1 mol A ⎠
The pieces of information that were not needed were the molecular mass of B and the number of molecules
of A in a mole of A
3.19
2H2O2 J 2H2O + O2
3.20
Their formula weights must be identical.
3.21
First write the balanced equation for the reaction of NH4NO3 as an explosive:
2NH4NO3 (s) J 2N2 (g) + O2 (g) + 4H2O (g)
Then find the molecular mass of NH4NO3 (80.04 g/mol).
Then calculate the number of moles of NH4NO3 is in 1.00 kg of NH4NO3:
Finally, using the stoichiometric ratio of N2 to NH4NO3 calculate the number of moles of N2 and then
multiply by Avagadro's number:
⎛ 1000 g NH 4 NO3 ⎞ ⎛ 1 mol NH 4 NO3 ⎞
molecules of N2 = (1.00 kg NH4NO3) ⎜
⎟⎜
⎟×
⎝ 1 kg NH 4 NO3 ⎠ ⎝ 80.06 g NH 4 NO3 ⎠
⎛ 1 mol N 2
⎞
⎜
⎟
⎝ 1 mol NH 4 NO3 ⎠
⎛ 6.022 × 1023 molecules N
2
⎜
⎜
1
mol
N
2
⎝
⎞
⎟ = 7.53 × 1024 molecules of N2
⎟
⎠
3.22
(a)
(b)
The balanced equation describes the stoichiometry.
The scale of the reaction is determined by the number of moles used as reactants in the
experiment.
3.23
To determine the number of grams of sulfur that would react with a gram of arsenic, the stoichiometric
ratio of the arsenic to the sulfur in the compound is needed, as well as the molecular masses of sulfur and
arsenic.
39
Chapter 3
3.24
Reaction 1
Reaction 2
Review Problems
3.25
(a)
(b)
(c)
(d)
6 atom C:11 atom H
12 mole C:11 mole O
12 atom H:11 atom O
12 mole H:11 mole O
3.26
(a)
(b)
(c)
(d)
2 atom C:1 atom O
2 mole C:1 mole O
1 atom C:2 atom H
1 mole C:2 mole H
3.27
1:2,
2 mol N to 4 mol O or in the smallest whole number ratio 1 mol N to 2 mol O
3.28
1:4,
1 mol C:4 mol H
3.29
⎛ 2 mol Bi ⎞
mol Bi = (1.58 mol O) ⎜
⎟ = 1.05 mol Bi
⎝ 3 mol O ⎠
3.30
⎛ 2 mol V ⎞
mol V = (0.565 mol O) ⎜
⎟ = 0.226 mol V
⎝ 5 mol O ⎠
3.31
⎛
⎞
1 mol Ta
–3
1.56 × 1021 atoms Ta ⎜
⎟ = 2.59 × 10 mole Ta
23
⎝ 6.022 × 10 atoms Ta ⎠
3.32
⎛
1 mol I 2
1.80 × 1024 molecules of I2 ⎜
⎜ 6.022 × 1023 molecules I
2
⎝
3.33
(a)
(b)
(c)
3.34
⎞
⎟ = 2.99 mole I2
⎟
⎠
⎛ 2 mol Al ⎞
⎛ 3 mol S ⎞
⎜ 3 mol S ⎟ or ⎜ 2 mol Al ⎟
⎝
⎠
⎝
⎠
⎛
⎞
3 mol S
⎛ 1 mol Al2 (SO4 )3 ⎞
⎜
⎟ or ⎜
⎟
3 mol S
⎝
⎠
⎝ 1 mol Al2 (SO4 )3 ⎠
⎛ 2 mol Al ⎞
mol Al = ( 0.900 mol S) ⎜
⎟ = 0.600 mol Al
⎝ 3 mol S ⎠
(d)
⎛
⎞
3 mol S
mol S = (1.16 mol Al2 (SO 4 )3 ) ⎜
⎟ = 3.48 mol S
⎝ 1 mol Al2 (SO 4 )3 ⎠
(a)
⎛ 3 mol Fe ⎞
⎛ 1 mol Fe3O 4 ⎞
⎜
⎟ or ⎜
⎟
⎝ 3 mol Fe ⎠
⎝ 1 mol Fe3O4 ⎠
40
Chapter 3
(b)
(c)
(d)
⎛ 3 mol Fe ⎞
⎛ 4 mol O ⎞
⎜ 4 molO ⎟ or ⎜ 3 mol Fe ⎟
⎝
⎠
⎝
⎠
⎛ 3 mol Fe ⎞
mol Fe = (2.75 mol Fe3O4) ⎜
⎟ = 8.25 mol Fe
⎝ 1 mol Fe3O 4 ⎠
⎛ 3 mol Fe ⎞ ⎛ 1 mol Fe2 O3 ⎞
mol Fe2 O3 = ( 4.50 mol Fe3O 4 ) ⎜
⎟⎜
⎟ = 6.75 mol Fe 2 O3
⎝ 1 mol Fe3O 4 ⎠ ⎝ 2 mol Fe ⎠
3.35
Number of C, H, and O atoms in glucose = 6 atoms C + 12 atoms H + 6 atoms O = 24 atoms
⎛ 6.022 × 1023 molecules glucose ⎞ ⎛
⎞
24 atoms
Number of atoms = (0.260 mol glucose) ⎜
⎟⎜
⎜
⎟ ⎝ 1 molecule glucose ⎟⎠
1
mol
glucose
⎝
⎠
= 3.76 × 1024 atoms
3.36
Number of N, H, and O atoms in glucose = 2 atoms N + 4 atoms H + 3 atoms O = 9 atoms
⎛ 6.022 × 1023 molecules NH NO ⎞ ⎛
⎞
9 atoms
4
3⎟
Number of atoms = (0.356 mol NH4NO3) ⎜
⎜
⎜
⎟ ⎝ 1 molecule glucose ⎟⎠
1 mol NH 4 NO3
⎝
⎠
= 1.93 × 1024 atoms
3.37
⎛ 4 mol F ⎞ ⎛ 1 mol UF6 ⎞
mol UF6 = (1.25 mol CF4) ⎜
⎟⎜
⎟ = 0.833 mol CF4
⎝ 1 mol CF4 ⎠ ⎝ 6 mol F ⎠
3.38
⎛ 2 mol Fe ⎞ ⎛ 1 mol Fe3O 4 ⎞
mol Fe3O4 = (0.260 mol Fe2O3) ⎜
⎟⎜
⎟ = 0.173 mol Fe3O4
⎝ 1 mol Fe2 O3 ⎠ ⎝ 3 mol Fe ⎠
3.39
⎛ 2 mol Cr ⎞
mol Cr = (2.16 mol Cr2O3) ⎜
⎟ = 4.32 mol Cr
⎝ 1 mol Cr2 O3 ⎠
3.40
⎛ 3 mol O ⎞
mol O = (4.25 mol CaCO3) ⎜
⎟ = 12.8 mol O
⎝ 1 mol CaCO3 ⎠
3.41
Based on the balanced equation:
2 NH3(g) J N2(g) + 3H2(g)
From this equation the conversion factors can be written:
⎛ 1 mol N 2 ⎞
⎛ 3 mol H 2 ⎞
⎜
⎟ and ⎜
⎟
⎝ 2 mol NH3 ⎠
⎝ 2 mol NH3 ⎠
To determine the moles produced, simply convert from starting moles to end moles:
⎛ 1 mol N 2 ⎞
mole N2 = 0.145 mol NH3 ⎜
⎟ = 0.0725 mol N 2
⎝ 2 mol NH3 ⎠
The moles of hydrogen are calculated similarly:
⎛ 3 mol H 2 ⎞
mole H2 = 0.145 mol NH3 ⎜
⎟ = 0.218 mol H 2
⎝ 2 mol NH3 ⎠
3.42
Based on the balanced equation:
2 Al(s) + 3 S(g) J Al2S3(s)
41
Chapter 3
From this equation the conversion factor can be written:
⎛ 3 mol S ⎞
⎜ 2 mol Al ⎟
⎝
⎠
To determine the moles of S needed, simply convert from the moles of Al2S3 produced:
⎛ 3 mol S ⎞
mol S = (0.225 mol Al) ⎜
⎟ = 0.338 mol S
⎝ 2 mol Al ⎠
3.43
⎛ 1 mol C3 H8
atoms C = (4.13 mol H) ⎜
⎝ 8 mol H
23
⎞ ⎛ 6.022 × 10 molecules C3 H8
⎜
⎟⎜
1 mol C3 H8
⎠⎝
⎞⎛
⎞
3 atoms C
⎟⎜
⎟
⎟
⎠ ⎝ 1 molecule C3 H8 ⎠
= 9.33 × 1023 atoms C
⎞ ⎛ 6.022 × 1023 atoms H ⎞
⎟ = 1.11 × 1025 atoms H
⎟ ⎜⎜
⎟
1
mol
H
⎠⎝
⎠
3.44
⎛ 8 mol H
atom H = (2.31 mol C3H8) ⎜
⎝ 1 mol C3 H8
3.45
⎛ 1 mol Ni ⎞
mol Ni = 17.7 g Ni ⎜
⎟ = 0.302 mol Ni
⎝ 58.69 g Ni ⎠
3.46
⎛ 1 mol Cr ⎞
mol Cr = 85.7 g Cr ⎜
⎟ = 1.65 mol Cr
⎝ 52.00 g Cr ⎠
3.47
⎛
⎞ ⎛ 39.10 g K ⎞
1 mol K
–10
g K = 2.00 × 1012 atoms K ⎜
⎟⎜
⎟ = 1.30 × 10 g K
23
1
mol
K
⎠
⎝ 6.022 × 10 atoms K ⎠ ⎝
3.48
⎛
⎞ ⎛ 22.99 g Na ⎞
1 mol Na
–5
g Na = 4.00 × 1017 atoms Na ⎜
⎟⎜
⎟ = 1.53 × 10 g Na
23
1
mol
Na
⎠
⎝ 6.022 × 10 atoms Na ⎠ ⎝
3.49
(a)
⎛ 55.85 g Fe ⎞
g Fe = (1.35 mol Fe) ⎜
⎟ = 75.4 g Fe
⎝ 1 mole Fe ⎠
(b)
⎛ 16.0 g O ⎞
g O = (24.5 mol O) ⎜
⎟ = 392 g O
⎝ 1 mole O ⎠
(c)
⎛ 40.08 g Ca ⎞
g Ca = (0.876) ⎜
⎟ = 35.1 g Ca
⎝ 1 mole Ca ⎠
(a)
⎛ 32.07 g S ⎞
g S = (0.546 mol S) ⎜
⎟ = 17.5 g S
⎝ 1 mole S ⎠
(b)
⎛ 14.01 g N ⎞
g N = (3.29 mol N) ⎜
⎟ = 46.1 g N
⎝ 1 mole N ⎠
(c)
⎛ 26.98 g N ⎞
g Al = (8.11 mol Al) ⎜
⎟ = 219 g Al
⎝ 1 mole N ⎠
3.50
3.51
⎛ 1 mol C-12 ⎞
mol C–12 = 6 g × ⎜
⎟ = 0.5 mol C–12
⎝ 12.00 g C-12 ⎠
⎛ 6.022 × 1023 atoms C-12 ⎞
Number of atoms C–12 = 0.5 mol ⎜
⎟ = 3.01 × 1023 atoms C–12
⎜
⎟
1
mol
C-12
⎝
⎠
42
Chapter 3
3.52
3.53
⎛ 6.022 × 1023 atoms C-12 ⎞
Number of atoms of C–12 = 1.5 mol C–12 ⎜
⎟ = 9.033 × 1023 atoms C–12
⎜
⎟
1
mol
C-12
⎝
⎠
⎛ 12.00 g C-12 ⎞
g C–12 = 1.5 mol C–12 × ⎜
⎟ = 18 g C–12
⎝ 1 mol C-12 ⎠
Note: all masses are in g/mole
=
(a)
NaHCO3
=
=
=
(b)
(NH4)2CO3
=
=
(c)
CuSO4·5H2O
=
=
=
=
(d)
K2Cr2O7
=
=
(e)
Al2(SO4)3
=
=
=
3.54
(a)
(b)
(c)
(d)
(e)
3.55
(a)
(b)
1Na + 1H + 1C + 3O
(22.98977) + (1.00794) + (12.0107) + (3 × 15.9994)
84.00661 g/mole = 84.0066 g/mol
2N + 8H + C + 3O
(2 × 14.0067) + (8 × 1.00794) + (12.0107) + (3 × 15.9994)
96.08582 g/mole = 96.0858 g/mol
1Cu + 1S +9O + 10H
63.546 + 32.065 + (9 × 15.9994) + (10 × 1.00794)
249.685 g/mole
2K + 2Cr + 7O
(2 × 39.0983) + (2 × 51.9961) + (7 × 15.9994)
294.1846 g/mole
2Al + 3S + 12O
(2 × 26.98154) + (3 × 32.065) + (12 × 15.9994)
342.15088 g/mole = 342.151 g/mol
Note: all masses are in g/mole
=
1Ca + 2N + 6O
Ca(NO3)2
=
(40.078) + (2 × 14.0067) + (6 × 15.9994)
=
164.0878 g/mole = 164.088 g/mol
Pb(C2H5)4
=
1Pb + 8C + 20H
=
(207.2) + (8 × 12.0107) + (20 × 1.00794)
=
323.4 g/mole (Since the mass of Pb is known exactly.)
