Transcript Document 7477111

Chapter Calculations with Chemical
Formulas and Equations
Chemistry 1061: Principles of Chemistry I
Andy Aspaas, Instructor
Molecular weight and formula weight
• Molecular weight: sum of atomic weights for all
atoms in a substance
– MW of H2O = 2(1.008 amu) + (16.00 amu) =
18.016 amu
• Formula weight: same as above, but for any
compound, molecule or not
– FW of NaCl = (22.99 amu) + (35.45 amu) = 58.44
amu
Mass and moles
• Mole: quantity of a substance that contains as many
molecules or formula units as the number of atoms
in exactly 12 grams of carbon-12
– That number = Avogadro’s number, 6.02 x 1023
– Simply a quantity like pair, dozen, gross
• Molar mass: mass of 1 mole of a substance
– Equal to formula weight of that substance!
– One mole of oxygen has a mass of 16.00 g
– One mole of H2O has a mass of 18.016 g
Mass and moles
• Mole calculations
– Use atomic, molecular, or formula weight as a
conversion factor in dimensional analysis to
convert between mass and moles, and vice-versa
– How many moles in 100.0 g Fe? 100g of H2O?
– What is the mass of 4.72 moles of Mg? 3.25
moles of glucose (C6H12O6)?
Molecules and moles
• Use Avogadro’s number as a conversion factor to
convert between number of particles and moles, and
vice versa
• 6.02 x 1023 particles = 1 mole
• Combine the two methods to convert mass to
number of particles.
Determining mass percentage from a formula
• Mass percentage: mass of certain constituent divided by
mass of whole, expressed as percentage
• Mass % from formula:
– Assume you have 1 mol of the substance, what mass do
you have?
– Use formula to determine number of moles of each
element present
– Convert moles of element to mass
– Divide mass of element by mass of compound and
multiply by 100%
Elemental analysis
• Determines amount of C, H, and O in a compound
that contains only those elements
• Compound is burned in an oven, amounts of CO2
and H2O given off are recorded
• Since 1 mol of CO2 contains 1 mol C atoms, the
mass of carbon in the substance can be found
• 1 mol H2O contains 2 mol H atoms - mass of H can
be found
• Mass % of C and H found
• O mass % is what’s left over
Determining empirical formulas from % composition
• Empirical formula: simplest formula of a substance that has
all integer whole-number subscripts
• If given % compositions of elements, assume you have 100.0
g of the substance
– 14.0% of an element becomes 14.0 g of that element
• Convert grams to moles for each element
• Divide all mole numbers by the smallest one
• If not all answers are integers (within experimental error),
multiply to make them all the smallest possible integers
Molecular formula from empirical formula
• Molecular formula is some multiple of the empirical
(simplest) formula
– Empirical formulas of acetylene (C2H2) and
benzene (C6H6) are both CH
• Molecular weight = n x empirical formula weight
• n = molecular weight / empirical formula weight
• Multiply all subscripts in empirical formula by n
Stoiciometry
• Chemical equations can be interpreted as either
individual molecules or numbers of moles
• N2(g) + 3H2(g) ---> 2NH3(g)
– 1 N2 molecule reacts with 3 H2 molecules to
make 2 NH3 molecules
– Or… 1 mol N2 reacts with 3 mol H2 to make 2 mol
NH3
• Use coefficients as conversion factors when
converting moles of any constituent in a chemical
reaction to moles of any other constituent
Masses in chemical equations
• How many grams of a product will be yielded from a
given mass of reactant?
• Do same process as before, but convert masses to
moles first, and moles back to masses
– Mass NH3 --> mol NH3 --> mol H2 --> mass H2
Limiting reactant
• Often times, one reactant will be in excess of the
other
• Number of moles of product is determined by
starting moles of limiting reactant
• Determine number of moles of each reactant, and
calculate how many moles of product each would
make
• The reactant that makes the least amount of moles
of product is the limiting reactant
Theoretical yield
• Theoretical yield: maximum amount of product that
can be obtained by a reaction from given amounts
of reactants
• In actuality, less product is often obtained
– Losses due to side-reactions
– Losses due to incomplete separations
• Percentage yield
= (actual yield / theoretical yield) x 100%