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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. 2H 2 + O 2 → 2H 2 O. 2 molecules 2 moles 0.84 moles - PowerPoint PPT Presentation
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REACTION STOICHIOMETRY
1792 JEREMIAS RICHTERThe amount of substances produced or consumed in chemical reactions can be quantified
4F-1 (of 14)
INFORMATION FROM CHEMICAL EQUATIONS
2H2 + O2 → 2H2O
2 molecules2 moles
0.84 moles0.028 moles
1 molecule1 moles
0.42 moles0.014 moles
2 molecules2 moles
0.84 moles0.028 moles
4F-2 (of 14)
The moles that react and form do so in the ratio of the balanced equation
INFORMATION FROM CHEMICAL EQUATIONS
2H2 + O2 → 2H2O
0.60 moles-0.60 moles0.00 moles
0.40 moles-0.30 moles0.10 moles
0.00 moles+0.60 moles
0.60 moles
4F-3 (of 14)
startingreacting
ending
0.50 moles-0.40 moles0.10 moles
0.20 moles-0.20 moles0.00 moles
0.00 moles+0.40 moles
0.40 moles
startingreacting
ending
2H2 + O2 → 2H2O
The moles that react and form do so in the ratio of the balanced equation
MASS CALCULATIONS
Calculate the mass of oxygen needed to burn 5.00 g of propanol, C3H8O.
C3H8O + O2 →
CO2 + H2O3 44½
4F-4 (of 14)
MASS CALCULATIONS
Calculate the mass of oxygen needed to burn 5.00 g of propanol, C3H8O.
C3H8O + O2 →
CO2 + H2O6 85.00 g x g2 mol 9 mol
9 2
2 mol C3H8O = 9 mol O2
5.00 g C3H8O
x 9 mol O2 _________________
2 mol C3H8O
= 12.0 g O2
x 1 mol C3H8O ____________________
60.094 g C3H8O
x 32.00 g O2 ______________
1 mol O2
4F-5 (of 14)
Calculate the mass of carbon dioxide produced from the 5.00 g of propanol.
C3H8O + O2 →
CO2 + H2O6 85.00 g x g2 mol 6 mol
9 2
2 mol C3H8O = 6 mol CO2
5.00 g C3H8O
x 6 mol CO2 _________________
2 mol C3H8O
= 11.0 g CO2
x 1 mol C3H8O ____________________
60.094 g C3H8O
x 44.01 g CO2 ________________
1 mol CO2
4F-6 (of 14)
C3H8O + O2 →
CO2 + H2O6 85.0 g 6.0 g
9 2 12.0 g 11.0 g
4F-7 (of 14)
LIMITING REACTANT CALCULATIONS
LIMITING REACTANT – The reactant that is completely used up in a reaction
4F-8 (of 14)
x
12 pieces
x 1 sandwich _______________
1 piece
= 12 sandwiches
22 slices
x 1 sandwich _______________
2 slices
= 11 sandwiches actual amount produced
limiting reactant
12 pieces 22 slices
4F-9 (of 14)
Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride.
Ba + WF3 →
BaF2 + W 25.0 g 26.0 g3 mol 2 mol
323x g
2 mol
2
4F-10 (of 14)
25.0 g Ba
x 2 mol W ____________
3 mol Ba
= 22.3 g W
x 1 mol Ba _______________
137.3 g Ba
x 183.8 g W ______________
1 mol W
Ba + WF3 → 25.0 g 26.0 g3 mol 2 mol
23 BaF2 + W 3x g
2 mol
2
Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride.
4F-11 (of 14)
25.0 g Ba
x 2 mol W ____________
3 mol Ba
= 22.3 g W
x 1 mol Ba _______________
137.3 g Ba
x 183.8 g W _______________
1 mol W
Ba + WF3 → 25.0 g 26.0 g3 mol 2 mol
23 BaF2 + W 3x g
2 mol
2
Calculate the mass of tungsten metal produced when 25.0 g of barium reacts with 26.0 g of tungsten (III) fluoride.
