REACTION STOICHIOMETRY
1792 JEREMIAS RICHTER
The amount of substances produced or consumed in chemical
reactions can be quantified
4F-1 (of 14)
INFORMATION FROM CHEMICAL EQUATIONS
2H2
+
O2
→
2H2O
2 molecules
1 molecule
2 molecules
2 moles
1 moles
2 moles
0.84 moles
0.42 moles
0.84 moles
0.028 moles
0.014 moles
0.028 moles
The moles that react and form do so in the ratio of the balanced equation
4F-2 (of 14)
INFORMATION FROM CHEMICAL EQUATIONS
2H2
+
O2
→
2H2O
0.60 moles
0.40 moles
0.00 moles
starting
-0.60 moles
-0.30 moles
+0.60 moles
reacting
0.00 moles
0.10 moles
0.60 moles
ending
2H2
+
O2
→
2H2O
0.50 moles
0.20 moles
0.00 moles
starting
-0.40 moles
-0.20 moles
+0.40 moles
reacting
0.10 moles
0.00 moles
0.40 moles
ending
The moles that react and form do so in the ratio of the balanced equation
4F-3 (of 14)
MASS CALCULATIONS
Calculate the mass of oxygen needed to burn 5.00 g of propanol, C3H8O.
C3H8O
4F-4 (of 14)
+ 4½ O2
→
3 CO2
+
4 H2 O
MASS CALCULATIONS
Calculate the mass of oxygen needed to burn 5.00 g of propanol, C3H8O.
2 C3H8O
+
→
9 O2
5.00 g
xg
2 mol
9 mol
6 CO2
+
8 H2 O
2 mol C3H8O = 9 mol O2
5.00 g C3H8O x
1 mol C3H8O
____________________
60.094 g C3H8O
= 12.0 g O2
4F-5 (of 14)
x
9 mol O2
_________________
2 mol C3H8O
x 32.00 g O2
______________
1 mol O2
Calculate the mass of carbon dioxide produced from the 5.00 g of propanol.
2 C3H8O
+
→
9 O2
6 CO2
5.00 g
xg
2 mol
6 mol
+
8 H2 O
2 mol C3H8O = 6 mol CO2
5.00 g C3H8O x
1 mol C3H8O
____________________
60.094 g C3H8O
= 11.0 g CO2
4F-6 (of 14)
x
6 mol CO2
_________________
2 mol C3H8O
x 44.01 g CO2
________________
1 mol CO2
2 C3H8O
5.0 g
4F-7 (of 14)
+
9 O2
12.0 g
→
6 CO2
11.0 g
+
8 H2O
6.0 g
LIMITING REACTANT CALCULATIONS
LIMITING REACTANT – The reactant that is completely used up in a reaction
4F-8 (of 14)
x
12 pieces
12 pieces x 1 sandwich
_______________
22 slices
= 12 sandwiches
1 piece
22 slices
x 1 sandwich
_______________
2 slices
limiting reactant
4F-9 (of 14)
= 11 sandwiches
actual amount produced
Calculate the mass of tungsten metal produced when 25.0 g of barium reacts
with 26.0 g of tungsten (III) fluoride.
3 Ba
4F-10 (of 14)
+
2 WF3
→
3 BaF2
+
2W
25.0 g
26.0 g
xg
3 mol
2 mol
2 mol
Calculate the mass of tungsten metal produced when 25.0 g of barium reacts
with 26.0 g of tungsten (III) fluoride.
3 Ba
25.0 g Ba x
2 WF3
→
3 BaF2
+
2W
25.0 g
26.0 g
xg
3 mol
2 mol
2 mol
1 mol Ba
_______________
137.3 g Ba
4F-11 (of 14)
+
x 2 mol W x 183.8 g W
____________
______________
3 mol Ba
1 mol W
=
22.3 g W
Calculate the mass of tungsten metal produced when 25.0 g of barium reacts
with 26.0 g of tungsten (III) fluoride.
