Transcript Mole Relationships and Percent Composition

CH3 Mass Relationships
in Chemical Reactions
C(s) + O2(g)
1 atom C + 1 molecule O2
CO2(g)
1 molecule carbon dioxide
1 atom C = 12.0107 amu
Use ATOMIC MASS UNITS – “amu”
1 amu = 1.66 x 10-24g
Mass of 1 carbon atom = 1.99 x 10-23g
Mass of 1000. carbon atoms in amu
(1000. C atoms) 12.01 amu = 12,010 amu
1 C atom
Pile of carbon weighs 3.00 x 1020 amu
(3.00 x 1020 amu) 1 C atom = 2.50 x 1019 carbon atoms
12.01 amu
Calculating AMU and atoms
Calculating the mass of 1000
carbon atoms
(12,010 amu)
1.66 x 10-24g = 1.99 x 10-20g
1 amu
Too difficult to weigh!!
What is the mass in grams of 6.022 x 1023
atoms of carbon?
Scientists determined how many carbons
were in 12.01g carbon.
Found to be 6.022 x 1023 atoms
We now call this
Avogadro’s number!
The periodic tables gives use the mass of each
element in terms of amu/atom and g/mol!!

For ease, scientists renamed Avogadro’s
number of atoms or molecules a mole
Aluminum
1 mole
6.02 x 1023
atoms
26.98g
Sulfur
1 mole
6.02 x 1023
atoms
32.07g
Water
1 mole
6.02 x 1023
molecules
18.02g
The MOLE
You have 0.500g Na. How many moles of sodium
do you have?
0.500g Na
x
1 mol Na
22.99g Na
= 0.0217 mol Na
How many atoms of sodium are in 0.500g Na?
23
0.0217mol Na x 6.022 x 10 atoms
1 mol Na
= 1.31 x 1022 atoms Na

You have 5.00 x 1020 Cr atoms. How
many moles do you have? What is the
mass (in grams)?
5.00 x 1020 Cr atoms x (
1 mol Cr
)
6.022 x 1023 Cr atoms
= 8.30 x 10-4 mol Cr
8.30 x 10-4 mol Cr x (51.996 g Cr)
1 mol Cr
= 0.0432 g Cr

A molecule or chemical compound is a collection of
atoms
CH4 - contains 1 carbon and 4 hydrogen
5 CH4 – contains 5 carbon and 20 hydrogen
Therefore….
Avogadro’s
Number
of CH4
And…
Avogadro’s
Number
of C atoms
4X Avogadro’s
Number of
H atoms
1 mol CH4 = 1 mol C + 4 mol H
Molar Mass
1 mol C + 4 mol H = 1 mol CH4
12.01 g C
mol C
+ 4 x (1.008g H) = 16.04g CH4
mol H
mol CH4
Molar Mass – mass of 1 mol of any
compound (in grams)
Molar Mass continued

What is the molar mass of SO2?
1 mol S
1 mol SO2
2 mol O
1 mol SO2
x (32.06 g S) = 32.06 g S/mol SO2
mol S
x (15.999 g O) = 31.998 g O/mol SO2
mol O
64.062 g SO2
mol SO2
.

Determine the molar mass of C2H5Cl
2 mol C
x (12.01 g C) = 24.02 g C/mol C2H5Cl
1 mol C2H5Cl
mol C
5 mol H
x (1.008 g H) = 5.04 g H/mol C2H5Cl
1 mol C2H5Cl
mol H
1 mol Cl x (35.45 g Cl) = 35.45 g Cl/mol C2H5Cl
1 mol C2H5Cl
mol Cl
64.51 g C2H5Cl
mol C2H5Cl
.