26.0 g WF3
x 2 mol W _____________
2 mol WF3
= 19.8 g W
x 1 mol WF3 _________________
240.8 g WF3
x 183.8 g W _______________
1 mol W
WF3 is the limiting reactant19.8 g W are produced
4F-12 (of 14)
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 + HCl →
MgCl2 + H2
x g 14.5 g1 mol 2 mol
2
1 mol Mg = 2 mol HCl
14.5 g HCl
x 1 mol Mg ______________
2 mol HCl
= 4.834 g Mg
x 1 mol HCl ________________
36.458 g HCl
x 24.31 g Mg ______________
1 mol Mg
4F-13 (of 14)
4.834 g Mg x 100 _______________
6.50 g alloy
= 74.4% Mg 100% - 74.4% = 25.6% Ag
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+ + I- →
PbI2 x g
2 mol
21.42 g1 mol
1.42 g PbI2
x 2 mol I-
_____________
1 mol PbI2
= 0.7818 g I-
x 1 mol PbI2 _________________
461.0 g PbI2
x 126.9 g I-
_____________
1 mol I-
0.7818 g I-
___________________
1.17 g sample
= 66.8% I- in the sample
x 100
4F-14 (of 14)
Determine the percentage by mass of iodide in a solid unknown.
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
x 0.275 L solution = 0.0825 mol K2CO3
0.300 mol K2CO3______________________
L solution
4G-1 (of 12)
Find the moles of each ion in the 0.300 M potassium carbonate solution.
0.0825 mol K2CO3 = 0.165 mol K+
x 2
0.0825 mol K2CO3 = 0.0825 mol CO32-
x 1
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.
x 0.0100 L solution = 0.004500 mol BaCl2
0.450 mol BaCl2______________________
L solution 0.00450 mol Ba2+ and 0.00900 mol Cl-
x 0.0200 L solution = 0.006000 mol K2SO4
0.300 mol K2SO4______________________
L solution 0.0120 mol K+ and 0.00600 mol SO4
2-
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- → BaSO4
Initial molesReacting molesFinal moles
0.00450 0.00600 0 -0.00450 -0.00450 +0.00450
0 0.00150 0.00450
0 mol Ba2+ 0.00150 mol SO42-
0.00900 mol Cl- 0.0120 mol K+
4G-4 (of 12)
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 M Ba2+ 0 mol Ba2+ ______________________
0.0300 L solution
= 0.0500 M SO42- 0.00150 mol SO4
2- _______________________
0.0300 L solution = 0.300 M Cl- 0.00900 mol Cl-
_______________________
0.0300 L solution
= 0.400 M K+ 0.0120 mol K+ _______________________
0.0300 L solution
4G-5 (of 12)
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.
x 0.0200 L solution = 0.007000 mol HCl
0.350 mol HCl___________________
L solution 0.00700 mol H+ and 0.00700 mol Cl-
x 0.0300 L solution = 0.007500 mol NaOH
0.250 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- → H2O
Initial molesReacting molesFinal moles
0.00700 0.00750 0 -0.00700 -0.00700 +0.00700
0 0.00050 0.00700
0 mol H+ 0.00050 mol OH-
0.00700 mol Cl- 0.00750 mol Na+
4G-7 (of 12)
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 M H+ 0 mol H+ ______________________
0.0500 L solution
= 0.010 M OH- 0.00050 mol OH- _______________________
0.0500 L solution
= 0.140 M Cl- 0.00700 mol Cl- _______________________
0.0500 L solution
= 0.150 M Na+ 0.00750 mol Na+ _______________________
0.0500 L solution
4G-8 (of 12)
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.