3 Ba
25.0 g Ba x
+
2 WF3
→
3 BaF2
2W
25.0 g
26.0 g
xg
3 mol
2 mol
2 mol
1 mol Ba
_______________
x 2 mol W x 183.8 g W
137.3 g Ba
____________
3 mol Ba
_______________
_________________
_____________
240.8 g WF3
2 mol WF3
WF3 is the limiting reactant
19.8 g W are produced
=
22.3 g W
=
19.8 g W
1 mol W
26.0 g WF3 x 1 mol WF3 x 2 mol W x 183.8 g W
4F-12 (of 14)
+
_______________
1 mol W
Determine the percentage of magnesium and silver in an alloy of the two
metals.
A 6.50 gram sample of the alloy reacts with 14.5 grams of hydrogen chloride.
Mg
+
2 HCl
xg
14.5 g
1 mol
2 mol
→
MgCl2
+
H2
1 mol Mg = 2 mol HCl
14.5 g HCl x
1 mol HCl
________________
36.458 g HCl
x
1 mol Mg
______________
2 mol HCl
4.834 g Mg x 100 = 74.4% Mg
_______________
6.50 g alloy
4F-13 (of 14)
x
24.31 g Mg
______________
= 4.834 g Mg
1 mol Mg
100% - 74.4% = 25.6% Ag
Determine the percentage by mass of iodide in a solid unknown.
A 1.17 gram sample of the unknown is dissolved in water, treated with lead
(II) ions, and 1.42 grams of precipitate are collected.
Pb2+
+
1.42 g PbI2 x 1 mol PbI2
_________________
461.0 g PbI2
0.7818 g I-
___________________
1.17 g sample
4F-14 (of 14)
x 100
=
2 I-
→
PbI2
xg
1.42 g
2 mol
1 mol
x 2 mol I-
_____________
1 mol PbI2
x 126.9 g I-
_____________
1 mol I-
66.8% I- in the sample
=
0.7818 g I-
MOLES FROM SOLUTION DATA
Find the moles of potassium carbonate contained in 275 mL of a 0.300 M
potassium carbonate solution.
M
= n
___
V
MV = n
0.300 mol K2CO3
______________________
L solution
4G-1 (of 12)
x 0.275 L solution = 0.0825 mol K2CO3
Find the moles of each ion in the 0.300 M potassium carbonate solution.
0.0825 mol K2CO3 x 2 = 0.165 mol K+
0.0825 mol K2CO3 x 1 = 0.0825 mol CO32-
4G-2 (of 12)
10.0 mL of 0.450 M BaCl2 are mixed with 20.0 mL of 0.300 M K2SO4.
(a) Find the moles of each ion in the solution.
0.450 mol BaCl2
______________________
x 0.0100 L solution = 0.004500 mol BaCl2
L solution
 0.00450 mol Ba2+ and 0.00900 mol Cl0.300 mol K2SO4
______________________
x 0.0200 L solution = 0.006000 mol K2SO4
L solution
 0.0120 mol K+ and 0.00600 mol SO42-
4G-3 (of 12)
10.0 mL of 0.450 M BaCl2 are mixed with 20.0 mL of 0.300 M K2SO4.
(b) Find the moles of each ion after any reaction.
Ba2+
+
SO42-
0.00450
0.00600
-0.00450
0
-0.00450
0.00150
→
BaSO4
0
+0.00450
0.00450
 0 mol Ba2+
0.00150 mol SO420.00900 mol Cl- 0.0120 mol K+
4G-4 (of 12)
Initial moles
Reacting moles
Final moles
10.0 mL of 0.450 M BaCl2 are mixed with 20.0 mL of 0.300 M K2SO4.
(c) Find the final molarities of each ion in the solution.
0 mol Ba2+
______________________
= 0 M Ba2+
0.0300 L solution
0.00150 mol SO42-
_______________________
= 0.0500 M SO42-
0.0300 L solution
0.00900 mol Cl-
_______________________
= 0.300 M Cl-
0.0300 L solution
0.0120 mol K+
_______________________
0.0300 L solution
4G-5 (of 12)
= 0.400 M K+
20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH.
