Formulas, Reactions, and Amounts PHYS 1090 Unit 10.
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Transcript of Formulas, Reactions, and Amounts PHYS 1090 Unit 10.
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
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
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
Count Atoms
• Activity I
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
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−
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−
Balance Charges
• Activity II
Identify and Balance
• Activity III
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
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
Count Atoms
• Activity IV
Balance Equations
• Adjust coefficients to balance equations
• Activities V, VI
Moles
How many?
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
Avogadro’s Number
• Why 6.02 × 1023?
• NA of Carbon-12 atoms has a mass of exactly 12 g.
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
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
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
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
Find Moles
• Activity IX
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”
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
Find Formula Masses
• Activity X
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
Find Moles
• Activity XI
Reactions
Recipes and equivalents
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
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
Find Equivalent Moles
• Activity XII
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
Find Masses from Equivalents
• Activity XIII
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
Finding Equivalent Masses
• Find moles of all reactants and products
• Convert to masses
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
Find Equivalent Masses
• Activity XIV
Limiting Reagents
• Reactants may not be present in equivalent amounts!
• The one with the fewest equivalents limits the outcome.
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
Find the Limiting Reagent and Yields
• Activity XV