: also called molecular weights, formula weights, or atomic masses
(however normally the units for atomic masses are in Daltons (Da) or amu)
Calculating molecular weights of compounds and formula weights of ionic compounds:
Using the number of atoms present in the compound and the molecular weight of the
individual atoms or ions, we can calculate an overall molecular weight for the entire compound.
Molecular weight of the compound = sum of the atomic mass of the elements present in the
compound.
Example: Molecular weight of C2H6
2C atoms = 2 * 12.01g/mole = 24.02g/mole
6H atoms = 6* 1.008g/mole = 6.048g/mole
Then the individual atomic masses are summed 24.02+6.048 = 30.07g/mole
Example: Molecular weight of Al(NO3)3
1 Al = 1* 26.98g/mole = 26.98g/mole
3N = 3* 14.01g/mole = 42.03g/mole
9O (remember 3 sets of NO3 means 9 oxygens total!!) =
9*15.99g/mole = 143.91g/mole
Then take the sum: 26.98+42.03+143.91 = 212.92g/mole
Molar Mass: the mass of a substance per 1 mole of its entities (atoms, molecules, ions, or
formula units). Units for molar mass are grams/mole. Once again, the periodic table is used to
calculate the molar mass (previously we called this molecular mass – it’s the same thing!!)
1.) to find the molar mass of an element, simply look at the atomic mass of an element on
the periodic table.
a.) monatomic elements: Cu, Fe, Mg etc . . . Their molar mass value is the same as
their amu value – thus the molar mass of Cu is 63.55 grams/mole ; for Fe it’s 55.85
grams/mole ; and for Mg it’s 24.31 grams/mole.
b.) diatomic elements: X2: H2, N2, O2, F2, Cl2, Br2, I2: have to take into account that
there are 2 things present! So the molar mass of O 2 is 2xmolar mass of 1
oxygen:
O2 molar mass
2 x 15.9994 = 31.9988 grams/mole
c.) the same applies to sulfur, whose major form exists as S8
S8 molar mass
8 x 32.066 = 256.528 grams/mole
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2.) to find the molar mass/molecular mass of a compound, we perform the same
calculation . . .
a.) for ionic compounds we call this the formula weight: Cr2(SO4)3
this species results from Cr+3 + SO4-2
it would be named: chromium (III) sulfate
The formula weight would be determined by taking the sum of
all the atoms present in the compound:
Cr2(SO4)3
Cr = 2 x 51.9961 = 103.9922
S = 3 x 32.066 = 96.198
O = 12 x 15.9994 = 191.9928
Adding the masses of the atoms together we get:
103.9922 + 96.198 + 191.9928 = 392.183 grams/mole
The KEY point is that the subscript in the formula refers to the number of
atoms/ions present in the compound – and the parentheses indicate
multiplication – remember (SO4)3 indicates 3 SO4-2 groups: so we have SO4-2 +
SO4-2 + SO4-2 adding up the atoms present we see the 3 sulfurs and the 4+4+4 = 12
oxygens. The subscripts also indicate the number of moles of atoms or ions
present.
The mole is useful as it allows one to calculate the mass (g, kg, µg) of a substance in a
particular sample if you know the moles of the substance (we’ll use this when we talk
about solutions!!!). If you know the mass of a substance you can inversely calculate the
number of moles:
moles
grams
OR grams
moles USE molecular weight!
Remember the units for molecular weight are grams/mole
So it can be used as the conversion factor between the two
species!!