2Na + 1S +14O + 20H
Na2SO4·10H2O =
=
(2 × 22.98977) + 32.065 + (14 × 15.9994) + (20 × 1.00794)
=
322.19494 g/mole = 322.195 g/mol
Fe4[Fe(CN)6]3
=
7Fe + 18C + 18N
=
(7 × 55.845) + (18 × 12.0107) + (18 × 14.0067)
=
859.2282 g/mole = 859.228 g/mol
Mg3(PO4)2
=
3Mg + 2P + 8O
=
(3 × 24.3050) + (2 × 30.97376) + (8 × 15.9994)
=
262.85772 g/mole = 262.8577 g/mol
⎛ 1 mole CaCO3 ⎞
moles CaCO3 = ( 21.5 g CaCO3 ) ⎜
⎟ = 0.215 moles CaCO3
⎝ 100.09 g CaCO3 ⎠
⎛ 1 g NH
⎞ ⎛ 1 mole NH ⎞
3
3
⎟
moles NH3 = (1.56 ng NH3 ) ⎜
= 9.16 × 10−11 moles NH3
9
⎜ 1× 10 ng NH ⎟ ⎜⎝ 17.03 g NH3 ⎟⎠
3⎠
⎝
(c)
⎛ 1 mole Sr ( NO3 )
2
moles Sr ( NO3 )2 = 16.8 g Sr ( NO3 )2 ⎜
⎜ 211.6 g Sr ( NO3 )
2
⎝
(d)
⎛ 1 g Na CrO
2
4
moles Na 2 CrO4 = ( 6.98 µg Na 2 CrO4 ) ⎜
⎜ 106 µg Na CrO
2
4
⎝
(
)
= 4.31 × 10−8 moles Na 2 CrO 4
43
⎞
⎟ = 7.94 × 10−2 moles Sr ( NO3 )2
⎟
⎠
⎞ ⎛ 1 mole Na CrO ⎞
2
4
⎟⎜
⎟ ⎝ 162.0 g Na 2 CrO4 ⎟⎠
⎠
Chapter 3
3.56
(a)
(b)
(c)
(d)
3.57
(a)
(b)
(c)
(d)
3.58
(a)
(b)
(c)
(d)
3.59
⎛ 1 mole Ca(OH) 2 ⎞
mol Ca(OH)2 = (9.36 g Ca(OH)2) ⎜
⎟ = 0.126 mol Ca(OH)2
⎝ 74.10 g Ca(OH) 2 ⎠
⎛ 1000 g PbSO 4 ⎞ ⎛ 1 mole PbSO 4 ⎞
mol PbSO4 = (38.2 kg PbSO4) ⎜
⎟⎜
⎟ = 126 mol PbSO4
⎝ 1 kg PbSO 4 ⎠ ⎝ 303.3 g PbSO 4 ⎠
⎛ 1 mole H 2 O 2 ⎞
mol H2O2 = (4.29 g H2O2) ⎜
⎟ = 0.126 mol H2O2
⎝ 34.01 g H 2 O 2 ⎠
⎛ 1 g NaAuCl4 ⎞ ⎛ 1 mol NaAuCl4 ⎞
mol NaAuCl4 = 4.65 mg NaAuCl4 ⎜
⎟⎜
⎟
⎝ 1000 mg NaAuCl4 ⎠ ⎝ 361.8 g NaAuCl4 ⎠
= 1.29 × 10–5 mol NaAuCl4
⎛ 310.18 g Ca 3 ( PO 4 ) ⎞
2 ⎟ = 388 g Ca (PO )
g Ca3(PO4)2 = (1.25 mol Ca3(PO4)2) ⎜
3
4 2
⎜ 1 mol Ca 3 ( PO 4 ) ⎟
2 ⎠
⎝
⎛ 241.86 mg Fe ( NO3 ) ⎞
3⎟
mg Fe(NO3)3 = (0.625 mmol Fe(NO3)3) ⎜
= 151 mg Fe(NO3)3
⎜ 1 mmol Fe ( NO3 ) ⎟
3 ⎠
⎝
= 0.151 g Fe(NO3)3
⎛ 58.12 µg C 4 H10 ⎞
–5
µg C4H10 = (0.600 µmol C4H10) ⎜
⎟ = 34.9 µg C4H10 = 3.49 × 10 g C4H10
1
µ
mol
C
H
4 10 ⎠
⎝
⎛ 96.09 g ( NH 4 ) CO3 ⎞
2
⎟ = 139 g (NH4)2CO3
g (NH4)2CO3 = (1.45 mol (NH4)2CO3) ⎜
⎜ 1 mol ( NH 4 ) CO3 ⎟
2
⎝
⎠
⎛ 136.31 g ZnCl2 ⎞
g ZnCl2 = (0.754 mol ZnCl2) ⎜
⎟ = 103 g ZnCl2
⎝ 1 mol ZnCl2 ⎠
⎛ 1 mol KIO
⎞
3 ⎟ ⎛ 214.00 g KIO3 ⎞ ⎛ 1000 µg KIO3 ⎞
µg KIO3 = (0.194 µmol KIO3) ⎜
⎟⎜
⎟
⎜ 6
⎟⎜
⎝ 10 µmol KIO3 ⎠ ⎝ 1 mol KIO3 ⎠ ⎝ 1 g KIO3 ⎠
= 4.15 × 10–5 g KIO3
⎛ 1 mol POCl3 ⎞ ⎛ 153.33 g POCl3 ⎞
⎟
g POCl3 = (0.322 mmol POCl3) ⎜
= 0.0494 g POCl3
⎜ 103 mmol POCl ⎟ ⎜⎝ 1 mol POCl3 ⎟⎠
3⎠
⎝
⎛ 132.1 g ( NH 4 ) HPO 4
2
g (NH4)2HPO4 = (4.31 ×10–3 mol (NH4)2HPO4) ⎜
⎜ 1 mol ( NH 4 ) HPO 4
2
⎝
= 0.569 g (NH4)2HPO4
⎞
⎟
⎟
⎠
⎛ 1000 g N ⎞ ⎛ 1 mol N ⎞ ⎛ 1 mol ( NH 4 )2 CO3 ⎞
⎟
kg fertilizer = (1 kg N) ⎜
⎟⎜
⎟⎜
⎟
2 mol N
⎝ 1 kg N ⎠ ⎝ 14.01 g N ⎠ ⎜⎝
⎠
⎛ 96.09 g ( NH 4 ) CO3 ⎞⎛ 1 kg ( NH 4 ) CO3 ⎞
2
2
⎟⎜
⎟ = 3.43 kg fertilizer
×⎜
⎜ 1 mol ( NH 4 ) CO3 ⎟⎜ 1000 g ( NH 4 ) CO3 ⎟
2
2
⎝
⎠⎝
⎠
3.60
⎛ 1000 g P ⎞ ⎛ 1 mol P ⎞ ⎛ 1 mol P2 O5 ⎞ ⎛ 141.94 g P2 O5 ⎞⎛ 1 kg P2 O5 ⎞
kg P2O5 = (1 kg P) ⎜
⎟⎜
⎟ = 2.3 kg P2O5
⎟⎜
⎟⎜
⎟⎜
⎝ 1 kg P ⎠ ⎝ 30.97 g P ⎠ ⎝ 2 mol P ⎠ ⎝ 1 mol P2 O5 ⎠⎝ 1000 g P2 O5 ⎠
44
Chapter 3
3.61
3.62
The formula CaC2 indicates that there is 1 mole of Ca for every 2 moles of C. Therefore, if there are 0.150
moles of C there must be 0.0750 moles of Ca.
⎛ 40.078 g Ca ⎞
g Ca = (0.075 mol Ca) ⎜
⎟ = 3.01 g Ca
⎝ 1 mole Ca ⎠
⎛
⎞
2 moles I
⎟ = 1.00 moles I
mol I = 0.500 mol Ca ( IO3 )2 ⎜
⎜ 1 mole Ca ( IO3 ) ⎟
2 ⎠
⎝
⎛ 389.9 g Ca ( IO3 ) ⎞
2 ⎟
g Ca ( IO3 )2 = 0.500 mol Ca ( IO3 )2 ⎜
= 195 g Ca ( IO3 )2
⎜ 1 mole Ca ( IO3 ) ⎟
2
⎝
⎠
(
)
(
3.63
3.64
3.65
)
⎛
⎞
2 moles N
⎟ = 1.30 mol N
mol N = (0.650 mol (NH4)2CO3) ⎜
⎜ 1 mole ( NH 4 ) CO3 ⎟
2
⎝
⎠
⎛ 96.09 g (NH 4 )2 CO3 ⎞
g (NH4)2CO3 = (0.650 mol (NH4)2CO3) ⎜
⎟ = 62.5 g (NH4)2CO3
⎝ 1 mole (NH 4 )2 CO3 ⎠
⎛
⎞
2 moles N
mol N = (0.556 mol NH4NO3) ⎜
⎟ = 1.11 mol N
1
mole
NH
NO
4
3⎠
⎝
⎛ 80.04 g NH 4 NO3 ⎞
g NH4NO3 = (0.556 mol NH4NO3) ⎜
⎟ = 44.5 g NH4NO3
⎝ 1 mole NH 4 NO3 ⎠
Assume one mole total for each of the following.
(a)
The molar mass of NaH2PO4 is 119.98 g/mol.
23.0 g Na
× 100% = 19.2%
% Na =
119.98 g NaH 2 PO4
%H=
2.02 g H
× 100% = 1.68%
119.98 g NaH 2 PO4
%P=
31.0 g P
× 100% = 25.8%
119.98 g NaH 2 PO4
64.0 g O
× 100 % = 53.3 %
119.98 g NaH 2 PO4
The molar mass of NH4H2PO4 is 115.05 g/mol.
14.0 g N
× 100% = 12.2%
%N=
115.05 g NH 4 H 2 PO 4
%O=
(b)
%H=
6.05 g H
× 100% = 5.26%
115.05 g NH 4 H 2 PO 4
%P=
31.0 g P
× 100% = 26.9%
115.05 g NH 4 H 2 PO4
%O=
64.0 g O
× 100 % = 55.6 %
115.05 g NH 4 H 2 PO 4
45
Chapter 3
(c)
(d)
The molar mass of (CH3)2CO is 58.08 g/mol
36.0 g C
%C=
× 100% = 62.0%
58.08 g ( CH3 )2 CO
%H=
6.05 g H
× 100% = 10.4%
58.08 g ( CH3 )2 CO
%O=
16.0 g O
× 100% = 27.6%
58.08 g ( CH3 )2 CO
The molar mass of calcium sulfate dihydrate is 172.2 g/mol.
40.1 g Ca
% Ca =
× 100% = 23.3%
172.2 g CaSO 4 • 2H 2 O
%S=
32.1 g S
× 100% = 18.6%
172.2 g CaSO 4 • 2H 2 O
%O=
96.0 g O
× 100% = 55.7%
172.2 g CaSO 4 • 2H 2 O
4.03 g H
× 100 % = 2.34 %
172.2 g CaSO 4 • 2H 2 O
The molar mass of CaSO4•2H2O is 172.2 g/mol.
%H=
(e)
% Ca =
3.66
(a)
40.1 g Ca
× 100% = 23.3%
172.2 g CaSO 4 • 2H 2 O
%S=
32.1 g S
× 100% = 18.6%
172.2 g CaSO 4 • 2H 2 O
%O=
96.0 g O
× 100% = 55.7%
172.2 g CaSO 4 • 2H 2 O
%H=
4.03 g H
× 100 % = 2.34 %
172.2 g CaSO 4 • 2H 2 O
The molar mass of (CH3)2N2H2 is 60.12 g/mol.
%C=
24.02 g C
× 100% = 40.0%
60.12 g (CH3 ) 2 N 2 H 2
%H=
8.06 g H
× 100% = 13.4%
60.12 g (CH3 )2 N 2 H 2
28.0 g N
× 100% = 46.6%
60.12 g (CH3 )2 N 2 H 2
The molar mass of CaCO3 is 100.1 g/mol.
%N=
(b)
% Ca =
40.08 g Ca
× 100% = 40.0%
100.1 g CaCO3
%C=
12.01 g C
× 100% = 12.0%
100.1 g CaCO3
%O=
48.00 g O
× 100% = 48.0%
100.1 g CaCO3
46
Chapter 3
(c)
(d)
The molar mass of Fe(NO3)3 is 241.9 g/mol.
55.85 g Fe
× 100% = 23.1%
% Fe =
241.9 g Fe ( NO3 )3
%N =
42.03 g N
× 100% = 17.4%
241.9 g Fe ( NO3 )3
%O =
144.00 g O
× 100% = 59.5%
241.9 g Fe ( NO3 )3
The molar mass of C3H8 is 44.11 g/mol.
