Ch 3 Calculations (Chemical Formulas and Eqns)
Molar Masses
- Molecular Mass or Weight (MM or MW) is the sum of
atomic weights for all atoms in a molecular compound.
For example, MM for H2O is (2 × 1.01) + 16.00 = 18.02 amu
- Formula Mass or Weight (FM or FW) is similar, but the sum
uses the formula unit of a non-molecular (ionic) compound.
For example, FM for NaCl is 23.0 + 35.5 = 58.5 amu
- A mole is the quantity of atoms in exactly 12 g of 12C.
This is Avogadro’s number: NA = 6.022 × 1023
-
amu
g
molecules
(or
mole
).
This quantity represents a mole for all atoms, molecules, and formula units.
NA allows us to convert between amu and g so that
g
amu
Molar Mass in (mole) = Molecular Mass in (atom)
Ex 3.3 Mass of a Cl atom (using NA as a conversion factor)
g
mole
atoms
6.022 x 1023
mole
35.5
= 5.90 × 10-23 g/atom
Ex 3.4 Converting moles into grams
- For ZnI2, the FM is 65.4 + 2(126.9) = 319.2 g/mol
- The mass of ZnI2 is 319.2 g/mol × 0.0654 mol = 20.9 g
Ex 3.5 Converting grams into moles
- For PbCrO4, the FM is 207.2 + 52.0 + 4(16.0) = 323.2 g/mol
-
The number of moles is
45.6 g
323.2
g
mol
= 0.141 mol
Mass Percentages in a Compound
-
Mass % of a component = [
mass of component in compound
total mass of compound
] × 100%
Ex 3.7 Finding Mass %’s
- For formaldehyde, CH2O, the MM is 12.0 + 2(1.01) + 16.0 = 30.0 g/mol
- The component masses in one mole of CH2O are:
C = 12.0 g
H = 2 × 1.01 = 2.02 g
O = 16.0 g
- The mass %’s are:
%C=
12.0
30.0
= 40.0 %
%H=
2.02
30.0
= 6.73 %
%O=
16.0
30.0
= 53.3 %
Ex 3.8 Finding mass of C in a sample of CH2O
- (83.5 g CH2O) × (40.0 % by mass C in CH2O) = 33.4 g C
Total Carbon / Hydrogen Measurement (Fig 3.6)
m
-
CnHmOq is combusted with O2 to create nCO2 and ( 2 )H2O
-
Masses of CO2 and H2O created are converted into masses of C and H,
by using mass %’s, similar to examples 3.7 and 3.8.
After subtracting masses of C and H from the sample mass,
the remaining mass, if any, in the sample is due to O.
See example 3.9.
-
Empirical Formula
- Written with the smallest possible set of integer subscripts.
- Same as formula unit for most ionic compounds (except Hg22+).
- But is not necessarily same as formula for molecular compounds.
- C2H2 (acetylene) and C6H6 (benzene) both have the same empirical formula: CH.
Ex 3.10 Finding the Empirical Formula from the Element Masses
- 1.587 g of a nitrogen oxide has 0.483 g N and 1.104 g O
-
0.483 g N
= 0.0345 mol N
g
mol
14.0
1.104 g O
= 0.06900 mol O
g
mol
16.00
moles O
moles N
=
0.06900
0.0345
= 2.00
The empirical formula is NO2. So, the compound can be NO2 or N2O4.
Ex 3.11 Finding the Empirical Formula from Mass Percentages
- 100.0 g sodium dichromate has 17.4 g Na, 39.7 g Cr, and 42.8 g O
1 mol
(17.5 g)(
-
(39.7 g)(
) = 0.763 mol Cr
-
(42.8 g)(
) = 2.68 mol O
-
Double these numbers to get all integer values: Na2Cr2O7
23.0 g
1 mol
52.0 g
1 mol
16.0 g
) = 0.761 mol Na
0.761
-
0.761
0.763
0.761
2.68
0.761
= 1.000
= 1.003
= 3.52
Relating empirical formula mass to molecular mass
- For molecular compounds, MM is an integer multiple of empirical formula mass
- n =
molecular mass
empirical formula mass
3.12 Finding Molecular Mass from Empirical Formula
- Acetic acid mass %’s are 39.9 % C, 6.7 % H, and 53.4 % O.
