For online help with
this topic, see the chemcal module "Stoichiometry"
which deals with moles; balancing equations; stoichiometric calculations;
molarity and solution stoichiometry.
MOLE CONCEPT (Advanced Questions 1)
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Shortcut to Questions
Q: 1 2 3
4 5
6 7 8 9
10 11

1

(1.)
What mass of methanol (CH_{3}OH) must be added to 2.5 mole
of ethanol (C_{2}H_{5}OH) in order to produce a solution
containing equal numbers of molecules of the two species?
(2.)
Assuming no chemical reaction occurs, what is the total number of
molecules present in the resulting solution?


2 
The diatomic molecule of an element Z has a mass of 6.31 x 10^{23}g.
What is the atomic weight of the element Z?


3 
Hydrogen chloride gas (7.3 g) is in a container at a pressure of
2.5 atmosphere and a temperature of 300 K.
(1.)
How many moles of gas are present?
(2.)
How many molecules of gas are present?
(3.)
If all the gas were dissolved in 100 cm^{3} of water, what
mass of silver nitrate would need to be added to precipitate all
the chloride ion produced.


4 
What is the mass in g of each of the following at 273 K and 101
kPa?
(1.)
neon (22.4 litre).
(2.)
ice (22.4 litre).
(3.)
mercury (0.10 mol).


5 
Formic acid (HCOOH) is a liquid monoprotic acid.
(1.) What
amount (in mole) of formic acid is present in 1.00 litre of the
pure substance at 298 K?
(2.) What
mass of formic acid should be taken and diluted to obtain 1.00 litre
of a 0.075 M solution?


6 
How many atoms are in the following at 273 K and 101 kPa?
(1.)
neon (40.4 g)
(2.)
silver (54 g)
(3.)
fluorine (22.4 litre)
(4.)
iodine (127 g)


7 
Hydrogen iodide is a gas at room conditions. If 5.0 litre of hydrogen
iodide at 298 K and 760 mmHg is dissolved in water and the volume
made up to 1.000 litre, what is the molarity of the solution?


8 
A certain hydrate of sodium carbonate (2.433 g) was dried in an
oven at 383 K. The resulting anhydrous sodium carbonate was found
to have a constant weight of 1.112 g. What is the formula of the
hydrated salt?


9 
Magnesium oxide (2.00 g) is added to 10.0 cm^{3} of 2.00
M hydrochloric acid. Calculate which reactant is in excess and the
number of mole by which it is in excess.


10 
Pure formic acid (HCOOH), is a liquid monoprotic acid decomposed
by heat to carbon dioxide and hydrogen.
(1.)
What mass of the pure acid should be diluted with water to produce
1.00 litre of 2.00 M aqueous solution of formic acid?
(2.)
What volume of 0.250 M sodium hydroxide would be required to react
exactly with 30.0 cm^{3} of this dilute solution of formic
acid?
(3.)
What is the maximum volume of carbon dioxide at 273 K and 1.00 atmosphere
that could be obtained by heating 1.00 mole of formic acid?
(4.)
How many molecules of carbon dioxide would it contain?


11 
Barium carbonate (5.50 g) is added to 100 cm^{3} of 0.60
M nitric acid. Calculate which reactant is in excess and the number
of mole by which it is in excess.



MOLE
CONCEPT (Advanced Answers 1)


1

In order to produce a
solution containing equal numbers of molecules, there needs to be
the same number of moles of each species, therefore 2.5 moles methanol
will need to be added to the ethanol.
Mass CH_{3}OH
required = moles CH_{3}OH x molar mass CH_{3}OH
= 2.5 mol x 32.04 g
mol^{1} = 80 g.
No. molecules in the solution = no. moles of all species x N_{A}
= (moles CH_{3}CH_{2}OH
+ moles CH_{3}OH) x N_{A}
= (2.5 mol + 2.5 mol)
x 6.022 x 10^{23}
= 3.0 x 10^{24}
molecules.


2 
Since the element is
in diatomic form, Z_{2}, each atom will weigh half the molecular mass
of the diatomic molecule,
ie mass of 1 atom of
Z = 0.5 x 6.31 x 10^{23} g
= 3.16 x 10^{23}
g.
The gram atomic mass
(molar mass) of the element is the mass of one mole of Z atoms
= mass of elemental
atom x N_{A}
= 3.16 x 10^{23}
x 6.022 x 10^{23}
= 19.0 g mol^{1}
and the atomic weight
of Z = 19.0.