(a) Find the moles of each ion in the solution.
0.350 mol HCl
___________________
x 0.0200 L solution
= 0.007000 mol HCl
L solution
 0.00700 mol H+ and 0.00700 mol Cl0.250 mol NaOH x 0.0300 L solution = 0.007500 mol NaOH
______________________
L solution
 0.00750 mol Na+ and 0.00750 mol OH-
4G-6 (of 12)
20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH.
(b) Find the moles of each ion after any reaction.
H+
+
OH-
0.00700
0.00750
-0.00700
0
-0.00700
0.00050
→
H2O
0
+0.00700
0.00700
 0 mol H+
0.00050 mol OH0.00700 mol Cl- 0.00750 mol Na+
4G-7 (of 12)
Initial moles
Reacting moles
Final moles
20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH.
(c) Find the final molarities of each ion in the solution.
0 mol H+
______________________
= 0 M H+
0.0500 L solution
0.00050 mol OH-
_______________________
= 0.010 M OH-
0.0500 L solution
0.00700 mol Cl-
_______________________
= 0.140 M Cl-
0.0500 L solution
0.00750 mol Na+
_______________________
0.0500 L solution
4G-8 (of 12)
= 0.150 M Na+
20.0 mL of 0.350 M HCl are mixed with 30.0 mL of 0.250 M NaOH.
(d) Find the mass of water produced by the reaction.
0.007000 mol H2O x 18.016 g H2O
__________________
1 mol H2O
4G-9 (of 12)
= 0.126 g H2O
DILUTION CALCULATIONS
When a solution is diluted, only solvent is added , so the moles of solute
are unchanged
mol solute (concentrated)
=
mol solute (diluted)
MCVC
=
MDVD
4G-10 (of 12)
Calculate the volume of 6.00 M ammonia needed to prepare 250. mL of a
0.100 M ammonia solution.
MCVC = MDVD
MC = 6.00 M
MD = 0.100 M
VC = ?
VD = 250. mL
VC = MDVD = (0.100 M)(250. mL)
_______
________________________
MC
(6.00 M)
4G-11 (of 12)
= 4.17 mL
Calculate the volume of water that must be added to 5.00 mL of
concentrated hydrochloric acid (12.1 M) to make the acid 3.00 M.
MCVC = MDVD
MC = 12.1 M
MD = 3.00 M
VC = 5.00 mL
VD = ?
MCVC = VD = (12.1 M)(5.00 mL)
_______
_______________________
MD
(3.00 M)
= 20.2 mL
20.2 mL
- 5.00 mL
Volume of dilute solution
Volume of concentrated solution
15.2 mL
Water that must be added
____________
4G-12 (of 12)
REACTIONS IN SOLUTION
TITRATION – A technique in which one solution is used to analysis another
Buret: a solution of 1 reactant of
known concentration
Flask: another reactant of unknown
concentration, mass, etc.
STANDARD SOLUTION – A solution of known concentration
4H-1 (of 13)
The mass of sodium bicarbonate in an antacid tablet is to be determined.
ACID-BASE INDICATOR – A weak organic acid or base that changes color in
acidic or basic solutions
The tablet is dissolved in water, an acid-base indicator added, and 21.5 mL
of a 0.300 M hydrochloric acid solution produces a color change.
NaHCO3
+
HCl
xg
21.5 mL
0.300 M
1 mol
1 mol
→
NaCl
+
H2O
+
CO2
0.300 mol HCl x 0.0215 L solution x 1 mol NaHCO3 x 84.008 g NaHCO3
__________________
_________________
_______________________
L solution
1 mol HCl
mol NaHCO3
= 0.542 g NaHCO3
4H-2 (of 13)
A sodium hydroxide solution is to be standardized.
