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