= 0.126 g H2O 0.007000 mol H2O
x 18.016 g H2O __________________
1 mol H2O
4G-9 (of 12)
DILUTION CALCULATIONS
When a solution is diluted, only solvent is added , so the moles of soluteare 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
VC = MDVD _______
MC
= (0.100 M)(250. mL) ________________________
(6.00 M)
= 4.17 mL
MC =VC =
6.00 M?
MD =VD =
0.100 M250. mL
4G-11 (of 12)
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
MCVC = VD_______
MD
= (12.1 M)(5.00 mL) _______________________
(3.00 M)
= 20.2 mL
MC =VC =
12.1 M5.00 mL
MD =VD =
3.00 M?
20.2 mL- 5.00 mL____________
Volume of dilute solutionVolume of concentrated solution
15.2 mL Water that must be added
4G-12 (of 12)
TITRATION – A technique in which one solution is used to analysis another
Buret: a solution of 1 reactant of known concentration
REACTIONS IN SOLUTION
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 solutionsThe 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 →
NaCl + H2O + CO2 x g 21.5 mL
0.300 M1 mol 1 mol
x 1 mol NaHCO3 _________________
1 mol HCl
0.300 mol HCl x 0.0215 L solution __________________
L solution
= 0.542 g NaHCO3
x 84.008 g NaHCO3 _______________________
mol NaHCO3
4H-2 (of 13)
x 1 mol O.A.D. ____________________
126.08 g O.A.D.
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).
NaOH + H2C2O4 → .2H2O Na2C2O4 + H2O 34.2 mL
x M0.619 g
2 mol 1 mol
x 2 mol NaOH _________________
1 mol O.A.D.
0.619 g O.A.D.
= 0.287 M NaOH
x 1 _______________________
0.0342 L solution
2 24
_
4H-3 (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.
4H-4 (of 13)
( ) x 3
( ) x 5
MnO4- (aq) + V2+ (aq) →
MnO4- → Mn2+
+7 -2 +2
V2+ → V5+
+ 4H2O 8H+ +5e- +
+ 3e-
Mn2+ + V5+
+2 +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.
4H-5 (of 13)
3MnO4- → 3Mn2+
5V2+ → 5V5+
+ 12H2O 24H+ +15e- +
+ 15e-
15e- + 24H+ + 3MnO4- + 5V2+ →
3Mn2+ + 12H2O + 5V5+ + 15e-
MnO4- (aq) + V2+ (aq) →
+7 -2 +2 +2 +5Mn2+ + V5+
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.
10.0 mL x M
32.7 mL0.115 M
5 mol3 mol
x 5 mol V2+
________________
3 mol MnO4-
0.115 mol MnO4- x 0.0327 L solution
_____________________
L solution
= 0.627 M V2+
x 1 _______________________
0.0100 L solution
4H-6 (of 13)
3MnO4- (aq) + 5V2+(aq) → 3Mn2+ (aq) + 12H2O (l) + 5V5+ (aq) 24H+
(aq) +
molar mass H2X = grams H2X_______________
moles H2X
molar mass H2X = 0.109 grams H2X ______________________
? moles H2X
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.
4H-7 (of 13)
H2X + NaOH →
H(OH) + Na2X 0.409 gx mol
19.50 mL0.287 M
1 mol 2 mol
2
x 1 mol H2X _______________
2 mol NaOH
0.287 mol NaOH x 0.01950 L sol’n _____________________
L sol’n
= 0.002798 mol H2X
2
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.
0.409 g H2X________________________
0.002798 mol H2X
= 146 g/mol
4H-8 (of 13)
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 + HCl →
AlCl3 + H(OH) 10.0 mL
x M12.5 mL0.300 M
1 mol 3 mol
3
x 1 mol Al(OH)3 _________________
3 mol HCl
0.300 mol HCl x 0.0125 L solution __________________
L solution
= 0.125 M Al(OH)3
x 1 ____________
0.0100 L
3
4H-9 (of 13)
( ) x 8
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 Cr2O7
2- solution. Determine the percentage of S2- in cinnabar ore.
4H-10 (of 13)
( ) x 3
K2Cr2O7 (aq) + S2- (aq) →
K+ (aq) + Cr2O7
2- (aq) + S2- (aq) →
Cr2O72- → Cr3+
+1 +6 -2 -2
S2- → S8
2
8
+ 7H2O14H+ +6e- +
+ 16e-
Cr3+ + S8
+3 0
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 Cr2O7
2- solution. Determine the percentage of S2- in cinnabar ore.
