For on-line help with this topic, see the chemcal module "Stoichiometry" which deals with moles; balancing equations; stoichiometric calculations; molarity and solution stoichiometry.

THE MOLE CONCEPT (Advanced Questions 2)

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Q: 1 2 3 4 5 6 7 8 9 10

1

What volume of 0.100 M sulfuric acid would be needed to react completely with a mixture of 0.500 g of sodium hydroxide and 0.800 g of potassium hydroxide?
2

Calcium carbonate (15.0 g) is quantitatively decomposed by strong heat to calcium oxide and carbon dioxide gas. The solid oxide residue reacts completely with a certain amount of hydrochloric acid. What mass of sodium hydroxide would also react exactly with this amount of hydrochloric acid?

3

Petrol is a mixture of many hydrocarbon compounds but can be considered to be equivalent to about 25 mole of octane, C8H18.

In 4.55 litres of petrol,
(1) How many mole of oxygen must be used to burn this petrol, assuming the only products to be carbon dioxide and water?
(2) How many mole of carbon dioxide are formed?
(3) What mass (kg) of carbon dioxide is formed ?
(4) What volume of carbon dioxide at 273 K and 1.00 atmosphere is released into the atmosphere when a car consumes 45.5 litre of petrol?

4

A compound is composed of the elements X and Y combined so that every two atoms of X require three atoms of Y. Element X has an atomic weight of 75. If 0.0333 mole of X are combined with Y (1.60 g), calculate:
(1) The atomic weight of Y and,
(2) The molecular weight of the compound.

5

A water solution of silver nitrate (50.0 cm3) was treated with a slight excess of a water solution of sodium chloride. The precipitate when filtered, washed, and dried had a mass of 0.497 g. What was the molarity of the silver nitrate solution?

6

Fluorine gas was passed over tin(II) sulfide at 773 K and the products were gaseous sulfur hexafluoride, and a solid. The solid contained by weight 61.0 per cent tin and 39.0 per cent fluorine. A molar weight determination gave the value 198. What is the molecular formula of the solid? Write the equation for the reaction between fluorine and tin(II) sulfide.

7 A sample (0.210 g) of a gaseous compound containing only hydrogen and carbon was burnt. Carbon dioxide (0.660 g) was recovered.
(1) What is the empirical formula of the compound?
(2) A determination of the density of this hydrocarbon gave a value of 1.87 gram litre-1 at 273 K and 101 kPa. What is the molecular formula of the substance?
8

A sample (1.37 g) of an organic compound containing only carbon, hydrogen and oxygen was burnt in an excess of oxygen to yield water (1.64 g) and carbon dioxide (1.53 litre), measured at 273 K and 1.00 atm. What is the empirical formula of the compound?

9 A certain carbohydrate of molar weight 120 contains by weight 40.0% carbon; 6.7% hydrogen and 53.3% oxygen.
(1) What is its molecular formula?
(2) What weight of the carbohydrate will, after complete combustion, produce dry carbon dioxide (1.12 litre) at 273 K and 1.00 atmosphere?
10

When solid lead(IV) oxide (PbO2) is heated, it forms solid lead(II) oxide (PbO) and oxygen gas. Heating solid barium peroxide (BaO2) yields solid barium oxide (BaO) and oxygen gas. A mixture of lead(IV) oxide and barium peroxide was heated until both decompositions were complete.
If the initial mass of the mixture were 15.00 g and the final mass were 13.80 g, what mass of lead(IV) oxide was present in the original mixture?

 

MOLE CONCEPT 2 (advanced answers)

 

1

An acid reacts with a hydroxide to give a salt and water. This type of reaction is sometimes called "neutralisation". The balanced equations for the neutralisation reactions here are:
H2SO4 + 2NaOH
® 2H2O + Na2SO4

H2SO4 + 2KOH ® 2H2O + K2SO4

In each reaction, one mole of sulfuric acid reacts with two moles of the relevant hydroxide, so

moles of sulfuric acid required = (moles of NaOH + moles of KOH).

Moles of NaOH = (mass NaOH / molar mass NaOH )

= (0.500 g / 40.00 g mol-1)

= 0.0125 mol.

Moles of KOH = (mass KOH / molar mass KOH)

= (0.800 g / 56.11 g mol-1)

= 0.0143 mol.


Total moles (NaOH + KOH) = (0.0125 mol + 0.0143 mol)

= 0.0268 mol.

From the equations, 1.00 mol H2SO4 reacts with 2.00 mol of either hydroxide.

Therefore moles H2SO4 required = 0.500 x total moles hyroxides

= 0.500 x 0.0268 mol = 0.0134 mol.

The volume of 0.100 M H2SO4 required

= moles H2SO4 / molarity of H2SO4 solution

= 0.0134 mol / 0.100 M L-1 = 0.134 L.
 

