Transcript No Slide Title

1) A sample of a compound
contains 12.15 g of Mg and 19 g of F.
What is the percent composition of
this compound (SHOW WORK)?
Period 3 Day 4 11-20
Calculating Empirical Formulas
In an unknown compound you find
4.04 g of N and 11.46 g O. Empirical
formula?
Periods 1 and 3 have a portal assignment due NO
later than Saturday (11-21) night at midnight.
1) A sample of a compound
contains 12.15 g of Mg and 19 g of F.
What is the percent composition of
this compound (SHOW WORK)?
Period 8 Day 4 11-20
We will have a TIMED ATB momentarily.
Get out your chemistry materials.
day 5 12-1
1) Earth’s population is about 6.9
billion. If every person on Earth
participates in counting identical
particles at a rate of 2 particles per
second, how many years would it take to
count 6.0 X 1023 particles? Assume that
there are 365 days in every year.
1.38 X
6
10
years
day 5 12-1
2) What is the mass in grams of 1.00 X
1012 lead atoms?
3.44 X
-10
10
g
day 5 12-1
3) Tin exists in Earth’s crust as SnO2.
Name this compound! Calculate the
percent composition by mass?
79% Sn and 21% O
day 5 12-1
1) A sample of a compound
contains 12.15 g of Mg and 19 g of F.
What is the percent composition of
this compound (SHOW WORK)?
Period 1 Day 5 12-1
Periods 1 and 3 have a portal assignment due NO
later than Saturday (11-21) night at midnight.
46 g of Sodium (Na) = __________ atoms?
46 g Na
X
1 mol Na
23 g Na
X
6.022 * 1023
atoms Na
1 mol Na
=
NOTES PAGE 3
46 g of Sodium (Na) = __________ atoms?
46 g Na
X
1 mol Na
23 g Na
X
6.022 * 1023
atoms Na
1 mol Na
=
NOTES PAGE 3
2 moles of sulfuric acid = _______ atoms of H
6.022 * 1023
2 moles H2SO4
X
molecules H2SO4
1 mol H2SO4
X
2 atoms of H
1 molecule
H2SO4
=
NOTES PAGE 3
Heavy
Light
Heavy
Light
Mass Spectrometer
Mass Spectrum of Ne
15
Empirical Formulas vs.
Molecular Formulas
Empirical formula – smallest wholenumber mole ratio for a compound
Sometimes
the same.
But not always!
Molecular formula – actual # of atoms
of each ele. in a molecular compound
NOTES PAGE 7
Molecular Formulas
Molecular formula – actual # of atoms
of each ele. in a molecular compound
To Calculate:
Compare the molar mass of the
empirical formula to the molar
mass of the molecular formula
Determining Molecular Formulas
Practice # 1 Determine the molecular
formula of the compound with an
empirical formula of CH and a formula
mass of 78.110 amu
Empirical mass = 13.019 g/mol
x=6
Molecular formula = C6H6
Complete the section on Mass
Spectrometer in your notes on page 5.
Be ready to discuss it tomorrow!
How about a quest on Friday 12-4!!!
Pages 107-112: #s 3.5, 3.14, 3.16, 3.20,
3.26, 3.30, 3.40, 3.44, 3.48, 3.52, 3.96
The answers for all except 3.5 are in the
back of the book – CHECK them! This
is your review for the quest!!! YOU
MUST SHOW YOUR WORK!!!
1) Allicin is the compound responsible
for the smell of garlic. An analysis of
the compound gives the following
percent composition by mass: C =
44.4% H = 6.21% S = 39.5% O = 9.86%.
Calculate its empirical formula. What is
its molecular formula given that its
molar mass is about 162 g?
Period 1 Day 6 12-2
Determining Molecular Formulas
In an unknown compound you find
4.04 g of N and 11.46 g O. Empirical
formula? This unknown compound
has a molar mass of 108.0 g/mol.
What is the molecular formula?
Empirical formula = N2O5
Empirical mass = 108.009 g/mol
Molecular formula = N2O5
Practice
A sample of a compound contains
3.003 g of C, 0.504 g of H, and 4.000 g
of O. If the molar mass is 180.157
g/mol what is the molecular formula.
Empirical
formula
Molecular
Formula
3.11
Question of the Day
A sample of a compound contains 30.46 percent nitrogen and
69.54 percent oxygen by mass, as determined by a mass
spectrometer.
In a separate experiment, the molar mass of the compound is
found to be between 90 g and 95 g.
Determine the molecular formula and the accurate molar mass
of the compound.
Period 1 Day 1 12-3
3.11
Strategy
To determine the molecular formula, we first need to determine
the empirical formula. Comparing the empirical molar mass to
the experimentally determined molar mass will reveal the
relationship between the empirical formula and molecular
formula.
Solution
We start by assuming that there are 100 g of the compound.
Then each percentage can be converted directly to grams; that
is, 30.46 g of N and 69.54 g of O.
3.11
Let n represent the number of moles of each element so that
Thus, we arrive at the formula N2.174O4.346, which gives the
identity and the ratios of atoms present. However, chemical
formulas are written with whole numbers.