36.03 g C
%C =
× 100% = 81.7%
44.11 g C3 H8
8.08 g H
× 100% = 18.3%
44.11 g C3 H8
The molar mass of Al2(SO4)3 is 342.2 g/mol.
54.0 g Al
% Al =
× 100% = 15.8%
342.2 g Al2 ( SO 4 )3
%H =
(e)
%S =
96.2 g S
× 100% = 28.1%
342.2 g Al2 ( SO 4 )3
%O =
192.0 g O
× 100% = 56.1%
342.2 g Al2 ( SO4 )3
( 7.14 × 10
21
)
⎛
⎞ ⎛ 5 mol O ⎞⎛ 16.0 g O ⎞
1 mol N
atoms N ⎜
⎟⎜
⎟⎜
⎟ = 0.474 g O
23
⎝ 6.02 × 10 atoms N ⎠ ⎝ 2 mol N ⎠⎝ 1 mol O ⎠
3.67
gO =
3.68
⎛
⎞ ⎛ 5 mol C ⎞ ⎛ 12.01 g C ⎞
1 mol H
g C = (4.25 × 1023 atoms H) ⎜
⎟⎜
⎟⎜
⎟ = 3.53 g C
23
⎝ 6.02 × 10 atoms H ⎠ ⎝ 12 mol H ⎠ ⎝ 1 mol C ⎠
3.69
% O in morphine =
48.00 g O
× 100% = 16.82% O
285.36 g C17 H19 NO3
80.00 g O
× 100% = 21.65% O
369.44 g C21H 23 NO5
Therefore heroin has a higher percentage oxygen.
% O in heroin =
3.70
% N in carbamazepine =
28.02 g N
× 100% = 11.9% N
236.29 g C15 H12 N 2 O
14.01 g N
× 100% = 4.20% N
333.52 g C20 H31 NO3
Therefore, carbamazepine has a higher percentage of nitrogen.
% N in carbetapentane =
47
Chapter 3
3.71
For C17H25N, the molar mass (17C + 25H + 1N) equals 243.43 g/mole, and the three theoretical values for
% by weight are calculated as follows:
%C=
204.2 g C
× 100% = 83.89%
243.4 g C17 H 25 N
%H=
25.20 g H
× 100% = 10.35%
243.4 g C17 H 25 N
%N=
14.01 g N
× 100% = 5.76%
243.4 g C17 H 25 N
These data are consistent with the experimental values cited in the problem.
3.72
For C20H25N3O, the molar mass (20C + 25H + 3N + O) equals 323.44 g/mole, and the theoretical values for
% by weight are calculated as follows:
240.22 g C
%C=
× 100% = 74.27%
323.44 g C20 H 25 N3O
%H=
25.20 g H
× 100% = 7.791%
323.44 g C20 H 25 N3O
%N=
42.02 g N
× 100% = 12.99%
323.44 g C20 H 25 N3O
%O=
16.00 g O
× 100% = 4.947%
323.44 g C20 H 25 N3O
(a)
(b)
3.73
The % by mass oxygen in the suspected sample may be determined by difference:
100% – (74.07 + 7.95 + 9.99)% = 7.99 %.
These data are not consistent with the theoretical formula for LSD.
70.90 g Cl
× 100% = 58.63% Cl
120.92 g CCl2 F2
% Cl in Freon-12 =
% Cl in Freon 141b =
70.9 g Cl
× 100% = 60.62% Cl
116.95 g C2 H3Cl2 F
Therefore Freon 141b has a higher percentage chlorine.
3.74
% F in Freon-12 =
38.00 g F
× 100% = 31.43% F
120.92 g CCl2 F2
% F in Freon 113 =
57.00 g F
× 100% = 30.42% F
187.37 g C2 Cl3 F3
Therefore Freon-12 has a higher percentage chlorine.
3.75
%P =
0.539 g P
× 100% = 22.9%
2.35 g compound
% Cl = 100% − 22.9% = 77.1%
3.76
1.47 g N
× 100% = 25.9%
5.67 g compound
%O = 100% − 25.9% = 74.1%
%N =
48
Chapter 3
3.77
The molecular formula is some integer multiple of the empirical formula. This means that we can divide
the molecular formula by the largest possible whole number that gives an integer ratio among the atoms in
the empirical formula.
(a)
SCl
(b)
CH2O (c)
NH3
(d)
AsO3 (e)
HO
3.78
(a)
3.79
This type of combustion analysis takes advantage of the fact that the entire amount of carbon in the original
sample appears as CO2 among the products. Hence the mass of carbon in the original sample must be equal
to the mass of carbon that is found in the CO2.
CH3O
(b)
HSO4
(c)
C2H5
(d)
⎛ 1 mole CO2 ⎞ ⎛ 1 mole C
g C = (19.73 × 10–3 g CO2) ⎜
⎟⎜
⎝ 44.01 g CO2 ⎠ ⎝ 1 mole CO2
BH3
(e)
C2H6O
⎞ ⎛ 12.011 g C ⎞
–3
⎟⎜
⎟ = 5.385 × 10 g C
1
mole
C
⎝
⎠
⎠
Similarly, the entire mass of hydrogen that was present in the original sample ends up in the products as
H2O:
⎛ 1 mole H 2 O ⎞ ⎛ 2 mole H ⎞ ⎛ 1.008 g H ⎞
= 7.150 × 10–4 g H
g H = (6.391 × 10–3 g H2O) ⎜
⎟⎜
⎟⎜
⎟
⎝ 18.02 g H 2 O ⎠ ⎝ 1 mole H 2 O ⎠ ⎝ 1 mole H ⎠
The mass of oxygen is determined by subtracting the mass due to C and H from the total mass:
6.853 mg total – (5.385 mg C + 0.7150 mg H) = 0.753 mg O.
Now, convert these masses to a number of moles:
⎛ 1 mol C ⎞
–4
mol C = (5.385 × 10–3 g C) ⎜
⎟ = 4.484 × 10 mol C
12.011
g
C
⎝
⎠
⎛ 1 mol H ⎞
–4
mol H = (7.150 × 10–4 g H) ⎜
⎟ = 7.094 × 10 mol H
1.0079
g
H
⎝
⎠
⎛ 1 mol O ⎞
–5
mol O = (7.53 × 10–4 g O) ⎜
⎟ = 4.71 × 10 mol H
⎝ 15.999 g O ⎠
The relative mole amounts are:
for C, 4.483 × 10–4 mol / 4.71 × 10–5 mol = 9.52
for H, 7.094 × 10–4 mol / 4.71 × 10–5 mol = 15.1
for O, 4.71 × 10–5 mol / 4.71 × 10–5 mol = 1.00
The relative mole amounts are not whole numbers as we would like. However, we see that if we double the
relative number of moles of each compound, there are approximately 19 moles of C, 30 moles of H and 2
moles of O. If we assume these numbers are correct, the empirical formula is C19H30O2, for which the
formula weight is 290 g/mole.
In most problems where we attempt to determine an empirical formula, the relative mole amounts should
work out to give a “nice” set of values for the formula. Rarely will a problem be designed that gives very
odd coefficients. With experience and practice, you will recognize when a set of values is reasonable.
3.80
This type of combustion analysis takes advantage of the fact that the entire amount of carbon in the original
sample appears as CO2 among the products. Hence the mass of carbon in the original sample must be equal
to the mass of carbon that is found in the CO2.
⎛ 1 mole CO 2 ⎞ ⎛ 1 mole C ⎞ ⎛ 12.0107 g C ⎞
–3
g C = (18.490 × 10–3 g CO2) ⎜
⎟⎜
⎟⎜
⎟ = 5.0461 × 10 g C
⎝ 44.0095 g CO 2 ⎠ ⎝ 1 mole CO 2 ⎠ ⎝ 1 mole C ⎠
49
Chapter 3
Similarly, the entire mass of hydrogen that was present in the original sample ends up in the products as
H2O:
⎛ 1 mole H 2 O ⎞ ⎛ 2 mole H ⎞ ⎛ 1.0079 g H ⎞
–4
g H = (6.232 × 10–3 g H2O) ⎜
⎟⎜
⎟⎜
⎟ = 6.973 × 10 g H
18.015
g
H
O
1
mole
H
O
1
mole
H
⎝
⎠
2
2
⎝
⎠⎝
⎠
The mass of oxygen is determined by subtracting the mass due to C and H from the total mass:
5.983 mg total – (5.0462 mg C + 0.6972 mg H) = 0.240 mg O.
Now, convert these masses to a number of moles:
⎛ 1 mol C ⎞
–4
mol C = (5.0461 × 10–3 g C) ⎜
⎟ = 4.2013 × 10 mol C
⎝ 12.0107 g C ⎠
⎛ 1 mol H ⎞
–4
mol H = (6.973× 10–4 g H) ⎜
⎟ = 6.918 × 10 mol H
⎝ 1.0079 g H ⎠
⎛ 1 mol O ⎞
–5
mol O = (2.40 × 10–4 g O) ⎜
⎟ = 1.50 × 10 mol H
15.999
g
O
⎝
⎠
The relative mole amounts are:
for C, 4.2013 × 10–4 mol / 1.50 × 10–5 mol = 28.0
for H, 6.918 × 10–4 mol / 1.50 × 10–5 mol = 46.1
for O, 1.50 × 10–5 mol / 1.50 × 10–5 mol = 1.00
and the empirical formula is C28H46O.
3.81
All of the carbon is converted to carbon dioxide so,
⎛ 1 mol CO 2 ⎞ ⎛ 1 mol C ⎞ ⎛ 12.01 g C ⎞
g C = (1.312 g CO2) ⎜
⎟⎜
⎟⎜
⎟ = 0.358 g C
⎝ 44.01 g CO2 ⎠ ⎝ 1 mol CO2 ⎠ ⎝ 1 mol C ⎠
⎛ 1 mol C ⎞
–2
mol C = (0.358 g C) ⎜
⎟ = 2.98 × 10 mol C
12.01
g
C
⎝
⎠
All of the hydrogen is converted to H2O, so
⎛ 1 mol H 2 O ⎞⎛ 2 mol H ⎞ ⎛ 1.008 g H ⎞
g H = (0.805 g H2O) ⎜
⎟⎜
⎟⎜
⎟ = 0.0901 g H
⎝ 18.02 g H 2 O ⎠⎝ 1 mol H 2 O ⎠ ⎝ 1 mol H ⎠
⎛ 1 mol H ⎞
–2
mol H = (0.0901 g H) ⎜
⎟ = 8.93 × 10 mol H
⎝ 1.008 g H ⎠
The amount of O in the compound is determined by subtracting the mass of C and the mass of H from the
sample.
g O = 0.684 g – 0.358 g – 0.0901 g = 0.236 g O
⎛ 1 mol O ⎞
–2
mol O = (0.236 g O) ⎜
⎟ = 1.48 × 10 mol O
⎝ 16.00 g O ⎠
The relative mole ratios are:
for C, 0.0298 moles / 0.0148 moles = 2.01
for H, 0.0893 moles/ 0.0148 moles = 6.03
for O, 0.0148 moles / 0.0148 moles = 1.00
The relative mole amounts give the empirical formula C2H6O
50
Chapter 3
3.82
All of the carbon is converted to carbon dioxide so,
⎛ 1 mol CO 2 ⎞ ⎛ 1 mol C ⎞ ⎛ 12.01 g C ⎞
g C = (2.01 g CO2) ⎜
⎟⎜
⎟⎜
⎟ = 0.549 g C
⎝ 44.01 g CO2 ⎠ ⎝ 1 mol CO2 ⎠ ⎝ 1 mol C ⎠
⎛ 1 mol C ⎞
mol C = (0.549 g C) ⎜
⎟ = 0.0457 mol C
⎝ 12.01 g C ⎠
All of the hydrogen is converted to H2O, so
⎛ 1 mol H 2 O ⎞⎛ 2 mol H ⎞ ⎛ 1.008 g H ⎞
g H = (0.827 g H2O) ⎜
⎟⎜
⎟⎜
⎟ = 0.0925 g H
⎝ 18.02 g H 2 O ⎠⎝ 1 mol H 2 O ⎠ ⎝ 1 mol H ⎠
⎛ 1 mol H ⎞
mol H = (0.0925 g H) ⎜
⎟ = 0.0918 mol H
⎝ 1.008 g H ⎠
The amount of O in the compound is determined by subtracting the mass of C and the mass of H from the
sample.
g O = 0.822 g – 0.549 g – 0.0925 g = 0.181 g
⎛ 1 mol O ⎞
mol O = (0.181 g O) ⎜
⎟ = 0.0113 mol O
⎝ 16.00 g O ⎠
The relative mole ratios are:
for C, 0.0457 moles / 0.0113 moles = 4.04
for H, 0.0918 moles/ 0.0113 moles = 8.12
for O, 0.0113 moles / 0.0113 moles = 1.00
The relative mole amounts give the empirical formula C4H8O.
3.83
We begin by realizing that the mass of oxygen in the compound may be determined by difference:
0.896 g total – (0.111 g Na + 0.477 g Tc) = 0.308 g O.
