Percent composition and empirical formulas

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.


• Chlorine atoms are heavier than sodium atoms.

An example.


• Chlorine atoms are heavier than sodium atoms. By mass:

• Na=22.99g/mol

• Cl=35.45g/mol

An example.



• %Na=22.99g/58.44 g x 100%

• %Cl=35.45g/58.44 g x 100%

An example.



• %Na=22.99g/58.44 g x 100%

• %Cl=35.45g/58.44 g x 100%FM of NaCl !

An example.



• %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.


(FM=199.88g/mol)

Try it.


(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.


(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) CaO

2) Na3N

3) Al2(SO4)3

4) NaNO3

5) NaNO2

Practice

• ? %comp of:

1) CaO 71.47%Ca 28.53%O

2) Na3N 83.12%Na 16.88%N

3) Al2(SO4)3 15.77%Al 28.11%S 56.12%O

4) NaNO3 27.05%Na 16.48%N56.47%O

5) NaNO2 33.32%Na 20.30%N46.38%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?




Uhh… 16.09/(16.09 +6.91) x 100%=…

So what?




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?



• 8.39 g Ti

• 5.61 g O



• 8.39 g Ti x 1mol Ti/47.90 g

• 5.61 g O x 1mol O/16.00 g



• 8.39 g Ti x 1mol Ti/47.90 g=.175 mol Ti

• 5.61 g O x 1mol O/16.00 g=.350 mol O

• Ti.175O.350




• 5.61 g O x 1mol O/16.00 g=.350 mol O

• Ti.175O.350 Ti.175/.175O.350/.175




• 5.61 g O 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.


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?


• 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



• 1.15 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



• 1.15 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




Na.05S.05O.075 NaSO1.5




Na.05S.05O.075 NaSO1.5

Don’t try to round the decimal away!




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.255AlO1.5

• C.89H1.18 CH1.33

• C1.58H 4.22 O1.58 CH2.67O

• C.190H2.38Cl2.38 CH1.25Cl1.25

• Al.17O.255AlO1.5 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.


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.




• FM(CH2O)=30g/mol x 2=60 g/mol

• CH2O x 2= C2H4O2

Try it.




C2H4O2

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Percent composition and empirical formulas