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Moles

When someone says they want a dozen donuts, roses, or eggs we know they mean they want 12. A mole in chemistry also tells us the amount of something.


  • Just like how a dozen of something = 12, 1 mole of something =6.02x1023 molecules
  • 6.02x1023 is referred to as Avagadro's number (NA)

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Wize Tip
You should memorize Avagadro's number since it is not always provided on exams! We'll soon see that you need to know it to solve problems :)
NA=6.02×1023  molecules/mole\boxed{\bold{N_A=6.02\times 10^{23}\ \bold{\text{ molecules/mole}}}}

We use the unit "moles" to help make very small amounts more measurable. A mole is just like any other unit!

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There are 2 Equations Related to Moles:


n=mM\boxed{n=\frac{m}{M}} n=NNA\boxed{n=\frac{N}{N_{A_{ }}}}

n=# of moles
m=mass (g)
M=molar mass (g/mol)
N=# of molecules or atoms
NA=Avagadro's number=6.02x1023 molecules

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Example #1:

If we are told that a sample of CO2(s) weighs 11g, how many moles are present?

m=11g, n=?
n=m/M
We are given m but need to solve for M:
M=12g/mol + 2(16g/mol)
M=44g/mol

n=m/M
n=11g/44g/mol
n=0.25moles!


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Example #2

We have 2 moles of CO2 present in our sample, how many molecules are there?

n=2 moles, N=? NA=6.02x1023
n=N/NA
N=n(NA)
N=2(6.02x1023)
N=1.2x1024 molecules

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Example #3

How many oxygen atoms there in 1 mole of CO2?

1) Find Number of Molecules In the Sample

n=N/NA N=? (how many molecules are there)
N=n(NA)
N=1(6.02x1023)
N=6.02x1023 molecules



2) Find the Number of O atoms in the Sample

In each molecule there are
2
O atoms.
Therefore to find the number of O atoms:
# of molecules in sample x
2
O atoms/molecule =# of O atoms
# of O atoms=
6.02x1023 x2=1.2x1024 atoms!


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Example: Calculate the Number of Molecules of Ethyl Mercaptan


The volatile liquid ethyl mercaptan,C2H6S,{\small\text{C}_2\text{H}_6\text{S},}is one of the most odoriferous substances known. It is added to natural gas to make gas leaks detectable. (d = 0.84 g/mL; MW = 62.1 g/mol)
  1. How manyC2H6S,{\small\text{C}_2\text{H}_6\text{S},}molecules are contained in a 3.0 μL{\small \mu L} sample?
N=? V=3 uL d=0.84 g/mL M=62.1g/mol NA=6.02x1023

n=N/NA
N=n(NA)
Need to solve for moles first before we can use this equation.

n=m/M need the mass before we can solve for moles
d=m/V
m=dV
density is given in g/mL so the V needs to be in mL so the units cross out and we have a mass in g.
1uL=1x10-6L
3uL=3x10-6L
Now we need to convert into mL:
1000mL=1L
V=3x10-3mL
m=dV
m=(0.84g/mL)(3x10-3mL)
m=2.52x10-3 g

n=m/M now solve for moles
n=2.52x10-3g / 62.1 g/mol
n=4.05797x10-5 moles

Finally since we have the total moles in the sample, we can solve for the total molecules:
N=n(NA)
N=4.05797x10-5 moles (6.02x1023 molecules/mole)
N= 2.4x1019 molecules

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  1. In the same 3.0 µL sample, how many C and H atoms are there?

#C atoms=2.4×1019molecules×2 C atomsmolecules=4.8×1019 C atoms\# \text{C atoms} = 2.4\times10^{19} \text{molecules}\times\dfrac{2\text{ C atoms}}{\text{molecules}} = 4.8\times10^{19}\text{ C atoms}

#H atoms=2.4×1019molecules×6 H atomsmolecules=1.4×1020 H atoms\# \text{H atoms} = 2.4\times10^{19} \text{molecules}\times\dfrac{6\text{ H atoms}}{\text{molecules}} = 1.4\times10^{20}\text{ H atoms}

Practice: Converting Mass to Number of Atoms

Calculate the number of nitrogen atoms in 2.25 g of bismuth(III) nitrate.

Practice: Finding the Number of Moles of Iron

Calculate the number of moles of iron atoms in 14.1 g of iron oxide, Fe2O3
Extra Practice