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Los ensayos triaxiales de corte se realizan en especímenes cilíndricos y sólidos de suelo. La
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SHS STEM General Chemistry 1: Atoms, Moles, Equations, StoichiometryPaula Marie Llido
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By mahasiswa yang masih belajar untuk menggapai cita cita dan kesuksesan menuju masa depan yang gemilang
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2. 2
Atomic Masses
• Balanced equation tells us the relative numbers
of molecules of reactants and products
C + O2 → CO2
1 atom of C reacts with 1 molecule of O2
to make 1 molecule of CO2
• If I want to know how many O2 molecules I will
need or how many CO2 molecules I can make, I
will need to know how many C atoms are in the
sample of carbon I am starting with
3. 3
Atomic Masses
• Dalton used the percentages of elements in
compounds and the chemical formulas to deduce
the relative masses of atoms
• Unit is the amu.
– atomic mass unit
– 1 amu = 1.66 x 10-24
g
• We define the masses of atoms in terms of atomic
mass units
– 1 Carbon atom = 12.01 amu,
– 1 Oxygen atom = 16.00 amu
– 1 O2 molecule = 2(16.00 amu) = 32.00 amu
4. 4
Atomic Masses
• Atomic masses allow us to convert weights
into numbers of atoms
If our sample of carbon weighs 3.00 x 1020
amu we will
have 2.50 x 1019
atoms of carbon
atomsC10x2.50
amu12.01
atomC1
xamu10x3.00 1920
=
Since our equation tells us that 1 C atom reacts
with 1 O2 molecule, if I have 2.50 x 1019
C
atoms, I will need 2.50 x 1019
molecules of O2
5. 5
Example #1
• Determine the mass of 1 Al atom
1 atom of Al = 26.98 amu
• Use the relationship as a conversion factor
amu2024
atomAl1
amu26.98
xatomsAl75 =
Calculate the Mass (in amu) of 75 atoms of Al
6. 6
Chemical Packages - Moles
• We use a package for atoms and molecules
called a mole
• A mole is the number of particles equal to the
number of Carbon atoms in 12 g of C-12
• One mole = 6.022 x 1023
units
• The number of particles in 1 mole is called
Avogadro’s Number
• 1 mole of C atoms weighs 12.01 g and has 6.02 x 1023
atoms
7. 7
Figure 8.1: All these samples of pure
elements contain the same number (a
mole) of atoms: 6.022 x 1023
atoms.
8. 8
Figure 8.2: One-mole samples of iron
(nails), iodine crystals, liquid mercury, and
powdered sulfur.
9. 9
Example #2
Use the Periodic Table to determine the mass
of 1 mole of Al
1 mole Al = 26.98 g
Use this as a conversion factor for grams-to-
moles
Almol0.371
g26.98
Almol1
xAlg10.0 =
Compute the number of moles
and number of atoms in 10.0 g of Al
10. 10
Use Avogadro’s Number to determine the
number of atoms in 1 mole
1 mole Al = 6.02 x 1023
atoms
Use this as a conversion factor for moles-to-
atoms
atomsAl10x2.23
Almol1
atoms10x6.02
xAlmol0.371 23
23
=
Example #2
Compute the number of moles
and number of atoms in 10.0 g of Al
11. 11
Use Avogadro’s Number to determine the
number of atoms in 1 mole
1 mole Al = 6.02 x 1023
atoms
Use this as a conversion factor for atoms-to-
moles
Almol0.370
atoms10x6.02
Almol1
xatomsAl10x2.23 23
23
=
Compute the number of moles
and mass of 2.23 x 1023
atoms of Al
Example #3
12. 12
Use the Periodic Table to determine the mass
of 1 mole of Al
1 mole Al = 26.98 g
Use this as a conversion factor for moles-to-
grams
Alg9.99
Almol1
g26.98
xAlmol0.370 =
Compute the number of moles
and mass of 2.23 x 1023
atoms of Al
Example #3
13. 13
Molar Mass
• The molar mass is the mass in grams of one
mole of a compound
• The relative weights of molecules can be
calculated from atomic masses
water = H2O = 2(1.008 amu) + 16.00 amu
= 18.02 amu
• 1 mole of H2O will weigh 18.02 g, therefore the
molar mass of H2O is 18.02 g