Given: 2.50 moles of Fe want to know ? grams
2.50 moles Fe x
55.847 grams Fe
= 1.40 x 102 grams Fe
1 mole Fe
2
Given: 45.789 grams of Fe want to know ? moles
45.789 grams of Fe x
1 mole Fe
= 8.1990 x 10-1 moles
55.847 grams Fe
Given: 25.12 moles of Al2(SO4)3 want to know ? grams
Step 1: calculate molecular weight of Al2(SO4)3
Step 2: convert moles to grams
following steps above:
Al2(SO4)3
Al: 2 x 26.9815 = 53.9630
S: 3 x 32.066 = 96.198
O: 12 x 15.9994= 191.9928
Sum = 53.9630+96.198+191.9928 = 342.154 grams/mole
25.12 moles Al2(SO4)3 x
342.154 grams Al 2 (SO 4 ) 3
= 8.595 x 103 grams Al2(SO4)3
1 mole Al 2 (SO 4 ) 3
Given: 500.23 grams of Al2(SO4)3 want to know ? moles
500.23 grams Al2(SO4)3 x
1 mole Al 2 (SO 4 ) 3
= 1.4620 grams Al2(SO4)3
342.154 grams Al 2 (SO 4 ) 3
But – what exactly IS this mole thing??
: the central concept of chemistry.
Typically, we perform a chemical reaction, and we take grams of substance A and react it with
grams of substance B. But if we examine what is really going on, we are talking about this
chemical reaction as each atom or molecule reacts with other atoms or molecules. HOW then
can we talk about quantities of things that are just soooooo small. We are not asked to go into
the balance room and weigh out 5 molecules of A . . . but we can, using this concept of the mole,
determine how many molecules/atoms are involved in chemical reactions.
The mole (n): a unit derived by chemists to count up the chemical entities involved in chemical
reactions which can be determined by weighing the substance.
-defined as the amount of a substance that contains the same number of entities as there
are atoms in exactly 12 grams of 12C.
- the mole specifies the number of objects in a fixed mass, therefore 1 mole of a substance
represents a fixed number of chemical entities
3
1 mole of substance = 6.022 x 1023 entities
Think about the size of this number. 109 is a billion. 1018 is a billion billion. 1023 is one hundred
thousand billion billion. If you had 6.022 x 1023 dollars you could spend a billion dollars a
second for your entire lifetime and still have used less than 0.001% of your money. If C atoms
were the size of peas, 6.022 x 1023 of them would cover the entire surface of our planet to a depth
of over 100m!!
those entities could be atoms, molecules, formula units, ions
1 mole 12C = 6.022 x 1023 carbon atoms
1 mole H2O = 6.022 x 1023 H2O molecules
1 mole NaCl = 6.022 x 1023 formula units
1 mole of electrons = 6.022 x 1023 electrons
1 mole of Na+1 ions = 6.022 x 1023 Na+1 ions
1 mole of electrons = 6.022 x 1023 electrons
There are several relationships to be familiar with: the first is:
1Da = 1u = 1 amu = 1.6606 x 10-24 grams
example: Calculate the number of grams in one 1H atom
1.6606 x10 24 g
1H = 1.0078 amu x
 1.67x 10-24 grams
1amu
Calculating the grams of a molecule is treated the same way – except – you must sum the amu
values for each element present (just like we will do for calculating molecular masses)
Example: calculate the number of grams in one molecule of H2O
1H = 1.0078 amu x 2H atoms = 2.0156 amu
16O = 15.9994 amu x 1O atom = 15.9994 amu
total amu = 15.9994 + 2.0156 = 18.0150 amu
1.6606 x10 24 g
 2.9916 x 10-23 grams
H2O = 18.0150 amu x
1amu
The central relationship between the mass of 1 atom and the mass of 1 mole of those atoms
comes from the periodic table. The atomic mass is expressed in amu (u, or Da)– which is
numerically equivalent to the mass of one mole of atoms of the element expressed in g/mole.
- Thus, 1 Na atom has a mass of 22.99 amu = 1 mole of Na atoms with a mass of 22.99 g
- 1 Si atom has a mass of 28.09 amu = 1 mole of Si atoms with a mass of 28.09 g
- Therefore, our units can be g/mole. This means that we can say that 1.0078 amu =
1.0078 grams/mole.
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“the monkey wrench”
Another relationship exists that you should be aware of. It is DIFFERENT than the relationship
between grams and moles and the two should NOT be confused. Avogadro’s Number (NA) is
used to calculate the number of entities – e.g. atoms, molecules, formula units, ions, etc in 1
mole of substance. NOTICE it is not used for grams!! Grams to moles and moles to grams
only uses the molecular weight!!
