Chem 020 Tutorial Test 1, Wednesday September 26, 07/08
1. 3.00 g of a compound composed of C, H and O is burned in excess oxygen to
yield CO2 and H2O. Calculate both the empirical and molecular formulae for the
compound given that 2.93 g of CO2 and 0.60 g of H2O were produced and that
the molar mass of the compound is estimated to be 90 g. ( 3 MARKS)
CxOyHz + aO2 → x CO2 + z/ CO2
Moles C in sample = moles CO2 produced = 2.93/44 = 0.0666
Moles H in sample = 2 x moles H2O produced = 2 x ( 0.60/18) = 0.067
Mass O in sample = Mass sample – mass C – mass H
=Mass sample – (Molar mass C)x moles C -(Molar Mass H)x molar mass H
= 3.00 – 12 x 0.0666 – 1x 0.067 = 2.133
Therefore moles O in sample = mass/molar mass= 2.133/16 = 0.1334
Now: moles C in sample = 0.0666, moles H in sample = 0.0667, moles O in sample =
0.1334
Dividing by the lowest # yields the empirical formula: CO2H
The molar mass of the empirical formula is 45, given the molecular mass is 90, the
molecular formula must be 2 x the empirical formula or C2O4H2.
2. A metal alloy is 80% Fe and has a density of 6.6 g/cc. 100 cc of this alloy were
subject to chemical analysis to yield Fe as FeCl3. If 996.7g of FeCl3 are
produced, calculate the % yield for the reaction. ( 3 MARKS)
Density = mass/volume = 6.6 g/cc = mass Fe/volume
Mass alloy = (6.6g/cc) x 100cc= 660g in alloy sample.
Mass Fe in alloy = (mass alloy) x 0.8 = 528.0g
Now, for mass Fe in FeCl3:
996.7 g of FeCl3 contain (55.85/162.2) x 996.7g = 343.6 g Fe
Yield = (mass Fe in FeCl3 produced/ Mass Fe in alloy) x 100 = (343.6/ 528.0) x 100
= 65%
3. NO is produced by the following process:
4 NH3 + 5 O2 → 4 NO + 6 H2O
What is the maximum mass of H2O that could be obtained from reacting 60 g of NH3
with 150 g O2? ( 2 MARKS)
Moles NH3 = (Mass/molar mass) = 60/17 = 3.53
Moles O2 required = (5/4) x 3.53 = 4.41
Moles O2 available = 150/32 = 4.69
NH3 is limiting and moles H2O produced = (6/4) x Moles NH3 = (6/4) x 3.53 = 5.295
Mass H2O produced = 18 x 5.295 = 95.3 g
4. There are three stable isotopes of oxygen: 16O mass = 15.999 amu, abundance =
99.76%; 17O mass = 16.999 amu, abundance = 0.04%; 18O mass = 17.999 amu,
abundance = 0.2%.
What would the average atomic mass be for O if the abundance of all three isotopes
were equal? ( 2 MARKS)
Assume any equal # of atoms of each isotope, say 300.
Then # 16O atoms = 17O atoms = 18O atoms = 100
Average mass = (total mass)/(total # of atoms)
Mass 100 16O atoms = 100 x 15.999 amu = 1599.9 amu
Mass 100 17O atoms = 100 x 16.999 amu = 1699.9 amu
Mass 100 18O atoms = 100 x 17.999 amu = 1799.9 amu
Total mass for all 300 O atoms = 5099.7 amu
Average mass = 5099.7 amu/300 = 16.999 amu
Chem 020 Tutorial Test 1, Tuesday, October2, + Wednesday Oct 32007
1. 3.00 g of a compound composed of C, H and O is burned in excess oxygen to
yield CO2 and H2O. Calculate both the empirical and molecular formulae for the
compound given that 7.76 g of CO2 and 1.59 g of H2O were produced and that the
molar mass of the compound is estimated to be 136 g.
CxOyHz + aO2 →
x CO2 + z/ CO2
Moles C in sample = moles CO2 produced = 7.765/44 = 0.1765
Moles H in sample = 2 x moles H2O produced = 2 x (1.59/18) = 0.1767
Mass O in sample = Mass sample – mass C – mass H
=Mass sample – (Molar mass C)x moles C -(Molar Mass H)x molar mass H
= 3.00 – 12 x 0.1765 – 1x 0.1767 = 0.7053
Therefore moles O in sample = mass/molar mass= 0.7053/16 = 0.0.0441
Now: moles C in sample = 0.1765, moles H in sample = 0.1767, moles O in sample =
0.0441
Dividing by the lowest # yields the empirical formula: C4OH4
The molar mass of the empirical formula is 68, given the molecular mass is 136, the
molecular formula must be 2 x the empirical formula or C8O2H8.
2. A metal alloy is 85 % Fe and has a density of 7.6 g/cc. 100 cc of this alloy were
subject to chemical analysis to yield Fe as FeCl3. If 996.7g of FeCl3 are produced,
calculate the % yield for the reaction.
Density = mass/volume = 7.6 g/cc = mass Fe/volume
Mass alloy = (7.6g/cc) x 100cc= 760g in alloy sample.