Try to convert to whole numbers by dividing the subscripts by
the smaller subscript (2.174). After rounding off, we obtain NO2
as the empirical formula.
3.11
The molecular formula might be the same as the empirical
formula or some integral multiple of it (for example, two, three,
four, or more times the empirical formula).
Comparing the ratio of the molar mass to the molar mass of the
empirical formula will show the integral relationship between the
empirical and molecular formulas.
The molar mass of the empirical formula NO2 is
empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g
3.11
Next, we determine the ratio between the molar mass and the
empirical molar mass
The molar mass is twice the empirical molar mass. This means
that there are two NO2 units in each molecule of the compound,
and the molecular formula is (NO2)2 or N2O4. The actual molar
mass of the compound is two times the empirical molar mass,
that is, 2(46.01 g) or 92.02 g, which is between 90 g and 95 g.
3.11
Check Note that in determining the molecular formula from the
empirical formula, we need only know the approximate molar
mass of the compound. The reason is that the true molar mass
is an integral multiple (1×, 2×, 3×, . . .) of the empirical molar
mass. Therefore, the ratio (molar mass/empirical molar mass)
will always be close to an integer.
Practice # 3 (expect a similar test question)
A sample of a compound contains 6.0
g of C and 16 g of O. The total sample
is 22 g. The molecular mass is 44 g.
% comp
27% C 73% O
Empirical
formula
CO2
Molecular
Formula
CO2
NOTES PAGE 8
3.11
Question of the Day
A sample of a compound contains 30.46 percent nitrogen and
69.54 percent oxygen by mass, as determined by a mass
spectrometer.
In a separate experiment, the molar mass of the compound is
found to be between 90 g and 95 g.
Determine the molecular formula and the accurate molar mass
of the compound.
Period 3 Day 1 12-3
3.11
Strategy
To determine the molecular formula, we first need to determine
the empirical formula. Comparing the empirical molar mass to
the experimentally determined molar mass will reveal the
relationship between the empirical formula and molecular
formula.
Solution
We start by assuming that there are 100 g of the compound.
Then each percentage can be converted directly to grams; that
is, 30.46 g of N and 69.54 g of O.
3.11
Let n represent the number of moles of each element so that
Thus, we arrive at the formula N2.174O4.346, which gives the
identity and the ratios of atoms present. However, chemical
formulas are written with whole numbers.
Try to convert to whole numbers by dividing the subscripts by
the smaller subscript (2.174). After rounding off, we obtain NO2
as the empirical formula.
3.11
The molecular formula might be the same as the empirical
formula or some integral multiple of it (for example, two, three,
four, or more times the empirical formula).
Comparing the ratio of the molar mass to the molar mass of the
empirical formula will show the integral relationship between the
empirical and molecular formulas.
The molar mass of the empirical formula NO2 is
empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g
3.11
Next, we determine the ratio between the molar mass and the
empirical molar mass
The molar mass is twice the empirical molar mass. This means
that there are two NO2 units in each molecule of the compound,
and the molecular formula is (NO2)2 or N2O4. The actual molar
mass of the compound is two times the empirical molar mass,
that is, 2(46.01 g) or 92.02 g, which is between 90 g and 95 g.
3.11
Check Note that in determining the molecular formula from the
empirical formula, we need only know the approximate molar
mass of the compound. The reason is that the true molar mass
is an integral multiple (1×, 2×, 3×, . . .) of the empirical molar
mass. Therefore, the ratio (molar mass/empirical molar mass)
will always be close to an integer.
Heavy
Light
Heavy
Light
Mass Spectrometer
Mass Spectrum of Ne
37
Example 3.7
How many hydrogen atoms
are present in 25.6 g of
urea [(NH2)2CO], which is
used as a fertilizer, in
animal feed, and in the
manufacture of polymers?
The molar mass of urea is
60.06 g.
urea
Example 3.7
Strategy
We are asked to solve for atoms of hydrogen in 25.6 g of urea.
We cannot convert directly from grams of urea to atoms of
hydrogen.
How should molar mass and Avogadro’s number be used in
this calculation?
How many moles of H are in 1 mole of urea?
Example 3.7
Solution
To calculate the number of H atoms, we first must convert
grams of urea to moles of urea using the molar mass of urea.
This part is similar to Example 3.2.
The molecular formula of urea shows there are four moles of H
atoms in one mole of urea molecule, so the mole ratio is 4:1.
Finally, knowing the number of moles of H atoms, we can
calculate the number of H atoms using Avogadro’s number.
We need two conversion factors: molar mass and Avogadro’s
number.
Example 3.7
We can combine these conversions
into one step:
= 1.03 × 1024 H atoms
Check Does the answer look reasonable?
How many atoms of H would 60.06 g of urea contain?
Example 3.11
Question of the Day
A sample of a compound contains 30.46 percent nitrogen and
69.54 percent oxygen by mass, as determined by a mass
spectrometer.
In a separate experiment, the molar mass of the compound is
found to be between 90 g and 95 g.
Determine the molecular formula and the accurate molar mass
of the compound.