Avogadro’s Number is used to convert:
Moles to atoms
Moles to ions
Moles to molecules
Moles to formula units
Avogadro’s Number = 6.022 x 1023
entities X
moles X
Given: 4 moles of carbon want to know ? atoms
4 moles Carbon x
6.022x10 23 atoms of carbon
= 2.409 x 1024 atoms of carbon
1 mole of carbon
Given: 16 grams of carbon want to know ? atoms
Step 1: convert grams to moles (using MW)
Step 2: convert moles to atoms (using NA)
16 grams x
1 mole C
= 1.33 moles of Carbon
12.00 grams C
1.33 moles C x
6.022x10 23 atoms of carbon
= 8.00 x 1023 atoms of carbon
1 mole of carbon
Given: ethylene which consists of C2H4 molecules. How many atoms of C are
in 2.50 moles of C2H4 and how many atoms of H are in 2.50 moles of C2H4?
Note: 1 mole of C2H4 contains 2 moles of C and 4 moles of H
2.50 moles C2H4 x
2 mole C
6.022x10 23 atoms of carbon
x
= 3.01 x 1024 C atoms
1 mole C 2 H 4
1 mole of carbon
2.50 moles C2H4 x
4 moles H
6.022x10 23 atoms of H
x
= 6.02 x 1024 H atoms
1 mole C 2 H 4
1 mole of Hydrogen
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Given: How many Ca ions are there in 4.56 grams of calcium chloride? How
many chloride ions are there in 4.56 grams of calcium chloride?
Step 1: write molecular formula for calcium chloride
Step 2: calculate FW (formula weight) for calcium chloride
Step 3: convert grams to moles using answer from Step 2
Step 4: convert moles to atoms using NA
Calcium chloride = Ca+2 + Cl-1 = CaCl2
CaCl2 = Ca: 1 x 40.078 = 40.078
Cl: 2 x 35.453 = 70.906
FW = 40.078 + 70.906 = 110.984 grams/mole
4.56 grams CaCl2 x
1 mole CaCl 2
1 mole Ca
6.022x10 23 Ca 2 ions
x
x
= 2.47 x 1022 Ca+2 ions
1 mole of Ca
110.984 grams CaCl 2 1 mole CaCl 2
4.56 grams CaCl2 x
1 mole CaCl 2
2 mole Cl
6.022x10 23 Cl -1 ions
x
x
= 4.95 x 1022 Cl-1 ions
1
mole
CaCl
1
mole
of
Cl
110.984 grams CaCl 2
2
Given: 4.56 g of calcium chloride ? formula units do you have
6.022x10 23 formula units of CaCl 2
1 mole CaCl 2
4.56 grams CaCl2 x
x
= 2.47 x1022 formula
1 mole of CaCl 2
110.984 grams CaCl 2
units of CaCl2
Each element in a compound makes up its own percentage of that compound. A certain % of
the mass of CaCl2 is attributed to the calcium and a certain % of the mass of CaCl2 is attributed
to the chlorine (chloride). As with any % - the sum of the parts should add up to 100!
The mass % of a particular element in the compound can be calculated by the following
equation:
moles X in formula x molar mass of X (grams
mole)
Mass % of element X =
mass (grams) of 1 mole of compound
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Given: A compound with the molecular formula C3H4O is known as acrolein. ? mass
percent of each element in acrolein. ? grams of oxygen in 144.5 mg of acrolein
There are 3 moles of C for every 4 moles of H for every 1 mole of O in acrolein.
