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
1 mol Na
400 g Na
= 17.40 mol Na
22.99 g 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 =
– N: 14.01 × 2 =
– O: 16.00 × 6 =
• Ca(NO3)2:
40.08
28.02
96.00
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
1 mol Ca(NO3)2
164.10 g Ca(NO3)2
= 0.609 mol Ca(NO3)2
Find Moles
• Activity XI
Reactions
Recipes and equivalents
Mole Equivalents
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
1 eq Ca(OH)2
1 eq HCl
1 eq CaCl2
1 eq H2O
= 1 mol
= 2 mol
= 1 mol
= 2 mol
Equivalent moles
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
If we use 1.80 mol Ca(OH)2, that is
(1.80 mol)(1 eq/1 mol) = 1.80 eq
1.80 eq HCl∙(2 mol/1 eq) = 3.60 mol HCl
1.80 eq CaCl2∙(1 mol/1 eq) = 1.80 mol CaCl2
1.80 eq H2O∙(2 mol/1 eq) = 3.60 mol H2O
Find Equivalent Moles
• Activity XII
Masses from Equivalent Moles
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
If we use 1.80 mol Ca(OH)2 = 133.37g
that is (1.80 mol)(1 eq/1 mol) = 1.80 eq
1.80 eq HCl
= 3.60 mol
1.80 eq CaCl2 = 1.80 mol
1.80 eq H2O
= 3.60 mol
= 131.26 g
= 199.77 g
= 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