34.2 mL of the sodium hydroxide solution are required _to neutralize a
solution made with 0.619 grams of solid H2C2O4.2H2O (m = 126.08 g/mol).
2 NaOH
+
H2C2O4 .2H2O
34.2 mL
xM
0.619 g
2 mol
1 mol
0.619 g O.A.D. x
1 mol O.A.D.
____________________
126.08 g O.A.D.
= 0.287 M NaOH
4H-3 (of 13)
x
→
Na2C2O4
2 mol NaOH x
2 H2O
+ 4
1
_________________
_______________________
1 mol O.A.D.
0.0342 L solution
A 10.0 mL aliquot of a solution containing V2+ ions is acidified, and 32.7 mL
of a 0.115 M MnO4- solution produces a light purple color. If the V2+ was
oxidized to V5+, determine the molarity of the V2+ in the original solution.
MnO4- (aq) + V2+ (aq) → Mn2+ + V5+
+7 -2
( 5e- +
+2
8H+ +
+2
MnO4-
→
( V2+ →
4H-4 (of 13)
+5
Mn2+ + 4H2O ) x 3
V5+ + 3e- ) x 5
A 10.0 mL aliquot of a solution containing V2+ ions is acidified, and 32.7 mL
of a 0.115 M MnO4- solution produces a light purple color. If the V2+ was
oxidized to V5+, determine the molarity of the V2+ in the original solution.
MnO4- (aq) + V2+ (aq) → Mn2+ + V5+
+7 -2
+2
+2
15e- + 24H+ + 3MnO4-
+5
→ 3Mn2+ + 12H2O
5V2+ → 5V5+ + 15e15e- + 24H+ + 3MnO4- + 5V2+ → 3Mn2+ + 12H2O + 5V5+ + 15e-
4H-5 (of 13)
A 10.0 mL aliquot of a solution containing V2+ ions is acidified, and 32.7 mL
of a 0.115 M MnO4- solution produces a light purple color. If the V2+ was
oxidized to V5+, determine the molarity of the V2+ in the original solution.
24H+ (aq) + 3MnO4- (aq) + 5V2+(aq) → 3Mn2+ (aq) + 12H2O (l) + 5V5+ (aq)
32.7 mL
0.115 M
10.0 mL
xM
3 mol
5 mol
0.115 mol MnO4- x 0.0327 L solution x
5 mol V2+
_____________________
________________
L solution
3 mol MnO4-
= 0.627 M V2+
4H-6 (of 13)
x
1
_______________________
0.0100 L solution
Calculate the molar mass of a diprotic acid if 0.409 grams of it are
neutralized by 19.50 mL of a 0.287 M sodium hydroxide solution.
molar mass H2X
= grams H2X
_______________
moles H2X
molar mass H2X
= 0.109 grams H2X
______________________
? moles H2X
4H-7 (of 13)
Calculate the molar mass of a diprotic acid if 0.409 grams of it are
neutralized by 19.50 mL of a 0.287 M sodium hydroxide solution.
H2X
+
2 NaOH
0.409 g
x mol
19.50 mL
0.287 M
1 mol
2 mol
→
2 H(OH)
0.287 mol NaOH x 0.01950 L sol’n x 1 mol H2X
_____________________
_______________
L sol’n
2 mol NaOH
0.409 g H2X
________________________
0.002798 mol H2X
4H-8 (of 13)
=
146 g/mol
+
Na2X
= 0.002798 mol H2X
The molarity of an aluminum hydroxide solution is to be determined.
An acid-base indicator is added to 10.0 mL of the aluminum hydroxide
solution, and 12.5 mL of 0.300 M hydrochloric acid produces a color change.
Al(OH)3
+
3 HCl
10.0 mL
xM
12.5 mL
0.300 M
1 mol
3 mol
→
AlCl3
+ 3 H(OH)
0.300 mol HCl x 0.0125 L solution x 1 mol Al(OH)3
__________________
_________________
L solution
3 mol HCl
= 0.125 M Al(OH)3
4H-9 (of 13)
x
1
____________
0.0100 L
Cinnabar ore contains S2- ions. A 1.534 g sample of the ore is dissolved in
acid, then all the S2- is oxidized by 20.4 mL of a 0.110 M Cr2O72- solution.