Next we can convert each mass of an element into the corresponding number of moles of that element as
follows:
⎛ 1 mol Na ⎞
−3
mol Na = ( 0.111 g Na ) ⎜
⎟ = 4.83 × 10 mol Na
23.00
g
Na
⎝
⎠
⎛ 1 mol Tc ⎞
−3
mol Tc = ( 0.477 g Tc ) ⎜
⎟ = 4.82 × 10 mol Tc
98.9
g
Tc
⎝
⎠
⎛ 1 mol O ⎞
−2
mol O = ( 0.308 g O ) ⎜
⎟ = 1.93 × 10 mol O
16.0
g
O
⎝
⎠
Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the
simplest mole ratio among the three elements in the compound:
for Na, 4.83 × 10–3 moles / 4.82 × 10–3 moles = 1.00
for Tc, 4.82 × 10–3 moles / 4.82 × 10–3 moles = 1.00
for O, 1.93 × 10–2 moles / 4.82 × 10–3 moles = 4.00
These relative mole amounts give us the empirical formula: NaTcO4.
51
Chapter 3
3.84
⎛ 1 mol C ⎞
mol C = (0.423 g C) ⎜
⎟ = 0.0352 mol C
⎝ 12.01 g C ⎠
⎛ 1 mol Cl ⎞
mol Cl = (2.50 g Cl) ⎜
⎟ = 0.0705 mol Cl
⎝ 35.45 g Cl ⎠
⎛ 1 mol F ⎞
mol F = (1.34 g F) ⎜
⎟ = 0.0705 mol F
⎝ 19.00 g F ⎠
Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the
simplest mole ratio among the three elements in the compound:
for C, 0.0352 moles / 0.0352 moles = 1.00
for Cl, 0.0705 moles / 0.0352 moles = 2.000
for F, 0.0705 moles / 0.0352 moles = 2.00
These relative mole amounts give us the empirical formula CCl2F2
3.85
Assume a 100 g sample:
⎛ 1 mol C ⎞
mol C = (14.5 g C) ⎜
⎟ = 1.21 mol C
⎝ 12.01 g C ⎠
⎛ 1 mol Cl ⎞
mol Cl = (85.5 g Cl) ⎜
⎟ = 2.41 mol Cl
⎝ 35.45 g Cl ⎠
Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the
simplest mole ratio among the three elements in the compound:
for C, 1.21 moles / 1.21 moles = 1.00
for Cl, 2.41 moles /1.21 moles = 2.000
These relative mole amounts give us the empirical formula CCl2
3.86
To solve this problem we will assume that we have a 100 g sample. This implies that we have 77.26 g Hg, 9.25 g
C, 1.17 g H and 12.32 g O. The amount of oxygen was determined by subtracting the total amounts of the other
three elements from the total assumed mass of 100 g. Convert each of these masses into a number of moles:
⎛ 1 mole Hg ⎞
moles Hg = ( 77.26 g Hg ) ⎜
⎟ = 0.3852 moles Hg
⎝ 200.59 g Hg ⎠
⎛ 1 mole C ⎞
moles C = ( 9.25 g C ) ⎜
⎟ = 0.770 moles C
⎝ 12.011 g C ⎠
⎛ 1 mole H ⎞
moles H = (1.17 g H ) ⎜
⎟ = 1.16 moles H
⎝ 1.008 g H ⎠
⎛ 1 mole O ⎞
moles O = (12.32 g O ) ⎜
⎟ = 0.7700 moles O
⎝ 15.999 g O ⎠
The relative mole amounts are determined as follows:
for Hg, 0.3852 moles / 0.3852 moles = 1.000
for C, 0.770 moles / 0.3852 moles = 2.00
for H, 1.16 moles / 0.3852 moles = 3.01
for O, 0.7700 moles / 0.3852 moles = 1.999
and the empirical formula is HgC2H3O2. The empirical formula weight is 259.6 g/mole, which must be
multiplied by 2 in order to obtain the molecular weight. This means that the molecular formula is twice the
empirical formula, or Hg2C4H6O4.
52
Chapter 3
3.87
Assume a 100 g sample:
⎛ 1 mol C ⎞
mol C = (72.96 g C) ⎜
⎟ = 6.074 mol C
⎝ 12.011 g C ⎠
⎛ 1 mol H ⎞
mol H = (5.40 g H) ⎜
⎟ = 5.36 mol H
⎝ 1.008 g H ⎠
To find the number of moles of O, first we have to find the number of grams of O:
100 g total = (72.96 g C) + (5.40 g H) + (x g O)
g O = 21.64 g O
⎛ 1 mol O ⎞
mol O = (21.64 g O) ⎜
⎟ = 1.353 mol O
⎝ 15.999 g O ⎠
Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the
simplest mole ratio among the three elements in the compound:
for C, 6.074 moles / 1.353 moles = 4.49
for H, 5.36 moles / 1.353 moles = 3.96
for O, 1.353 moles / 1.353 moles = 1.00
These relative mole amounts give us the empirical formula C4.5H4O
Since we cannot have decimals as subscripts, multiply all of the subscripts by 2 to get the formula: C9H8O2
3.88
Assume a 100 g sample:
⎛ 1 mol C ⎞
mol C = (63.2 g C) ⎜
⎟ = 5.26 mol C
⎝ 12.01 g C ⎠
⎛ 1 mol H ⎞
mol H = (5.26 g H) ⎜
⎟ = 5.22 mol H
⎝ 1.008 g H ⎠
⎛ 1 mol O ⎞
mol O = (31.6 g O) ⎜
⎟ = 1.98 mol O
⎝ 16.00 g O ⎠
Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the
simplest mole ratio among the three elements in the compound:
for C, 5.26 moles / 1.98 moles = 2.66 = (8/3)
for H, 5.22 moles / 1.98 moles = 2.64 = (8/3)
for O, 1.98 moles / 1.98 moles = 1.00 = (3/3)
These relative mole amounts give us the empirical formula C8H8O3
3.89
The formula mass for the compound C19H30O2 is 290 g/mol. Thus, the empirical and molecular formulas
are equivalent.
3.90
The formula mass for the compound C28H46O is 399 g/mol. Thus, the empirical and molecular formulas
are equivalent.
3.91
(a)
Formula mass = 135.1 g
270.4 g/mol
= 2.001
135.1 g/mol
The molecular formula is Na2S4O6
(b)
Formula mass = 73.50 g
147.0 g/mol
= 2.000
73.50 g/mol
53
Chapter 3
The molecular formula is C6H4Cl2
3.92
3.93
(c)
Formula mass = 60.48 g
181.4 g/mol
= 2.999
60.48 g/mol
The molecular formula is C6H3Cl3
(a)
Formula mass = 122.1 g
732.6 g/mol
= 6.000
122.1 g/mol
The molecular formula is Na12Si6O18
(b)
Formula mass = 102.0 g
305.9 g/mol
= 2.999
102.0 g/mol
The molecular formula is Na3P3O9
(c)
Formula mass = 31.03 g
62.1 g/mol
= 2.00
31.03 g/mol
The molecular formula is C2H6O2
First, determine the amount of oxygen in the sample by subtracting the masses of the other elements from
the total mass:
0.6216 g – (0.1735 g C + 0.01455 g H + 0.2024 g N) = 0.2312 g O.
Now, convert these masses into a number of moles for each element:
⎛ 1 mole C ⎞
–2
mol C = (0.1735 g C) ⎜
⎟ = 1.445 × 10 mol C
12.011
g
C
⎝
⎠
⎛ 1 mole H ⎞
–2
mol H = (0.01455 g H) ⎜
⎟ = 1.444× 10 mol H
1.0079
g
H
⎝
⎠
⎛ 1 mole N ⎞
–2
mol N = (0.2024 g N) ⎜
⎟ = 1.444 × 10 mol N
14.007
g
N
⎝
⎠
⎛ 1 mole O ⎞
–2
mol O = (0.2312 g O) ⎜
⎟ = 1.444 × 10 mol O
⎝ 15.999 g O ⎠
These are clearly all the same mole amounts, and we deduce that the empirical formula is CHNO, which
has a formula weight of 43. It can be seen that the number 43 must be multiplied by the integer 3 in order
to obtain the molar mass (3 × 43 = 129), and this means that the empirical formula should similarly be
multiplied by 3 in order to arrive at the molecular formula, C3H3N3O3.
3.94
To solve this problem we will assume that we have a 100 g sample. This implies that we have 75.42 g C, 6.63 g H,
8.38 g N and 9.57 g O. The amount of oxygen was determined by subtracting the total amounts of the other three
elements from the total assumed mass of 100 g. Convert each of these masses into a number of moles:
⎛ 1 mole C ⎞
mol C = (75.42 g C) ⎜
⎟ = 6.279 mol C
⎝ 12.011 g C ⎠
⎛ 1 mole H ⎞
mol H = (6.63 g H) ⎜
⎟ = 6.58 mol H
⎝ 1.008 g H ⎠
54
Chapter 3
⎛ 1 mole N ⎞
mol N = (8.38 g N) ⎜
⎟ = 0.598 mol N
⎝ 14.01 g N ⎠
⎛ 1 mole O ⎞
mol O = (9.57 g O) ⎜
⎟ = 0.598 mol O
⎝ 16.00 g O ⎠
The relative mole amounts are determined as follows:
for C, 6.279 mol / 0.598 mol = 10.5
for H, 6.58 mol / 0.598 mol = 11.0
for N, 0.598 mol / 0.598 mol = 1.00
for O, 0.598 mol / 0.598 mol = 1.00
In order to obtain whole numbers, each of these values is multiplied by 2 and we determine the empirical
formula is C21H22N2O2. The empirical formula weight is 334 g/mole. This means that the molecular
formula is the same as the empirical formula, or C21H22N2O2.
3.95
From the information provided, we can determine the mass of mercury as the difference between the total
mass and the mass of bromine:
g Hg = 0.389 g compound – 0.111 g Br = 0.278 g Hg
To determine the empirical formula, first convert the two masses to a number of moles.
⎛ 1 mole Hg ⎞
–3
mol Hg = (0.278 g Hg) ⎜
⎟ = 1.39 × 10 mol Hg
200.59
g
Hg
⎝
⎠
⎛ 1 mole Br ⎞
–3
mol Br = (0.111 g Br) ⎜
⎟ = 1.39 × 10 mol Br
⎝ 79.904 g Br ⎠
Now, we would divide each of these values by the smaller quantity to determine the simplest mole ratio
between the two elements. By inspection, though, we can see there are the same number of moles of Hg
and Br. Consequently, the simplest mole ratio is 1:1 and the empirical formula is HgBr.
To determine the molecular formula, recall that the ratio of the molecular mass to the empirical mass is
equivalent to the ratio of the molecular formula to the empirical formula. Thus, we need to calculate an
empirical mass:
(1 mole Hg)(200.59 g Hg/mole Hg) + (1 mole Br)(79.904 g Br/mole Br) = 280.49 g/mole HgBr.
The molecular mass, as reported in the problem is 561 g/mole. The ratio of these is:
561 g/mole
= 2.00
280.49 g/mole
So, the molecular formula is two times the empirical formula or Hg2Br2.
3.96
From the information provided, the mass of sulfur is the difference between the total mass and the mass of
antimony:
g S = 0.6662 g compound – 0.4017 g Sb = 0.2645 g S
To determine the empirical formula, first convert the two masses to a number of moles.
⎛ 1 mole S ⎞
–3
mol S = (0.2645 g S) ⎜
⎟ = 8.249 × 10 mol S
32.065
g
S
⎝
⎠
55
Chapter 3
⎛ 1 mole Sb ⎞
–3
mol Sb = (0.4017 g Sb) ⎜
⎟ = 3.299 × 10 mol Sb
121.76
g
Sb
⎝
⎠
Now, divide each of these values by the smaller quantity to determine the simplest mole ratio between the
two elements:
For Sb: 3.299 × 10–3 moles/3.299 × 10–3 moles = 1.000 mol Sb
For S: 8.249 × 10–3 moles/3.299 × 10–3 moles = 2.500 mol S
Hence the empirical formula is Sb2S5, and the empirical mass is (2 × Sb) + (5 × S) = 403.85 g/mol. Since
the molecular mass reported in the problem is the same as the calculated empirical mass, the empirical
formula is the same as the molecular formula.