Period 8 Day 1 12-3
Example 3.11
Strategy
To determine the molecular formula, we first need to determine
the empirical formula. Comparing the empirical molar mass to
the experimentally determined molar mass will reveal the
relationship between the empirical formula and molecular
formula.
Solution
We start by assuming that there are 100 g of the compound.
Then each percentage can be converted directly to grams; that
is, 30.46 g of N and 69.54 g of O.
Example 3.11
Let n represent the number of moles of each element so that
Thus, we arrive at the formula N2.174O4.346, which gives the
identity and the ratios of atoms present. However, chemical
formulas are written with whole numbers.
Try to convert to whole numbers by dividing the subscripts by
the smaller subscript (2.174). After rounding off, we obtain NO2
as the empirical formula.
Example 3.11
The molecular formula might be the same as the empirical
formula or some integral multiple of it (for example, two, three,
four, or more times the empirical formula).
Comparing the ratio of the molar mass to the molar mass of the
empirical formula will show the integral relationship between the
empirical and molecular formulas.
The molar mass of the empirical formula NO2 is
empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g
Example 3.11
Next, we determine the ratio between the molar mass and the
empirical molar mass
The molar mass is twice the empirical molar mass. This means
that there are two NO2 units in each molecule of the compound,
and the molecular formula is (NO2)2 or N2O4. The actual molar
mass of the compound is two times the empirical molar mass,
that is, 2(46.01 g) or 92.02 g, which is between 90 g and 95 g.
Example 3.11
Check Note that in determining the molecular formula from the
empirical formula, we need only know the approximate molar
mass of the compound. The reason is that the true molar mass
is an integral multiple (1×, 2×, 3×, . . .) of the empirical molar
mass. Therefore, the ratio (molar mass/empirical molar mass)
will always be close to an integer.
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
1Na
NaCl
22.99 amu
1Cl + 35.45 amu
NaCl
58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
48
Percent composition of an element in a compound =
n x molar mass of element
x 100%
molar mass of compound
n is the number of moles of the element in 1 mole
of the compound
2 x (12.01 g)
x 100% = 52.14%
46.07 g
6 x (1.008 g)
%H =
x 100% = 13.13%
46.07 g
1 x (16.00 g)
%O =
x 100% = 34.73%
46.07 g
%C =
C2H6O
52.14% + 13.13% + 34.73% = 100.0%
49
Example 3.8
Phosphoric acid (H3PO4) is a
colorless, syrupy liquid used in
detergents, fertilizers,
toothpastes, and in carbonated
beverages for a “tangy” flavor.
Calculate the percent
composition by mass of H, P,
and O in this compound.
Example 3.8
Strategy
Recall the procedure for calculating a percentage.
Assume that we have 1 mole of H3PO4.
The percent by mass of each element (H, P, and O) is given by
the combined molar mass of the atoms of the element in 1 mole
of H3PO4 divided by the molar mass of H3PO4, then multiplied
by 100 percent.
Example 3.8
Solution The molar mass of H3PO4 is 97.99 g. The percent by
mass of each of the elements in H3PO4 is calculated as follows:
Check Do the percentages add to 100 percent? The sum of
the percentages is (3.086% + 31.61% + 65.31%) = 100.01%.
The small discrepancy from 100 percent is due to the way we
rounded off.
Percent Composition and Empirical Formulas
53
Example
Pages 107-112: #s 3.5, 3.14, 3.16, 3.20,
3.26, 3.30, 3.40, 3.44, 3.48, 3.52, 3.96
The answers for all except 3.5 are in the
back of the book – CHECK them! This
is your review for the quest!!! YOU
MUST SHOW YOUR WORK!!!
1) Earth’s population is about 6.9
billion. If every person on Earth
participates in counting identical
particles at a rate of 2 particles per
second, how many years would it take to
count 6.0 X 1023 particles? Assume that
there are 365 days in every year.
day 5 12-1
2) What is the mass in grams of 1.00 X
1012 lead atoms?
day 5 12-1
3) Tin exists in Earth’s crust as SnO2.
Name this compound! Calculate the
percent composition by mass?
day 5 12-1
1) Earth’s population is about 6.9
billion. If every person on Earth
participates in counting identical
particles at a rate of 2 particles per
second, how many years would it take to
count 6.0 X 1023 particles? Assume that
there are 365 days in every year.
day 5 12-1
2) What is the mass in grams of 1.00 X
1012 lead atoms?
day 5 12-1
3) Tin exists in Earth’s crust as SnO2.
Name this compound! Calculate the
percent composition by mass?
day 5 12-1
1) Earth’s population is about 6.9
billion. If every person on Earth
participates in counting identical
particles at a rate of 2 particles per
second, how many years would it take to
count 6.0 X 1023 particles? Assume that
there are 365 days in every year.
day 5 12-1
2) What is the mass in grams of 1.00 X
1012 lead atoms?
day 5 12-1
3) Tin exists in Earth’s crust as SnO2.
Name this compound! Calculate the
percent composition by mass?
day 5 12-1
Converting:
_________
=
_________
Mass to moles
Moles to mass
Moles to atoms /
molecules / particles
_________
= _________
Atoms / molecules /
partilces to moles
Changing a
substance
=
_________
_________