- Find moles of C, H, and O in 100 g of acetic acid as follows:
C=
H=
O=
-
39.9 g
g
12.0 mol
6.7 g
g
1.01 mol
53.4 g
16.0
g
mol
= 3.33 moles
= 6.6 moles
= 3.34 moles
Divide the results by 3.33 to get empirical formula, which is CH2O.
Empirical formula mass = 12.0 + 2(1.01) + 16.0 = 30.0 g
Acetic acid MM = 60.0 g/mol = (n)(30.0 g) and n = 2.
Actual Formula is (CH2O)2 = C2H4O2 (or CH3COOH).
Stoichiometry
- Stoichiometry is the calculation of quantities of reactants and products
involved in a chemical reaction.
- The reaction equation can use either molecules or moles as units.
The ratios for the substances are the same, regardless of which units are used.
1 molecule N2 + 3 molecules H2  2 molecules NH3
- If the equation is in moles, then we can convert the moles into masses.
(1 mole N2) × (28.0 g/mol) =
28.0 g N2
(3 moles H2) × (2.02 g/mol) =
6.06 g H2
(2 moles NH3) × (17.0 g/mol) =
34.0 g NH3
- Moles are proportional to stoichiometric coefficients. So, we can use the coefficients to
create a conversion factor, and then find the amount of H2 needed to create 3.2 mol NH3.
(3.2 mol NH3)(
3 mol H2
2 mol NH3
-
) = 4.8 moles H2
We can find the mass of H2 needed to create 1 ton of NH3 by using the same
conversion factor along with conversions between moles and masses.
454 g
1 mol NH3
First, convert to moles of NH3:
(2000 lb)(
)(
1 lb
Then, find the moles of H2:
(5.34×104 moles NH3)(
Finally, convert to mass of H2:
(8.01×104 moles H2)(
)=
17.0 g
3 mol H2
2 mol NH3
2.02 g H2
mol
5.34 × 104 moles NH3
) = 8.01 × 104 moles H2
)=
1.62 × 105 g H2
Ex 3.13 Determining the Mass of a Reactant
Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g)
-
Convert mass of product into moles:
(1000 g Fe2O3) (
-
Convert into amount of reactant: (6.25 moles Fe2O3) (
1 mol
160 g
2 mol Fe
1 mol Fe2 O3
) = 6.25 moles Fe2O3
55.85 g
) ( 1 mol Fe) = 698 g Fe
Limiting Reactant
- A substance that will be entirely consumed by a complete reaction is limiting.
- A substance that cannot be entirely consumed by a complete reaction is present in excess.
- The maximum moles of product possible are determined by the
starting moles of the limiting reactant. See Figures 3.14 and 3.15.
Ex 3.15 Finding the Amount of Product Possible from the Limiting Reactant
- Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
- We have 0.30 mol Zn and 0.52 mol HCl.
1 mol H2
-
(0.30 mol Zn) (
-
(0.52 mol HCl) (
-
HCl is the limiting reactant because it makes less H2 than Zn.
Only 0.26 moles of H2 are produced when the reaction is complete.
Once 0.26 moles of H2 are created, there will be no HCl left, and the reaction stops.
Mass of H2 produced in a complete reaction is (0.26 mol H2)(2.02 g/mol) = 0.525 g
-
) = 0.30 mol H2
1 mol Zn
1 mol H2
2 mol HCl
)= 0.26 mol H2
Theoretical Yield
- Maximum amount of product that can be obtained from a complete reaction.
- The amount produced when the limiting reactant is completely consumed.
- In Ex 3.15 above, the theoretical yield of H2 is 0.26 moles.
Actual Yield
- The actual yield is usually less than the theoretical yield for several reasons.
∙ Products or reactants can escape or be lost.
∙ Other competing reactions can deplete reactants.
∙ Reactions typically stop before 100% completion.
Actual Yield Percentage
(Actual Yield)
-
Actual Yield % = [
-
Suppose the theoretical yield is 27.3 g, and 23.8 g are actually created.
(Theoretical Yield)
23.8 g
Then, the actual yield % = (
27.3 g
-
] x 100%
) × 100% = 87.2%.
The same value will be found if the units are in moles instead of g (because units cancel).