3.97
(a)
(b)
(c)
(d)
(e)
Mg(OH)2 + 2HBr J MgBr2 + 2H2O
2HCl + Ca(OH)2 J CaCl2 + 2H2O
Al2O3 + 3H2SO4 J Al2(SO4)3 + 3H2O
2KHCO3 + H3PO4 J K2HPO4 + 2H2O + 2CO2
C9H20 + 14O2 J 9CO2 + 10H2O
3.98
(a)
(b)
(c)
(d)
(e)
CaO + 2HNO3 → Ca(NO3)2 + H2O
Na2CO3 + Mg(NO3)2 J MgCO3 + 2NaNO3
(NH4)3PO4 + 3NaOH J Na3PO4 + 3NH3 + 3H2O
2LiHCO3 + H2SO4 J Li2SO4 + 2H2O + 2CO2
C4H10O + 6O2 J 4CO2 + 5H2O
3.99
(a)
(b)
(c)
(d)
(e)
Ca(OH)2 + 2HCl J CaCl2 + 2H2O
2AgNO3 + CaCl2 J Ca(NO3)2 + 2AgCl
Pb(NO3)2 + Na2SO4 J PbSO4 + 2NaNO3
2Fe2O3 + 3C J 4Fe + 3CO2
2C4H10 + 13O2 J 8CO2 + 10H2O
3.100
(a)
(b)
(c)
(d)
(e)
2SO2 + O2 J 2SO3
2NaHCO3 + H2SO4 J Na2SO4 + 2H2O + 2CO2
P4O10 + 6H2O J 4H3PO4
Fe2O3 + 3H2 J 2Fe + 3H2O
2Al + 3H2SO4 J Al2(SO4)3 + 3H2
3.101
4Fe(s) + 3O2(g) J 2Fe2O3(s)
3.102
2NO(g) + O2(g) J 2NO2(g)
3.103
36 mol H
3.104
24 moles of O
3.105
2FeCl3 + SnCl2 J 2FeCl2 + SnCl4
3.106
AlCl3(aq) + 3AgNO3(aq) J 3AgCl(s) + Al(NO3)3(aq)
3.107
(a)
(b)
(c)
4P + 5O2 J P4O10
⎛ 1 mol P ⎞ ⎛ 5 mol O 2 ⎞ ⎛ 32.0 g O 2 ⎞
g O2 = (6.85 g P) ⎜
⎟ = 8.85 g O2
⎟⎜
⎟⎜
⎝ 30.97 g P ⎠ ⎝ 4 mol P ⎠ ⎝ 1 mol O 2 ⎠
⎛ 1 mol O 2 ⎞ ⎛ 1 mol P4 O10 ⎞ ⎛ 283.9 g P4 O10 ⎞
g P4O10 = (8.00 g O2) ⎜
⎟ = 14.2 g P4O10
⎟⎜
⎟⎜
⎝ 32.00 g O2 ⎠ ⎝ 5 mol O 2 ⎠ ⎝ 1 mol P4 O10 ⎠
56
Chapter 3
(d)
3.108
(a)
(b)
(c)
(d)
3.109
(a)
(b)
(c)
(d)
3.110
(a)
(b)
(c)
(d)
3.111
3.112
⎛ 1 mol P4 O10 ⎞ ⎛ 4 mol P
g P = (7.46 g P4O10) ⎜
⎟⎜
⎝ 283.9 g P4 O10 ⎠ ⎝ 1 mol P4 O10
2C4H10 + 13O2 J 8CO2 + 10H2O
⎛ 1 mol H 2 O ⎞ ⎛ 2 mol C4 H10 ⎞ ⎛ 58.12 g C4 H10 ⎞
g C4H10 = (4.46 g H2O) ⎜
⎟ = 2.88 g C4H10
⎟⎜
⎟⎜
⎝ 18.02 g H 2 O ⎠ ⎝ 10 mol H 2 O ⎠ ⎝ 1 mol C4 H10 ⎠
⎛ 1 mol C4 H10 ⎞⎛ 13 mol O2 ⎞ ⎛ 32.0 g O 2 ⎞
g O2 = (2.88 g C4H10) ⎜
⎟⎜
⎟⎜
⎟ = 10.3 g O2
⎝ 58.12 g C4 H10 ⎠⎝ 2 mol C4 H10 ⎠ ⎝ 1 mol O2 ⎠
⎛ 8 mol CO2 ⎞⎛ 44.01 g CO 2 ⎞
g CO2 = (0.0496 mol C4H10) ⎜
⎟⎜
⎟ = 8.73 g CO2
⎝ 2 mol C4 H10 ⎠ ⎝ 1 mol CO 2 ⎠
⎛ 1 mole Na 2S2 O3 ⎞
mol Na2S2O3 = (0.12 mol Cl2) ⎜
⎟ = 0.030 mol Na2S2O3
⎝ 4 mole Cl2 ⎠
⎛ 8 mole HCl ⎞
mol HCl = (0.12 mol Cl2) ⎜
⎟ = 0.24 mol HCl
⎝ 4 mole Cl2 ⎠
⎛ 5 mole H 2 O ⎞
mol H2O = (0.12 mol Cl2) ⎜
⎟ = 0.15 mol H2O
⎝ 4 mole Cl2 ⎠
⎛ 5 mole H 2 O ⎞
mol H2O = (0.24 mol HCl) ⎜
⎟ = 0.15 mol H2O
⎝ 8 mole HCl ⎠
⎛ 25 mole O 2 ⎞
mol O2 = (6 mol C8H18) ⎜
⎟ = 80 mol O2
⎝ 2 mole C8 H18 ⎠
(Note: This calculation is limited due to sig figs.)
⎛ 16 mole CO 2 ⎞
mol CO2 = (0.5 mol C8H18) ⎜
⎟ = 4 mol CO2
⎝ 2 mole C8 H18 ⎠
⎛ 18 mole H 2 O ⎞
mol H2O = (8 mol C8H18) ⎜
⎟ = 70 mol H2O
⎝ 2 mole C8 H18 ⎠
⎛ 25 mole O 2 ⎞
mol O2 = ( 6.00 mol CO2) ⎜
⎟ = 9.38 mol O2
⎝ 16 mole CO 2 ⎠
⎛ 2 mole C8 H18 ⎞
mol C8H18 = (6.00 mol CO2) ⎜
⎟ = 0.750 mol C8H18
⎝ 16 mole CO 2 ⎠
⎛ 1000 g H 2 O 2 ⎞ ⎛ 1 mol H 2 O 2 ⎞ ⎛ 1 mol O 2 ⎞ ⎛ 32.00 g O 2 ⎞ ⎛ 1 kg O 2 ⎞
kg O2 = 1.0 kg H2O2 ⎜
⎟⎜
⎟⎜
⎟⎜
⎟⎜
⎟
⎝ 1 kg H 2 O 2 ⎠ ⎝ 34.01 g H 2 O 2 ⎠ ⎝ 2 mol H 2 O 2 ⎠ ⎝ 1 mol O2 ⎠ ⎝ 1000 g O 2 ⎠
= 0.47 kg O2
⎛ 1000 g KClO3 ⎞⎛ 1 mol KClO3 ⎞ ⎛ 3 mol O 2 ⎞ ⎛ 32.00 g O2 ⎞⎛ 1 kg O2 ⎞
kg O2 = 1.0 kg KClO2 ⎜
⎟⎜
⎟⎜
⎟⎜
⎟⎜
⎟
⎝ 1 kg KClO3 ⎠⎝ 122.6 g KClO3 ⎠ ⎝ 2 mol KClO3 ⎠ ⎝ 1 mol O 2 ⎠⎝ 1000 g O 2 ⎠
= 0.39 kg O2
⎛
3.113
⎞ ⎛ 30.97 g P ⎞
⎟⎜
⎟ = 3.26 g P
⎠ ⎝ 1 mol P ⎠
(a)
–⎜
1 mol Zn
0.11 mol Au(CN)2 ⎜
−
⎝ 2 mol Au(CN)2
⎞ ⎛ 65.39 g Zn ⎞
⎟⎜
⎟ ⎝ 1 mol Zn ⎟⎠ = 3.6 g Zn
⎠
57
Chapter 3
⎛
3.114
(b)
⎞ ⎛ 197.0 g Au ⎞
⎟
⎟ = 22 g Au
0.11 mol Au(CN)2 ⎜
− ⎟⎜
⎝ 2 mol Au(CN)2 ⎠ ⎝ 1 mol Au ⎠
(c)
⎛ 2 mol Au(CN) −
2
0.11 mol Zn ⎜
⎜
1
mol
Zn
⎝
(a)
(b)
(c)
–⎜
2 mol Au
⎞ ⎛ 249.0 g Au(CN) −
2
⎟⎜
⎟ ⎜ 1 mol Au(CN) −
2
⎠⎝
⎞
⎟ = 55 g Au(CN)2–
⎟
⎠
⎛ 5 mol O2 ⎞ ⎛ 32.00 g O 2 ⎞
3 mol C3H8 ⎜
⎟⎜
⎟ = 500 g O2
⎝ 1 mol C3 H8 ⎠ ⎝ 1 mol O2 ⎠
⎛ 3 mol CO 2 ⎞ ⎛ 44.01 g CO 2 ⎞
0.1 mol C3H8 ⎜
⎟⎜
⎟ = 13 g CO2
⎝ 1 mol C3 H8 ⎠ ⎝ 1 mol CO 2 ⎠
⎛ 4 mol H 2 O ⎞ ⎛ 18.01 g H 2 O ⎞
4 mol C3H8 ⎜
⎟⎜
⎟ = 300 g H2O
⎝ 1 mol C3 H8 ⎠ ⎝ 1 mol H 2 O ⎠
⎞ ⎛ 63.013 g HNO3 ⎞
⎟ = 30.28 g HNO3
⎟⎜
⎠ ⎝ 1 mol HNO3 ⎠
3.115
⎛ 1 mol Cu ⎞ ⎛ 8 mol HNO3
g HNO3 = (11.45 g Cu) ⎜
⎟⎜
⎝ 63.546 g Cu ⎠ ⎝ 3 mol Cu
3.116
⎛ 1 mol N 2 H 4 ⎞⎛ 7 mol H 2 O 2 ⎞ ⎛ 34.02 g H 2 O 2 ⎞
g H2O2 = (852 g N2H4) ⎜
⎟⎜
⎟⎜
⎟ = 6330 g H2O2
⎝ 32.05 g N 2 H 4 ⎠⎝ 1 mol N 2 H 4 ⎠ ⎝ 1 mol H 2 O2 ⎠
3.117
3AgNO3 + FeCl3 → 3AgCl + Fe(NO3)3
Calculate the amount of FeCl3 that are required to react completely with all of the available silver nitrate:
⎛ 1 mol AgNO3 ⎞ ⎛ 1 mol FeCl3 ⎞⎛ 162.21 g FeCl3 ⎞
g FeCl3 = (18.0 g AgNO3) ⎜
⎟⎜
⎟⎜
⎟
⎝ 169.87 g AgNO3 ⎠ ⎝ 3 mol AgNO3 ⎠⎝ 1 mol FeCl3 ⎠
= 5.73 g FeCl3
Since more than this minimum amount is available, FeCl3 is present in excess, and AgNO3 must be the
limiting reactant.
We know that only 5.73 g FeCl3 will be used. Therefore, the amount left unused is:
32.4 g total – 5.73 g used = 26.7 g FeCl3
3.118
First, calculate the amount of H2O needed to completely react with the available ClO2;
⎛ 1 mol ClO2 ⎞ ⎛ 3 mol H 2 O ⎞ ⎛ 18.02 g H 2 O ⎞
g H2O = 142.0 g ClO2 ⎜
⎟⎜
⎟⎜
⎟ =18.97 g H2O
⎝ 67.45 g ClO 2 ⎠ ⎝ 6 mol ClO 2 ⎠ ⎝ 1 mol H 2 O ⎠
So, there is excess H2O present. The amount that remains is 38.0 g – 18.97 g = 19.0 g H2O.
3.119
First calculate the number of moles of water that are needed to react completely with the given amount of
NO2:
⎛ 1 mol NO2 ⎞ ⎛ 1 mol H 2 O ⎞ ⎛ 18.02 g H 2 O ⎞
–4
g H2O = 0.0010 g NO2 ⎜
⎟⎜
⎟⎜
⎟ = 1.3 × 10 g H2O
46.01
g
NO
3
mol
NO
1
mol
H
O
2 ⎠⎝
2 ⎠⎝
2
⎝
⎠
Since this is less than the amount of water that is supplied, the limiting reactant must be NO2.
Therefore, to calculate the amount of HNO3:
⎛ 1 mol NO2 ⎞ ⎛ 2 mol HNO3 ⎞ ⎛ 63.02 g HNO3 ⎞
g HNO3 = 0.0010 g NO2 ⎜
⎟ = 0.913 mg HNO3
⎟⎜
⎟⎜
⎝ 46.01 g NO 2 ⎠ ⎝ 3 mol NO 2 ⎠ ⎝ 1 mol HNO3 ⎠
58
Chapter 3
3.120
(a)
First calculate the number of moles of water that are needed to react completely with the given
amount of PCl5:
⎛ 4 mole H 2 O ⎞
mol H2O = (0.360 mol PCl5) ⎜
⎟ = 1.44 mol H2O
⎝ 1 moles PCl5 ⎠
Since this is less than the amount of water that is supplied, the limiting reactant must be PCl5.
This can be confirmed by the following calculation:
⎛ 1 mol PCl5 ⎞
mol PCl5 = (2.88 mol H2O) ⎜
⎟ = 0.720 mol PCl5
⎝ 4 mol H 2 O ⎠
(b)
3.121
(a)
which also demonstrates that the limiting reactant is PCl5.
