Download CH. 3 - STOICHIOMETRY: CHEMICAL CALCULATIONS I. Molecular

CH. 3 - STOICHIOMETRY: CHEMICAL CALCULATIONS
I. Molecular Masses and Formula Masses
A. molecular mass - sum of masses of atoms represented in a molecular formula
B. formula mass - sum of masses of atoms or ions present in a formula unit
II. The Mole and Avogadro’s Number
A. mole (mol) - amount of substance that contains as many elementary entities
as there are atoms in exactly 12g of the carbon-12 isotope
B. Avogadro’s number (NA) - 6.022 x 1023
C. molar mass - mass of 1 mol of that substance
D. mass percent composition - proportions of constituent elements in a
compound as the number of grams of each element per 100g of the compound
E. molar amounts cannot be measured directly, they must be related to
quantities that we can measure
Calculate the mass percent of each element in ammonium nitrate, NH4NO3.
1. Determine the molar mass of ammonium nitrate.
(2 x 14.01) + (4 x 1.008) + (3 x 16.00) = 80.05g/mol NH4NO3
2. For one mole of compound, determine mass ratios and percentages.
%N =
28.02g N__ x 100 = 35.00% N
80.05g NH4NO3
%H =
4.032g H
x 100 = 5.04% H
80.05g NH4NO3
%O =
48.00g O x 100 = 59.96% O
80.05g NH4NO3
Find the empirical formula for butane, C4H10.
1. Convert the percent of each element to a mass.
(4 x 12.011) + (10 x 1.00794) = 58.123g C4H10
%C =
48.044g C x 100 = 82.66% = 82.66g C
58.123g C4H10
%H = 10.0794g H x 100 = 17.341% = 17.34g H
58.123g C4H10
2. Convert the mass of each element to an amount in moles.
? mol C = 82.66g C x 1mol C_ = 6.882mol C
12.011g C
? mol H = 17.34g H x 1mol H = 17.20mol H
1.0079g H
(cont.)
(cont.)
3. Use the numbers of moles of elements as subscripts in tentative formula.
C6.882H17.20
4. Attempt to get integers as subscripts by dividing each of the subscripts in step 3 by the
smallest subscript.
C6.882H17.20  CH2.500
6.882
6.882
5. If any subscripts obtained after step 4 are fractional quantities, multiply each of the
subscripts by the smallest integer that will convert all the subscripts to integers. The result is
an empirical formula.
2.500 = 2/5
C2 x 1H2 x 2.500 = C2H5
III. Stoichiometric Equivalence and Reaction Stoichiometry
A. stoichiometric coefficients - numbers placed in front of formulas to balance
the equation, thereby indicating the combining ratios of reactants to products
B. stoichiometric factors (mole ratios) - conversion factors formed from
stoichiometric coefficients in a chemical equation
When 0.105mol propane is burned in excess of oxygen, how many moles of oxygen are
consumed?
C3H8 + 5O2  2CO2 + 4H2O
1. Write the stoichiometric equivalence.
1mol C3H8  5mol O2
2. Derive the conversion factors from the equivalence.
5mol O2
and 1mol C3H8
1mol C3H8
5mol O2
3. Find the number of moles of oxygen.
? mol O2 = 0.105mol C3H8 x 5mol O2 = 0.525mol O2
1mol C3H8
IV. Limiting Reactants
A. limiting reactant (reagent) - reactant that is completely consumed in a
reaction and limit’s the amount of product formed
Magnesium nitride can be formed by the reaction of magnesium metal with nitrogen
gas. (a) How many grams of magnesium nitride can be made through the reaction of
35.00g of magnesium and 15.00g of nitrogen? (b) How many grams of the excess
reactant remain after the reaction?
(a)
1. Write the balanced equation.
3Mg + N2  Mg3N2
2. Find the limiting reactant.
Assuming that Mg is the limiting reactant:
? mol Mg3N2 = 35.00g Mg x 1mol Mg x 1 mol Mg3N2 = 0.4800mol Mg3N2
24.305g Mg
3mol Mg
Assuming that N2 is the limiting reactant:
? mol Mg3N2 = 15.00g N2 x 1mol N2 x 1mol Mg3N2 = 0.5355mol Mg3N2
28.013g N2
1mol N2
0.4800mol < 0.5355mol, so magnesium is the limiting reactant.
3. Find the mass of Mg3N2 produced when 0.4800mol Mg3N2 has been formed.
? g Mg3N2 = 0.4800mol Mg3N2 x 100.93g Mg3N2 = 48.45g Mg3N2
1mol Mg3N2
(b)
1. Calculate how much N2 must have been consumed by 0.4800mol Mg3N2.
? g N2 = 0.4800mol Mg3N2 x 1mol N2 x 28.013g N2 = 13.45g N2
1mol Mg3N2
1mol N2
2. Find the mass of N2 present in excess.
15.00g N2 (initially) - 13.45g N2 (consumed) = 1.55g N2 (excess)
V. Yields of Chemical Reactions
A. theoretical yield - calculated/expected quantity of product
B. actual yield - measured/actual quantity of product
C. percent yield - ratio of actual yield to theoretical times 100
What mass of acetic acid is needed to prepare 252g ethyl acetate if the expected
percent yield is 85.0%? Assume that the other reactant, ethanol, is present in excess. The
equation for the reaction, carried out in the presence of H 2SO4, is
CH3COOH + HOCH2CH3  CH3COOCH2CH3 + H2O.
(ACETIC ACID)
(ETHANOL)
(ETHYL ACETATE)
1. Calculate the theoretical yield.
Percent yield = actual yield
x 100%
Theoretical yield
Theoretical yield = actual yield x 100%
Percent yield
= 252g ethyl acetate x 100% = 296g ethyl acetate
85.0%
2. Determine the mass of acetic acid required to produce 296g ethyl acetate.
296g CH3COOCH2CH3 x 1mol CH3COOCH2CH3 x
1mol CH3COOH
x
88.11g CH3COOCH2CH3 1mol CH3COOCH2CH3
x 60.05g CH3COOH = 202g CH3COOH
1mol CH3COOH
VI. Molar concentration
A. substances enter into chemical reactions according to their molar ratios
B. volumes of solutions are more convenient to measure than their masses
C. molarity/molar concentration (M) - amount of solute per liter of solution
What is the molarity of a solution in which 333g potassium hydrogen carbonate is
dissolved in enough water to make 10.0L of solution?
1. Convert mass to number of moles of KHCO3.
333g KHCO3 x 1mol KHCO3_
100.1g KHCO3
2. Use the expression as the numerator and solution volume, 10.0L, as the denominator.
333g KHCO3 x 1mol KHCO3_
Molarity =
100.1g KHCO3 = 0.333M KHCO3
10.0L solution
VII. Dilution
A. dilution - process of preparing a more dilute solution by adding solvent
B. addition of solvent does not change the amount of solute in a solution but
does change the solution concentration
How many milliliters of a 2.00M CuSO4 stock solution are needed to prepare 0.250L of
0.400M CuSO4?
1. ? mL = 0.250L x 0.400mol CuSO4 x
1L
x 1000mL = 50.0mL
1L
2.00mol CuSO4
1L