Average Atomic Mass
Since atoms are so small and the mass of individual atoms is also very small, it is not useful to use the units of grams or kilogram.
A new unit called the atomic mass unit (amu) was developed to deal with the very small units of mass for particles like the atom.
1 amu = 1.66 x 10-24 g
It has been found to be useful that instead of using absolute masses, it is best compare the relative masses of atoms using a reference isotope as a standard.
The carbon – 12 isotope has been set as this standard.
The carbon – 12 isotope has been assigned a mass of exactly 12 amu.
For example, the oxygen atom has a relative atomic mass of 15.999 amu.
The Mole
Another way to measure the amount of a
substance is to count the number of particles in that substance.
When talking about the number atoms, the number of individual atoms in the sample is ridiculously large.
Counting the individual atoms is NOT PRACTICAL.
Just as a dozen represents 12, 1 MOLE represents 6.02 x 1023 particles of a substance.
23
23
1002.6
1
1
1002.6
x
moleor
mole
x
This number (Avogadro’s Number) is named in honor of the Italian scientist Amedo Avogadro di Quarenga from whose work the concept was based.
How can we use Avogadro’s Number in a calculation?
Example #1 How many moles of magnesium make up 1.25 x 1023 atoms of magnesium? Answer: In order to solve this problem, we must realize that the conversion needed is
23
23
1002.6
1
1
1002.6
x
moleor
mole
x
Which form do we use???
Remember the techniques that we used when we did conversions factors
“What you want goes on top.”
1.25 x 1023 atoms Mg Mgatomsx
Mgmole231002.6
1
= .208 moles Mg
Example #2 How many atoms of sulfur are present in 0.75 moles of sulfur? Answer: In order to solve this problem, we must again realize that the conversion needed is
23
23
1002.6
1
1
1002.6
x
moleor
mole
x
Which form do we use???
0.75 moles S Smole
Satomsx
1
1002.6 23
= 4.5 x 1023 atoms S
Molar Mass
One atom of aluminum has a mass of 4.48 x 10-23 g.
This mass is determined by adding up the masses of the protons, neutrons and electrons within the aluminum atom.
What would be the mass of 1 mole of aluminum atoms (or 6.02 x 1023 atoms of aluminum)?
6.02 x 1023 atoms 4.48 x 10-23 g
1 atoms=
= 26.98 g 26.98 g is the mass of 6.02 x 1023 atoms
of aluminum OR
26.98 g is the mass of 1 mole of aluminum
THEREFORE,
26.98 g is the mass of 1 mole of Aluminum
More commonly stated, 26.98 g/mole is the MOLAR MASS of aluminum.
The Atomic Weights of the elements on
the Periodic Table also represent the Molar Masses of every element. Let’s find the molar masses of some atoms: Molar Mass O = ? Molar Mass W = ? Molar Mass Mg = ? How do we use Molar Mass?
Example #3 How many grams of iron are present in 1.78 moles of iron? Answer: In order to solve this problem, we must realize that the conversion needed is the molar mass of iron:
g
moleor
mole
g
85.55
1
1
85.55
Which form do we use???
Femole
FegFemoles
1
85.5578.1
= 99.4 g Fe Example #4 How many moles of calcium are present in 5.78 g of calcium?
Answer: In order to solve this problem, we must realize that the conversation needed is the molar mass of calcium:
g
moleor
mole
g
08.40
1
1
08.40
Which form do we use???
Cag
CamoleCag
08.40
178.5
= 0.0144 moles Ca
How can we determine the molar mass of a compound? 1) You must know the chemical formula of the compound. 2) Add the molar masses of the individual atoms in the molecule. Example: SO3 The molar mass of sulfur is
32.06 g/mole. The molar mass of oxygen is
16.00 g/mole. Therefore, molar mass of SO3 =
+ 3(16.00 g/mole)1(32.06 g/mole)
80.06 g/mole) Try this: What is the molar mass of BaBr2?
Mole – Mass Conversions The molar mass of an element or
compound can be used to the convert mass of a substance into moles. Example #1: How many moles are in 6.59 g of Nickel? 1. Determine the molar mass of Ni.
mole
g69.58
2. Calculate the moles of Ni.
Nig
NimolesNig
69.5859.6
= .112 moles Ni
Example #2: How many moles are in 92.2 g of FeO? 1. Determine the molar mass of FeO.
+ 16.00 g/mole)55.85 g/mole
71.85 g/mole
2. Calculate the moles of FeO.
FeOg
FeOmolesFeg
85.712.92
= 1.28 moles FeO
The molar mass of an element or compound is used to convert moles of a substance into mass.
Example #3: How many grams are in 3.84 moles of NO2? 1. Determine the molar mass of NO2.
+ 2(16.00) g/mole)14.01 g/mole
46.01g/mole 2. Calculate the moles of NO2.
2
22
01.4684.3
NOmoles
NOgNOmoles
= 177 g NO2
Percent Composition
All substances are composed of elements.
Those substances that contain several elements, do so with specific amounts of each component element.