3 
(1.)
Moles HCl = mass HCl / molar mass HCl
= 7.3 g / 36.46 g mol^{1} = 0.20 mol.
(2.)
No. molecules HCl = moles HCl x N_{A}
= 0.20 mol x 6.022 x 10^{23} = 1.20
x 10^{23} mlecules
.
(3.)
Balanced equation for the reaction:
AgNO_{3} + HCl → AgCl +
HNO_{3
}Silver chloride is insoluble and precipitates as a solid.
From the above equation, 1 mol HCl reacts with 1 mol AgNO_{3},
therefore moles AgNO_{3} required =
moles HCl
= 0.20 mol.
Mass AgNO_{3} required = moles AgNO_{3}
x molar mass AgNO_{3}
= 0.20 mol x 169.91 g mol^{1} = 34
g.


4

(1.)
Using the ideal gas approximation:
moles Ne (273 K, 101 kPa) = volume Ne / molar volume
= 22.4 L / 22.4 L mol^{1} = 1.00 mol.
Mass Ne = moles Ne x molar mass Ne
= 1.00 mol x 20.18 g mol^{1} = 20.2
g.
(2.)
Note that ice is a solid, not a gas  do not be mislead into using
the molar volume of an ideal gas which is 22.4 L at these conditions.
While the molar volume of all ideal gases is independent of the
gas used, this is not true of liquids or solids. Instead, the density
of each particular solid or liquid substance must be available in
order to calculate any mass  volume relationship.
Using the density of ice at 273 K:
Mass H_{2}O (at 273 K) = volume ice x density ice
= 22,400 cm^{3} x 0.917 g cm^{3}
= 2.05 x 10^{4} g.
(3.)
Mass Hg = moles Hg x molar mass Hg
= 0.10 x 200.6 = 20 g.


5 
(1.)
Mass HCOOH (298 K) = density
HCOOH x volume HCOOH
= 1.220 g cm^{3}
x 1000 mL
= 1.22 kg (note that
1 cm^{3} = 1 mL, 1000 cm^{3} = 1 L).
Moles HCOOH = mass HCOOH
/ molar mass HCOOH
= 1.22 kg / 46.03 g mol^{1}
= 26.5 mol.
(2.)
0.075 M means 0.075 moles
of solute per litre of solution.
Therefore 0.075 mol is
HCOOH required.
Mass HCOOH = moles HCOOH
x molar mass HCOOH
= 0.075 mol x 46.03 g
mol^{1} = 3.5 g.


6 
(1.)
Moles Ne = mass Ne / molar mass Ne
= 40.4 g / 20.18 g mol^{1}
= 2.00 mol.
No. atoms Ne = moles Ne x N_{A}
= 2.00 mol x 6.022 x 10^{23}
= 1.20 x 10^{24} atoms.
(2.)
Moles Ag = mass Ag / molar mass Ag
= 54 g / 107.9 g mol^{1} = 0.50 mol.
No. atoms Ag = moles Ag x N_{A}
= 0.50 mol x 6.022 x 10^{23}
= 3.0 x 10^{23} atoms.
(3.)
Note that fluorine exists as diatomic F_{2} (not F) in it's
ordinary gaseous state.
Moles F_{2} (273 K, 101 kPa) = volume F_{2} / molar
volume
= 22.4 L / 22.4 L mol^{1}
= 1.00 mol.
No. molecules F_{2} = moles F_{2}
x N_{A}
= 1.00 mol x 6.022 x 10^{23}
= 6.02 x 10^{23} atoms.
No. atoms F = no. molecules F_{2} x
2
= 6.02 x 10^{23} x 2 = 1.20 x 10^{24}
atoms.
(4.)
Iodine is a diatomic solid, I_{2},
Moles I_{2}
= mass I_{2} / molar mass I_{2}
= 127 g / (126.9 X 2)g mol^{1}
= 0.500 mol.
As iodine is diatomic, no. of I atoms = 2 x
moles I_{2} x N_{A}
= 2 x 0.500 x 6.022 x 10^{23}
= 6.02 x 10^{23} atoms.


