Chemistry I-Honors
Class notes
Mole Concept / Molarity / Empirical & Molecular Formulas
Mole Concept
-
three basic quantities – molar mass, molar volume, number of particles
-
mass units on the periodic table are actually dimensionless
– listed as amu (atomic mass units)
– most commonly, if the “mole” being used is actually a “gram-mole”, then the mass on the periodic
table is in units of grams. The quantity for that mass is 6.02 x 1023 atoms (known as Avogadro’s #).
-
molar volume has a defined temperature and pressure, known as STP – stands for standard temperature
and pressure – defined as 0oC or 273 K) and 1.0 atmosphere. The molar volume at STP = 22.4 liters.
-
the number of particles is based on Avogadro’s number – may be any type of particle – atoms, ions,
people, eyeballs, etc.
-
For most mole concept problems, you need to go through three steps:
#1. Convert the given quantity (whatever it is) into moles
#2. Use the mole ratio between the substance given and the desired substance
#3. Get out of moles and into the desired unit.
molar volume
moles
# of particles
molecular mass
-2Example problems:
1. What is the mass of a 256.0-ml sample of carbon dioxide gas at STP conditions?
0.2560 liters
x
1 mole
----------22.4 liters
x
44.0 grams
------------1 mole
= 0.503 grams
2. How many carbon atoms are in the sample described in #1?
0.2560 liters
x
1 mole
----------22.4 liters
x
1 mole C
------------x
1 mole CO2
6.02 x 1023 atoms of C
----------------= 6.88 x 1021 atoms
1 mole C
3. How many atoms total are there in the gas sample of #1?
0.2560 liters
x
1 mole
----------22.4 liters
x
3 moles atoms
------------x
1 mole CO2
6.02 x 1023 atoms
----------------1 mole of atoms
= 2.06 x 1022 atoms
4. What is the mass, in grams, of a single molecule of carbon dioxide gas?
1 molecule of CO2
1 mole CO2
44.0 grams
x -----------------------x --------------6.02 x 1023 molecules
1 mole CO2
= 7.30 x 10-23 grams
5. If a sample of cupric sulfate pentahydrate has 8.85 x 1025 atoms of oxygen in it, what is the mass (in kg)
of this sample? CuSO4 . 5 H2O Molecular Wt. = 249.6 g/mole
8.85 x 1025 atoms of O
1 mole of O
1 mole of cpd.
249.6 grams
x ------------------------- x ----------------- x --------------- = 4.08 kilograms
6.02 x 1023 atoms of O
9 moles of O
1 mole of cpd.
6. What is the percentage of metal in a 5.00-gram sample of potassium dichromate? K2Cr2O7
-
note: the 5.00-gram sample does not matter; the percentage of metal in any size sample is the same.
% metal =
2(39.1 g/mole) + 2(52.0 g/mole)
------------------------------------------------------------- x 100 = 61.93 %
2(39.1 g/mole) + 2(52.0 g/mole) + 7(16.0 g/mole)
-3Molarity
-
defined as the moles of solute per liter of solution; use a capital “M” or [ ] for molarity;
moles of solute may be the entire compound or only one of the ions
-
for the most part, in dilute aqueous solutions, the volume of water used to dissolve the solute and the
volume of the solution are the nearly the same
the denominator must be in liters
-
When a concentrated solution is diluted with water, the number of moles of solute does not change, but
the ratio of moles of solute to volume of solution does change.
For dilution problems, the formula M1V1 = M2V2 can be used.
-
Using the density of a solution and its percent by weight composition, the molarity can be found.
Example problems:
1. What is the molarity of a solution if 44.0 grams of solid sodium carbonate is added to enough water to
make 250.0 ml of solution?
44.0 grams / (106.0 g/mole)
M = -------------------------------- = 1.66 M = [Na2CO3]
0.2500 liter
2. What is the [CO3-2] in the solution? What is the [Na+1] in the solution?
1.66 M = [Na2CO3]; then [CO3-2] = 1.66 M;
[Na+1] = 3.32 M
3. If the above solution has 450.0 ml of water added to it, what is the molarity of the final, diluted
solution?
