Percent composition and
empirical formulas
No matter how great and destructive your
problems may seem now, remember, you've
probably only seen the tip of them
An example.
• You could claim that NaCl is half sodium
and half chlorine—one of each.
An example.
• You could claim that NaCl is half sodium
and half chlorine—one of each.
• Chlorine atoms are heavier than sodium
atoms.
An example.
• You could claim that NaCl is half sodium
and half chlorine—one of each.
• Chlorine atoms are heavier than sodium
atoms. By mass:
•
•
Na=22.99g/mol
Cl=35.45g/mol
An example.
• You could claim that NaCl is half sodium
and half chlorine—one of each.
• Chlorine atoms are heavier than sodium
atoms. By mass:
• %Na=22.99g/58.44 g x 100%
• %Cl=35.45g/58.44 g x 100%
An example.
• You could claim that NaCl is half sodium
and half chlorine—one of each.
• Chlorine atoms are heavier than sodium
atoms. By mass:
• %Na=22.99g/58.44 g x 100%
• %Cl=35.45g/58.44 g x 100%
FM of NaCl !
An example.
• You could claim that NaCl is half sodium
and half chlorine—one of each.
• Chlorine atoms are heavier than sodium
atoms. By mass:
• %Na=22.99g/58.44 g x 100%=39.34% and
• %Cl=35.45g/58.44 g x 100%=60.66%
Definition
• % composition of a compound:
• % A=
mass A in the compound x 100%
mass of the compound
PS
• All of the %’s add up to 100%
• The %’s are constant, no matter how
much of the substance!
• (AKA: the law of definite proportions)
Try it.
• What is the percent composition of CaBr2?
Try it.
• What is the percent composition of CaBr2?
(FM=199.88g/mol)
Try it.
• What is the percent composition of CaBr2?
(FM=199.88g/mol)
• %Ca=40.08g/199.88 g x 100%
and
• %Br=2 x 79.90g/199.88 g x 100%
Try it.
• What is the percent composition of CaBr2?
(FM=199.88g/mol)
• %Ca=40.08g/199.88 g x 100%=20.05%
and
• %Br=2 x 79.90g/199.88 g x 100%=79.95%
Practice
•
? %comp of:
1)
2)
3)
4)
5)
CaO
Na3N
Al2(SO4)3
NaNO3
NaNO2
Practice
•
? %comp of:
1)
2)
3)
4)
5)
CaO
Na3N
Al2(SO4)3
NaNO3
NaNO2
71.47%Ca
83.12%Na
15.77%Al
27.05%Na
33.32%Na
28.53%O
16.88%N
28.11%S 56.12%O
16.48%N 56.47%O
20.30%N 46.38%O
So what?
• Iron (II) oxide is 77.73% iron.
• Iron (III) oxide is 69.94% iron.
• An iron oxide that contains 16.09 g of iron
and 6.91 g of oxygen has to be…
So what?
• Iron (II) oxide is 77.73% iron.
• Iron (III) oxide is 69.94% iron.
• An iron oxide that contains 16.09 g of iron
and 6.91 g of oxygen has to be…
Uhh… 16.09/(16.09 +6.91) x 100%=…
So what?
• Iron (II) oxide is 77.73% iron.
• Iron (III) oxide is 69.94% iron.
• An iron oxide that contains 16.09 g of iron
and 6.91 g of oxygen has to be…
Iron (III) oxide
On the other hand…
• A sample contains 8.39 g Ti and 5.61 g O.
What is the formula of this compound?
On the other hand…
• A sample contains 8.39 g Ti and 5.61 g O.
What is the formula of this compound?
• 8.39 g Ti
• 5.61 g O
On the other hand…
• A sample contains 8.39 g Ti and 5.61 g O.
What is the formula of this compound?
• 8.39 g Ti
• 5.61 g O
x 1mol Ti/47.90 g
x 1mol O/16.00 g
On the other hand…
• A sample contains 8.39 g Ti and 5.61 g O.
What is the formula of this compound?
• 8.39 g Ti
• 5.61 g O
• Ti.175O.350
x 1mol Ti/47.90 g=.175 mol Ti
x 1mol O/16.00 g=.350 mol O
On the other hand…
• A sample contains 8.39 g Ti and 5.61 g O.
What is the formula of this compound?
• 8.39 g Ti
• 5.61 g O
x 1mol Ti/47.90 g=.175 mol Ti
x 1mol O/16.00 g=.350 mol O
• Ti.175O.350 Ti.175/.175O.350/.175
On the other hand…
• A sample contains 8.39 g Ti and 5.61 g O.
