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Week 2 CHEM 1310 - Sections L and M 1
Wed | Aug 29, 2007
Chapter 3: Stoichiometry– Focus on Sections 3.2 - 3.9– Chemical Formulas– Moles and Related Calculations
Bring PRS units on Friday, Aug 31
WebAssign HW due Thurs before midnight
Week 2 CHEM 1310 - Sections L and M 2
Mole
Avogadro’s Number, N0The number of atoms inexactly 12 grams of 12C.
6.022 x 1023 entities x mol-1
Mole (mol)The amount of a substance
containing as manyelementary particles asthere are in 12 g of 12C.
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Week 2 CHEM 1310 - Sections L and M 3
Counting in Chemistry
1 mole = different masses of elements
Week 2 CHEM 1310 - Sections L and M 4
Molar MassGiven: Molecular FormulaFind: Molecular Weight (i.e. Molar Mass)
Sulfuric Acid: H2SO4
Solution:(1) Multiply the atomic mass by the # of each atom type(2) Sum the resulting weights
H (1.0079 g/mol) x 2 = 2.0158 g/molS (32.065 g/mol) x 1 = 32.065 g/molO (15.999 g/mol) x 4 = 63.996 g/mol
SUM: 98.077 g/mol
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Week 2 CHEM 1310 - Sections L and M 5
Calculations
Given: Moles of an element or compoundFind: Weight of element or compound
EXAMPLE What is the weight of 0.03 mol of gold?
Atomic Mass of Au: 196.97 g/mol
0.03 mol x 196.97 gmol
= 6 g of Au
Week 2 CHEM 1310 - Sections L and M 6
Calculations Involving Moles
Given: Weight of a moleculeFind: Number of atoms
EXAMPLE How many atoms are contained in 1.5 g of NaOH?
FW of NaOH: 40 g/mol
1.5 g x 1 mol40 g
x 6.022 x 1023 atoms1 mol
= 2.3 x 1022 atoms
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Week 2 CHEM 1310 - Sections L and M 7
Chemical Formulas
Empirical Formula: a chemical formula of a compoundthat expresses the relative chemical amounts of itselements in terms of the smallest integers.
Molecular Formula: a chemical formula that specifiesthe actual number of atoms of each element in onemolecule.
Empirical FormulaCH2O
Molecular FormulaC6H12O6
(Glucose)
Week 2 CHEM 1310 - Sections L and M 8
Given: Mass CompositionFind: Empirical Formula
EXAMPLEA sample of a tin and chlorine compound with a
mass of 2.57g was found to contain 1.17 g of tin. What is the compound’s empirical formula
Atomic Mass of Tin: 118.710 g/molAtomic Mass of Chlorine: 35.453 g/mol
Calculations with Formulas
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Week 2 CHEM 1310 - Sections L and M 9
Given: Empirical Formula and Molecular MassFind: Molecular Formula
EXAMPLEThe empirical formula of squalene is C3H5.
Its molar mass is 410.7 g/mol.
What is the molecular formula of squalene?
Calculations with Formulas
Week 2 CHEM 1310 - Sections L and M 10
More Calculations with Moles
Given: Amount of a compoundFind: Amount of an element in compound
Compute the mass (in grams) oflithium in 65.4 g of Li2CO3
(lithium carbonate).
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Week 2 CHEM 1310 - Sections L and M 11
More Complexity…
Given: Amount of an element within a compoundFind: Amount of a different element in compound
EXAMPLEChlorophyll, the green pigment in plants,
has the formula C55H72MgN4O5
If 0.0011g of Mg is available to a plant cell forchlorophyll synthesis, how many grams of carbon
will be required to completely use up themagnesium?
Week 2 CHEM 1310 - Sections L and M 12
Density & MW
Given: Density and Molecular WeightFind: Molar Concentration (mol/L)
Density of NaCl: 2.16 g/cm3
FW of NaCl: 58.443 g/molcm3 = mL = 1 x 10-3 L
1 mol58.443 g
x 2.16 gmL
x 1000 mLL
= 36.96 M NaCl
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Week 2 CHEM 1310 - Sections L and M 13
Dilutions, Proportions
M1 x V1 = M2 x V2
What volume of 0.1M NaCl is needed to make0.2 L of 10 mM NaCl?
(0.1M) x V1 = (10 x 10-3M) x (0.2L)
V1 = 0.02L or 20 mL
Remember this!
Week 2 CHEM 1310 - Sections L and M 14
On Friday
Complete Chapter 3
– Chemical Equations
– How to balance chemical equations
– Calculations involving reactants and products
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