2

Balanced equation for the thermal decomposition of CaCO3 to give calcium oxide and carbon dioxide:
CaCO3
® CaO + CO2
Moles CaCO3 = mass CaCO3 / molar mass CaCO3

= 15.0 g / 100.19 g mol-1 = 0.150 mol.

From above equation, 0.150 mol CaCO3 will produce 0.150 mol CaO.

Oxides of metals react with acids to form a salt and water.

Reaction of CaO with hydrochloric acid:
CaO + 2HCl
® CaCl2 + H2O

From this equation moles HCl required = 2 x moles CaO

= 2 x 0.150 mol = 0.300 mol.

Acids also react with hydroxides of metals to form a salt and water.

Balanced equation for the reaction between HCl hydrochloric acid and sodium hydroxide::
HCl + NaOH
® NaCl + H2O
From this equation, 1.00 mol NaOH reacts with 1.00 molHCl

Therefore moles NaOH = moles HCl = 0.300 mol.

Mass NaOH = mol NaOH x molar mass NaOH

= 0.300 mol x 40.00 g mol-1 = 12.0 g.
 

3

Balanced equation for combustion of octane:
2C8H18 + 25O2
® 16CO2 + 18H2O


(1.) From the equation, 1 mol C8H18 requires 12.5 mol O2 for combustion. Therefore moles O2 required to burn 25 moles C8H18 = 25 x 12.5 mol

= 3.1 x 102 mol.


(2.) From the equation, 1 mole C8H18 produces 8 mole CO2.

Hence moles CO2 produced from 25 mol C8H18 = 25 x 8 mol

= 2.0 x 102 mol.


(3.) Mass CO2 = moles CO2 x molar mass CO2

= 2 x 102 mol x 44.01 g mol-1 = 8.8 kg.


(4.) Using the ideal gas approximation:
Volume CO2 (273 K, 1.00 atm) = moles CO2 x molar volume

= 200 mol x 22.4 L mol-1 = 4.5 x 104 L

4

The empirical formula of the compund is X2Y3.

If 0.0333 mol X are combined with Y, then mol Y present

= 3/2 x 0.0333 mol = 0.0500 mol.

The weight of Y is 1.60 g, therefore molar mass of Y

= mass Y / moles Y = 1.60 g / 0.0500 mol = 32.0 g mol-1and the atomic weight of Y = 32.0

Molecular weight of X2Y3

= (2 x atomic weight of X) + (3 x atomic weight of Y)

= 2 x 75 + 3 x 32.0 = 246.

5

Balanced equation:
AgNO3 + NaCl
® AgCl + NaCl
Moles AgCl = mass AgCl / molar mass AgCl

= 0.497 g / 143.35 g mol-1 = 0.00347 mol.


From the balanced equation, moles AgNO3 = moles AgCl

= 0.00347 mol.

Since this amount of AgNO3 was contained in the original 50.0 cm3 of silver nitrate solution, the molarity of the AgNO3 solution

= moles AgNO3 x (1000 cm3 / 50.0 cm3)

= 0.00347 x 20.0 = 0.0694 M.

6

In 100.0 g of compound, moles Sn = mass Sn / molar mass of Sn atoms

= 61.0 g / 118.7 g mol-1 = 0.514 mol.

and moles F = mass F / molar mass of F atoms

= 39.0 g / 19.00 g mol-1 = 2.05 mol.

Thus the ratio of Sn atoms to F atoms = 0.514 : 2.05

Dividing through by 0.514:
Molar ratio Sn:F = 1:3.99
Allowing for errors, the empirical formula is SnF4.
The formula weight of SnF4 is 195, which is very close to the molar mass determination of 198, therefore the molecular formula is SnF4. This allows the balanced equation to be determined:
SnS + 5F2
® SnF4 + SF6

7

(1 ) Moles CO2 = mass CO2 / molar mass CO2

= 0.660 g / 44.01 g mol-1 = 0.0150 mol.

Since all the C present in the CO2 must have been produced from C in the compound, there must have been 0.0150 mol C in the original sample. Mass C in sample = moles C x molar mass C atoms

= 0.0150 x 12.01 = 0.180 g.

Therefore mass of H in original sample (since it only contains C and H)

= mass of sample - mass C in sample

= 0.210 g - 0.180 g = 0.030 g.

Moles H in sample = mass of H in sample / molar mass H atoms

= 0.030 g / 1.008 g mol-1 = 0.030 mol.

Mole ratio of C : H = 0.0150 : 0.030

Dividing through by 0.0150 gives the empirical formula CH2.


(2 ) As the substance is a gas, the ideal gas approximation can be used:

1 mole of any gas at 273K, 1 atm will occupy 22.4 L.

Mass of 1 mole gas = density of gas x volume of 1 mole

= 1.87 g L-1 x 22.4 L = 41.9 g.