4H-11 (of 13)
K2Cr2O7 (aq) + S2- (aq) →
K+ (aq) + Cr2O7
2- (aq) + S2- (aq) →
8Cr2O72- → 16Cr3+
+1 +6 -2 -2
24S2- → 3S8
+ 56H2O112H+ +48e- +
+ 48e-
48e- + 112H+ + 8Cr2O72- + 24S2- → 16Cr3+ + 56H2O + 3S8 + 48e-
Cr3+ + S8
+3 0
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 Cr2O7
2- solution. Determine the percentage of S2- in cinnabar ore.
x g20.4 mL0.110 M
x 24 mol S2-
_________________
8 mol Cr2O72-
0.110 mol Cr2O72- x 0.0204 L solution
_______________________
L solution
= 0.2159 g S2-
x 32.07 g S2- _____________
mol S2-
0.2159 g S2- x 100 _______________
1.534 g ore
= 14.1% S2- in the ore
4H-12 (of 13)
8Cr2O72- (aq) + 24S2-(aq) → 16Cr3+ (aq) + 56H2O (l) + 3S8
(s) 112H+ (aq) +
8 mol 24 mol
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 + KOH →
H(OH) + K2SO4 25.0 mL
x M33.5 mL0.240 M
1 mol 2 mol
2
x 1 mol H2SO4 _________________
2 mol KOH
0.240 mol KOH x 0.0335 L solution ___________________
L solution
= 0.161 M H2SO4
x 1 ____________
0.0250 L
2
4H-13 (of 13)
MOLAR MASSES AND STOICHIOMETRIC CONVERSIONS
2 mol C (12.01 g/mol) = 24.02 g6 mol H (1.008 g/mol) = 6.048 g
1 mol O (16.00 g/mol) = 16.00 g46.068 g
Calculate the molar mass of ethanol, C2H5OH
4I-1 (of 8)
46.068 g C2H5OH = 1 mol C2H5OH
x 1 mol C2H5OH ______________________
46.068 g C2H5OH
Calculate the number of ethanol molecules in 25.0 mL of pure ethanol. The density of the ethanol is 0.789 g/mL.
25.0 mL C2H5OH
x 6.022 x 1023 molecules C2H5OH ________________________________________
1 mol C2H5OH
= 2.58 x 1023 molecules C2H5OH
x 0.789 g C2H5OH ____________________
1 mL C2H5OH
0.789 g C2H5OH = 1 mL C2H5OH
4I-2 (of 8)
x 0.789 g C2H5OH ____________________
1 mL C2H5OH
x 1 mol C2H5OH ______________________
46.068 g C2H5OH
Calculate the number of carbon atoms in a 10.0 mL sample of pure ethanol.
10.0 mL C2H5OH
x 2 mol C __________________
1 mol C2H5OH
= 2.06 x 1023 atoms C
x 6.022 x 1023 atoms C ___________________________
1 mol C
4I-3 (of 8)
x 40. mL C2H5OH ____________________
100 mL vodkax 1 mol C2H5OH ______________________
46.068 g 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.
45.0 mL vodka
x 6.022 x 1023 molecules C2H5OH ________________________________________
1 mol C2H5OH
= 1.9 x 1023 molecules C2H5OH
x 0.789 g C2H5OH ____________________
1 mL C2H5OH
80. Proof vodka = 40.% C2H5OH by volume
100 mL vodka = 40. mL C2H5OH
4I-4 (of 8)
x 46.068 g C2H5OH _______________________
mol C2H5OH
Calculate the mass of one ethanol molecule, in grams.
= 7.650 x 10-23 g
x mol C2H5OH ________________________________________
6.022 x 1023 molecules C2H5OH
4I-5 (of 8)
1 molecule C2H5OH
x 1 mol O ____________
16.00 g O
A metal oxide with the formula M2O3 is 29.0% oxygen by mass. Calculate the molar mass of metal M.