⎛ 5 mol HCl ⎞ ⎛ 36.46 g HCl ⎞
g HCl = (0.360 mol PCl5) ⎜
⎟⎜
⎟ = 65.6 g HCl
⎝ 1 mol PCl5 ⎠ ⎝ 1 mol HCl ⎠
First determine the amount of Fe2O3 that would be required to react completely with the given
amount of Al:
⎛ 1 mol Fe2 O3 ⎞
mol Fe2O3 = (4.20 mol Al) ⎜
⎟ = 2.10 mol Fe2O3
⎝ 2 mol Al ⎠
Since only 1.75 mol of Fe2O3 are supplied, it is the limiting reactant. This can be confirmed by
calculating the amount of Al that would be required to react completely with all of the available Fe2O3:
⎛ 2 mol Al ⎞
mol Al = (1.75 mol Fe2O3) ⎜
⎟ = 3.50 mol Al
⎝ 1 mol Fe 2 O3 ⎠
Since an excess (4.20 mol – 3.50 mol = 0.70 mol) of Al is present, Fe2O3 must be the limiting
reactant, as determined above.
(b)
3.122
(a)
⎛ 2 mol Fe ⎞ ⎛ 55.847 g Fe ⎞
g Fe = (1.75 mol Fe2O3) ⎜
⎟⎜
⎟ = 195 g Fe
⎝ 1 mol Fe 2 O3 ⎠ ⎝ 1 mol Fe ⎠
First determine the amount of C2H4 that would be required to react completely with the given
amount of H2O:
⎛ 1000 g C2 H 4 ⎞ ⎛ 1 mol C2 H 4 ⎞ ⎛ 1 mol H 2 O ⎞
g H2O = 1.0 kg C2H4 ⎜
⎟⎜
⎟⎜
⎟ ×
⎝ 1 kg C2 H 4 ⎠ ⎝ 28.05 g C2 H 4 ⎠ ⎝ 1 mol C2 H 4 ⎠
⎛ 18.02 g H 2 O ⎞ ⎛ 1 kg H 2 O ⎞
⎜
⎟⎜
⎟ = 0.64 kg H2O
⎝ 1 mol H 2 O ⎠ ⎝ 1000 g H 2 O ⎠
Since only 0.010 kg of H2O are supplied, it is the limiting reactant. This can be confirmed by
calculating the amount of C2H4 that would be required to react completely with all of the available H2O:
⎛ 1000 g H 2 O ⎞ ⎛ 1 mol H 2 O ⎞ ⎛ 1 mol C2 H 4 ⎞
g C2H4 = 0.01 kg H2O ⎜
⎟⎜
⎟⎜
⎟
⎝ 1 kg H 2 O ⎠ ⎝ 18.02 g H 2 O ⎠ ⎝ 1 mol H 2 O ⎠
⎛ 28.05 g C2 H 4 ⎞ ⎛ 1 kg C2 H 4 ⎞
⎜
⎟⎜
⎟ = 0.016 kg C2H4
⎝ 1 mol C2 H 4 ⎠ ⎝ 1000 g C2 H 4 ⎠
59
Chapter 3
(b)
3.123
⎛ 1000 g H 2 O ⎞ ⎛ 1 mol H 2 O ⎞
g C2H5OH = (0.010 kg H2O) ⎜
⎟⎜
⎟
⎝ 1 kg H 2 O ⎠ ⎝ 18.02 g H 2 O ⎠
⎛ 1 mol C2 H5 OH ⎞ ⎛ 46.08 g C2 H5OH ⎞
× ⎜
⎟ = 26 g C2H5OH
⎟⎜
⎝ 1 mol H 2 O ⎠ ⎝ 1 mol C2 H5 OH ⎠
First, determine how much H2SO4 is needed to completely react with the AlCl3
⎛ 1 mol AlCl3 ⎞ ⎛ 3 mol H 2SO 4 ⎞ ⎛ 98.08 g H 2SO 4 ⎞
g H2SO4 = 25 g AlCl3 ⎜
⎟⎜
⎟⎜
⎟
⎝ 133.34 g AlCl3 ⎠ ⎝ 2 mol AlCl3 ⎠ ⎝ 1 mol H 2SO 4 ⎠
= 27.58 g H2SO4
There is an excess of H2SO4 present.
Determine the theoretical yield:
⎛ 1 mol AlCl3 ⎞ ⎛ 1 mol Al2 (SO 4 )3 ⎞ ⎛ 342.17 g Al2 (SO 4 )3 ⎞
g Al2(SO4)3 = 25.00 g AlCl3 ⎜
⎟⎜
⎟⎜
⎟
⎝ 133.33 g AlCl3 ⎠ ⎝ 2 mol AlCl3 ⎠ ⎝ 1 mol Al2 (SO 4 )3 ⎠
= 32.08 g Al2(SO4)3
actual yield
28.46 g
% yield =
× 100 =
× 100 = 88.72%
theoretical yield
32.08 g
3.124
Assume there is excess oxygen present and determine the theoretical yield of carbon dioxide.
⎛ 1 mol CH3OH ⎞ ⎛ 2 mol CO 2 ⎞ ⎛ 44.01 g CO2 ⎞
g CO 2 = (6.40 g CH3OH) ⎜
⎟⎜
⎟⎜
⎟ = 8.79 g CO2
⎝ 32.04 g CH3OH ⎠ ⎝ 2 mol CH3OH ⎠ ⎝ 1 mol CO2 ⎠
6.12 g
% yield =
× 100% = 69.6%
8.79g
3.125
First determine the theoretical yield:
⎛ 1 mol Ba(NO3 )2 ⎞ ⎛ 1 mol BaSO4 ⎞⎛ 233.39 g BaSO 4 ⎞
g BaSO4 = (75.00 g Ba(NO3)2 ⎜
⎟⎜
⎟⎜
⎟
⎝ 261.34 g Ba(NO3 ) 2 ⎠ ⎝ 1 mol Ba(NO3 )2 ⎠ ⎝ 1 mol BaSO4 ⎠
= 66.98 g BaSO4
Then calculate a % yield:
actual yield
64.45 g
× 100 =
× 100 = 96.22%
% yield =
theoretical yield
66.98 g
3.126
The theoretical yield is
⎛ 1 mole NaCl ⎞ ⎛ 1 mol NaHCO3
g Na2CO3 = 120 g NaCl ⎜
⎟⎜
⎝ 58.44 g NaCl ⎠ ⎝ 1 mol NaCl
= 108.8 g Na2CO3
% yield =
3.127
⎞ ⎛ 1 mol Na 2 CO3 ⎞⎛ 105.99 g Na 2 CO3 ⎞
⎟⎜
⎟
⎟⎜
⎠ ⎝ 2 mol NaHCO3 ⎠⎝ 1 mol Na 2 CO3 ⎠
actual yield
85.4 g
× 100 =
× 100 = 78.5%
theoretical yield
108.8 g
If the yield for this reaction is only 71% and we need to have 11.5 g of product, we will attempt to make 16
g of product. This is determined by dividing the actual yield by the percent yield. Recall
actual yield
× 100 . If we rearrange this equation we can see
that; % yield =
theoretical yield
60
Chapter 3
actual yield
× 100 . Substituting the values from this problem gives the 16 g of
% yield
product mentioned above.
that theoretical yield =
⎛ 1 mol KC7 H5 O 2 ⎞ ⎛ 1 mol C7 H8 ⎞⎛ 92.14 g C7 H8 ⎞
g C7 H8 =16 g KC7 H5 O2 ⎜
⎟⎜
⎟⎜
⎟ = 9.2 g C7 H8
⎝ 160.21 g KC7 H5O 2 ⎠ ⎝ 1 mol KC7 H5O 2 ⎠⎝ 1 mol C7 H8 ⎠
3.128
First, determine how much MnI2 is needed to completely react with the F2
⎛ 1 mol F2 ⎞ ⎛ 2 mol MnI2 ⎞ ⎛ 308.75 g MnI 2 ⎞
g MnI2 = 10.0 g F2 ⎜
⎟⎜
⎟⎜
⎟ = 12.5 g MnI2
38.00
g
F
13
mol
F
1
mol
MnI
2
2
2
⎝
⎠⎝
⎠⎝
⎠
There is an excess of F2 present.
Note that MnI2 is the limiting reactant and that MnI2 and MnF3 are in a 1:1 ratio, so the number of mole of
MnI2 equals the number of moles of MnF3. According to the problem statement, we will only prepare 56%
of this number of moles of MnF3.
⎛ 1 mol MnI2 ⎞ ⎛ 2 mol MnF3 ⎞ ⎛ 111.93 g MnF3 ⎞
g MnF3 = 10.0 g MnI2 ⎜
⎟ ( 0.56 ) = 2.03 g MnF3
⎟⎜
⎟⎜
⎝ 308.75 g MnI2 ⎠ ⎝ 2 mol MnI2 ⎠ ⎝ 1 mol MnF3 ⎠
Additional Exercises
BU
3.129
Assume the hydrogen is the limiting reactant.
⎞
1 lb O2
⎛ 453.59237 g ⎞ ⎛ 1 mol H 2 ⎞ ⎛ 1 mol O 2 ⎞⎛ 31.9988 g O 2 ⎞ ⎛
lb O2 = 227,641 lb H2 ⎜
⎜
⎟⎜
⎟⎜
⎟⎜
⎟
⎟
1 lb
⎝
⎠ ⎝ 2.01588 g H 2 ⎠ ⎝ 2 mol H 2 ⎠⎝ 1 mol O2 ⎠ ⎝ 453.59237 g O 2 ⎠
=1,806,714 lb O2
Since this is more than the amount of O2 that is supplied, the limiting reactant must be O2. Next calculate
the amount of H2 needed to react completely with all of the available O2.
⎞
1 lb H 2
⎛ 453.59237 g ⎞ ⎛ 1 mol O2 ⎞ ⎛ 2 mol H 2 ⎞ ⎛ 2.01588 g H 2 ⎞ ⎛
lb H2 = 1,361,936 lb O2 ⎜
⎟⎜
⎟
⎟ ⎜ 31.9988 g O ⎟ ⎜ 1 mol O ⎟ ⎜ 1 mol H
1
lb
453.59237
g
H
⎝
⎠⎝
2 ⎠⎝
2 ⎠⎝
2
2⎠
⎠⎝
=171,600 lb H2
Since only 171,600 lb. of H2 reacted, there are 227,641 lb. – 171,600 lb. = 56,041 lb. of unreacted H2.
3.130
Since 5.00 g represents 86.0% of the required amount, we can solve for the amount that should be made:
5.00 g = 86.0 % of X; X = 5.81 g Pb(NO3)2.
⎛ 1 mol Pb(NO3 )2 ⎞ ⎛ 1 mol PbO ⎞ ⎛ 223.2 g PbO ⎞
g PbO = (5.81 g Pb(NO3)2) ⎜
⎟⎜
⎟⎜
⎟ = 3.92 g PbO
⎝ 331.21 g Pb(NO3 ) 2 ⎠ ⎝ 1 mol Pb(NO3 ) 2 ⎠ ⎝ 1 mol PbO ⎠
3.131
⎛ 1 mol N ⎞ ⎛ 1 mol (NH 2 ) 2 CO ⎞ ⎛ 60.06 g (NH 2 ) 2 CO ⎞
g (NH2)2CO = 6.00 g N ⎜
⎟⎜
⎟ ⎜ 1 mol (NH ) CO ⎟ = 12.9 g (NH2)2CO
2 mol N
⎠⎝
⎝ 14.007 g N ⎠ ⎝
2 2
⎠
3.132
⎛ 2000 lb Hg ⎞ ⎛ 1 kg Hg ⎞ ⎛ 1000 g Hg ⎞ ⎛ 1 mol Hg ⎞
lb (CH3)2Hg = (263 tons Hg)(0.010) ⎜
⎟⎜
⎟⎜
⎟⎜
⎟
⎝ 1 ton Hg ⎠ ⎝ 2.205 lb Hg ⎠ ⎝ 1 kg Hg ⎠ ⎝ 200.59 g Hg ⎠
⎛ 1 mol (CH3 ) 2 Hg ⎞ ⎛ 230.66 g (CH3 ) 2 Hg ⎞
6
⎟ = 2.7 × 10 g (CH3)2Hg
⎜
⎟⎜
1
mol
Hg
1
mol
(CH
)
Hg
⎝
⎠⎝
3 2
⎠
61
Chapter 3
⎛ 1k g (CH3 )2 Hg ⎞ ⎛ 2.205 lb (CH3 ) 2 Hg ⎞
2.7 × 106 g (CH3)2Hg ⎜
⎟⎜
⎟ = 5950 lb (CH3)2Hg
⎝ 1000 g (CH3 )2 Hg ⎠ ⎝ 1 kg (CH3 ) 2 Hg ⎠
3.133
Calculate the cost of one mole nitrogen from each compound:
⎛
⎞⎛ 1 kg NH 4 NO3 ⎞ ⎛ 80.04 g NH 4 NO3 ⎞ ⎛ 1 mol NH 4 NO3 ⎞
$625
(a)
$ per mol N = ⎜
⎟⎜
⎟⎜
⎟⎜
⎟
2 mol N
⎠
⎝ 25 kg NH 4 NO3 ⎠⎝ 1000 g NH 4 NO3 ⎠ ⎝ 1 mol NH 4 NO3 ⎠ ⎝
(b)
= $1.00 per mol N
⎛
$55
$ per mol N = ⎜
⎜ 1 kg ( NH 4 ) HPO4
2
⎝
⎛ 1 mol ( NH 4 ) HPO 4
2
⎜
⎜
2
mol
N
⎝
⎞⎛ 1 kg ( NH 4 ) HPO 4
2
⎟⎜
⎟⎜ 1000 g ( NH 4 ) HPO 4
2
⎠⎝
⎞ ⎛ 132.1 g ( NH 4 ) HPO 4
2
⎟⎜
⎟ ⎜ 1 mol ( NH 4 ) HPO 4
2
⎠⎝
⎞
⎟
⎟
⎠
⎞
⎟ = $3.66 per mol N
⎟
⎠
⎛
⎞ ⎛ 1 kg CH 4 ON 2 ⎞ ⎛ 60.06 g CH 4 ON 2 ⎞ ⎛ 1 mol CH 4 ON 2 ⎞
$60
$ per mol N = ⎜
⎟⎜
⎟⎜
⎟⎜
⎟
2 mol N
⎠
⎝ 5 kg CH 4 ON 2 ⎠ ⎝ 1000 g CH 4 ON 2 ⎠ ⎝ 1 mol CH 4 ON 2 ⎠ ⎝
= $0.36 per mol N
⎛ $128 ⎞⎛ 1 kg NH3 ⎞ ⎛ 17.03 g NH3 ⎞ ⎛ 1 mol NH3 ⎞
(d)
$ per mol N = ⎜
⎟⎜
⎟⎜
⎟⎜
⎟
⎝ 50 kg NH3 ⎠⎝ 1000 g NH3 ⎠ ⎝ 1 mol NH3 ⎠ ⎝ 1 mol N ⎠
= $0.04 per mol N
NH3 is the cheapest and could be the most economical.