Determine the percentage of S2- in cinnabar ore.
K2Cr2O7 (aq) + S2- (aq) →
K+ (aq) + Cr2O72- (aq) + S2- (aq) →
+1
+6 -2
-2
Cr3+ + S8
+3
0
( 6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O ) x 8
( 8 S2- →
4H-10 (of 13)
S8 + 16e- ) x 3
Cinnabar ore contains S2- ions. A 1.534 g sample of the ore is dissolved in
acid, then all the S2- is oxidized by 20.4 mL of a 0.110 M Cr2O72- solution.
Determine the percentage of S2- in cinnabar ore.
K2Cr2O7 (aq) + S2- (aq) →
K+ (aq) + Cr2O72- (aq) + S2- (aq) →
+1
+6 -2
-2
Cr3+ + S8
+3
0
48e- + 112H+ + 8Cr2O72- → 16Cr3+ + 56H2O
24S2- → 3S8 + 48e48e- + 112H+ + 8Cr2O72- + 24S2- → 16Cr3+ + 56H2O + 3S8 + 48e-
4H-11 (of 13)
Cinnabar ore contains S2- ions. A 1.534 g sample of the ore is dissolved in
acid, then all the S2- is oxidized by 20.4 mL of a 0.110 M Cr2O72- solution.
Determine the percentage of S2- in cinnabar ore.
112H+ (aq) + 8Cr2O72- (aq) + 24S2-(aq) → 16Cr3+ (aq) + 56H2O (l) + 3S8 (s)
20.4 mL
0.110 M
xg
8 mol
24 mol
0.110 mol Cr2O72- x 0.0204 L solution x
24 mol S2-
_______________________
_________________
L solution
8 mol Cr2O72-
= 0.2159 g S20.2159 g S2- x 100
_______________
1.534 g ore
4H-12 (of 13)
= 14.1% S2- in the ore
x 32.07 g S2_____________
mol S2-
Calculate the molarity of a sulfuric acid solution if 25.0 mL of it are
neutralized by 33.5 mL of a 0.240 M potassium hydroxide solution.
H2SO4
+
2 KOH
25.0 mL
xM
33.5 mL
0.240 M
1 mol
2 mol
→
2 H(OH)
+
0.240 mol KOH x 0.0335 L solution x 1 mol H2SO4
___________________
L solution
= 0.161 M H2SO4
4H-13 (of 13)
_________________
2 mol KOH
K2SO4
x
1
____________
0.0250 L
MOLAR MASSES AND STOICHIOMETRIC CONVERSIONS
Calculate the molar mass of ethanol, C2H5OH
2 mol C (12.01 g/mol)
6 mol H (1.008 g/mol)
1 mol O (16.00 g/mol)
=
=
=
24.02 g
6.048 g
16.00 g
46.068 g
46.068 g C2H5OH = 1 mol C2H5OH
4I-1 (of 8)
Calculate the number of ethanol molecules in 25.0 mL of pure ethanol.
The density of the ethanol is 0.789 g/mL.
0.789 g C2H5OH = 1 mL C2H5OH
25.0 mL C2H5OH x 0.789 g C2H5OH x
____________________
1 mol C2H5OH
______________________
1 mL C2H5OH
46.068 g C2H5OH
x 6.022 x 1023 molecules C2H5OH
________________________________________
1 mol C2H5OH
4I-2 (of 8)
= 2.58 x 1023 molecules C2H5OH
Calculate the number of carbon atoms in a 10.0 mL sample of pure
ethanol.