7 
Using the ideal gas approximation:
moles HI (298 K, 760 mmHg) = volume HI / molar volume
= 5.0 L / 24.5 L mol^{1}
= 0.20 mol.
Therefore there will
be 0.20 moles HI dissolved in 1.000 L H_{2}O, so the solution
will be 0.20 M.


8 
The formula of anhydrous
sodium carbonate is Na_{2}CO_{3}. The difference
in mass between the original hydrated salt, and the dry anhydrous
salt is due solely to lost water.
Mass
of H_{2}O lost = mass hydrated salt  mass anhydrous salt
= 2.433 g  1.112 g =
1.321 g.
Moles
H_{2}O lost = mass H_{2}O / molar mass H_{2}O
= 1.321 g / 18.02 g mol^{1}
= 0.07331 mol.
Moles Na_{2}CO_{3} present in dried sample
= mass Na_{2}CO_{3}
/ molar mass Na_{2}CO_{3}
= 1.112 g / 105.99 g
mol^{1} = 0.01049 mol.
The ratio of H_{2}O to Na_{2}CO_{3} = moles
H_{2}O / moles Na_{2}CO_{3}
= 0.07331 / 0.01049 =
6.989 / 1
Taking into consideration
the error implicit in the given masses, as well as the possibility
of rounding errors, this figure which must be an integer is taken
as 7, hence providing the hydrated salt formula Na_{2}CO_{3}.7H_{2}O.


9 
The oxides of metals
react with acids to form a salt and water. The stoichiometric ratios
of the reactants is found from the balanced equation:
MgO + 2HCl →
MgCl_{2} + H_{2}O
Moles MgO = mass MgO / molar mass MgO
= 2.00 g / 40.31 g mol^{1}
= 0.0496 mol.
Moles HCl = molarity HCl x volume HCl solution (L)
= 2.00 M x ((10.0 cm^{3})
/ 1000 cm^{3}) = 0.0200 mol.
From the equation, 1 mol MgO reacts with 2 mol HCl, therefore 0.0496
mol MgO requires 2 x 0.0496 = 0.0992 mol HCl for complete reaction.
Since only 0.0200 mol
HCl is present, MgO is present in excess.


10 
(1.)
2.00 M means there are 2.00 moles HCOOH dissolved per litre of solution.
Mass HCOOH = moles HCOOH x molar mass HCOOH
= 2.00 mol x 46.03 g mol^{1} = 92.1
g.
(2.)
Balanced equation:
HCOOH + NaOH → HCOONa + H_{2}O
From this equation, 1 mol NaOH will react with 1 mol HCOOH.
Moles HCOOH = volume solution x molarity
= 0.0300 L x 2.00 M = 0.0600 mol.
Therefore 0.0600 mol NaOH is required for reaction.
Volume of 0.250 M NaOH solution required
= moles NaOH required / molarity of NaOH solution
= 0.0600 mol / 0.250 mol L^{1} = 0.240
L.
(3.)
Balanced equation for thermal decomposition:
HCOOH → H_{2} + CO_{2
}From the equation, 1 mol HCOOH produces 1 mol gaseous CO_{2}.
From the ideal gas approximation, 1 mol gas
occupies 22.4 L at 273 K, 1.00 atm.
(4.)
1 mol CO_{2} contains N_{A} molecules
= 6.022 x 10^{23} molecules.


11 
The stoichiometric ratios
of the reactants is found from the balanced equation:
BaCO_{3} + 2HNO_{3} →
Ba(NO_{3})_{2}
+ H_{2}O + CO_{2
}Moles BaCO_{3} = mass BaCO_{3} / molar mass
BaCO_{3}
= 5.50 g / 197.31 g mol^{1}
= 0.0279 mol.
Moles HNO_{3} = molarity HNO_{3} x volume HNO_{3}
solution (L)
= 0.60 M x ((100 cm^{3})
/ 1000 cm^{3}) = 0.060 mol.
From the equation, 1 mol BaCO_{3} reacts with 2 mol HNO_{3},
therefore 0.0279 mol BaCO_{3} requires 2 x 0.0279 = 0.0558
mol HNO_{3} for complete reaction.
There is 0.060 mol HNO_{3}
present,
so HNO_{3} is
in excess by (0.060  0.0558) = 0.004 mol.


