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Homework #
Chapter 3
Stoichiometry
- Need to find average molar mass of X (M X
𝑋
46
𝑋
46 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋
47
𝑋
47 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋
48
𝑋
48 )
49
𝑋
49 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋
50
𝑋
𝑋
Find the atomic weight on the periodic table that matches 47.88 amu.
Ti titanium
- Need to find the mass present of
151
Eu and
153
Eu
Know
M
Eu
= 151.96 amu
fraction of
151
Eu = x
𝐸𝑢
fraction of
153
Eu = 1-x
𝐸𝑢
𝐸𝑢
151
𝐸𝑢
153
𝐸𝑢
Therefore
151
Eu and 52%
153
Eu
- Need to find the molar mass of
185
Ru(𝑀
𝑅𝑢
𝑅𝑢
187
𝑅𝑢
187 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑅𝑢
185
𝑅𝑢
𝑅𝑢
185
𝑅𝑢
- Need to find the average molar mass of X (M X
𝑋
𝑎
𝑋
𝑎 ) + (𝐹𝑟𝑎𝑐𝑟𝑖𝑜𝑛 𝑜𝑓 𝑋
𝑏
𝑋
𝑏
𝑐
𝑋
𝑐 )
𝑑
𝑋
𝑋
Find the atomic weight on the periodic table that matches 207.22 amu.
Lead (Pb)
Note: Grey information is given
a) Moles of Sample
6
6
1 𝑚𝑜𝑙 𝐶
6
𝐻
6
- 12 𝑔 𝐶
6
𝐻
6
6
6
Molecules in Sample
6
6
- 02214 × 10
23
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐶
6
𝐻
6
1 𝑚𝑜𝑙 𝐶
6
𝐻
6
) = 3. 27 × 10
22
6
6
Atoms in Sample
3. 27 × 10
22
6
6
12 𝑎𝑡𝑜𝑚𝑠
1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶
6
𝐻
6
) = 3. 92 × 10
23
b) Mass of Sample
2
- 02 𝑔 𝐻
2
𝑂
1 𝑚𝑜𝑙 𝐻
2
𝑂
2
Molecules in Sample
2
- 02214 × 10
23
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝐻
2
𝑂
1 𝑚𝑜𝑙 𝐻
2
𝑂
) = 1. 35 × 10
23
2
Total Atoms in Sample
1. 35 × 10
23
2
3 𝑎𝑡𝑜𝑚𝑠
1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐻 2
𝑂
) = 4. 04 × 10
23
c) Moles of Sample
2. 71 × 10
22
2
1 𝑚𝑜𝑙 𝐶𝑂
2
- 02214 × 10
23
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶𝑂
2
2
Mass of Sample
2
- 01 𝑔 𝐶𝑂 2
1 𝑚𝑜𝑙 𝐶𝑂
2
2
Total Atoms in Sample
2. 71 × 10
22
2
3 𝑎𝑡𝑜𝑚𝑠
1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶𝑂
2
) = 8. 13 × 10
22
