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%  255.8 g 100%  27.9% Fe
399.9 g
Fe (SO )
2
43
332.1 g
3S
% S 
100% 
100%  24.1% S
399.9 g
Fe (SO )
2
43
1216.0 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