Moles  # of ATOMS
1. A Canadian penny contains 0.106 mol of copper. How many atoms of copper are in a
Canadian penny?
0.106 mol of copper x 6.02 x 1023 atoms per mole = 6.38 x 1022 atoms of copper
2. A sample of sucrose, C12H22O11, contains 0.16 moles of sucrose. How many atoms of
carbon are in this sample?
Since there are 12 atoms of carbon per molecule, there will be 12 moles of C atoms per
mole of sucrose molecules.
Use a ratio: 12 mol C/ mol of sucrose = x mol C / 0.16 mol of sucrose
So, there are 12 x 0.16 = 1.92 mol of C atoms in 0.16 mol of sucrose molecules
Moles  # of MOLECULES < # of atoms
1. Hydrazine, N2H4,(l), is used in pharmaceuticals, rocket fuels and airbags.
a) How many molecules of hydrazine are present in a 3.65 mol sample of hydrazine?
3.65 mol x NA = 3.65 mol x 6.02 x 1023 molecules per mole = 2.20 x 1024 molecules of
hydrazine
b) How many nitrogen atoms would be present in this same sample of hydrazine?
There are 2 nitrogen atoms for every molecule, so 2 x 2.20 x 1024 = 4.40 x 1024 atoms of
nitrogen are in this sample
Moles  # of FORMULA UNITS  # of ions  mass
A recipe calls for half a teaspoon of salt, which contains 5.23 X 10-2 moles of sodium
chloride, NaCl(s).
a) How many formula units of sodium chloride are needed?
5.23 x 10-2 moles x 6.02 x 1023 formula units per mole = 3.15 x 1024 formula units
b) How many sodium ions are present?
Every formula unit has 1 ion of sodium so there are 3.15 x 1024 sodium ions
c) How many grams of sodium does this represent?
The molar mass of a sodium ion is the same as the mass of a sodium atom, 22.99 g per
mole
.0523 moles x 22.99 g per mole = 1.20 g of sodium
# of MOLECULES  MOLES
A cleaning solution contains 7.9 x 1026 molecules of ammonia, NH3(aq). What amount
in moles of ammonia is in the solution?
7.9 x 1026 molecules divided by 6.02 x 1023 molecules per mole =
1.3 x 103 moles of ammonia.
# of ATOMS  MOLES
A sample of rubbing alcohol solution contains ethanol, C2H5OH,(l). If the sample
contains 1.25 x 1023 atoms of hydrogen in the ethanol, what amount in moles of ethanol
is in the sample?
There are 6 atoms of hydrogen per molecule of ethanol
1.25 x 1023 divided by 6 = 0.208 x 1023 molecules of ethanol
0.208 x 1023 molecules divided by NA = 0.0346 moles
FINDING MOLAR MASS
1. Determine the molar mass of calcium phosphate, Ca3(PO4)2(s).
3 (40.08) + 2 (30.97) + 8 (16.00) = 310.18 g mol-1
2. Determine the molar mass of barium hydroxide octahydrate, Ba(OH)2•8 H2O (s)
137.33 + 2(16.00) + 2 (1.01) + 16 (1.01) + 8 (16.00) = 315.51 g mol-1
MASS  MOLES
What amount in moles is 15.3 g of sulfur dioxide, SO2(g)?
Mass divided by molar mass = # of moles m/M = n
Molar mass of SO2 is 2(16.00) + 32.07 = 64.07
15.3 g divided by 64.07 g mol-1 = 0.239 moles of sulfur dioxide
MASS  MOLES  PARTICLES
How many atoms of xenon, Xe, are present in 22 mg of xenon tetrafluoride, XeF4 (s)?
22 mg = 0.022 g
Molar mass of XeF4 = 131.29 + 4(19.00) = 207.29
0.022 g divided by 207.20 g per mole = 0.000106 moles
There is 1 atom of Xe in every molecule, so 0.000106 mol x NA = 6.39 x 1019 atoms of
Xe are present
# OF PARTICLES  MASS
What is the mass of 4.72 x 1023 formula units of chromium (III) iodide, CrI3(s) ?
How many moles? 4.72 x 1023 divided by NA = 0.784 moles
Molar mass of CrI3 = 52.00 + 3(126.90) = 432.7 g per mole
0.784 x 432.7 = 339.2368 g or 339 g
SOLUTION STOICHIOMETRY (with LR)
3. A student mixes 15.0 cm3 of 0.250 mol dm-3 aqueous sodium hydroxide, NaOH, with
20.0 cm3 of 0.400 mol dm-3 aqueous aluminum nitrate, Al(NO3)3. This results in the
formation of an insoluble precipitate of aluminum hydroxide.
a) Write the balanced chemical equation for this reaction.
3 NaOH(aq) + Al(NO3)3 (aq)  Al(OH)3 (s) + 3 NaNO3(aq)
b) Write the two dissociation equations that describe the dissociation of the two aqueous
reactants.[2]
3 NaOH(aq)  3 Na+ (aq) + 3 OH- (aq)
Al(NO3)3 (aq) ==> Al3+ (aq) + 3 NO3-(aq)
d) Write the net ionic equation for this reaction. [1]
Al (aq)+ 3 OH- (aq)  Al(OH)3 (s)
3+
d) Calculate the maximum mass of precipitate that forms.[4]
nNaOH = C x V = 0.250 x 0.015 = 0.00375 moles
3:1 = NaOH : Al(OH)3 so 3 x 0.00375 mol = 0.01125 mol of aluminum hydroxide
forms from this amount of sodium hydroxide
n Al(NO3)3 = C x V = .400 x .020 = .008 moles of aluminum nitrate
1:1 ratio so 0.008 moles of aluminum hydroxide forms from this amount of
aluminumnitrate
THEREFORE the Limiting Reactant is aluminum nitrate because it forms the least
product
0.008 mol of aluminum hydroxide are formed
The molar mass of aluminum hydroxide is 26.98 + 3 (1.01 + 16.00) = 78.01
.008 mol x 78.01 g per mole = 0.624 g of the precipitate are formed
e) What concentration of Al3+ remains after the reaction is complete?[3]
1. a) 16.20 x 10-3 dm3 of 0.1020 mol dm-3 aqueous silver nitrate, AgNO3, is added to
14.80 x 10-3 dm3 of 0.1250 mol dm-3 aqueous sodium chloride, NaCl. Calculate the
maximum mass (g) of silver chloride which could be obtained from this reaction.
AgNO3(aq) + NaCl (aq)  AgCl(s) + NaNo3 (aq)
AgNO3: n = C x V = .01620 x .1020 = 0.0016524 moles
1:1 so 0.0016524 mol of AgCl would be formed
NaCl: n = C x V = 0.01480 x .1250 = .00185 moles
1:1 so 0.00185 mol of AgCl would be formed
So AgNO3 is the LR because it produces the least product
So 0.0016524 mol of AgCl will form molar mass of AgCl is 107.87 + 35.45 = 143.32
0.0016524 mol x 143.32 g per mol = 0.237 g of AgCl forms
b) If 0.225 g of silver chloride were actually produced, what is the percent yield?
0.225 / 0.237 x 100 = 94.9 per cent yield