Knowing the moles of each element, calculate grams
3 moles C x
12.000 g C
= 36.000 grams C in 1 mole
1 moles C
4 moles H x
1.0078 g H
= 4.0312 grams H in 1 mole
1 moles H
1 mole O x
15.999 g O
= 15.999 grams O in 1 mole
1 moles O
Total mass of acrolein = 36.000+4.0312+15.999 = 56.030 grams/1mole
Mass % of carbon =
36.000 g C
x100 = 64.251% C
56.030g acrolein
Mass % of hydrogen =
Mass % of oxygen =
4.0312 g H
x 100 = 7.195% H
56.030g acrolein
15.999 g O
x 100 = 28.554% O
56.030g acrolein
Math check: 64.251%+7.195%+28.554% = 100.000%
Sample of acrolein 144.5 mg ? grams of O:
1 grams
= 0.1445 grams
1000 mg
Acrolein is composed of 28.554% O, so 28.554% of 0.1445 grams is oxygen
0.1445 grams x 0.28554 = 0.04126 or 4.126 x 10-2 grams of oxygen in 144.5 mg of
acrolein
Convert 144.5mg to grams: 144.5 mg x
7
When given the chemical formula for a compound, you can see how easy it is to determine
mass percent of each element present in the compound. What if only the mass percents were
given? Could you use that information in reverse to predict the molecular formula?
We are going to determine the subscripts (the number of moles) of each atom present in the
compound. Chemists will take a sample compound, break it down into its components,
calculate the number of grams, convert that to moles, and then convert the number of moles into
a whole number ratio. This is a valid representation of the formula, however, this is the
empirical formula.
The “problem” with the empirical formula is that it is the smallest whole number ratio of the
elements present in the compound, and therefore it may not be an accurate representation of
the compound itself. (C2H2 is NOT C6H6 even though they have the same empirical formula!!)
But nonetheless, they are useful and can be used to help determine the molecular formula.
Calculating the Empirical Formula:
1. If given grams, convert grams to moles
2. Divide through by the lowest number of moles
3. These numbers are the subscripts, but remember – they must be whole numbers (no
partial atoms exist!!) you therefore MAY have to multiply through by the smallest integer
that will turn all numbers into whole numbers. What you do to one number – you must
must must must must do to all the others.
For example: if you calculated that you have 2 moles of A, 2.5 moles of B, and 1 mole of C,
you will notice that 2.5 is not a whole number. We need to make it so, therefore we
multiply by the SMALLEST number that will make it a whole number. Multiplying by 1
gets us nowhere. What about 2? Yes, that works. BUT we must multiply everything by 2!
Therefore we will have 4 moles of A, 5 moles of B, and 2 moles of C.
1.
2.
3.
4.
If given % of atoms, assume 100% and just turn the % into grams
Covert the grams to moles
Divide through by the lowest number of moles
These numbers are the subscripts, but remember – they must be whole numbers (no
partial atoms exist!!) you therefore MAY have to multiply through by the smallest integer
that will turn all numbers into whole numbers. What you do to one number – you must
must must must must do to all the others.
How do empirical formulas tie into molecular formulas?? They each contain the same atoms
(e.g. if we have an empirical formula of CH, the molecular formula will also have CH in it – and
nothing else!!) but the difference is, molecular formulas are some whole number multiple of the
empirical formula: CH could mean C2H2 or C4H4 or C6H6. Thus, if we know (are given) the
molar mass of the compound, we can compare our empirical formula mass to the given molar
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mass. If they are the same, then we have an empirical formula that is the molecular formula. If
they are different, then we must find the whole number multiple!
Determining the Whole Number Multiple:
grams
(given)
mole
whole-number multiple =
grams
empirical formula mass in
mole
molar mass in
Remember that the whole number multiple applies to ALL the elements in the compound!
Another type of chemical reaction to be aware of is the combustion reaction. Combustion is
used to measure the amount of hydrogen and carbon in an organic substance as it burns in the
presence of oxygen. All of the hydrogen is converted or H 2O, while all of the carbon is
converted to CO2. If the organic compound contains a halogen, nitrogen, or oxygen, simple
math (!!) can be used to determine its mass. We will know the mass of the components at the
end of the combustion reaction and we know the mass of the starting compounds.
Conservation of mass tells us that the mass must be conserved. Using addition/subtraction, the
mass of the missing component can be found.
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