d) Molecules in Sample
3. 35 × 10
22
3
1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶𝐻
3
𝑂𝐻
6 𝑎𝑡𝑜𝑚𝑠
= 5. 58 × 10
21
3
Moles of Sample
5. 58 × 10
21
3
1 𝑚𝑜𝑙 𝐶𝐻 3
𝑂𝐻
- 02214 × 10
23
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶𝐻
3
𝑂𝐻
3
Mass of Sample
3
- 05 𝑔 𝐶𝐻 3
𝑂𝐻
1 𝑚𝑜𝑙 𝐶𝐻
3
𝑂𝐻
3
- a) 20. 0 𝑚𝑔 𝐶 8
10
4
2
1 𝑔
1000 𝑚𝑔
1 𝑚𝑜𝑙 𝐶
8
𝐻
10
𝑁
4
𝑂
2
- 22 𝑔 𝐶
8
𝐻
10
𝑁
4
𝑂
2
) = 1. 03 × 10
− 4
8
10
4
2
b) 2. 72 × 10
21
2
5
1 𝑚𝑜𝑙 𝐶
2
𝐻
5
𝑂𝐻
- 02214 × 10
23
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝐶𝐻
3
𝑂𝐻
2
5
c) 1. 50 𝑔 𝐶𝑂
2
1 𝑚𝑜𝑙 𝐶𝑂 2
- 01 𝑔 𝐶𝑂 2
2
Mass of
Sample
Moles of Sample Molecules in Sample
Total Atoms in
Sample
a) 4.24 g C 6
H
6
0.0543 mol 3.27×
22
molec 3.92×
23
atoms
b) 4.04 g 0.224 mol H 2
O 1.35×
23
molec 4.04×
23
atoms
c) 1.98 g 0.0450 mol 2.71×
22
molec CO 2 8.13×
22
atoms
d) 0.297 g 0.00927 mol 5.58×
21
molec 3.35×
22
atoms
in CH 3
OH
44. C
8
H
10
N
4
O
2
Assume 1 mol C 8
H
10
N
4
O
2
8
10
4
2
- 22 𝑔 𝐶 8
𝐻 10
𝑁 4
𝑂 2
1 𝑚𝑜𝑙 𝐶
8
𝐻
10
𝑁
4
𝑂
2
8
10
4
2
8
10
4
2
8 𝑚𝑜𝑙 𝐶
1 𝑚𝑜𝑙 𝐶
8
𝐻
10
𝑁
4
𝑂
2
- 01 𝑔 𝐶
1 𝑚𝑜𝑙 𝐶
C
12
H
22
O
11
Assume 1 mol C 12
H
22
O
11
12
22
11
- 34 𝑔 𝐶
12
𝐻
22
𝑂
11
1 𝑚𝑜𝑙 𝐶 12
𝐻 22
𝑂 11
12
22
11
12
22
11
12 𝑚𝑜𝑙 𝐶
1 𝑚𝑜𝑙 𝐶
12
𝐻
22
𝑂
11
- 01 𝑔 𝐶
1 𝑚𝑜𝑙 𝐶
C
2
H
5
OH
Assume 1 mol C 2
H
5
OH
2
5
- 08 𝑔 𝐶
2
𝐻
5
𝑂𝐻
1 𝑚𝑜𝑙 𝐶
2
𝐻
5
𝑂𝐻
2
5
2
5
2 𝑚𝑜𝑙 𝐶
1 𝑚𝑜𝑙 𝐶
2
𝐻
5
𝑂𝐻
- 01 𝑔 𝐶
1 𝑚𝑜𝑙 𝐶
In order of increase mass % of carbon
C
12
H
22
O
11
< C
8
H
10
N
4
O
2
< C
2
H
5
OH
or
sucrose < caffeine < ethanol
- Assume 1 mol of fungal laccase
In 1 fungal laccase molecule there are 4 Cu atoms
Therefore, in 1 mol of fungal laccase there are 4 moles of Cu atoms
Find the mass of 4 mol Cu.
- 55 𝑔 𝐶𝑢
1 𝑚𝑜𝑙 𝐶𝑢
Find mass of fungal laccase
𝐶𝑢
𝑡𝑜𝑡𝑎𝑙
𝑡𝑜𝑡𝑎𝑙
𝐶𝑢
The mass total was for 1 mol of the fungal laccase, therefore, in order to get the molar mass one
needs to divide by 1 mol.
𝑓𝑢𝑛𝑔𝑎𝑙 𝑙𝑎𝑐𝑐𝑎𝑠𝑒
𝑔
𝑚𝑜𝑙
- Compound 1
