Percent Composition Percent Composition the percentage by mass of each element in a compound Percent Part Whole x 100 So Percent composition of a compound or molecule ID: 591004
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Slide1
Percent Composition, Empirical Formulas, Molecular FormulasSlide2
Percent Composition
Percent Composition –
the percentage by mass of each element in a compound
Percent =
_______
Part
Whole
x 100%
So…
Percent compositionof a compound or =molecule
Mass of element in 1 mol
____________________
Mass of 1 mol
x 100%Slide3
Percent Composition
Example:
What is the percent composition of Potassium Permanganate (KMnO
4
)?
Molar Mass of KMnO4
K = 1(39.1) = 39.1
Mn = 1(54.9) = 54.9
O = 4(16.0) = 64.0
MM = 158 gSlide4
Percent Composition
Example:
What is the percent composition of Potassium Permanganate (KMnO
4
)?
= 158 g
% K
Molar Mass of KMnO4
39.1 g K
158 gx 100 =
24.7 %% Mn
54.9 g Mn
158 g
x 100 =
34.8 %
% O
64.0 g O
158 g
x 100 =
40.5 %
K = 1(39.10) = 39.1
Mn = 1(54.94) = 54.9
O = 4(16.00) =
64.0
MM = 158Slide5
Percent Composition
Determine the percentage composition of sodium carbonate (
Na
2CO3)?
Molar Mass
Percent Composition
% Na
=
46.0 g
106 gx 100% =
43.4 %% C =
12.0 g
106 g
x 100% =
11.3 %
% O
=
48.0 g
106 g
x 100% =
45.3 %
Na = 2(23.00) = 46.0
C = 1(12.01) = 12.0
O = 3(16.00) =
48.0
MM= 106 gSlide6
Percent Composition
Determine the percentage composition of ethanol (C
2
H5OH)?
% C = 52.13%, % H = 13.15%, % O = 34.72%
_______________________________________________
Determine the percentage composition of sodium oxalate(Na
2C2O
4)?
% Na = 34.31%, % C = 17.93%, % O = 47.76%Slide7
Percent Composition
Calculate the mass of bromine in 50.0 g of Potassium bromide
.
1. Molar Mass of KBr
K = 1(39.10) = 39.10
Br =1(79.90) =79.90
MM = 119.0
79.90 g
___________
119.0 g = 0.6714
3. 0.6714 x 50.0g = 33.6 g Br
2.Slide8
Percent Composition
Calculate the mass of nitrogen in 85.0 mg of the amino acid lysine, C
6
H14N2O
2.
1. Molar Mass of C
6H14N
2O2
C = 6(12.01) = 72.06
H =14(1.01) = 14.14MM = 146.2
28.02 g
___________
146.2 g
= 0.192
3. 0.192 x 85.0 mg = 16.3 mg N
2.
N = 2(14.01) = 28.02
O = 2(16.00) =
32.00Slide9
Hydrates
Hydrated salt –
salt that has water molecules trapped within the crystal lattice
Examples:
CuSO4•5H2
O , CuCl2•2H
2O
Anhydrous salt – salt without water molecules
Examples: CuCl
2Can calculate the percentage of water in a hydrated salt.Slide10
Percent Composition
Calculate the percentage of water in sodium carbonate decahydrate, Na
2
CO3•10H2O.
1. Molar Mass of Na
2CO
3•10H2O
Na = 2(22.99) = 45.98
C = 1(12.01) = 12.01
MM = 286.2H = 20(1.01) = 20.2
O = 13(16.00)= 208.00
H = 20(1.01) = 20.2
Water
O = 10(16.00)= 160.00
MM = 180.2
2.
3.
180.2 g
_______
286.2 g
67.97 %
x 100%=
or
H = 2(1.01) = 2.02
O = 1(16.00) =
16.00
MM H2O = 18.02
So…
10 H
2
O = 10(18.02) = 180.2Slide11
Percent Composition
Calculate the percentage of water in Aluminum bromide hexahydrate, AlBr
3
•6H2O.
1. Molar Mass of AlBr3
•6H2O
Al = 1(26.98) = 26.98
Br = 3(79.90) = 239.7
MM = 374.8
H = 12(1.01) = 12.12O = 6(16.00) =
96.00H = 12(1.01) = 12.1
Water
O = 6(16.00)= 96.00
MM = 108.1
2.