• 1 mole of H2O will contain 16.00 g of oxygen and
2.02 g of hydrogen
14. 14
Percent Composition
• Percentage of each element in a compound
– By mass
• Can be determined from
the formula of the compound or
the experimental mass analysis of the compound
• The percentages may not always total to 100%
due to rounding
100%
whole
part
Percentage ×=
15. 15
Determine the mass of each element in 1
mole of the compound
2 moles C = 2(12.01 g) = 24.02 g
6 moles H = 6(1.008 g) = 6.048 g
1 mol O = 1(16.00 g) = 16.00 g
Determine the molar mass of the compound
by adding the masses of the elements
1 mole C2H5OH = 46.07 g
Determine the Percent Composition
from the Formula C2H5OH
Example #4
16. 16
Divide the mass of each element by the molar
mass of the compound and multiply by 100%
52.14%C100%
46.07g
24.02g
=×
13.13%H100%
46.07g
6.048g
=×
34.73%O100%
46.07g
16.00g
=×
Determine the Percent Composition
from the Formula C2H5OH
Example #4
17. 17
Empirical Formulas
• The simplest, whole-number ratio of atoms in a
molecule is called the Empirical Formula
– can be determined from percent composition or
combining masses
• The Molecular Formula is a multiple of the
Empirical Formula
% A mass A (g) moles A
100g MMA
% B mass B (g) moles B
100g MMB
moles A
moles B
18. 18
Find the greatest common factor (GCF) of the
subscripts
factors of 20 = (10 x 2), (5 x 4)
factors of 12 = (6 x 2), (4 x 3)
GCF = 4
Divide each subscript by the GCF to get the empirical
formula
C20H12 = (C5H3)4
Empirical Formula = C5H3
Determine the Empirical Formula of
Benzopyrene, C20H12
Example #5
19. 19
Convert the percentages to grams by
assuming you have 100 g of the compound
– Step can be skipped if given masses
47gC
100g
47gC
100g =×
47gO
100g
47gO
100g =× 6.0gH
100g
6.0gH
100g =×
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Example #6
20. 20
Convert the grams to moles
Cmol3.9
12.01g
Cmol1
C47g =×
Omol2.9
16.00g
Omol1
Og47 =×
Hmol6.0
1.008g
Hmol1
Hg6.0 =×
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Example #6
21. 21
Divide each by the smallest number of moles
1.32.9Cmol3.9 =÷
12.9Omol2.9 =÷
22.9Hmol6.0 =÷
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Example #6
22. 22
If any of the ratios is not a whole number, multiply all
the ratios by a factor to make it a whole number
– If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then
multiply by 3; if ?.25 or ?.75 then multiply by 4
43x1.32.9Cmol3.9 ==÷
33x12.9Omol2.9 ==÷
63x22.9Hmol6.0 ==÷
Multiply all the
Ratios by 3
Because C is 1.3
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Example #6
23. 23
° Use the ratios as the subscripts in the
empirical formula
43x1.32.9Cmol3.9 ==÷
33x12.9Omol2.9 ==÷
63x22.9Hmol6.0 ==÷
C4H6O3
Determine the Empirical Formula of
Acetic Anhydride if its Percent Composition is
47% Carbon, 47% Oxygen and 6.0% Hydrogen
Example #6
24. 24
Molecular Formulas
• The molecular formula is a multiple of the
empirical formula
• To determine the molecular formula you need to
know the empirical formula and the molar mass
of the compound
25. 25
Determine the empirical formula
• May need to calculate it as previous
C5H3
Determine the molar mass of the empirical
formula
5 C = 60.05 g, 3 H = 3.024 g
C5H3 = 63.07 g
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
Example #7
26. 26
Divide the given molar mass of the
compound by the molar mass of the
empirical formula
– Round to the nearest whole number
4
07.63
252
=
g
g
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
Example #7
27. 27
Multiply the empirical formula by the calculated
factor to give the molecular formula
(C5H3)4 = C20H12
Determine the Molecular Formula of Benzopyrene
if it has a molar mass of 252 g and an
empirical formula of C5H3
Example #7
