percent composition, empirical formulas, molecular formulas
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Percent Composition, Empirical Formulas, Molecular Formulas
Percent Composition• Percent Composition – the percentage by mass of
each element in a compound
Percent = _______PartWhole
x 100%
Percent compositionof a compound or =molecule
Mass of element in 1 mol____________________Mass of 1 mol
x 100%
Percent CompositionExample: What is the percent composition of Potassium Permanganate (KMnO4)?
Molar Mass of KMnO4K = 1(39.1) =
39.1Mn = 1(54.9) = 54.9O = 4(16.0) = 64.0
MM = 158 g
Percent CompositionExample: What is the percent composition of Potassium Permanganate (KMnO4)? = 158 g
% K
Molar Mass of KMnO4
39.1 g K158 g
x 100 =24.7 %
% Mn54.9 g Mn158 g
x 100 =34.8 %
% O64.0 g O158 g
x 100 = 40.5 %K = 1(39.10) = 39.1
Mn = 1(54.94) = 54.9
O = 4(16.00) = 64.0MM = 158
Percent CompositionDetermine the percentage composition of sodium carbonate (Na2CO3)?
Molar Mass Percent Composition
% Na =46.0 g106 g
x 100% =43.4 %
% C =12.0 g106 g
x 100% =11.3 %
% O =48.0 g106 g
x 100% =45.3 %
Na = 2(23.00) = 46.0C = 1(12.01) = 12.0O = 3(16.00) = 48.0 MM= 106 g
Percent CompositionDetermine the percentage composition of ethanol (C2H5OH)?
% C = 52.13%, % H = 13.15%, % O = 34.72%
_______________________________________________
Determine the percentage composition of sodium oxalate(Na2C2O4)?
% Na = 34.31%, % C = 17.93%, % O = 47.76%
Percent CompositionCalculate 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.
Percent CompositionCalculate the mass of nitrogen in 85.0 mg of the amino acid lysine, C6H14N2O2.
1. Molar Mass of C6H14N2O2
C = 6(12.01) = 72.06 H =14(1.01) = 14.14
MM = 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) = 28.02O = 2(16.00) = 32.00
HydratesHydrated salt – salt that has water molecules trapped within the crystal lattice
Examples: CuSO4•5H2O , CuCl2•2H2O
Anhydrous salt – salt without water molecules
Examples: CuCl2
Can calculate the percentage of water in a hydrated salt.
Percent CompositionCalculate the percentage of water in sodium carbonate decahydrate, Na2CO3•10H2O.
1. Molar Mass of Na2CO3•10H2ONa = 2(22.99) = 45.98
C = 1(12.01) = 12.01
MM = 286.2
H = 20(1.01) = 20.2O = 13(16.00)= 208.00
H = 20(1.01) = 20.2
Water
O = 10(16.00)= 160.00MM = 180.2
2.
3.180.2 g _______286.2 g
67.97 % x 100%=
orH = 2(1.01) = 2.02O = 1(16.00) = 16.00
MM H2O = 18.02
So… 10 H2O = 10(18.02) = 180.2
Percent CompositionCalculate the percentage of water in Aluminum bromide hexahydrate, AlBr3•6H2O.
1. Molar Mass of AlBr3•6H2OAl = 1(26.98) = 26.98
Br = 3(79.90) = 239.7
MM = 374.8
H = 12(1.01) = 12.12O = 6(16.00) = 96.00
H = 12(1.01) = 12.1
Water
O = 6(16.00)= 96.00MM = 108.1
2.
3.108.1 g _______374.8 g
28.85 % x 100%=
orMM = 18.02For 6 H2O = 6(18.02) = 108.2
Percent CompositionIf 125 grams of magnesium sulfate heptahydrate is completely dehydrated, how many grams of anhydrous magnesium sulfate will remain?MgSO4 . 7 H2O1. 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 g
H = 2 x 1.01 = 2.02 gO = 1 x 16.00 = 16.00 g
MM = 18.02 g
MM H2O =7 x 18.02 g = 126.1 gTotal MM = 120.4 g + 126.1 g = 246.5 g
2. % MgSO4
120.4 g246.5 g
X 100 =48.84 %
3. Grams anhydrous MgSO4
0.4884 x 125 = 61.1 g
Percent CompositionIf 145 grams of copper (II) sulfate pentahydrate is completely dehydrated, how many grams of anhydrous copper sulfate will remain?CuSO4 . 5 H2O1. 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 g
H = 2 x 1.01 = 2.02 gO = 1 x 16.00 = 16.00 g
MM = 18.02 g
MM H2O =5 x 18.02 g = 90.1 gTotal MM = 159.6 g + 90.1 g = 249.7 g
2. % CuSO4
159.6 g249.7 g
X 100 =63.92 %
3. Grams anhydrous CuSO4
0.6392 x 145 = 92.7 g
Percent CompositionA 5.0 gram sample of a hydrate of BaCl2 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 hydratex 100 = 14 %
Percent CompositionA 7.5 gram sample of a hydrate of CuCl2 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 hydratex 100 = 29 %
Percent CompositionA 5.0 gram sample of Cu(NO3)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 =4.0
Percent CompositionA 7.5 gram sample of CuSO4•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 =5.0
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:
C4H10 - molecular
C2H5 - empirical
C6H12O6 - molecular
CH2O - empirical
Formulas
Is H2O2 an empirical or molecular formula?
Molecular, it can be reduced to HOHO = empirical formula
Calculating Empirical FormulaAn 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 O
Calculating Empirical FormulaAn 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, Al2O3
Calculating Empirical FormulaA 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 Co1 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
CoCl2
Calculating Empirical FormulaWhen 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
FeO
Calculating Empirical FormulaA 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 Pb1 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 O
Calculating Empirical FormulaA 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
PbHAsO4
Calculating Empirical FormulaThe 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 O
Calculating Empirical FormulaThe 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
C6NH11O
Calculating Molecular FormulaA 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 = 61.94gO = 5 x 16.00g = 80.00 g
141.94 g
Step 2: Divide MM by Empirical Formula Mass
238.88 g141.94g
= 2
Step 3: Multiply
(P2O5)2 =
P4O10
Calculating Molecular FormulaA compound has an experimental molar mass of 78 g/mol. Its empirical formula is CH. What is its molecular formula?
C = 12.01 gH = 1.01 g 13.01 g
78 g/mol
13.01 g/mol= 6
(CH)6 =
C6H6