29.0 g O x 2 mol M __________
3 mol O
= 58.8 g/mol 71.0 g M_________________
1.208 mol M
4I-6 (of 8)
g Mmol M
Molar Mass of M = = 71.0 g M ? mol M
= 1.208 mol M
1 mol Na (22.99 g/mol) = 22.99 g1 mol N (14.01 g/mol) =
14.01 g3 mol O (16.00 g/mol) =48.00 g
85.00 g
THEORETICAL PERCENT COMPOSITION OF COMPOUNDS BY MASS
NaNO3
Calculate the percent composition by mass of sodium nitrate
22.99 g Na 100___________________
85.00 g NaNO3
% Na = = 27.05 % Na
48.00 g O 100___________________
85.00 g NaNO3
% O = = 56.47 % O
14.01 g N 100___________________
85.00 g NaNO3
% N = = 16.48 % N
4I-7 (of 8)
BaCl2.2H2O
1 mol Ba (137.3 g/mol) = 137.3 g2 mol Cl (35.45 g/mol) = 70.90 g
2 mol H2O (18.016 g/mol) = 36.032 g244.232 g
Calculate the percentage by mass of water in barium chloride dihydrate
36.032 g H2O 100____________________________
244.232 g BaCl2.2H2O
% H2O = = 14.75 % H2O
4I-8 (of 8)
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 + O2 →
CO2 + H2O
8.45 g CO2
= 2.306 g C x 12.01 g C ________________
44.01 g CO2
1.73 g H2O
= 0.1936 g H x 2.016 g H ________________
18.016 g H2O
4J-1 (of 9)
x 1 mol C_____________
12.01 g C
= 0.1920 mol C
2.306 g C
x 1 mol H____________
1.008 g H
= 0.1921 mol H
0.1936 g H
0.1920 mol C_________________
0.1920
0.1921 mol H_________________
0.1920
= 1.00 mol C = 1.00 mol H
Empirical formula: CH
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.
4J-2 (of 9)
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 + O2 →
B2O3 + H2O
12.89 g B2O3
= 4.0029 g B x 21.62 g B ________________
69.62 g B2O3
12.27 g H2O
= 1.3730 g H x 2.016 g H ________________
18.016 g H2O
4J-3 (of 9)
x 1 mol B_____________
10.81 g B
= 0.37030 mol B
4.0029 g B
x 1 mol H_____________
1.008 g H
= 1.3621 mol H
1.3730 g H
0.37030 mol B___________________
0.37030
1.3621 mol H_________________
0.37030
= 1.000 mol B = 3.678 mol H
Empirical formula: B3H11
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.
4J-4 (of 9)
x 3 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 + O2 → CO2 + H2O + NaOb
7.34 g CO2
= 2.003 g C x 12.01 g C ________________
44.11 g CO2
2.51 g H2O
= 0.2809 g H 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)
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.
x 1 mol C_____________
12.01 g C
= 0.1668 mol C
2.003 g C
x 1 mol H____________
1.008 g H
= 0.2787 mol H
0.2809 g H
0.1668 mol C_________________
0.1111
0.2787 mol H_________________
0.1111= 1.50 mol C = 2.51 mol H
Empirical formula: C3H5N2
x 1 mol N_____________
14.01 g N
= 0.1111 mol N
1.5561 g N
0.1111 mol N________________
0.1111= 1.00 mol N
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 + O2 → CO2 + H2O
5.49 g CO2
= 1.498 g C x 12.01 g C ________________
44.11 g CO2
2.25 g H2O
= 0.2518 g H 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)
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.
x 1 mol C____________
12.01 g C
= 0.1247 mol C
1.498 g C
x 1 mol H____________
1.008 g H
= 0.2498 mol H
0.2518 g H
0.1247 mol C_________________
0.06251
0.2498 mol H_________________
0.06251= 1.99 mol C = 4.00 mol H
Empirical formula: C2H4O
x 1 mol O_____________
16.00 g O
= 0.06251 mol O
1.0002 g O
0.06251 mol O__________________
0.06251= 1.00 mol O
4J-8 (of 9)
C2H4O
2 mol C (12.01 g/mol) = 24.02 g4 mol H (1.008 g/mol) = 4.032 g1 mol O (16.00 g/mol) = 16.00 g
44.052 g
90 g/mol________________
44.052 g/mol
≈ 2
Molecular formula: C4H8O2
Calculate the molecular formula of the previous compound if it has a molar mass of about 90 g/mol.
4J-9 (of 9)