formulas, reactions, and amounts
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Formulas, Reactions, and Amounts. PHYS 1090 Unit 10. Why?. Fine structure of matter not obvious Formulas force us to mind atomic composition Materials react in definite proportions Simple ratios emphasized in formulas Explain quantitative results Understand the measurements. - PowerPoint PPT PresentationTRANSCRIPT
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Formulas, Reactions, and Amounts
PHYS 1090 Unit 10
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Why?
• Fine structure of matter not obvious– Formulas force us to mind atomic composition
• Materials react in definite proportions– Simple ratios emphasized in formulas
• Explain quantitative results– Understand the measurements
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Molecular Formulas
• Molecule: group of connected atoms of definite composition
• Formula: tells how many atoms of each element per molecule– Often more information is necessary to
unambiguously specify molecule
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Formulas
• Elements represented by symbols (One capital letter or one cap + one lowercase)– H, Li, Na, C, N, etc.
• Subscripts tell how many atoms of each– No subscript means “1”
LiBr: 1 Li+ + 1 Br−
SrF2: 1 Sr+2 + 2 F−
• Can subscript groups, e.g. B(OH)3
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Count Atoms
• Activity I
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Ionic Compounds
• Ion: electrically-charged object
• Ionic compound: composed of ions, each containing one or more atoms, connected only by electrostatic attraction– Charges balance to zero
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Charges of Ions
• Many atoms have one preferred charge– Na+, Ca+2, Br−
• Charges specified for others– Iron(II), lead(IV)
• Ions can be groups of atoms– CO3
−2, ClO4−
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Ionic Compound Formulas
• Formula unit: fewest positive + negative ions to balance charge
Li+ + Br−: 1 Li+ + 1 Br−
Sr+2 + F−: 1 Sr+2 + 2 F−
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Balance Charges
• Activity II
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Identify and Balance
• Activity III
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Reaction Equations
Reactants → Products
• “+” btw different reactants and products
• Coefficients: how many formula units of each species– No coefficient means “1”
2 C + O2 → 2 CO
• Conservation of atoms from reactants to products
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Counting Atoms
• Multiply number in each formula unit by coefficient
• Add together atoms of each type in all reactants• Add together atoms of each type in all products• These are the same in a balanced equation
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Count Atoms
• Activity IV
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Balance Equations
• Adjust coefficients to balance equations
• Activities V, VI
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Moles
How many?
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Mole
• A counting unit: 6.02 × 1023 items– Abbreviation “mol” (save one whole letter!)
– 6.02 × 1023 = “Avogadro’s number” = NA
• Compare to – dozen– pair– gross– score
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Avogadro’s Number
• Why 6.02 × 1023?
• NA of Carbon-12 atoms has a mass of exactly 12 g.
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Atomic Mass
• A sample of 1 mol of atoms of an element has a mass in grams equal to its atomic mass– More correctly “molar mass of the element”
• (Unstated) units = g/mol
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Finding Atomic Mass
• On the periodic table– After the atomic number
• It’s that number that I warned you is not the mass number– Now you know what it’s for
• Depends on isotopic abundances– Generally similar for different sources
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Find Masses
• Activity VII– Mass of 1 mole
• Activity VIII– Mass of arbitrary numbers of moles– Multiply atomic mass by moles– E.g. (2.0 mol B)(10.81 g B /mol B) = 21.62 g B
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Finding Moles
• Divide sample mass by molar mass• E.g. (400 g Na) / (22.99 g Na/mol Na) = 17.40 mol Na
• Or think of it as
400 g Na = 17.40 mol Na22.99 g Na1 mol Na
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Find Moles
• Activity IX
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Formula Mass
• Mass in grams of a mole of formula units– Mass of a mole of molecules
• Molar mass of compound– “molecular mass”– “molecular weight”– “formula weight”
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Finding Formula Mass
• Multiply each element’s molar mass by its number in the formula unit
• Add products together
• Example: Ca(NO3)2
– Ca: 40.08 × 1 = 40.08– N: 14.01 × 2 = 28.02– O: 16.00 × 6 = 96.00
• Ca(NO3)2: 164.10 g/mol
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Find Formula Masses
• Activity X
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Other Way Around
• Given mass, how many moles are there?
• Divide sample mass by molar mass – Just like atomic masses
• Example: 100 g Ca(NO3)2
100 g Ca(NO3)2 = 0.609 mol Ca(NO3)2164.10 g Ca(NO3)2
1 mol Ca(NO3)2
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Find Moles
• Activity XI
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Reactions
Recipes and equivalents
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Mole Equivalents
1 eq Ca(OH)2 = 1 mol
1 eq HCl = 2 mol
1 eq CaCl2 = 1 mol
1 eq H2O = 2 mol
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
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Equivalent moles
If we use 1.80 mol Ca(OH)2, that is (1.80 mol)(1 eq/1 mol) = 1.80 eq
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
1.80 eq HCl∙(2 mol/1 eq) = 3.60 mol HCl
1.80 eq CaCl2∙(1 mol/1 eq) = 1.80 mol CaCl21.80 eq H2O∙(2 mol/1 eq) = 3.60 mol H2O
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Find Equivalent Moles
• Activity XII
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Masses from Equivalent Moles
If we use 1.80 mol Ca(OH)2 = 133.37gthat is (1.80 mol)(1 eq/1 mol) = 1.80 eq
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
1.80 eq HCl = 3.60 mol = 131.26 g
1.80 eq CaCl2 = 1.80 mol = 199.77 g
1.80 eq H2O = 3.60 mol = 64.85 g
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Find Masses from Equivalents
• Activity XIII
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Equivalent Masses
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
1 mol Ca(OH)2 74.096 g/mol 74.096 g
2 mol HCl 36.458 g/mol 72.916 g
1 mol CaCl2 110.98 g/mol 110.98 g
2 mol H2O 18.016 g/mol 36.032 g
• 74.096 + 72.916 = 147.012• 110.98 + 36.032 = 147.012
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Finding Equivalent Masses
• Find moles of all reactants and products
• Convert to masses
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Finding Equivalent Masses
• Example: – Ca(OH)2 + 2HCl → CaCl2 + 2H2O (previous)
– 10 g Ca(OH)2
• N mol Ca(OH)2 = 10 g/74.096 g/mol = 0.13496 mol
– 2N mol HCl = 0.26992 mol = 9.84 g
– N mol CaCl2 = 0.13496 mol = 14.98 g
– 2N mol H2O = 0.26992 mol = 4.86 g
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Find Equivalent Masses
• Activity XIV
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Limiting Reagents
• Reactants may not be present in equivalent amounts!
• The one with the fewest equivalents limits the outcome.
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Limiting Reagents
Example:
Mg(OH)2 + 2HCl → MgCl2 + 2H2O; 50 g Mg(OH)2 + 50 g HCl
– Mg(OH)2: 58.326 g/mol 50 g = 0.857 mol = 0.857 eq
– HCl: 36.458 g/mol 50 g = 1.371 mol = 0.686 eq
• HCl is limiting– Mg(OH)2: use 0.686 mol × 58.326 g/mol = 40.01 g
– MgCl2: make 0.686 mol × 95.21 g/mol = 65.31 g
– H2O: make 1.371 mol × 18.016 g/mol = 24.70 g
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Find the Limiting Reagent and Yields
• Activity XV