Chapter 3 Lecture
1. Gallium consists of two isotopes of masses
68.95 amu and 70.95 amu with abundances of
60.16% and 39.84%, respectively. What is the
atomic mass of gallium?
Atomic mass = (mass I1)(% I1) + (mass I2)(% I2) + (continued)
(make sure that percents add up to 100 or approximately 100)
= (68.95 amu)(0.6016) + (70.95 amu)(0.3984)
= 69.75 amu
2. Magnesium exists as three isotopes in nature. One
isotope (25Mg) has a mass of 24.99 amu and a relative
abundance of 10.13%. The other two isotopes have
masses of 23.99 amu (24Mg) and 25.98 amu (26Mg). What
are their relative abundances?
24.31 = (24.99)(0.1013) + (23.99)(% I2) + (25.98)(% I3)
24.31 = (24.99)(0.1013) + (23.99)(x) + (25.98y)
0.1013 + x + y = 1 (all % abundances add up to 100%)
x = 0.8987 – y
24.31 = (24.99)(0.1013) + (23.99)(0.8987 - y) + (25.98y)
y = 10.99%, x = 78.88 %
3. Calculate the molar masses of the following
compounds:
a. Zr(SeO3)2
b. NH4OH c. Ca2Fe(CN)6.12H20
91.22
14.01
+ 78.96(2)
+ 1.01(5)
+ 16.00(6)
+ 16.00
345.14 g/mole 35.06 g/mole
40.08(2)
+ 55.85
+ 12.01(6)
+ 14.01(6)
+ 1.01(24)
+ 16.00(12)
508.37 g/mole
4. What are the mass percents of iron and
oxygen in Fe2O3?
Iron: (2 x 55.85 g/mole) x 100= 69.94 %
(159.7 g/mole)
Oxygen: (3 x 16.00 g/mole) x 100 = 30.06 %
(159.7 g/mole)
Two ways of solving the same problem
5. How many grams of iron can be extracted from 567
grams of iron(III) oxide?
567 g Fe2O3 ( 1 mole Fe2O3 ) ( 2 moles Fe) (55.85 g Fe) =
(159.7 g Fe2O3) (1 mole Fe2O3) (1 mole Fe)
OR
Use the strategy we just learned in #3!
mass of compound
in sample
x
% iron in the =
compound
mass iron in
the sample
(567 g Fe2O3)( 0.6994) = 397 g Fe
6. A substance contains 23.0 g sodium, 27.0 g
aluminum, and 114 g fluorine. How many grams of
sodium are there in a 120 gram sample of the
substance?
Total mass: 164 g
23.0 g Na = 0.140 or 14.0%
164g
(0.140)(120g) = 17 g Na
7. A 25.0 gram sample of a compound contains 6.64 grams
potassium, 8.84 g chromium, 9.52g oxygen. Find the empirical
formula of this compound.
Step 1: Convert each mass to moles
0.170 mole K
0.170 mole Cr
0.170
0.170
0.595 mole O
0.170
Step 2. Get ratio of moles (by dividing each # moles by smallest #
moles)
Mole ratio: 1 mole K: 1 mole Cr: 3.50 moles O
Step 3: If the numbers in the ratio are not integers, double, triple, etc
each number until they are all integers….you may want to round
slightly (ex. 2.98 can be 3 but 2.49 should be doubled to 4.98, then
rounded to 5)
New ratio: 2 : 2 : 7
K2Cr2O7, potassium dichromate
8. Phenol is a compound which contains 76.57% carbon, 6.43% hydrogen, and
17.0% oxygen. What is the empirical formula of phenol?
If you are given percents of elements instead of actual masses, turn each
percent into a gram amount out of a 100g sample
Step 1: Convert each mass to moles
76.57g C/12.01 g C
6.43 g H/1.01g H
17.0 g O/16.00 g O
6.376 mole C
1.06
1.06 mole O
1.06
6.366 mole H
1.06
Step 2. Get ratio of moles (by dividing each # moles by smallest # moles)
Mole ratio: 6 mole C: 6 mole H: 1 moles O
C6H6O = phenol!
9. The empirical formula of styrene is CH; its formula weight
(‘molar mass’) is 104.1 g/mole. What is the molecular
formula of styrene?
molecular formula = multiple of the empirical formula
* molar mass of the compound is needed
Molar mass styrene = 104.1 g/mole
Molar mass emp.form. = 13.02 g/mole
= 7.995391… = 8
molecular formula = C8H8
10. A 5.00 gram sample of an acid contains 2.00 g
carbon, 0.336 g hydrogen, and 2.66 g oxygen. Find the
molecular formula of this acid and give the proper
name if its molar mass = 60.06 g/mole
Try this yourself: answer: acetic acid
The Molar Highway
Finding Various Quantities using The Mole
mass of element
or ion in
compound
moles of element or
ion in the compound
# atoms or ions
in sample
Need:
Need:
# of that atom or
ion in compound
atomic mass of
element or ion
Need:
Mass
compound
in sample
Need:
molar mass of compound
# of that
atom or ion
in
compound
1 mole of a
compound
# molecules
compound in
sample
Need:
1 mole = 6.02 x 1023 particles
11. A sample of 3.9 x 1019 strontium
chloride molecules contains how many
milligrams of chlorine?
3.9 x 1019 molec. SrCl2 (2 atoms Cl ) (1 mole Cl)
( 35.45 g Cl) ( 1000 mg Cl) =
(1 molec SrCl2) (6.02 x 1023 atoms Cl) (1 mole Cl) ( 1 g Cl)
4.6 mg chlorine
2Sb(s) + 3I2(s) → 2SbI3(s
12a. What mass of antimony(III) iodide can form from
1.20g Sb ?
1.20 g Sb
(
(
1
mol Sb )
121.76 g Sb ) (
(
2
mol SbI3)
(
502.46
2
mol Sb)
(
1
g SbI3 )
mol SbI3 )
= 4.94 g SbI3
Wait! We know how much Sb we have. But
what if there is just a tiny, tiny amount of
iodine available? Hmmmm…
2Sb(s) + 3I2(s) → 2SbI3(s)
12b. Determine the theoretical yield (mass in
grams) of antimony(III) iodide formed when 1.20g
Sb and 2.40 g I2 are mixed.
2.40 g I2
(
(
1
253.8
= 3.17 g SbI3
mol I2 )
(
2
mol SbI3)
(
502.46
g I2 )
(
3
mol I2)
(
1
g SbI3 )
mol SbI3 )
wait but 1.20 g Sb predicts 4.94 g SbI3
5. Do another calculation with the other amount given
6. Choose the lower amount…the higher amount is not possible. The
theoretical yield is calculated based upon the limiting reactant.
3.17 g SbI3. I2 is the limiting reactant (even though there is a greater
amount of it) and is completely used up. Some Sb will be leftover.
12c. If the experiment actually produces 3.00g
SbI3, calculate the percent yield of the
experiment.
percent yield = actual or experiment yield x 100
theoretical yield
percent yield = 3.00 g x 100
3.17 g
94.6 % yield