Mass Fe in alloy = (mass alloy) x 0.85 = .85 x 760 = 646.0g
Now, for mass Fe in FeCl3:
996.7 g of FeCl3 contain (55.85/162.2) x 996.7g = 343.6 g Fe
Yield = (mass Fe in FeCl3 produced/ Mass Fe in alloy) x 100 = (343.6/ 646.0) x 100
= 53 %
3. NO is produced by the following process:
4 NH3 + 5 O2 → 4 NO + 6 H2O
What is the maximum mass of H2O that could be obtained from reacting 60 g of NH3
with 150 g O2?
Moles NH3 = (Mass/molar mass) = 60/17 = 3.53
Moles O2 required = (5/4) x 3.53 = 4.41
Moles O2 available = 150/32 = 4.69
NH3 is limiting and moles H2O produced = (6/4) x Moles NH3 = (6/4) x 3.53 = 5.295
Mass H2O produced = 18 x 5.295 = 95.3 g
4.
There are three stable isotopes of oxygen: 16O mass = 15.999 amu, abundance =
99.76%; 17O mass = 16.999 amu, abundance = 0.04%; 18O mass = 17.999 amu,
abundance = 0.2%.
What would the average atomic mass be for O if the abundance of all three isotopes were
equal?
Assume any equal # of atoms of each isotope, say 300.
Then # 16O atoms = 17O atoms = 18O atoms = 100
Average mass = (total mass)/(total # of atoms)
Mass 100 16O atoms = 100 x 15.999 amu = 1599.9 amu
Mass 100 17O atoms = 100 x 16.999 amu = 1699.9 amu
Mass 100 18O atoms = 100 x 17.999 amu = 1799.9 amu
Total mass for all 300 O atoms = 16.00339 amu
Average mass = 5099.7 amu/300 = 16.999 amu
Chem 020 Tutorial Test 1, September 24 and 25, 07/08
1. 3.00 g of a compound composed of C, H and O is burned in excess oxygen to
yield CO2 and H2O. Calculate both the empirical and molecular formulae for the
compound given that 7.20 g of CO2 and 1.47 g of H2O were produced and that the
molar mass of the compound is estimated to be 110 g.
CxOyHz + aO2 →
x CO2 + z/ CO2
Moles C in sample = moles CO2 produced = 7.2/44 = 0.1636
Moles H in sample = 2 x moles H2O produced = 2 x ( 1.47/18) = 0.1636
Mass O in sample = Mass sample – mass C – mass H
=Mass sample – (Molar mass C) x moles C -(Molar Mass H)x molar mass H
= 3.00 – 12 x 0.1636 – 1x 0.1636 = 0.8732
Therefore moles O in sample = mass/molar mass= 0.955/16 = 0.0546
Now: moles C in sample = 0.1636, moles H in sample = 0.1636, moles O in sample =
0.0546
Dividing by the lowest # yields the empirical formula: C3OH3
The molar mass of the empirical formula is 55, given the molecular mass is 110, the
molecular formula must be 2 x the empirical formula or C6O2H6.
2. A metal alloy is 80% Fe and has a density of 6.6 g/cc. 100 cc of this alloy were
subject to chemical analysis to yield Fe as Fe(NO3)3. If 796.7g of Fe(NO3)3 are
produced, calculate the % yield for the reaction.
Density = mass/volume = 6.6 g/cc = mass alloy/volume
Mass alloy = (6.6g/cc) x 100cc= 660g
Mass Fe = 0.8 x 660 = 528 g
796.7 g of Fe(NO3)3 contain (55.85/241.85) x 796.7g = 183.98 g Fe
But mass Fe expected = 528 g
% yield Fe = (mass Fe in product/Mass Fe in alloy) x 100 = (183.98/528) x 100 =
34.8 %
3. NO is produced by the following process:
4 NH3 + 5 O2 → 4 NO + 6 H2O
What is the maximum mass of H2O that could be obtained from reacting 60 g of NH3
with 150 g O2?
Moles NH3 = (Mass/molar mass) = 60/17 = 3.53
Moles O2 required = (5/4) x 3.53 = 4.41
Moles O2 available = 150/32 = 4.69
NH3 is limiting and moles H2O produced = (6/4) x Moles NH3 = (6/4) x 3.53 = 5.295
Mass H2O produced = 18 x 5.295 = 95.3 g
4.
There are three stable isotopes of oxygen: 16O mass = 15.999 amu, abundance =
99.76%; 17O mass = 16.999 amu, abundance = 0.04%; 18O mass = 17.999 amu,
abundance = 0.2%.
What would the average atomic mass be for O if the abundance of the three isotopes were
20% 16O, 30% 17O and 50% 18O?
Assume any total # of O atoms, say 100.
Average mass = (total mass)/(total # of atoms)
# of 16O atoms = 20
Mass 20 16O atoms = 20 x 15.999 amu = 319.98 amu
# of
17
O atoms =30
Mass 30 17O atoms = 30 x 16.999 amu = 509.97 amu
# of
18
O atoms = 50
Mass 50 18O atoms = 50 x 17.999 amu = 899.95 amu
Total mass for all 100 O atoms = 1729.9 amu
Average mass = 1729.9 amu/100 = 17.299 amu
Strong acids/bases