(c)
3.134
3.135
⎛ 1 mol C6 H 6 ⎞ ⎛ 6 mol C
g Na2C2O4 = (125 g C6H6) ⎜
⎟⎜
⎝ 78.11 g C6 H 6 ⎠ ⎝ 1 mol C6 H 6
= 643 g NaC2O4
⎞ ⎛ 1 mol Na 2 C2 O4
⎟⎜
2 mol C
⎠⎝
⎞ ⎛ 134.00 g Na 2 C2 O4 ⎞
⎟ ⎜ 1 mol Na C O ⎟
⎠⎝
2 2 4 ⎠
Only 27% of the paint is left in paint chip after 73% has evaporated.
The mass of the wet paint is:
0.15 g
= 0.56 g
0.27
g PbCr2O7 = 0.56 g sample × 14.5% PbCr2O7 = 0.0812 g PbCr2O7
⎛ 1 mol PbCr2 O7 ⎞ ⎛ 1 mol Pb
⎞ ⎛ 207.2 g Pb ⎞
g Pb = 0.0812 g PbCr2O7 ⎜
⎟⎜
⎟⎜
⎟ = 0.040 g Pb
⎝ 423.2 g PbCr2 O7 ⎠ ⎝ 1 mol PbCr2 O7 ⎠ ⎝ 1 mol Pb ⎠
3.136
First determine the percentage by weight of each element in the respective original samples. This is done
by determining the mass of the element in question present in each of the original samples. The percentage
by weight of each element in the unknown will be the same as the values we calculate.
⎛ 1 mol CaCO3 ⎞ ⎛ 1 mol Ca ⎞ ⎛ 40.1 g Ca ⎞
g Ca = (0.160 g CaCO3) ⎜
⎟⎜
⎟⎜
⎟ = 0.0641 g Ca
⎝ 100.09 g CaCO3 ⎠ ⎝ 1 mol CaCO3 ⎠ ⎝ 1 mol Ca ⎠
% Ca = (0.0641/0.250) × 100% = 25.6% Ca
⎛ 1 mol BaSO 4 ⎞⎛ 1 mol S
g S = (0.344 g BaSO4) ⎜
⎟⎜
⎝ 233.8 g BaSO 4 ⎠⎝ 1 mol BaSO 4
% S = (0.0472/0.115) × 100% = 41.0% S
62
⎞ ⎛ 32.07 g S ⎞
⎟⎜
⎟ = 0.0472 g S
⎠ ⎝ 1 mol S ⎠
Chapter 3
⎛ 1 mol NH3 ⎞ ⎛ 1 mol N ⎞ ⎛ 14.01 g N ⎞
g N = (0.155 g NH3) ⎜
⎟⎜
⎟⎜
⎟ = 0.128 g N
⎝ 17.03 g NH3 ⎠ ⎝ 1 mol NH3 ⎠ ⎝ 1 mol N ⎠
% N = (0.128/0.712) × 100% = 18.0% N
% C = 100.0 – (25.6 + 41.0 + 18.0) = 15.4% C. Next, we assume 100 g of the compound, and convert
these weight percentages into mole amounts:
⎛ 1 mol Ca ⎞
mol Ca = ( 25.6 g Ca ) ⎜
⎟ = 0.639 mol Ca
⎝ 40.08 g Ca ⎠
⎛ 1 mol S ⎞
mol S = ( 41.0 g S) ⎜
⎟ = 1.28 mol S
⎝ 32.07 g S ⎠
⎛ 1 mol N ⎞
mol N = (18.0 g N ) ⎜
⎟ = 1.28 mol N
⎝ 14.07 g N ⎠
⎛ 1 mole C ⎞
moles C = (15.4 g C ) ⎜
⎟ = 1.28 moles C
⎝ 12.01 g C ⎠
Dividing each of these mole amounts by the smallest, we have:
For Ca: 0.639 mol / 0.639 mol = 1.00
For S: 1.28 mol / 0.639 mol = 2.00
For N: 1.28 mol / 0.639 mol = 2.00
For C: 1.28 mol / 0.639 mol = 2.00
The empirical formula is therefore CaC2S2N2, and the mass of the empirical unit is Ca + 2S + 2N + 2C
= 156 g/mol. Since the molecular mass is the same as the empirical mass, the molecular formula is
CaC2S2N2.
3.137
(a)
One mole of N2, 2 moles of H2O and 1/2 mole of O2 for a total of 3 1/2 moles of gases.
(b)
⎛ 2000 lb ⎞ ⎛ 453.59 g ⎞
mol of gases = (1.00 ton NH4NO3) ⎜
⎟⎜
⎟
⎝ 1 ton ⎠ ⎝ 1 lb ⎠
⎛ 1 mol NH 4 NO3 ⎞ ⎛ 3.5 mol gas ⎞
4
× ⎜
⎟⎜
⎟ = 3.97 × 10 mol gas
80.04
g
NH
NO
1
mol
NH
NO
4
3 ⎠⎝
4
3⎠
⎝
3.138
Assume 100 g of magnesium boron compound, therefore there are 52.9 g of Mg and 47.1 g of B.
⎛ 1 mol Mg ⎞
mol of Mg = 52.9 g Mg ⎜
⎟ = 2.18 mol Mg
⎝ 24.305 g Mg ⎠
⎛ 1 mol B ⎞
mol of B = 47.1g B ⎜
⎟ = 4.36 mol Mg
⎝ 10.811 g B ⎠
Divide the number of moles of each element by the least number of moles:
2.18 mol Mg
=1
2.18 mol Mg
4.26 mol B
=2
2.18 mol Mg
Formula is MgB2.
63
Chapter 3
3.139
⎛ 1 mol Cl ⎞ ⎛ 1 mol F ⎞ ⎛ 18.998 g F ⎞
–10
g F = (1.0 × 10–9 g Cl) ⎜
⎟⎜
⎟ ⎜ 1 mol F ⎟ = 5.4 × 10 g F
35.453
g
Cl
1
mol
Cl
⎠⎝
⎠
⎝
⎠⎝
3.140
First, we determine the number of grams of chlorine in the original sample:
⎛ 1 mol AgCl ⎞⎛ 1 mol Cl ⎞ ⎛ 35.453 g Cl ⎞
g Cl = (0.3383 g AgCl) ⎜
⎟⎜
⎟⎜
⎟ = 0.08369 g Cl
⎝ 143.32 g AgCl ⎠⎝ 1 mol AgCl ⎠ ⎝ 1 mol Cl ⎠
The mass of Cr in the original sample is thus 0.1246 – 0.08369 g = 0.0409 g Cr. Converting to moles, we
have:
⎛ 1 mol Cl ⎞
–3
for Cl: 0.08369 g ⎜
⎟ = 2.361 × 10 mol Cl
35.453
g
Cl
⎝
⎠
⎛ 1 mol Cr ⎞
–4
for Cr: 0.0409 g Cr ⎜
⎟ = 7.866 × 10 mol Cr
51.996
g
Cr
⎝
⎠
The relative mole amounts are:
for Cl: 2.361 × 10–3 mol / 7.87 × 10–4 mol = 3.00
for Cr: 7.87 × 10–4 mol / 7.87 × 10–4 mol = 1.00
The empirical formula is thus CrCl3.
3.141
Overall percentage yield = (0.835)(0.714) × 100% = 59.6%
64
Bringing It Together: Chapters 1 – 3
1.
(a)
24.6 cm 3 sig. fig.
0.35140 m 5 sig fig
7,424 mm 4 sig. fig.
⎛ 100 cm ⎞
⎛ 1 cm ⎞
3
vol = 24.6 cm × 0.35410 m ⎜
⎟ × 7.424 mm ⎜ 10 mm ⎟ = 647 cm
⎝ 1m ⎠
⎝
⎠
3
⎛ 1 ft ⎞
3
3⎜
ft = 647 cm ⎝ 30.48 cm ⎟⎠ = 0.0228 ft3
(b)
(c)
⎛ 7.140 g ⎞ ⎛ 1 kg ⎞
kg = 647 cm3 ⎜⎝ 1 cm3 ⎟⎠ ⎜ 1000 g ⎟ = 4.62 kg
⎝
⎠
(d)
2.
An atom is the smallest representative sample of an element while a molecule is the smallest representative
sample of a compound. Molecules are made of atoms. A mole is a unit of measure for the amount of a
substance; 6.022 × 1023 things are in a mole.
3.
The atomic mass of element X is half the size of the atomic mass of element Y.
4.
According to Dalton's atomic theory, a chemical reaction is simply a reordering of atoms from one
combination to another. If no atoms are gained or lost and if the masses of the atoms cannot change, then
the mass after the reaction must be the same as the mass before.
For the law of definite proportions, the theory states a given compound always has atoms of the same
elements in the same numerical ratio. The same mass ratio would exist regardless of how many molecules
were in the sample.
5.
A and B do not necessarily need to be the same element. They could be two different elements with the
same number of neutrons by coincidence.
6.
⎛ 30.48 cm ⎞
cm = 3.14 ft ⎝ 1 ft ⎟⎠
7.
244
94 Pu
238
94 Pu
8.
Protons:
Neutrons
Electrons
28
32
28
9.
23
+
11 Na
10.
(a)
(b)
(c)
(d)
(e)
(f)
3
3⎜
3
A molar mass of iron
Possible
An atom of iron
Not possible – atoms cannot be seen by eye
A molecule of water
Not possible – water molecules are too small to be seen by eye
A mole of water
Possible
An ion of sodium
Not possible – Na+ is too small to be seen
A formula unit of sodium chloride
Not possible – one NaCl is too small to be seen.
65
Bringing It Together: Chapters 1 – 3
11.
NONMETALS
METALLOIDS
METALS
12.
Zr and Hf
13.
Calcium
Iron
Helium
Gadolinium
Iodine
Sodium
14.
Ductile is the ability to be drawn into wire.
Malleable is the ablility to be hammmered or rolled into thin sheets.
15.
Mercury (m.p. –39 °C)
16.
Metalloids are semiconductors.
17.
Ga, In, Sn, Tl, Pb, Bi
18.
(a)
(c)
(e)
(g)
(i)
(k)
(m)
KNO3
Co3(PO4)2
FeBr3
Al2Se3
BrF5
Sr(C2H3O2)2
Cu2S
19.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
sodium chlorate
calcium phosphate
sodium permanganate
aluminum phosphide
iodine trichloride
phosphorous trichloride
potassium chromate
calcium cyanide
manganese(II) chloride
sodium nitrite
iron(II) nitrate
20.
Empirical formula are written for ionc compounds since discrete molecules do not exist, the smallest set of
subscripts that specify the correct ratio of the ions is used.
Alkaline earth metal
Transition metal
Noble gas
Inner transition metal
Halogen
Alkali metal
Tungsten (m.p. 3400 °C)
(b)
(d)
(f)
(h)
(j)
(l)
CaCO3
MgSO3
Mg3N2
Cu(ClO4)2
N2O5
(NH4)2Cr2O7
66
Bringing It Together: Chapters 1 – 3
21.
Al2O3, MgO, NO2
22.
23
204 g
⎛
⎞ ⎛ 6.022 × 10 molecules ⎞
–1
MM = ⎜
⎟⎟ = 1230 g mol
⎟ ⎜⎜ 1 mol molecules
23
×
1.00
10
molecules
⎝
⎠⎝
⎠
23.
K4Fe(CN)6
24.