10.0 mL C2H5OH x 0.789 g C2H5OH x 1 mol C2H5OH
____________________
______________________
1 mL C2H5OH
46.068 g C2H5OH
x 6.022 x 1023 atoms C = 2.06 x 1023 atoms C
___________________________
1 mol C
4I-3 (of 8)
x
2 mol C
__________________
1 mol C2H5OH
Calculate the number of ethanol molecules in 45.0 mL of 80. proof vodka.
The density of the vodka is 0.92 g/mL.
80. Proof vodka = 40.% C2H5OH by volume
100 mL vodka = 40. mL C2H5OH
45.0 mL vodka x 40. mL C2H5OH x 0.789 g C2H5OH
x
____________________
____________________
100 mL vodka
1 mL C2H5OH
1 mol C2H5OH
______________________
x 6.022 x 1023 molecules C2H5OH
________________________________________
46.068 g C2H5OH
= 1.9 x 1023 molecules C2H5OH
4I-4 (of 8)
1 mol C2H5OH
Calculate the mass of one ethanol molecule, in grams.
1 molecule C2H5OH x
mol C2H5OH
________________________________________
6.022 x 1023 molecules C2H5OH
= 7.650 x 10-23 g
4I-5 (of 8)
x
46.068 g C2H5OH
_______________________
mol C2H5OH
A metal oxide with the formula M2O3 is 29.0% oxygen by mass. Calculate
the molar mass of metal M.
Molar Mass of M
=
gM
mol M
29.0 g O x
1 mol O
x 2 mol M
16.00 g O
3 mol O
____________
71.0 g M
_________________
1.208 mol M
4I-6 (of 8)
=
__________
58.8 g/mol
=
71.0 g M
? mol M
= 1.208 mol M
THEORETICAL PERCENT COMPOSITION OF COMPOUNDS BY MASS
Calculate the percent composition by mass of sodium nitrate
NaNO3
1 mol Na (22.99 g/mol)
1 mol N (14.01 g/mol)
3 mol O (16.00 g/mol)
= 22.99 g
= 14.01 g
= 48.00 g
85.00 g
% Na =
22.99 g Na
___________________
 100 =
27.05 % Na
 100 =
16.48 % N
 100 =
56.47 % O
85.00 g NaNO3
%N
=
14.01 g N
___________________
85.00 g NaNO3
%O
=
48.00 g O
___________________
85.00 g NaNO3
4I-7 (of 8)
Calculate the percentage by mass of water in barium chloride dihydrate
BaCl2.2H2O
1 mol Ba (137.3 g/mol)
=
2 mol Cl (35.45 g/mol)
=
2 mol H2O (18.016 g/mol) =
137.3
g
70.90 g
36.032 g
244.232 g
% H2O =
36.032 g H2O
____________________________
244.232 g BaCl2.2H2O
4I-8 (of 8)
 100
=
14.75 % H2O
EMPIRICAL AND MOLECULAR FORMULA CALCULATIONS
When a sample of a hydrocarbon is burned, 8.45 g CO2 and 1.73 g H2O
are produced. Calculate the empirical formula of the hydrocarbon.
CxHy
8.45 g CO2
x
+
O2
12.01 g C
________________
→
CO2
=
2.306 g C
=
0.1936 g H
44.01 g CO2
1.73 g H2O
x
2.016 g H
________________
18.016 g H2O
4J-1 (of 9)
+
H2 O
EMPIRICAL AND MOLECULAR FORMULA CALCULATIONS
When a sample of a hydrocarbon is burned, 8.45 g CO2 and 1.73 g H2O
are produced. Calculate the empirical formula of the hydrocarbon.
2.306 g C x
1 mol C
_____________
= 0.1920 mol C
12.01 g C
0.1936 g H x
1 mol H
____________
= 0.1921 mol H
1.008 g H
0.1920 mol C = 1.00 mol C
_________________
0.1920
Empirical formula: CH
4J-2 (of 9)
0.1921 mol H
_________________
0.1920
= 1.00 mol H
A borane is any compound composed of boron and hydrogen. When a
sample of a borane is burned, 12.89 g B2O3 and 12.27 g H2O are
produced. Calculate the empirical formula of the borane.
