Molecular Weight and Atomic Weight 22.990 Na Cl 35.452 Molecular Weight or Molar Mass 22.990 + 35.452 = 58.442
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Transcript Molecular Weight and Atomic Weight 22.990 Na Cl 35.452 Molecular Weight or Molar Mass 22.990 + 35.452 = 58.442
Molecular Weight and Atomic Weight
22.990
Na Cl
35.452
Molecular Weight or
Molar Mass
22.990 + 35.452 = 58.442
Take-Home Calculations
Molecular Weight = ∑ atomic weights
Mole = Weight / Molecular Weight
Number of atoms or molecules = Mole x
Avogadro’s number (6.022 x 1023)
Mass of one atom = grams of an
element / 6.022 X 1023
Molar Mass or Molecular Weight
Molar mass
mass in grams numerically equal to the
atomic weight of the element in grams.
Molar mass or Molecular weight of NaCl =
22.990 + 35.452 = 58.442 amu
amu = atomic mass unit
Molar Mass
How do we calculate the molar mass of
a compound?
• add atomic weights of each atom
The molar mass of propane, C3H8, is:
3 C 3 12.01 amu 36.03 amu
8 H 8 1.01 amu 8.08 amu
Molar mass
44.11 amu
Molar Mass
Calculate the molar mass of Ca(NO3)2?
1 Ca 1 40.08 amu 40.08 amu
2 N 2 14.01 amu 28.02 amu
6 O 6 16.00 amu 96.00 amu
Molar mass
164.10 amu
The Mole
A number of atoms, ions, or molecules
that is large enough to see and handle.
A mole = number of things
Avogadro’s number (NA) = 6.022 x 1023
• Just like a dozen = 12 things
• One mole = 6.022 x 1023 things
The number of moles
How many moles of Mg atoms are present in
73.4 g of Mg?
1 mol Mg atoms
? mol Mg 73.4 g Mg
24.30 g Mg
3.02 mol Mg
The Mole
One Mole of Cl2 Contains
• Cl2 or 70.90 g
• contains 6.022 x 1023 Cl2 molecules
2(6.022 x 1023 ) Cl atoms
One Mole of C3H8 Contains
• C3H8 or 44.11 g
• contains 6.022 x 1023 C3H8 molecules
3 (6.022 x 1023 ) C atoms
8 (6.022 x 1023 ) H atoms
The mass of a single atom
Calculate the mass of a single Mg atom,
in grams, to 3 significant figures.
1 mol Mg atoms
? g Mg 1 Mg atom
6.022 10 23 Mg atoms
24.30gMg
1 mol Mg atoms
23
4.04
10
g Mg
The number of atoms
How many atoms are contained in 1.67
moles of Mg?
6.022 1023 Mg atoms
? Mg atoms 1.67 mol Mg
1 mol Mg
1.00 10
24
Mg atoms
The number of molecules
Calculate the number of C3H8 molecules
in 74.6 g of propane.
? C3 H 8 molecules 74.6 g C3 H 8
1 mole C3 H 8
44.11 g C H
3
8
6.022 10 23 C3 H 8 molecules
44.11 g C3 H 8
1.02 10
24
molecules
The number of atoms
Calculate the number of O atoms in 26.5
g of Li2CO3.
? O atoms 26.5 g Li 2 CO3
6.022 10 form. units Li 2 CO3
1 mol Li 2 CO3
73.8 g Li 2 CO3
23
1 mol Li 2 CO3
6.49 10
3 O atoms
1 formula unit Li 2 CO3
23
O atoms
Percent Composition and Formulas of
Compounds
% composition = mass of an individual
element in a compound divided by the
total mass of the compound x 100%
Determine the percent composition of C
in C3H8.
mass C
%C
100%
mass C 3 H 8
3 12.01 g
44.11 g
81.68%
100%
Percent Composition and Formulas of
Compounds
What is the percent composition of H in C3H8?
mass H
%H
100%
mass C 3 H 8
8 H
100%
C3H 8
8 1.01 g
100% 18.32%
44.11 g
or
18.32% 100% 81.68%
Percent Composition and Formulas of
Compounds
Calculate the percent composition of
Fe2(SO4)3 to 3 significant figures
2 Fe 100% 255.8 g 100% 27.9% Fe
399.9 g
Fe (SO )
2
43
332.1 g
3S
% S
100%
100% 24.1% S
399.9 g
Fe (SO )
2
43
1216.0 g
% O 12 O 100%
100% 48.0% O
399.9 g
Fe (SO )
2
43
Total
100%
% Fe
Derivation of Formulas from Elemental
Composition
Empirical Formula - smallest whole-number ratio of atoms
present in a compound
• CH2 is the empirical formula for alkenes
(CH2CH2CH2CH2CH2)
Molecular Formula - actual numbers of atoms of each element
present in a molecule of the compound
• Ethene – CH2CH2 is C2H4
• Pentene – CH2CH2CH2CH2CH2 is C5H10
• percent composition is determined experimentally
Derivation of Formulas from Elemental
Composition
A compound contains 24.74% K, 34.76% Mn,
and 40.50% O by mass. What is its empirical
formula?
Make the simplifying assumption that we have
100.0 g of compound.
In 100.0 g of compound there are:
•
•
•
24.74 g of K
34.76 g of Mn
40.50 g of O
Derivation of Formulas from Elemental
Composition
1 mol K
24.74 g K
? mol K
0.6327 mol K
39.10 g K
? mol Mn 34.76 g Mn
1 mol Mn
0.6327 mol Mn
54.94 g Mn
? mol O
40.50 g O
1mol O
2.531 mol O
16.00 g O
obtain smallest whole number ratio
for K
0.6327
1K
for Mn
0.6327
for O
0.6327
0.6327
2.531
4O
0.6327
thus the chemical formula is KMnO 4
1 Mn
Derivation of Formulas from Elemental
Composition
A sample of a compound contains 6.541g
of Co and 2.368g of O. What is the
empirical formula for this compound?
? mol Co 6.541 g Co
1 mol Co
0.1110 mol Co
58.93 gCo
? mol O 2.368 g O
1mol O
0.1480 mol O
16.00 g O
find smallest whole number ratio
Derivation of Formulas from Elemental
Composition
A sample of a compound contains 6.541g
of Co and 2.368g of O. What is the
empirical formula for this compound?
for Co
0.1110
1 Co
for O
0.1110
0.1480
1.333O
0.1110
multipy both by 3 to turn fraction to whole number
1 Co 3 3 Co 1.333 O 3 4 O
Thus the compound' s formula is :
Co3O 4
Purity of Samples
The percent purity of a sample of a
substance is always represented as
% purity =
mass of pure substance
100%
mass of sample
mass of sample includes impurities
Purity of Samples
A bottle of sodium phosphate, Na3PO4, is
98.3% pure Na3PO4. What are the masses of
Na3PO4 and impurities in 250.0 g of this
sample of Na3PO4?
unit factor
98.3 g Na 3 PO 4
100.0 g sample
? g Na 3 PO 4 250.0 g sample
98.3 g Na 3 PO 4
100.0 g sample
= 246 g Na 3 PO 4
? g impurities = 250.0 g sample - 246 g Na 3 PO 4
= 4.00 g impurities