3.
108.1 g
_______
374.8 g
28.85 %
x 100%=
or
MM = 18.02
For 6 H2O = 6(18.02) = 108.2Slide12
Percent Composition
If 125 grams of magnesium sulfate
hepta
hydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain?
MgSO4
. 7 H
2O
1. Molar Mass
Mg = 1 x 24.31 = 24.31 g S = 1 x 32.06 = 32.06 g
O = 4 x 16.00 = 64.00 g MM = 120.37 gH = 2 x 1.01 = 2.02 g
O = 1 x 16.00 = 16.00 g MM = 18.02 g
MM H2O
=7 x 18.02 g = 126.1 g
Total MM =
120.4 g + 126.1 g =
246.5 g
2. % MgSO
4
120.4 g
246.5 g
X 100 =
48.84 %
3. Grams anhydrous MgSO
4
0.4884 x 125 =
61.1 g Slide13
Percent Composition
If 145 grams of copper (II) sulfate
penta
hydrate is completely dehydrated, how many grams of anhydrous copper sulfate will remain?
CuSO4
. 5 H
2O
1. Molar Mass
Cu = 1 x 63.55 = 63.55 g S = 1 x 32.06 = 32.06 g
O = 4 x 16.00 = 64.00 g MM = 159.61 gH = 2 x 1.01 = 2.02 g
O = 1 x 16.00 = 16.00 g MM = 18.02 g
MM H2O
=5 x 18.02 g = 90.1 g
Total MM =
159.6 g + 90.1 g =
249.7 g
2. % CuSO
4
159.6 g
249.7 g
X 100 =
63.92 %
3. Grams anhydrous CuSO
4
0.6392 x 145 =
92.7 g Slide14
Percent Composition
A 5.0 gram sample of a hydrate of BaCl
2
was heated, and only 4.3 grams of the anhydrous salt remained. What percentage of water was in the hydrate?1. Amount water lost
5.0 g hydrate
4.3 g anhydrous salt 0.7 g water
2. Percent of water
0.7 g water
5.0 g hydrate
x 100 = 14 %Slide15
Percent Composition
A 7.5 gram sample of a hydrate of CuCl
2
was heated, and only 5.3 grams of the anhydrous salt remained. What percentage of water was in the hydrate?1. Amount water lost
7.5 g hydrate
5.3 g anhydrous salt 2.2 g water
2. Percent of water
2.2 g water
7.5 g hydrate
x 100 = 29 %Slide16
Percent Composition
A 5.0 gram sample of Cu(NO
3
)2•nH2O is heated, and 3.9 g of the anhydrous salt remains. What is the value of n?
1. Amount water lost
5.0 g hydrate
3.9 g anhydrous salt 1.1 g water
2. Percent of water
1.1 g water
5.0 g hydratex 100 =
22 %
3. Amount of water
0.22 x 18.02 =
4.0Slide17
Percent Composition
A 7.5 gram sample of CuSO
4
•nH2O is heated, and 5.4 g of the anhydrous salt remains. What is the value of n?
1. Amount water lost
7.5 g hydrate 5.4 g anhydrous salt
2.1 g water
2. Percent of water
2.1 g water
7.5 g hydratex 100 =
28 %3. Amount of water
0.28 x 18.02 =
5.0Slide18
Formulas
Empirical Formula –
formula of a compound that expresses lowest whole number ratio of atoms.
Molecular Formula – actual formula of a compound showing the number of atoms present
Percent composition allow you to calculate the simplest ratio among the atoms found in compound.
Examples:
C4
H10
- molecular
C2H5
- empiricalC6
H12O6
- molecular
CH
2
O
- empiricalSlide19
Formulas
Is H
2
O2 an empirical or molecular formula?
Molecular, it can be reduced to HO HO = empirical formulaSlide20
Calculating Empirical Formula
An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula.
1. Determine the number of grams of each element in the compound.
4.151 g Al and 3.692 g O
2. Convert masses to moles.
4.151 g Al
1 mol Al
26.98 g Al
=
0.1539 mol Al
3.692 g O
1 mol O
16.00 g O
=
0.2308 mol OSlide21
Calculating Empirical Formula
An oxide of aluminum is formed by the reaction of 4.151 g of aluminum with 3.692 g of oxygen. Calculate the empirical formula.