Calcium Carbonate, CaCO3.
1.Calculate
the molar mass.
2.A certain sample contains 4.86 mol. What is
the mass in grams of this sample?
3.How many calcium carbonate molecules are
in this sample?
Calculating Number of Molecules



So far, we’ve learned to look at molecules
for the numbers/moles of constituent
atoms
It can be useful to know a compound’s
composition in terms of masses of
elements
We use mass percent to describe the mass
of an element in a mole of compound
Percent Composition
Mass Percent
Mass %
for a given
element
=
mass of the
element present
in 1 mol molecule
mass of 1 mol
molecules
x 100%
Calculating mass % of sulfur and oxygen
in SO2
1. Calculate the mass of 1 mole SO2
2. Calculate the mass of sulfur in 1 mole SO2
3. Calculate the mass of oxygen in 1 mole
SO2
4. Calculate the mass % of sulfur and oxygen
in SO2
1. Calculate the mass of 1 mole SO2
1 mol S
x (32.06 g S) = 32.06 g S/mol SO2
1 mol SO2
mol S
2 mol O
x (15.999 g O) = 31.998 g O/mol SO2
1 mol SO2
mol O
64.062
g SO2 .
mol SO2
2. Calculate the mass of sulfur in 1 mole SO2
1 mol S
1 mol SO2
x (32.06 g S) = 32.06 g S/mol SO2
mol S
3. Calculate the mass of oxygen in 1 mole
SO2
2 mol O x (15.999 g O) = 31.998 g O/mol SO2
1 mol SO2
mol O

4. Calculate the mass % of sulfur and
oxygen in SO2
32.06g S/mol SO2
x 100% =
64.062 g SO2/mol SO2
31.998 g O/mol SO2 x 100% =
64.062 g SO2/mol SO2
Calculating Mass %
50.04% S
49.96% O
Let’s say we have 0.2015g of an unknown sample.
We find that this sample contains 0.0806g carbon,
0.01353g H, and 0.1074g O.
First lets see how many moles of each element we
have.
0.0806g C x ( 1 mol C ) = 0.00671 mol C
12.01g C
0.01353g H x ( 1 mol H ) = 0.01342 mol H
1.0079g H
0.1074g O x ( 1 mol O ) = 0.00671 mol O
15.9994g O
Formulas of Compounds
Now let’s see how many atoms of each
element are in the sample.
0.00671 mol C x (6.022 x 1023 atoms C) = 4.04 x 1021 C atoms
1 mol C
0.01342 mol H x (6.022 x 1023 atoms H) = 8.08 x 1021 H atoms
1 mol H
0.00671 mol O x (6.022 x 1023 atoms O) = 4.04 x 1021 O atoms
1 mol O
What is the
relationship and
what does it tell
you??
Formulas of Compounds
The numbers of atoms of C, O, and H tell us
several things about our sample.
1.
2.
3.
There are the same number of carbon and
oxygen atoms.
There are twice as many hydrogen atoms
as carbon or oxygen atoms.
We can express the relative number of
atoms as CH2O
Formulas of Compounds
How do we know that CH2O is our compound in our
sample? Can our sample of molecules have C2H4O2 or
C6H12O6?
The answer is that we don’t know!
So, CH2O is known as the empirical formula. Empirical
formula expresses the smallest whole-number
ratio of the atoms present.
The molecular formula give the composition of the
molecules that are present in the actual formula.
C6H12O6 is the molecular formula for glucose while CH2O is
the molecular formula of formaldehyde.
Molecular vs. Empirical
1.
2.
3.
Calculate the number of moles of each
element.
Divide the number of moles by the
smallest number of moles.
Examine the ratios. Only whole numbers
are used in empirical formulas.
Calculating Empirical Formulas

An oxide of aluminum is formed by the
reaction of 4.151g of aluminum with 3.692g
of oxygen. Calculate the empirical formula
for this compound.
4.151g Al x ( 1 mol Al ) = 0.1539 mol Al
26.98g Al
3.692g O x ( 1 mol O ) = 0.2308 mol O
15.9994g O
0.1539 mol Al = 1.000 mol Al
0.1539
x 2 = 2 mol Al
0.2308 mol O = 1.500 mol O
0.1539
x 2 = 3 mol O
Al2O3
A white powder is analyzed and found to have an empirical
formula of P2O5. The compound has a molar mass of
283.88g. What is the compounds molecular formula?
2 mol P x 30.97g P = 61.94 g P/mol P2O5
mol P2O5
mol P
Molar Mass
P2O5 = 141.94g
5 mol O x 15.9994g O = 79.99 g O/mol P2O5
mol P2O5
mol O
141.94 x (n) = 283.88
n=2
Therefore, P4O10
is the molecular
formula
Calculating Molecular Formulas