What is the formula of this compound?
• 8.39 g Ti
• 5.61 g O
x 1mol Ti/47.90 g=.175 mol Ti
x 1mol O/16.00 g=.350 mol O
• Ti.175O.350 Ti.175/.175O.350/.175 TiO2
Try it.
• A sample contains 2.6366 g C, 0.4425 g H, and
3.5122 g O. What is the empirical formula of this
compound?
Try it.
• A sample contains 2.6366 g C, 0.4425 g H, and
3.5122 g O. What is the empirical formula of this
compound?
CH2O
Try it.
• A sample contains 35.378g C, 5.938 g H, and
31.418 g O. What is the formula of this
compound?
There are two things to watch out
for:
1) What if the smallest number is not 1?
2) What if the simplest whole number ratio
is smaller than the molecule?
1) What if the smallest number is not 1?
• A sample contains 1.15 g Na, 1.60 g S,
and 1.20 g O.
1) What if the smallest number is not 1?
• A sample contains 1.15 g Na, 1.60 g S,
and 1.20 g O.
• 1.15 g Na/22.99g/mol
• 1.60 g S /32.06 g/mol
• 1.20 g O/16 g/mol
1) What if the smallest number is not 1?
• A sample contains 1.15 g Na, 1.60 g S,
and 1.20 g O.
• 1.15 g Na/22.99g/mol=.0500 mol Na
• 1.60 g S /32.06 g/mol=.0500 mol S
• 1.20 g O/16 g/mol=
.0750 mol O
1) What if the smallest number is not 1?
• A sample contains 1.15 g Na, 1.60 g S,
and 1.20 g O.
• 1.15 g Na/22.99g/mol=.0500 mol Na
• 1.60 g S /32.06 g/mol=.0500 mol S
• 1.20 g O/16 g/mol=
.0750 mol O
Na.05S.05O.075
1) What if the smallest number is not 1?
• A sample contains 1.15 g Na, 1.60 g S,
and 1.20 g O.
• 1.15 g Na/22.99g/mol=.0500 mol Na
• 1.60 g S /32.06 g/mol=.0500 mol S
• 1.20 g O/16 g/mol=
.0750 mol O
Na.05S.05O.075 NaSO1.5
1) What if the smallest number is not 1?
• A sample contains 1.15 g Na, 1.60 g S,
and 1.20 g O.
• 1.15 g Na/22.99g/mol=.0500 mol Na
• 1.60 g S /32.06 g/mol=.0500 mol S
• 1.20 g O/16 g/mol=
.0750 mol O
Na.05S.05O.075 NaSO1.5 Don’t try to round
the decimal away!
1) What if the smallest number is not 1?
• A sample contains 1.15 g Na, 1.60 g S,
and 1.20 g O.
• 1.15 g Na/22.99g/mol=.0500 mol Na
• 1.60 g S /32.06 g/mol=.0500 mol S
• 1.20 g O/16 g/mol=
.0750 mol O
Na.05S.05O.075 NaSO1.5  Na2S2O3
• Al.17O.255
• C.89H1.18
• C1.58H 4.22 O1.58
• C1.90H2.38Cl2.38
• Al.17O.255AlO1.5
• C.89H1.18  CH1.33
• C1.58H 4.22 O1.58  CH2.67O
• C.190H2.38Cl2.38  CH1.25Cl1.25
• Al.17O.255AlO1.5  Al2O3
• C.89H1.18  CH1.33  C3H4
• C1.58H 4.22 O1.58  CH2.67O  C3H 8 O3
• C.190H2.38Cl2.38  CH1.25Cl1.25  C4H5Cl5
Try it.
• A sample contains 2.6366 g C, 0.4425 g H, and
3.5122 g O. What is the empirical formula of this
compound?
CH2O
• If we know that the FM of the compound is about
60g/mol, what is the molecular formula?
• FM(CH2O)=30g/mol
Try it.
• A sample contains 2.6366 g C, 0.4425 g H, and
3.5122 g O. What is the empirical formula of this
compound?
CH2O
• If we know that the FM of the compound is about
60g/mol, what is the molecular formula?
• FM(CH2O)=30g/mol x 2=60 g/mol
• CH2O x 2=
C2H4O2
Try it.
• A sample contains 2.6366 g C, 0.4425 g H, and
3.5122 g O. What is the empirical formula of this
compound?
CH2O
• If we know that the FM of the compound is about
60g/mol, what is the molecular formula?
C2H4O2