3. Find ratio by dividing each element by smallest amount of moles.
0.1539 moles Al
0.1539
= 1.000 mol Al
0.2308 moles O
0.1539
= 1.500 mol O
4. Multiply by common factor to get whole number. (cannot have fractions of atoms in compounds)
O = 1.500 x 2 = 3Al = 1.000 x 2 = 2
therefore,
Al2
O3Slide22
Calculating Empirical Formula
A 4.550 g sample of cobalt reacts with 5.475 g chlorine to form a binary compound. Determine the empirical formula for this compound.
4.550 g Co
1 mol Co
58.93 g Co
= 0.07721 mol Co
5.475 g Cl
1 mol Cl
35.45 g Cl
= 0.1544 mol Cl
0.07721 mol Co
0.1544 mol Cl
0.07721
0.07721
= 2
= 1
CoCl
2Slide23
Calculating Empirical Formula
When a 2.000 g sample of iron metal is heated in air, it reacts with oxygen to achieve a final mass of 2.573 g. Determine the empirical formula.
2.000 g Fe
1 mol Fe
55.85 g Fe
= 0.03581 mol Fe
0.573 g O
1 mol O
16.00 g
= 0.03581 mol Fe
Fe = 2.000 g
O = 2.573 g – 2.000 g = 0.5730 g
1 : 1
FeOSlide24
Calculating Empirical Formula
A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate.
1.3813 g Pb
1 mol Pb
207.2 g Pb
= 0.006667 mol Pb
0.00672 gH
1 mol H
1.008 g H
= 0.00667 mol H
0.4995 g As
1 mol As
74.92 g As
= 0.006667 mol As
0.4267g Fe
1 mol O
16.00 g O
= 0.02667 mol OSlide25
Calculating Empirical Formula
A sample of lead arsenate, an insecticide used against the potato beetle, contains 1.3813 g lead, 0.00672g of hydrogen, 0.4995 g of arsenic, and 0.4267 g of oxygen. Calculate the empirical formula for lead arsenate.
0.006667 mol Pb
0.00667 mol H
0.006667 mol As
0.02667 mol O
0.006667
0.006667
0.006667
0.006667
= 1.000 mol Pb
= 1.00 mol H
= 1.000 mol As
= 4.000 mol O
PbHAsO
4Slide26
Calculating Empirical Formula
The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6.
Step 1:
In 100.00g of Nylon-6 the masses of elements present are 63.38 g C, 12.38 g n, 9.80 g H, and 14.14 g O.
Step 2:
63.38 g C
1 mol C
12.01 g C
= 5.302 mol C
12.38 g N
1 mol N
14.01 g N
= 0.8837 mol N
9.80 g H
1 mol H
1.01 g H
= 9.72 mol H
14.14 g O
1 mol O
16.00 g O
= 0.8832 mol OSlide27
Calculating Empirical Formula
The most common form of nylon (Nylon-6) is 63.38% carbon, 12.38% nitrogen, 9.80% hydrogen and 14.14% oxygen. Calculate the empirical formula for Nylon-6.
Step 3:
5.302 mol C
0.8837
= 6.000 mol C
0.8837 mol N
0.8837
= 1.000 mol N
9.72 mol H
0.8837
= 11.0 mol H
0.8837 mol O
0.8837
= 1.000 mol O
6:1:11:1
C
6
NH
11
OSlide28
Calculating Molecular Formula
A white powder is analyzed and found to have an empirical formula of P2O5. The compound has a molar mass of 283.88g. What is the compound’s molecular formula?
Step 1: Molar Mass
P = 2 x 30.97 g = 61.94g
O = 5 x 16.00g =
80.00 g 141.94 g
Step 2: Divide MM by Empirical Formula Mass
238.88 g
141.94g
= 2
Step 3: Multiply
(P2O
5)2 =
P
4
O
10Slide29
Calculating Molecular Formula
A compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula?
C = 12.01 g
H = 1.01 g
13.01 g
78 g/mol
13.01 g/mol
= 6
(CH)
6 =
C6H
6