Stoichiometry © 2009, Prentice- Hall, Inc. Unit 10: Unit 10: Stoichiometry 1 Stoichiometry 1 Calculations with Calculations with Chemical Formulas Chemical Formulas
Dec 19, 2015
Stoichiometry
© 2009, Prentice-Hall, Inc.
Unit 10: Stoichiometry 1Unit 10: Stoichiometry 1
Calculations with Chemical Calculations with Chemical FormulasFormulas
Stoichiometry
Atomic Mass
Atoms are so small, it is difficult to discuss how much they weigh in grams.
• Use atomic mass unitsatomic mass units or amuamus– 1 amu is 1/12 the mass of a carbon-12 atom.
This gives us a basis for comparison.• The decimal numbers on the periodic table
are atomic masses in amu.
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Stoichiometry
Gram Atomic Mass
• Atomic mass in gramsgrams instead of amu’s.– Represents an amount that we can actually
measure in lab.
• Also known as a molemole, or molar mass, which we will come back to later…
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Stoichiometry
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Gram Formula Mass • The gram formula mass is the sum of the
atomic masses for the atoms in a chemical formula.
• So, the gram formula mass of calcium chloride, CaCl2, would be
Ca: 1 x 40.1 = 40.1
+ Cl: 2 x 35.5 = 71.0
111.1 amu
• Gram formula mass is generally used for either molecular or ionic compounds.
Stoichiometry
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Gram Formula Mass
• For the molecule, ethane (C2H6), the formula mass would be:
• For the ionic compound, (NH4)2CO3
N: 2 x 14.0 amu = 28.0
H: 8 x 1.0 amu = 8.0
C: 1 x 12.0 amu = 12.0
+O: 3 x 16.0 amu = 48.0
= 96.0 amu96.0 amu
C: 2 x 12.0 = 24.0
30.0 amu30.0 amu
+ H: 6 x 1.0 = 6.0
Stoichiometry
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Percent Composition
You can find the percentage (%) of the mass of a compound that comes from each of the elements in the compound by using these steps:
1.Calculate the formula mass of the compound.
2.Divide the mass of each element by the formula mass and multiply that fraction x 100.
Stoichiometry
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Percent Composition
So the percentage of carbon in ethane,C2H6, is…
C is 2 x 12.0 = 24.0
H is 6 x 1.0 = 6.0
30.030.024.0 amu
30.0 amu= x 100
= 80.0%
% Carbon
amu
Stoichiometry
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Avogadro’s Number
• 6.02 x 1023
• 1 mole of 12C has a mass of 12 g.
Stoichiometry
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Molar Mass
• By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol).– The molar mass of an element is the mass
number for the element that we find on the periodic table.
– The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).
Stoichiometry
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Using Moles
Moles provide a bridge from the molecular scale to the real-world scale.
Stoichiometry
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Mole Relationships
• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.
• One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.
Stoichiometry
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Calculating Empirical FormulasCalculating Empirical Formulas
• Empirical formula: Empirical formula: smallest, whole-number ratio of atoms of elements in a compound.
• You can calculate the empirical formula from the percent composition.
Stoichiometry
Steps for Empirical FormulaSteps for Empirical Formula
1. For each element, convert mass to moles. – If you have percents, use the percent as the
number of grams/100 grams of compound. Example: 67% would be 67 grams.
2. Find the mole ratio: Divide all the numbers by the smallest number of moles
3. Use the smallest, whole number ratio as the subscripts for the formula.
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Stoichiometry
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Calculating Empirical FormulasCalculating Empirical Formulas
Example:The compound para-aminobenzoic acid (you may have seen it listed as PABA on your bottle of sunscreen) is composed of:
carbon (61.31%)hydrogen (5.14%) nitrogen (10.21%)oxygen (23.33%)
Find the empirical formula of PABA.
Stoichiometry
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Calculating Empirical FormulasCalculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C: 61.31 g x = 5.105 mol C
H: 5.14 g x = 5.09 mol H
N: 10.21 g x = 0.7288 mol N0.7288 mol N
O: 23.33 g x = 1.456 mol O
1 mol12.01 g
1 mol14.01 g
1 mol1.01 g
1 mol16.00 g
Sunscreen PABAcarbon (61.31%)
hydrogen (5.14%) nitrogen (10.21%)oxygen (23.33%)
Stoichiometry
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Calculating Empirical FormulasCalculating Empirical FormulasCalculate the mole ratio by dividing by the smallest number of moles:
C: = 7.005 = 77
H: = 6.984 = 77
N: = 1.000 = 11
O: = 2.001 = 22
5.105 mol0.7288 mol
5.09 mol0.7288 mol
0.7288 mol0.7288 mol
1.458 mol0.7288 mol
Stoichiometry
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Calculating Empirical FormulasCalculating Empirical Formulas
These numbers are the subscripts for the empirical formula: C7H7NO2
The molecule is shown here.