Heating 0.6498 g of compound 1 leaves 0.6018 g Hg
𝑂
𝑡𝑜𝑡𝑎𝑙
𝐻𝑔
Determine moles of Hg
1 𝑚𝑜𝑙 𝐻𝑔
- 59 𝑔 𝐻𝑔
Determine moles of O
1 𝑚𝑜𝑙 𝑂
- 00 𝑔 𝑂
Divide through by smallest number of moles (0.00300 mol)
Mercury Oxygen
003000 𝑚𝑜𝑙
00300 𝑚𝑜𝑙
00300 𝑚𝑜𝑙
00300 𝑚𝑜𝑙
Empirical Formula
HgO
Compound 2
Heating 0.4172 g of compound 2 results in 0.016 g O being produced (mass of O is the
mass lost)
𝐻𝑔
𝑡𝑜𝑡𝑎𝑙
𝑂
Determine moles of Hg
1 𝑚𝑜𝑙 𝐻𝑔
- 59 𝑔 𝐻𝑔
Determine mol of O
1 𝑚𝑜𝑙 𝑂
- 00 𝑔 𝑂
Divide through by smallest number of moles (0.0010 0 mol)
Mercury Oxygen
002000 𝑚𝑜𝑙
00100 𝑚𝑜𝑙
00100 𝑚𝑜𝑙
00100 𝑚𝑜𝑙
Empirical Formula
Hg 2
O
- Assume 100 g sample therefore 56.79 g C, 6.56 g H, 28.37 g O, and 8.28 g N
Calculate the number of moles of C, H, O, and N
Carbon
1 𝑚𝑜𝑙 𝐶
- 01 𝑔 𝐶
Hydrogen
1 𝑚𝑜𝑙 𝐻
- 01 𝑔 𝐻
Oxygen
1 𝑚𝑜𝑙 𝑂
- 00 𝑔 𝑂
Nitrogen
1 𝑚𝑜𝑙 𝑁
- 01 𝑔 𝑁
Divide through by smallest mol amount (0.591 mol)
Carbon Hydrogen
729 𝑚𝑜𝑙
591 𝑚𝑜𝑙
50 𝑚𝑜𝑙
591 𝑚𝑜𝑙
Oxygen Nitrogen
773 𝑚𝑜𝑙
591 𝑚𝑜𝑙
591 𝑚𝑜𝑙
591 𝑚𝑜𝑙
Empirical Formula C 8
H
11
O
3
N
Divide through by smallest mole amount (3.64×
mol)
Carbon Hydrogen Oxygen
- 539 × 10
− 4
𝑚𝑜𝑙
- 77 × 10
− 4
𝑚𝑜𝑙
- 16 × 10
− 4
𝑚𝑜𝑙
- 77 × 10
− 4
𝑚𝑜𝑙
- 77 × 10
− 4
𝑚𝑜𝑙
- 77 × 10
− 4
𝑚𝑜𝑙
Multiple through by 2 to get whole numbers
Empirical Formula C
4
H
3
O
2
𝐶 4
𝐻 3
𝑂 2
𝑔
𝑚𝑜𝑙
Find molecular formula
𝑚
𝑛
5 𝑔
250 𝑚𝑜𝑙
𝑔
𝑚𝑜𝑙
166
𝑔
𝑚𝑜𝑙
- 07
𝑔
𝑚𝑜𝑙
= 2. 00 Multiply empirical formula by 2: C
8
H
6
O
4
61. C
x
H
y
O
z
+ O
2
CO
2
+ H
2
O (Combustion Reaction)
All of the C in the compound goes into forming CO
2
. Therefore, 𝑛
𝐶𝑂 2
𝐶
2
1 𝑔
1000 𝑚𝑔
1 𝑚𝑜𝑙 𝐶𝑂
2
- 01 𝑔 𝐶𝑂
2
1 𝑚𝑜𝑙 𝐶
1 𝑚𝑜𝑙 𝐶𝑂
2
) = 3. 638 × 10
− 4
All of the H in the compound goes into forming H
2
O. Therefore, 𝑛
𝐻 2
𝑂
𝐻
2
1 𝑔
1000 𝑚𝑔
1 𝑚𝑜𝑙 𝐻
2
𝑂
- 02 𝑔 𝐻
2
𝑂
2 𝑚𝑜𝑙 𝐻
1 𝑚𝑜𝑙 𝐻
2
𝑂
) = 4. 85 × 10
− 4
To find the mass of O, find the mass of C and H, and subtract them from the overall
mass of the compound.