V2 = 250.0 ml + 450.0 ml = 0.7000 liters; M1V1 = M2V2
1.66 M x .2500 liter
M2 = ------------------------- = 0.593 M
0.7000 liter
4. If an aqueous solution of hydrochloric acid is known to have a density of 1.15 grams per milliliter, and
is known to contain 28.5% by weight HCl, what is the molarity of the solution?
- the best way to start this problem is to assume that you have 1.00 liter of solution.
1150 grams soln.
-----------------------1 liter solution
x
28.5 grams HCl
1 mole HCl
------------------ x ------------------- = 8.98 M = [HCl]
100 grams soln.
36.5 grams HCl
-4Empirical & Molecular Formulas
-
the empirical formula of a substance is the simplest whole-number mole ratio of the elements that
make up the atom
the molecular formula is the actual number of elements that make up a compound – it will always be a
whole number multiple of the empirical formula
for example, the simple sugar – glucose – has a molecular formula of C6H12O6. Its empirical formula is
CH2O.
a special group of compounds called “hydrates” have a definite number of water molecules attached to
the “anhydrous” salt.
Analysis of the compounds may in several different units – grams, percentages, moles, number of
particles, etc. In all cases, the quantity needs to be converted into moles.
Once the number of moles of each substance is determined, divide all moles by the smallest mole
quantity. If fractions are produced, then multiply by a factor that will give whole numbers.
If percentages are given, assume that you have 100 grams; then the percentage will be equivalent to
grams.
Example Problems:
1.
A compound of boron and hydrogen contains 18.9% hydrogen and 81.1% boron. What is the
empirical formula of this compound?
for H: 18.9 g / 1.0 g/mole = 18.9 moles
2.
for B: 81.1 g / 10.8 g/mole = 7.51 moles
(18.9 moles H) / (7.51 moles B) = 2.52
(7.51 moles B) / (7.51 moles B) = 1.00
mole ratio is 2.5 : 1  x 2 = 5.0 : 2
 B2H5
A certain compound used for fertilizer analyzes 5.0% hydrogen, 35.0% nitrogen, and 60.0% oxygen.
What is its empirical formula?
for H: 5.0 g / 1.0 g/mole = 5.0 moles of hydrogen
for N: 35.0 g / 14.0 g/mole = 2.50 moles of nitrogen
for O: 60.0 g / 16.0 g/mole = 3.75 moles of oxygen
divide all three mole quantities by 2.50 
multiply all three quantities by 2 
H : N : O = 2.0 : 1.0 : 1.50
H:N:O = 4:2:3
Formula = H4N2O3
3.
If the compound in #2 is known to have a molecular weigh of 160.0 g/mole, what is the true, or
molecular, formula?
Molecular weight of the empirical formula = 80.0 g/mole
160.0 g/mole / 80.0 g/mole = 2; then true formula = (H4N2O3) x 2 = H8N4O6
-5-
4.
When a hydrate of calcium sulfate is heated, all of the water is driven off. If 34.0 grams of the
anhydrous salt remains when 43.0 grams of the hydrate is heated, what is the empirical formula of the
hydrate?
43.0 g – 34.0 g = 9.0 grams = mass of water; then 9.0 grams / 18.0 g/mole = 0.50 moles water
34.0 g CaSO4 / 136.2 g/mole = 0.250 moles
0.50 moles / 0.250 moles = 2; then the formula is CaSO4 . 2 H2O
5.
The complete combustion of a hydrocarbon (compound composed only of carbon & hydrogen) in
oxygen produced 176 grams of CO2 and 90.0 grams of H2O. What is the empirical formula of this
hydrocarbon?
CxHy + O2 
CO2 + H2O
176 grams CO2 / 44.0 g/mole = 4.00 moles of CO2 = 4.00 moles of carbon
90.0 grams H2O / 18.0 g/mole = 5.00 moles of H2O = 10.00 moles of hydrogen
C : H = 4.00 : 10.00  2 :5  C2H5
6.
If the compound in #5 is know to be a gas at STP conditions, and a 2.50-gram sample is known to
occupy 96.6 ml, what is the true, or molecular, formula of this compound?
(2.50 g / 0.0966 liters) (22.4 liters / 1 mole) = 60.0 g/mole = molecular weight
molecular weight of C2H5 = 29.0 g/mole
58.0 g/mole / 29.0 g/mole = 2; then (C2H5) x 2 = C4H10 = butane