Stoichiometry
Molecular FormulaMolecular Formula
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Empirical Formula is the smallest whole number ratio of atoms of elements in a compound.
Molecular Formula is the real formula of a compound. It is a multiple of the Empirical Formula. They may be the same!
1.1. CalculateCalculate the mass of the empirical formula.
2.2. DivideDivide the molecular mass given in the problem by the empirical formula mass.
3.3. MultiplyMultiply the subscripts in the empirical formula by the number you get to make new subscripts.
Molecular formula is a multiple Molecular formula is a multiple of the empirical formula! of the empirical formula!
Stoichiometry
Molecular Formula ProblemMolecular Formula Problem
• Analysis of a chemical used in photographic developing fluid indicates a chemical composition of:
65.4% C
5.45% H
29.09% O
• The molar mass is found to be 110.0 g/mol. Determine the empirical and molecular formulas.
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Stoichiometry
Find Empirical Formula First!Find Empirical Formula First!
65.4g C 1 mole C = 5.45 ÷ 1.82 = 33
12g C
5.45g H 1 mole H = 5.45 ÷1.82 = 33
1g H
29.09g O 1 mole O = 1.82 ÷ 1.82 = 11
16g O
CC33HH33OO = Empirical Formula
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Stoichiometry
Molecular FormulaMolecular Formula 1. Find molar mass C3H3O
C – 3 x 12 = 36
H – 3 x 1 = 3
O – 1 x 16 = +16
55 g/mole is Empirical Formula Mass
2. Divide the given Molecular Formula Mass (110g/mole) by the Empirical Formula Mass
110/55 = 2, so C3x2H3x2O1x2
3. Molecular Formula = CMolecular Formula = C66HH66OO22 © 2009, Prentice-Hall, Inc.
Stoichiometry
Molecular formula Molecular formula is some multiple of the empirical formulaempirical formula. It may be the same as the empirical formula!
Two samples of a compound must have the same percent composition to be the same compound or they would have different empirical formulas.
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Stoichiometry
HydratesHydrates
A hydratehydrate is an ionic compound that has a specific number of water molecules bound to the
atoms in its crystals. (This is NOT the same as being dissolved in water)
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Many compounds are found in nature as hydrates, such as protein crystals
Stoichiometry
Naming HydratesNaming Hydrates
To name a hydrate, give the name of the compound, and add the prefix for the number of waters + hydrate:
CuSOCuSO44·5H·5H22O O
copper(II) sulfate pentahydrate
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Stoichiometry
Formula of a HydrateFormula of a Hydrate
1.1. Weigh Weigh the hydrate, then heatheat it in a partly covered crucible to drive off the water.
2.2. Weigh it againWeigh it again to determine the amount of water lost from the anhydrous (no water) compound. Repeat until mass stops changing.
3.3. Calculate the empirical formula Calculate the empirical formula with the anhydrous (no water) compound as one unit and the water as the second.
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Stoichiometry
Hydrate ProblemHydrate ProblemProblem:
If 11.75 g of cobalt(II) chloride is heated, 9.25 g of anhydrous cobalt chloride, CoCl2, remains. What is the formula and name for this hydrate?
11.75 – 9.25 = 2.5 g water removed
9.25 g anhydrous CoCl2
Molar mass of CoCl2
Co – 1 x 59 = 59
Cl – 2 x 35.5 = +71
130 g/mol
9.25 g CoCl2 1 mole = .0712 ÷.0712 = 11
130 g
2.5 g H20 1 mole = .139 ÷ .0712 = 22
18 g
CoCl2·2H2O is cobalt(II) chloride dihydrate
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Stoichiometry
Hydrate ProblemHydrate ProblemA hydrate is found to have the following percent
composition: 48.8% MgSO4 and 51.22% H2O. What is the formula and name of this hydrate?
Molar mass of MgSO4 = 24.3 + 32 + 64 =120.3g/mol
Molar mass of H2O = 2 + 16 = 18 g/mol
48.8g 1 mole = .406 ÷ .406 = 11 120.3 g
51.22 g 1 mole = 2.85 ÷ .406 = 77 18 g
Magnesium sulfate heptahydrate: Magnesium sulfate heptahydrate: MgSOMgSO44· 7H· 7H22OO© 2009, Prentice-Hall, Inc.
Stoichiometry
HydratesHydrates
Sometimes hydration can have nifty color changes!
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