⎛ 241.6 g Cu(NO3 ) 2 ⋅ 3H 2 O ⎞
g Cu(NO3)2⋅3H2O = 0.118 mol ⎜
⎟ = 28.5 g Cu(NO3)2⋅3H2O
⎝ 1 mol Cu(NO3 ) 2 ⋅ 3H 2 O ⎠
25.
%C =
0.4343 g C
× 100% = 74.04%
0.5866 g nicotine
%H =
0.05103 g H
× 100% = 8.699%
0.5866 g nicotine
%N =
0.1013 g N
× 100% = 17.27%
0.5866 g nicotine
26.
=
=
=
4K + 1Fe + 6C + 6N
(4 × 39.10) + (1 × 55.85) + (6 × 12.01) + (6 × 14.01)
368.37 g/mole
⎛ 1 mol K ⎞
mol K = ( 37.56 g K ) ⎜
⎟ = 0.9607 mol K
⎝ 39.098 g K ⎠
⎛ 1 mol H ⎞
mol H = (1.940 g H ) ⎜
⎟ = 1.925 mol H
⎝ 1.00794 g H ⎠
⎛ 1 mol P ⎞
mol P = ( 29.79 g P ) ⎜
⎟ = 0.9618 mol P
⎝ 30.974 g P ⎠
Amount of O:
% O = 100% – 37.56% K – 1.940% H – 29.79% P = 30.71% O
⎛ 1 mol O ⎞
mol O = ( 30.71 g O ) ⎜
⎟ = 1.919 mol O
⎝ 15.9994 g O ⎠
Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the
simplest mole ratio among the three elements in the compound:
for K,
for H,
for H,
for O,
0.9607 moles / 0.9607 moles = 1.00
1.925 moles / 0.9607 moles = 2.00
0.9618 moles / 0.9607 moles = 1.00
1.919 moles / 0.9607 moles = 2.00
These relative mole amounts give us the empirical formula: KH2PO2.
27.
⎛ 29.6 mL C 2 H 5 OH ⎞ ⎛ 0.798 g C2 H5 OH ⎞
molecules C2H5OH = 1.00 fl. oz. ⎜
⎟⎜
⎟
⎝ 1 fl. oz. C2 H5 OH ⎠ ⎝ 1 mL C2 H5 OH ⎠
⎛ 1 mol C2 H5 OH ⎞ ⎛ 6.022 × 1023 molecules C2 H5 OH ⎞
23
×⎜
⎟⎟ = 3.08 × 10 molecules C2H5OH
⎟ ⎜⎜
1 mol C 2 H 5OH
⎝ 46.07 g C2 H 5 OH ⎠ ⎝
⎠
67
Bringing It Together: Chapters 1 – 3
28.
L C2H6O2 = 5.00 × 1024 molecules ×
⎛
⎞ ⎛ 62.07 g C 2 H 6 O 2 ⎞ ⎛ 1 mL C2 H 6 O 2 ⎞
1 mol C2 H 6 O 2
⎜⎜
⎟⎟ ⎜
⎟⎜
⎟
23
⎝ 6.022 × 10 molecules C2 H6 O2 ⎠ ⎝ 1 mol C2 H 6 O 2 ⎠ ⎝ 1.11 g C2 H 6 O 2 ⎠
⎛ 1 L C2 H6 O2 ⎞
×⎜
⎟ = 0.464 L
⎝ 1000 mL C2 H 6 O 2 ⎠
29.
30.
31.
32.
33.
⎛ 1 mol Cl2 ⎞ ⎛ 1 mol Cl2 O7 ⎞ ⎛ 7 mol O 2 ⎞
mol O2 = 2.56 g Cl2 ⎜
⎟ = 0.126 mol O2
⎟⎜
⎟⎜
⎝ 70.91 g Cl2 ⎠ ⎝ 1 mol Cl2 ⎠ ⎝ 2 mol Cl2 O7 ⎠
⎛ 32.00 g O 2 ⎞
g O2 = 0.126 mol O2 ⎜
⎟ = 4.03 g O2
⎝ 1 mol O 2 ⎠
(a)
(b)
2Fe2O3 + 12HNO3 J 4Fe(NO3)3 + 6H2O
2C21H30O2 + 55O2 J 42CO2 + 30H2O
⎛ 8 HNO3 ⎞
mol HNO3 = 2.56 mol Cu ⎜
⎟ = 6.83 mol HNO3
⎝ 3 mol Cu ⎠
⎛ 1 mol NH3 ⎞ ⎛ 5 mol O 2 ⎞
mol O2 = 56.8 g NH3 ⎜
⎟⎜
⎟ = 4.17 mol O2
⎝ 17.03 g NH3 ⎠ ⎝ 4 mol NH3 ⎠
⎛ 32.00 g O 2 ⎞
g O2 = 4.17 mol O2 ⎜
⎟ = 133 g O2
⎝ 1 mol O 2 ⎠
(a)
heat
→ CaO + CO2
CaCO3 ⎯⎯⎯
heat
MgCO3 ⎯⎯⎯
→ MgO + CO2
(b)
Let x = g CaCO3 and y = g MgCO3
x g CaCO3 + y g MgCO3 = 5.78 g sample
⎛ 1 mol CaCO3 ⎞⎛ 1 mol CaO ⎞ ⎛ 56.08 g CaO ⎞
g CaO = x g CaCO3 ⎜
⎟⎜
⎟⎜
⎟ = 0.560x g CaO
⎝ 100.09 g CaCO3 ⎠⎝ 1 mol CaCO3 ⎠ ⎝ 1 mol CaO ⎠
⎛ 1 mol MgCO3 ⎞ ⎛ 1 mol MgO ⎞ ⎛ 40.30 g MgO ⎞
g MgO = y g MgCO3 ⎜
⎟⎜
⎟⎜
⎟ = 0.478y g MgO
⎝ 84.31 g MgCO3 ⎠ ⎝ 1 mol MgCO3 ⎠ ⎝ 1 mol MgO ⎠
0.560x g CaO + 0.478y g MgO = 3.02 g total
x = 5.78 – y
0.560(5.78 – y) g CaO + 0.478y g MgO = 3.02 g
3.24 – 0.560y + 0.478y = 3.02
y = 2.68 g MgCO3
x = 5.78 – 2.68 = 3.10 g CaCO3
% CaCO3 =
% MgCO3 =
3.10 g CaCO3
× 100% = 53.6% CaCO3
5.78 g sample
2.68 g MgCO3
× 100% = 46.4% MgCO3
5.78 g sample
68
Bringing It Together: Chapters 1 – 3
34.
(a)
(b)
35.
100% yeild adipic acid would be
12.5 g adipic acid
= 18.2 g
0.686
Amount of cyclohexene needed:
⎛ 1 mol C6 H10 O 4 ⎞⎛ 3 mol C6 H10 ⎞⎛ 82.14 g C6 H10 ⎞
g C6H10 = 18.2 g ⎜
⎟⎜
⎟⎜
⎟ = 10.2 g C6H10
⎝ 146.1 g C6 H10 O 4 ⎠⎝ 3 mol C6 H10 O 4 ⎠⎝ 1 mol C6 H10 ⎠
⎛ 1 mol C6 H10 O 4 ⎞⎛ 4 mol Na 2 Cr2 O7 ⋅ 2H 2 O ⎞
g Na2Cr2O7⋅2H2O = 18.2 g ⎜
⎟⎜
⎟
3 mol C6 H10 O 4
⎝ 146.1 g C6 H10 O 4 ⎠⎝
⎠
⎛ 298.00 g Na 2 Cr2 O7 ⋅ 2H 2 O ⎞
×⎜
⎟ = 49.5 g Na2Cr2O7⋅2H2O
⎝ 1 mol Na 2 Cr2 O7 ⋅ 2H 2 O ⎠
⎛ 91.5 ton Fe possible ⎞ ⎛ 2000 lb Fe possible ⎞ ⎛ 454 g Fe ⎞ ⎛ 1 mol Fe ⎞
tons ore = 1.00 ton Fe ⎜
⎟⎜
⎟
⎟⎜
⎟⎜
⎝ 100 ton Fe recovered ⎠ ⎝ 1 ton Fe possible ⎠ ⎝ 1 lb Fe ⎠ ⎝ 55.8 g Fe ⎠
⎞
1 ton ore
⎛ 1 mol Fe2 O3 ⎞ ⎛ 160 g Fe 2 O3 ⎞ ⎛ 1 lb Fe2 O3 ⎞ ⎛ 1 ton Fe 2 O3 ⎞ ⎛
×⎜
⎜
⎟⎜
⎟⎜
⎟⎜
⎟ = 7.69 ton
⎟
⎝ 1 mol Fe ⎠ ⎝ 1 mol Fe2 O3 ⎠ ⎝ 454 g Fe2 O3 ⎠ ⎝ 2000 lb Fe2 O3 ⎠ ⎝ 0.341 ton Fe 2 O3 ⎠
ore
36.
(a)
(b)
37.
⎛
⎞ ⎛ 1000 mg Au ⎞ ⎛ 1 ton seawater ⎞
31.1 g Au
tons seawater = 1.0 troy ounce ⎜
⎟⎜
⎟⎜
⎟
⎝ 1.0 troy ounce Au ⎠ ⎝ 1 g Au ⎠ ⎝ 1.5 mg Au ⎠
⎛ 65 ton seawater ⎞
4
×⎜
⎟ = 1.3 × 10 tons seawater
⎝ 100 ton seawater ⎠
$625.10 per troy ounce
breakeven point =
= $0.048 per ton seawater
1.3 × 104 tons seawater
g O = 2.164 g – (0.5259 g Fe + 0.7345 g Cr) = 0.9036 g O
⎛ 1 mole Fe ⎞
mol Fe = 0.5259 g Fe ⎜
⎟ = 0.009416 mol Fe
⎝ 55.85 g Fe ⎠
⎛ 1 mole Cr ⎞
mol Cr = 0.7345 g Cr ⎜
⎟ = 0.01413 mol Cr
⎝ 51.9961 g Cr ⎠
⎛ 1 mole O ⎞
mol O = 0.9036 g O ⎜
⎟ = 0.05648 mol O
⎝ 15.9994 g O ⎠
Now we divide each of these numbers of moles by the smallest of the three numbers, in order to obtain the
simplest mole ratio among the three elements in the compound:
for Fe, 0.009416 mol / 0.009416 mol = 1.00
for Cr, 0.01413 mol / 0.009416 mol = 1.50
for O, 0.05648 mol / 0.009416 mol = 6.00
Multiply by 2 in order to have whole numbers.
These relative mole amounts give us the empirical formula: Fe2Cr3O12
To calculate the molecular mass, the molecular formula is needed.
69
Bringing It Together: Chapters 1 – 3
38.
The amount of water removed was
6.584 g sample – 2.889 g dry sample = 3.695 g H2O
⎛ 1 mol H 2 O ⎞
mol H2O = 3.695 g H2O ⎜
⎟ = 0.2051 mol H2O
⎝ 18.015 g H 2 O ⎠
⎛ 1 mol Na 2SO 4 ⎞
mol Na2SO4 = 2.889 g Na2SO4 ⎜
⎟ = 0.02034 mol Na2SO4
⎝ 142.04 g Na 2SO 4 ⎠
Determine the mole ratio:
0.02034 mol / 0.02034 mol = 1
For Na2SO4
0.2051 mol / 0.02034 mol = 10.08
For H2O
Formula Na2SO4⋅10H2O
39.
First determine the amount of NH3 that would be required to react completely with the given amount of O2:
⎛ 1 mol O 2 ⎞ ⎛ 4 mole NH3 ⎞ ⎛ 17.03 g NH3 ⎞
g NH3 = (58.0 g O 2 ) ⎜
⎟ = 24.7 g NH3
⎟⎜
⎟⎜
⎝ 32.0 g O 2 ⎠ ⎝ 5 moles O 2 ⎠ ⎝ 1 mole NH3 ⎠
Since 45.0 g of NH3 are supplied, O2 is the limiting reactant.
An excess (45.0 g – 24.7 g = 20.3 g) of NH3 is present.
The number of moles and grams of NO formed is:
⎛ 1 mol NH3 ⎞ ⎛ 4 mol NO ⎞
mol NO = 24.7 g NH3 ⎜
⎟⎜
⎟ = 1.45 mol NO
⎝ 17.03 g NH3 ⎠ ⎝ 4 mol NH3 ⎠
⎛ 30.01 g NO ⎞
g NO = 1.45 mol NO ⎜
⎟ = 43.5 g NO
⎝ 1 mol NO ⎠
40.
2Al(s) + Fe2O3(s) J 2Fe(l) + Al2O3(s)
3
⎛ 2.70 g Al ⎞ ⎛ 1 mol Al ⎞ ⎛ 2 mol Fe ⎞ ⎛ 55.85 g Fe ⎞ ⎛ 1 cm Fe ⎞
3
cm3 Fe = 14.0 cm3 Al ⎜
⎜
⎟ = 9.96 cm Fe
⎟⎜
⎟⎜
⎟⎜
⎟
⎝ 1 cm3 Al ⎠ ⎝ 26.98 g Al ⎠ ⎝ 2 mol Al ⎠ ⎝ 1 mol Fe ⎠ ⎜⎝ 7.86 g Fe ⎟⎠
70