Upload
godwin-daniels
View
227
Download
0
Embed Size (px)
Citation preview
Atomic MassAtomic Mass
Atoms are so small, it is difficult to weigh in grams (Use atomic mass units)Atomic mass is a weighted average of all masses for each isotope of an elementAre not whole numbers because they are averagesThese are the numbers found on the periodic table for each elementExample: page 80
Atoms are so small, it is difficult to weigh in grams (Use atomic mass units)Atomic mass is a weighted average of all masses for each isotope of an elementAre not whole numbers because they are averagesThese are the numbers found on the periodic table for each elementExample: page 80
The MoleThe MoleNumber used to count atomsIt is 6.022 X 1023 anything is one moleFound through the number of atoms in 12 grams of carbon-12Allows calculations/predictions for reactions to be madeExamples page 82, 83, 84
Number used to count atomsIt is 6.022 X 1023 anything is one moleFound through the number of atoms in 12 grams of carbon-12Allows calculations/predictions for reactions to be madeExamples page 82, 83, 84
Molar Mass (Molecular weight)
Molar Mass (Molecular weight)
Mass (grams) of one mole of a substanceFor a compound, add all the molar masses of the elements that make it upExamples
Mg3P2
Al2(Cr2O7)3
Ca(NO3)3
Mass (grams) of one mole of a substanceFor a compound, add all the molar masses of the elements that make it upExamples
Mg3P2
Al2(Cr2O7)3
Ca(NO3)3
Percent CompositionPercent Composition
Provides the contribution to a compound based on percentage To find
Have the mass of each elementAdd to total the mass of the
compoundDivide the mass of the element
by the mass of the compound Multiply by 100 (for a
percentage)
Provides the contribution to a compound based on percentage To find
Have the mass of each elementAdd to total the mass of the
compoundDivide the mass of the element
by the mass of the compound Multiply by 100 (for a
percentage)
Percent CompositionPercent Composition
Examples: Find the percent composition of each element in : Mg3P2, Al2(Cr2O7)3, and Ca(NO3)3
Mg: 54.1% P: 46.1%Al: 7.7% Cr: 44.5% O: 47.8%Ca: 17.8% N: 18.7% O: 63.6%
Examples: Find the percent composition of each element in : Mg3P2, Al2(Cr2O7)3, and Ca(NO3)3
Mg: 54.1% P: 46.1%Al: 7.7% Cr: 44.5% O: 47.8%Ca: 17.8% N: 18.7% O: 63.6%
Empirical formulasEmpirical formulasThe lowest ratio of atoms in a moleculeCan use percent compositions to determine empirical formulaBased on mole ratiosExamples
CH for C2H2 and C6H6
CH3 for C2H6
The lowest ratio of atoms in a moleculeCan use percent compositions to determine empirical formulaBased on mole ratiosExamples
CH for C2H2 and C6H6
CH3 for C2H6
Molecular FormulasMolecular Formulas
Molecular formula represents the number of each atom in a moleculeNeed molar mass to convert empirical formula to molecular formula
Uses a ratio of molar mass of empirical formula to known molar massRules for determining molecular formulas
1. Use mass percentages and molar mass to determine number of moles
Molecular formula represents the number of each atom in a moleculeNeed molar mass to convert empirical formula to molecular formula
Uses a ratio of molar mass of empirical formula to known molar massRules for determining molecular formulas
1. Use mass percentages and molar mass to determine number of moles
Molecular FormulasMolecular Formulas2. Determine moles of element
in one mole of compound3. The integers from step 2
are the subscripts for the elementsExample
Caffeine contains 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen and has a molar mass of 194.2 g/mol. What is the molecular formula for caffeine?
2. Determine moles of element in one mole of compound
3. The integers from step 2 are the subscripts for the elementsExample
Caffeine contains 49.48% carbon, 5.15% hydrogen, 28.87% nitrogen, and 16.49% oxygen and has a molar mass of 194.2 g/mol. What is the molecular formula for caffeine?
Caffeine AnswerCaffeine AnswerC8H10N4O2
http://en.wikipedia.org/wiki/File:Caffeine.svg
C8H10N4O2
http://en.wikipedia.org/wiki/File:Caffeine.svg
Chemical EquationsChemical Equations
Represents a reorganization of atoms in one or more substances (reactions)Reactant(s) --------> Product(s)Atoms must be equal on both sides of the equation (balancing equations)Cannot change formulas to balance equationsOccur as atoms, molecules, particles - not as grams!
Represents a reorganization of atoms in one or more substances (reactions)Reactant(s) --------> Product(s)Atoms must be equal on both sides of the equation (balancing equations)Cannot change formulas to balance equationsOccur as atoms, molecules, particles - not as grams!
Equation PracticeEquation PracticeCa(OH)2 + H3PO4 -------> H2O +
Ca3(PO4)2
Cr + S8 ------> Cr2S3
KClO3(s) -------->Cl2(g) + O2(g)
Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas.
Fe2O3(s) + Al(s) --------> Fe(s) +
Al2O3(s)
Ca(OH)2 + H3PO4 -------> H2O +
Ca3(PO4)2
Cr + S8 ------> Cr2S3
KClO3(s) -------->Cl2(g) + O2(g)
Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas.
Fe2O3(s) + Al(s) --------> Fe(s) +
Al2O3(s)
StoichiometryStoichiometry
Method of calculating amounts of reactants needed or products producedUses mole ratios (moles to moles) and molar mass (grams/mole) to perform calculationsFirst step is always to balance the equation!!How much carbon dioxide is absorbed by 1 kg of lithium hydroxide LiOH (s) + CO2 (g) -----> Li2CO3 (s) + H2O(l)
Answer is 920 grams (page 106)
Method of calculating amounts of reactants needed or products producedUses mole ratios (moles to moles) and molar mass (grams/mole) to perform calculationsFirst step is always to balance the equation!!How much carbon dioxide is absorbed by 1 kg of lithium hydroxide LiOH (s) + CO2 (g) -----> Li2CO3 (s) + H2O(l)
Answer is 920 grams (page 106)
Limiting Reactants (Reagents)
Limiting Reactants (Reagents)
The reactant controlling the amount of product made in a reactionCan use stoichiometric calculations to determine the reactant limiting the productsUsed to determine the theoretical yieldExample: N2 + H2 ------> NH3
What mass of ammonia can be produced from 100 grams of N and 500 grams of H? What is the limiting reactant? How much unreacted materials remain?
The reactant controlling the amount of product made in a reactionCan use stoichiometric calculations to determine the reactant limiting the productsUsed to determine the theoretical yieldExample: N2 + H2 ------> NH3
What mass of ammonia can be produced from 100 grams of N and 500 grams of H? What is the limiting reactant? How much unreacted materials remain?
Percent YieldPercent YieldComparison of the actual amount produced to the theoretical yieldRecorded as a percentagePercent yield = actual yield X 100%
theoretical yieldExample: Aluminum burns in bromine producing aluminum bromide. In a lab, 6.0 g of aluminum react with excess bromine. 50.3 g of aluminum bromide are produced. What are the values of the three types of yield?
Comparison of the actual amount produced to the theoretical yieldRecorded as a percentagePercent yield = actual yield X 100%
theoretical yieldExample: Aluminum burns in bromine producing aluminum bromide. In a lab, 6.0 g of aluminum react with excess bromine. 50.3 g of aluminum bromide are produced. What are the values of the three types of yield?