BxHy
12.89 g B2O3 x
+
O2
21.62 g B
________________
→
B2O3
=
4.0029 g B
=
1.3730 g H
69.62 g B2O3
12.27 g H2O
x
2.016 g H
________________
18.016 g H2O
4J-3 (of 9)
+
H2O
A borane is any compound composed of boron and hydrogen. When a
sample of a borane is burned, 12.89 g B2O3 and 12.27 g H2O are
produced. Calculate the empirical formula of the borane.
4.0029 g B x
1 mol B
_____________
= 0.37030 mol B
10.81 g B
1.3730 g H x
1 mol H
_____________
= 1.3621 mol H
1.008 g H
0.37030 mol B
___________________
= 1.000 mol B x 3
0.37030
Empirical formula: B3H11
4J-4 (of 9)
1.3621 mol H
_________________
0.37030
= 3.678 mol H x 3
When a 3.84 g sample of a compound containing C, H, and N is
combusted, 7.34 g CO2 and 2.51 g H2O are collected. Calculate the
empirical formula of the compound.
CxHyNz
7.34 g CO2
x
+
O2
12.01 g C
________________
→
CO2
+
=
2.003 g C
=
0.2809 g H
H2O
44.11 g CO2
2.51 g H2O
x
2.016 g H
_____+__________
18.016 g H2O
3.84 g CxHyNz
- 2.003 g C
- 0.2809 g H
________________________
1.5561 g N
4J-5 (of 9)
+
Na O b
When a 3.84 g sample of a compound containing C, H, and N is
combusted, 7.34 g CO2 and 2.51 g H2O are collected. Calculate the
empirical formula of the compound.
2.003 g C x
1 mol C
= 0.1668 mol C
_____________
12.01 g C
0.2809 g H x 1 mol H
= 0.2787 mol H
____________
1.008 g H
1.5561 g N x
1 mol N
= 0.1111 mol N
_____________
14.01 g N
0.1668 mol C
0.2787 mol H
0.1111 mol N
0.1111
0.1111
0.1111
= 1.50 mol C
= 2.51 mol H
= 1.00 mol N
_________________
_________________
Empirical formula: C3H5N2
4J-6 (of 9)
________________
When a 2.75 g sample of a compound containing C, H, and O is
combusted, 5.49 g CO2 and 2.25 g H2O are collected. Calculate the
empirical formula of the compound.
CxHyOz
5.49 g CO2
x
+
O2
12.01 g C
________________
→
CO2
=
1.498 g C
=
0.2518 g H
+
44.11 g CO2
2.25 g H2O
x
2.016 g H
_________________
18.016 g H2O
2.75 g CxHyOz
- 1.498 g C
- 0.2518 g H
________________________
1.0002 g O
4J-7 (of 9)
H2O
When a 2.75 g sample of a compound containing C, H, and O is
combusted, 5.49 g CO2 and 2.25 g H2O are collected. Calculate the
empirical formula of the compound.
1.498 g C x
1 mol C
= 0.1247 mol C
____________
12.01 g C
0.2518 g H x 1 mol H
= 0.2498 mol H
____________
1.008 g H
1.0002 g O x 1 mol O
= 0.06251 mol O
_____________
16.00 g O
0.1247 mol C
0.2498 mol H
0.06251 mol O
0.06251
0.06251
0.06251
= 1.99 mol C
= 4.00 mol H
_________________
_________________
Empirical formula: C2H4O
4J-8 (of 9)
__________________
= 1.00 mol O
Calculate the molecular formula of the previous compound if it has a
molar mass of about 90 g/mol.
C2H4O
2 mol C (12.01 g/mol)
=
24.02 g
4 mol H (1.008 g/mol)
1 mol O (16.00 g/mol)
=
=
4.032 g
16.00 g
44.052 g
90 g/mol
________________
≈ 2
44.052 g/mol
Molecular formula: C4H8O2
4J-9 (of 9)