The gram formula mass of the empirical CH2 unit is 14.03 g mol-1, and the molar mass of the gas is 41.9 g mol-1, so the number of empirical formula units in each molecule = 41.9 g mol-1 / 14.03 g mol-1 = 3, providing a molecular formula of C3H6.

8 Moles CO2 (273 K, 1.00 atm) = volume CO2 (1.53 L) / 22.4 L mol-1

= 0.0683 mol.

All C atoms present in the CO2 produced must be from C atoms in the sample, therefore there are 0.0683 mol C in the sample.

Mass C = moles C x molar mass C atoms

= 0.0683 mol x 12.01g mol-1 = 0.820 g.

Moles H2O = mass H2O / molar mass H2O

= 1.64 g / 18.02 g mol-1 = 0.0910 mol.

All H atoms present in the H2O produced must be from H atoms in the sample, each mol H2O molecules containing 2 mol H atoms.

Therefore moles H atoms in sample = 2 x moles H2O

= 2 x 0.0910 mol = 0.182 mol.

Mass H = moles H x molar mass H atoms

= 0.182 mol x 1.008 g mol-1 = 0.183 g.

The mass of O in the sample = total sample mass - mass C - mass H

= 1.37 g - 0.820 g - 0.183 g

= 0.37 g.

Therefore moles O in sample = mass O / molar mass O atoms

= 0.37 g / 16.00 g mol-1 = 0.023 mol.

Molar ratios: C : H : O = 0.0683 : 0.182 : 0.023
Dividing through by 0.023:

Molar ratios: C : H : O = 3.0 : 7.9 : 1.0

Allowing for errors, this gives the empirical formula C3H8O

9

(1) In 100.0 g of sample,

moles C atoms = mass C / molar mass C atoms

= 40.0 g / 12.01 g mol-1 = 3.33 mol

moles H atoms = mass H / molar mass H atoms

= 6.7 g / 1.008 g mol-1 = 6.6 mol

moles O atoms = mass O / molar mass O atoms

= 53.3 g / 16.00 g mol-1 = 3.33 mol

Molar ratios: C : H : O = 3.33 : 1.98 : 3.33


Dividing through by 3.33:
C : H : O = 1.00 : 1.98 : 1.00

Allowing for errors an empirical formula of CH2O can be assigned. (Note that all carbohydrates have the same empirical formula - hence the name "carbohydrate".)

The empirical formula unit CH2O has a formula weight = 30.0.

No. empirical units per molecule

= molar mass molecule / molar mass of empirical formula unit

= 120 / 30.0= 4.00.

This allows assignment of the molecular formula C4H8O4.


(2) Moles CO2 = volume CO2 (1.12 L) / 22.4 L mol-1= 0.0500 mol.

Combustion equation:
C4H8O4 + 4O2 ® 4CO2 + 4H2O

From the equation, 1 mole of C4H8O4 produces 4 moles of CO2

Therefore 0.0125 mole C4H8O4 is required to produce 0.0500 mole CO2.

Mass C4H8O4 = moles C4H8O4 x molar mass C4H8O4

= 0.0125 x 120 = 1.50 g.

10

Reaction equations:
2PbO2
® 2PbO + O2
2BaO2
® 2BaO + O2
The initial mass =15.00 g is composed of

(moles PbO2 x molar mass PbO2) + (moles BaO2 x molar mass BaO2).

As the oxygen gas has escaped, the final mass = 13.80 g is composed of

(moles PbO x molar mass PbO) + (moles BaO x molar mass BaO).


Note that the moles of PbO2 before reaction equals the moles of PbO after reaction and moles of BaO2 before reaction = moles of BaO after reaction (see reaction equations).

These two relationships can be used to form a pair of simultaneous equations:
Molar mass BaO2 = 169.3 g mol-1
Molar mass BaO = 153.3 g mol-1
Molar mass PbO2 = 239.2 g mol-1
Molar mass PbO = 223.3 g mol-1

Let P = moles PbO2 = moles PbO
Let B = moles BaO2 = moles BaO
Excluding units for the sake of clarity, the two equations are now:
(Eqn 1) 15 = 239.2 P + 169.3 B
(Eqn 2) 13.8 = 223.2 P + 153.3 B
From Eqn 2:
153.3 B = 13.80 - 223.2 P
B = (13.80 - 223.2 P) / 153.3
Substituting this into Eqn 1:
15.00 = 239.2 P + (169.3 / 153.3)(13.80 - 223.2 P)
15.00 = 239.2 P + (169.3 / 153.3)(13.80) - (169.3 / 153.3)(223.2 P)
15.00 = 239.2 P + 15.24 - 246.5 P
-0.24 = -7.3 P
P = 0.033,

Hence moles PbO2 = 0.033 mol.

Mass PbO2 = moles PbO2 x molar mass PbO2

= 0.033 mol x 239.2 g mol-1 = 7.9 g.