Calculate 𝑚
𝐶
𝐶
= 3. 638 × 10
− 4
- 01 𝑔 𝐶
1 𝑚𝑜𝑙 𝐶
Calculate 𝑚
𝐻
𝐻
= 4. 85 × 10
− 4
- 01 𝑔 𝐻
1 𝑚𝑜𝑙 𝐻
) = 4. 90 × 10
− 4
Calculate 𝑚
𝑂
𝑂
𝐶
𝑥
𝐻
𝑦
𝑂
𝑧
𝐶
𝐻
𝐶 𝑥
𝐻 𝑦
𝑂 𝑧
1 𝑔
1000 𝑚𝑔
𝑂
= 0. 01068 𝑔 − 0. 004369 𝑔 − 4. 90 × 10
− 4
Calculate 𝑛
𝑂
1 𝑚𝑜𝑙 𝑂
- 00 𝑔 𝑂
) = 3. 64 × 10
− 4
Divide through by smallest mole amount (3.638×
mol)
Carbon Hydrogen Oxygen
- 638 × 10
− 4
𝑚𝑜𝑙
- 638 × 10
− 4
𝑚𝑜𝑙
- 85 × 10
− 4
𝑚𝑜𝑙
- 638 × 10
− 4
𝑚𝑜𝑙
- 64 × 10
− 4
𝑚𝑜𝑙
- 638 × 10
− 4
𝑚𝑜𝑙
Multiple through by 3 to get whole numbers
Empirical Formula C
3
H
4
O
3
𝐶
3
𝐻
4
𝑂
8
𝑔
𝑚𝑜𝑙
Find molecular formula
- 1
𝑔
𝑚𝑜𝑙
- 07
𝑔
𝑚𝑜𝑙
= 2. 000 Multiply empirical formula by 2 C
6
H
8
O
6
63. X+2Y XY
2
- a) C 6
H
12
O
6
(s) + 6O 2
(g) 6CO 2
(g) + 6H 2
O(g)
b) Fe 2
S
3
(s) + 6HCl(g) 2FeCl 3
(s) + 3H 2
S(g)
c) CS 2
(l) + 2NH 3
(g) H 2
S(g) + NH 4
SCN(s)
- a) C 2
H
5
OH(l) + 3O 2
(g) 2CO 2
(g) + 3H 2
O(g)
b) 3 Pb(NO 3 ) 2 (aq) + 2 Na 3 PO 4 (aq) Pb 3 (PO 4 ) 2 (s) + 6 NaNO 3 (aq)
- a) 16Cr(s) + 3S 8
(s) 8Cr 2
S
3
(s)
b) 2NaHCO 3
(s)
heat
Na 2
CO
3
(s) + CO 2
(g) + H 2
O(g)
c) 2KClO 3
(s)
heat
2KCl(s) + 3O 2
(g)
d) 2Eu(s) + 6HF(g) 2EuF 3
(s) + 3H 2
(g)
1000 𝑔
1 𝑘𝑔
1 𝑚𝑜𝑙 𝐴𝑙
- 98 𝑔 𝐴𝑙
3 𝑚𝑜𝑙 𝑁𝐻
4
𝐶𝑙𝑂
4
3 𝑚𝑜𝑙 𝐴𝑙
- 50 𝑔 𝑁𝐻
4
𝐶𝑙𝑂
4
1 𝑚𝑜𝑙 𝑁𝐻
4
𝐶𝑙𝑂
4
- Use the mass of HNO 3
to calculate the moles of HNO 3
needed
1. 0 × 10
6
3
1000 𝑔
1 𝑘𝑔
1 𝑚𝑜𝑙 𝐻𝑁𝑂
3
- 02 𝑔 𝐻𝑁𝑂 3
) = 1. 6 × 10
7
3
Use moles of HNO 3
to calculate the moles of NH 3
needed
1. 6 × 10
7
3
3 𝑚𝑜𝑙 𝑁𝑂 2
2 𝑚𝑜𝑙 𝐻𝑁𝑂
3
2 𝑚𝑜𝑙 𝑁𝑂
2 𝑚𝑜𝑙 𝑁𝑂
2
4 𝑚𝑜𝑙 𝑁𝐻 3
4 𝑚𝑜𝑙 𝑁𝑂
) = 2. 4 × 10
7
3
Use the moles of NH 3
to calculate the kilograms of NH 3
produces
2. 4 × 10
7
3
- 04 𝑔 𝑁𝐻
3
1 𝑚𝑜𝑙 𝑁𝐻
3
1 𝑘𝑔
1000 𝑔
) = 4. 1 × 10
5
3
1 𝑚𝑜𝑙 𝐹𝑒
- 85 𝑔 𝐹𝑒
2 𝑚𝑜𝑙 𝐴𝑙
2 𝑚𝑜𝑙 𝐹𝑒
- 98 𝑔 𝐴𝑙
1 𝑚𝑜𝑙 𝐴𝑙
1 𝑚𝑜𝑙 𝐹𝑒
- 85 𝑔 𝐹𝑒
2 𝑚𝑜𝑙 𝐹𝑒 2
𝑂 3
2 𝑚𝑜𝑙 𝐹𝑒
- 70 𝑔 𝐹𝑒 2
𝑂 3
1 𝑚𝑜𝑙 𝐹𝑒
2
𝑂
3
2
3
1 𝑚𝑜𝑙 𝐹𝑒
- 85 𝑔 𝐹𝑒
2 𝑚𝑜𝑙 𝐴𝑙
2
𝑂
3
2 𝑚𝑜𝑙 𝐹𝑒
- 96 𝑔 𝐴𝑙
2
𝑂
3
1 𝑚𝑜𝑙 𝐴𝑙
2
𝑂
3
2
3
- 2NO(g) + O 2
(g) 2NO 2
(g)
The limiting reagent is NO.