For example, consider K2CrO4 and K2Cr2O7:
The relative amounts are element expressed can be expressed as the Percent Composition.
The percent composition will be determined as a function of mass of each element in a compound.
If a compound, AB, is made of
elements A & B, then the
100% xmass
massA
AB
A
Example #1: A compound is composed of 45.98 g of sodium and 70.90 g of chlorine. What is the percent composition? Answer: 45.98 g + 70.90 g 116.88 g
%34.3910088.116
98.45% x
g
gNa
%66.6010088.116
90.70% x
g
gCl
OR % Cl = 100% - 39.34% = 60.66%)
Example #2: An 8.20 g piece of magnesium combines with 5.40 g of oxygen to form a compound. What is the percent composition of the compound? 8.20 g + 5.40 g 13.60 g
%3.6010060.13
20.8% x
g
gMg
%7.3910060.13
40.5% x
g
gO
OR (% O = 100% - 60.3% = 39.7%) If information is not given about
the individual amounts for each element, the percent composition can be determined from the chemical formula.
Example #1: What is the percent composition of C3H8?
Answer: Step 1: Determine the molar mass of C3H8.
+ 3(12.01) g/mole)8(1.01) g/mole
44.11g/mole
Step 2: Determine the molar masses for each type of element. For C: 3(12.01 g/mole) = 36.03 g/mole For H: 8(1.01 g/mole) = 8.08 g/mole Step 3: Determine the percent composition for each element.
%68.8110011.44
03.36% x
g
gC
% H = 100% - 81.68% = 18.32%
Example #2: What is the percent composition of HCN? Answer: Step 1: Determine the molar mass of HCN.
+ 12.01 g/mole1.01 g/mole
27.03g/mole14.01 g/mole
Step 2: Determine the percent compositions for each element.
%43.4410003.27
01.12% x
g
gC
%83.5110003.27
01.14% x
g
gN
% H = 100% - 51.83% - 44.43%=3.74%
Example: What is the percent water in the hydrate cobalt (II) chloride dehydrate (CoC12 ● 2H2O)? Answer: Step 1: Determine the molar mass of CoC12 ● 2H2O.
1(58.93 g/mole) 2(35.45 g/mole) 4(1.01 g/mole) 2(16.00 g/mole) 165.87 g/mole
+
Step 2: Determine the percent water.
100/87.165
/)00.16(2)01.1(4% 2 x
moleg
molegOH
= 21.72 % water
EMPIRICAL FORMULA
The simplest ratio of the atoms in a molecule.
Sometimes the molecular formula is the same as the empirical formula.
Example:
NaCl → NaCl
Molecular Empirical Formula Formula
Sometimes they are not:
C3H9 → CH3
Molecular Empirical Formula Formula
Example #1: A compound has a composition 90.10 g P 8.90 g H What is the empirical formula? Answer: 1) Convert the mass to moles.
PmolePg
PmolePg 909.2
97.30
110.90
HmoleHg
HmoleHg 81.8
01.1
190.8
2) Select the atoms with the least number of moles. In this case it is the 2.909 mole P
3) Divide each of the number of moles calculated in step #1 by the # of moles determined in step #2.
0.1
909.2
909.2:PFor
00.302.3
909.2
81.8:HFor
Giving the ratio of P : H = 1 : 3
Empirical Formula PH3
Example #2: A compound has a % composition 65.2% Sc 34.8% O What is the empirical formula? 1) Convert the percent to mass.
65.2% Sc → 65.2 g Sc
+ 34.8% O → + 34.8 g O
100.0% 100.0 g 2) Convert the mass to moles.
ScmoleScg
ScmoleScg 45.1
96.44
12.65
OmoleOg
OmoleOg 18.2
00.16
18.34
3) Select the atoms with the least number of moles. In this case it is the 1.45 mole Sc
4) Divide each of the number of moles calculated in step #2 by the # of moles determined in step #3.
0.145.1
45.1:ScFor
50.145.1
18.2:OFor
Giving the ratio of Sc : O = 1 : 1.5
Empirical Formula Sc1O1.5 Is this formula possible? Can you have 1.5 atoms???
NO! Solution: Multiply the subscripts in order to get whole numbers.
→Multiply by two.
Empirical Formula Sc2O3
Example #3: A compound is composed of 0.59 g H and 9.40 g O. It has been determined that the molar mass of the compound is 34.0 g/mole. 1. What is the empirical formula? 2. What is the molecular formula? Answer: 1. Empirical Formula
OmoleOg
OmoleOg 590.0
00.16
140.9
HmoleHg
HmoleHg 58.0
01.1
159.0
0.158.0
58.0:HFor
0.158.0
59.0:OFor
Giving the ratio of H : O = 1 : 1
Empirical Formula HO 2. Molecular Formula The molar mass of the empirical formula: H O (1.01 g/mole) + (16.00 g/mole) = 17.01 g/mole for HO If the empirical molar mass is doubled, then it will equal the molar mass of the molecular compound. Therefore, the molecular formula must be double the amount of atoms. Giving: H2O2
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