- 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g)
NH
4
O
2
(L.R.) NO H
2
O
Initial 10 molec 10 molec 0 0
Change in terms of x
x=
+4x
+6x
Final 2 molec 0 8 molec 12 molec
The Reaction will go until either the NH 4
or O 2
is used up, therefore, x=2 and oxygen
is the limiting reagent
ICF tables can be done in molecules or moles.
22 total moleucles are present in the container once the reaction goes to completion.
The problem tells you that the coefficients in front of each of the species are 1
Determine the molecular formula of aspirin
C
x
H
y
O
z
+ O
2
CO
2
+ H
2
O (Combustion Reaction)
All of the carbon in the compound goes into forming CO
2
. Therefore, 𝑛
𝐶𝑂 2
𝐶
2
1 𝑚𝑜𝑙 𝐶𝑂
2
- 01 𝑔 𝐶𝑂
2
1 𝑚𝑜𝑙 𝐶
1 𝑚𝑜𝑙 𝐶𝑂
2
All of the hydrogen in the compound goes into forming H
2
O. Therefore, 𝑛
𝐻
2
𝑂
𝐻
2
1 𝑚𝑜𝑙 𝐻
2
𝑂
- 02 𝑔 𝐻
2
𝑂
2 𝑚𝑜𝑙 𝐻
1 𝑚𝑜𝑙 𝐻
2
𝑂
To find the mass of O, find the mass of C and H, and subtract them from the overall
weight of the compound.
Calculate 𝑚
𝐶
𝐶
- 01 𝑔 𝐶
1 𝑚𝑜𝑙 𝐶
Calculate 𝑚
𝐻
𝐻
- 01 𝑔 𝐻
1 𝑚𝑜𝑙 𝐻
Calculate 𝑚
𝑂
𝑂
𝐶
𝑥
𝐻
𝑦
𝑂
𝑧
𝐶
𝐻
𝑂
Calculate 𝑛
𝑂
1 𝑚𝑜𝑙 𝑂
- 00 𝑔 𝑂
Divide through by smallest mol amount (0. 0 22 mol)
Carbon Hydrogen Oxygen
0500 𝑚𝑜𝑙
022 𝑚𝑜𝑙
0444 𝑚𝑜𝑙
022 𝑚𝑜𝑙
022 𝑚𝑜𝑙
022 𝑚𝑜𝑙
Multiple through by 4 to get whole numbers
Empirical Formula C
9
H
8
O
4
𝐶
9
𝐻
8
𝑂
4
𝑔
𝑚𝑜𝑙
Since the molecular formula weight is between 170
𝑔
𝑚𝑜𝑙
and 190
𝑔
𝑚𝑜𝑙
the empirical and
molecular formula are the same.
Determine the formula or salicylic acid
Salicylic acid + C
4
H
6
O
3
aspirin (C
9
H
8
O
4
) + C
2
H
4
O
2
Salicylic acid: C
7
H
6
O
3
- You need to determine the amount of NO(g) produced.
You know that 2.00 mol of NH 3
are reacted with 10.00 mol of O 2
and 6.75 moles of O 2
remains
after the reaction goes to completion. This indicates that NH 3
is the limiting reagent
Determine the moles of O 2
reacted
𝑂
2
(𝑟𝑒𝑎𝑐𝑡𝑒𝑑)
𝑂
2
(𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙)
𝑂
2
(𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑)
NH 3 reacts by one of the following equations
4NH
3
(g) + 5O 2
(g) 4 NO(g) + 6H 2
O(g)
4NH
3
(g) + 7O 2
(g) 4NO 2
(g) + 6H 2
O(g)
Therefore for equation 1 𝑛
𝑂
2
5
4
𝑁𝑂
and for equation 2 𝑛
𝑂
2
7
4
𝑁𝑂
2
The total number of moles of O 2
used was 3.25 giving
5
4
𝑁𝑂
7
4
𝑁𝑂
2
Similarly for equation 1 𝑛
𝑁𝐻
3
𝑁𝑂
and for equation 2 𝑛
𝑁𝐻
3
𝑁𝑂
2
The total number of moles of NH 3
used was 2.00 giving
𝑁𝑂
𝑁𝑂
2
Combine equations and solve for mole of NO
𝑁𝑂 2
𝑁𝑂
5
4
𝑁𝑂
7
4
𝑁𝑂
𝑁𝑂
𝑁𝑂
𝑁𝑂
𝑁𝑂
