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

MOLE CONCEPT (General Questions 1)

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1

Which of the following has the greatest mass:
(1.) copper (100 g),
(2.) helium (6.00 mole),
(3.) 12.0 x 1023 atoms of silver?
Hydrogen gas (4.00 g) and chlorine gas (10.0 g) are allowed to react.
(1.) Which reactant gas is in excess?
(2.) What mass of hydrogen chloride is formed?
(1.) Name the compound of formula Na2SO4.10H2O.
(2.) What is its molar weight?
(3.) What is the percentage by weight of sulfur in this compound?
(4.) How many sodium ions are in one mole of this compound?
Calculate the volume of the following at 298 K and 101 kPa.
(1.) Water (2.00 mol)
(2.) Chlorine (1.00 mol)
(3.) Mercury (3.00 mol)
(4.) Neon (1.00 mol)
(5.) Nitrogen (1.20 x 1024 molecules)
Calculate the mass of the following at 298 K and 760 mmHg pressure.
(1.) Helium (49.0 litre)
(2.) Water (3.00 mol)
(3.) Ethanol (100 cm3)
(4.) Nitrogen (25.0 litre)
The price of chromium is 1.00 dollar gram-1. What is the price of one mole of chromium?
Calculate (1.) the molar weight, and (2.) the molar volume of liquid carbon tetrachloride at 298 K given the density of carbon tetrachloride at 298 K = 1.584 g cm-3.
What volume is occupied at 298 K and 101 kPa by 1.00 mole of each of the following:
(1.) hydrogen(g)
(2.) helium(g)
(3.) aluminium(s)
(4.) water(l)
What amount (in mole) of carbon is contained in 112 g of ethylene (C2H4)?
10  Which of the following has the greatest mass?
(1.) lead (3.00 g)
(2.) helium (22.4 litre) at 273 K and 101 kPa?
(3.) 3.01 x 1023 atoms of lithium.
11  Which of the following has the greatest mass?
(1.) lead (63.0 g)
(2.) fluorine (24.5 litre) at 298 K and 2.00 atmosphere.
(3.) 3.01 x 1023 atoms of caesium. 
12  In the chemical reaction requiring two atoms of aluminium for every three atoms of oxygen, how many mole of oxygen atoms is required by 2.7 g of aluminium? What mass of oxygen atoms are required?
13  Hydrogen gas reacts with chlorine gas to form hydrogen chloride. If hydrogen (2.00 mol) reacts with excess chlorine, what mass of hydrogen chloride can be obtained?
14  Nickel(II) sulfate-6-water (2.63 g) is converted to nickel(II) chloride-6-water. Calculate the theoretical yield in grams.
15 (1.) What mass of sodium hydroxide reacting with excess carbon dioxide is required to prepare 53.0 g of sodium carbonate?
(2.) What volume of carbon dioxide is released at 273 K and 1.00 atmosphere when this weight of sodium carbonate is treated with excess hydrochloric acid?
16 Calculate the mass of anhydrous sodium carbonate required to make 250 cm3 of 0.100 M solution.
17 A solution was prepared by dissolving nickel(II) nitrate-6-water (29.1 g) in some water and making the volume up to 1.000 litre with water. Assuming complete dissociation of the solid into ions, calculate the following for 100 cm3 of solution.
(1.) The number of mole of nickel(II) ion.
(2.) The number of mole of nitrate ion.
(3.) The number of individual nickel(II) ions.
(4.) The number of cm3 which theoretically contain one individual nickel (II) ion.
 
MOLE CONCEPT (General answers 1)   

The silver has the largest mass:
(1.) Mass of copper = 100 g (given).


(2.) Mass of helium = atomic mass He x no. moles He

= 4.003 x 6.00 = 24.0 g.

(3.) Moles silver = no. atoms Ag / NA

= 12.0 x 1023 / 6.022 x 1023 = 1.99 mol. 

Mass of silver = moles Ag x atomic weight Ag

= 1.99 mol x 107.9 g mol-1 = 215 g 

(1.) MolesH2 = mass H2 / molar mass H2

= 4.00 g / (2 x 1.008 g mol-1) = 2.00 mol.

 
Moles Cl2 = mass Cl2 / molar mass Cl2

= 10.0 g / (2 x 35.45 g mol-1) = 0.141 mol. 


From the balanced equation H2 + Cl2® 2HCl, one mole of hydrogen will react with one mole of chlorine, so hydrogen is in excess.


(2.) There are two moles HCl formed for each mole of Cl2, hence 2 x 0.141 = 0.282 moles HCl is produced. 
Mass HCl = no. mol HCl x molar mass HCl

= 0.282 x (1.008 + 35.45) = 10.3 g.

(1.) The name of Na2SO4.10H2O is sodium sulfate-10-water.


(2.) The molar weight of sodium sulfate-10-water is 322.21 g mol-1.


(3.) The percentage of sulfur

= (atomic mass S / molar mass Na2SO4.10H2O) x 100%
= (32.07 g mol-1 / 322.21 g mol-1) x 100%

= 9.95 %.


(4.) The formula of Na2SO4.10H2O contains two sodium ions, so one mole of Na2SO4.10H2O will contain two moles of sodium ions,

ie 2 x 6.022 x 1023 sodium ions

= 1.20 x 1024 sodium ions.

The volume of the gases is found from the ideal gas approximation of 24.5 L mol-1 (at 298K and 101 kPa), irrespective of what gas it is. There is no simple approximation for finding the density of solids and liquids, and known densities must be used for calculations.


(1.) Mass of water = moles H2O x molar mass H2O

= 2.00 mol x 18.02 g mol-1 = 36.0 g. 

Volume of water = mass H2O / density H2O

= 36.0 g / 1.00 g cm-3 = 36.0 cm3

(Note: this is equivalent to 36.0 mL as 1cm3 is the same volume as 1 mL)

(2.) Volume of chlorine = moles Cl2 / molar volume

= 1.00 mol x 24.5 L mol-1 = 24.5 L.

(3.) Mass of mercury = moles Hg x molar mass Hg

= 3.00 mol x 200.6 g mol-1 = 602 g. 

Volume of mercury = mass Hg / density Hg

= 602 g / 13.6 g cm-3 = 44.3 cm3.

(4.) Volume of neon = moles Ne / molar volume

= 1.00 mol x 24.5 L mol-1 = 24.5 L.

(5.) Moles N2 = no. molecules N2 / NA

= 1.20 x 1024 / 6.023 x 1023 = 1.99 mol. 

The volume of N2 = moles N2 / molar volume

= 1.99 mol x 24.5 L mol-1 = 48.8 L. 

The mass of gases is found using the same relationships used to obtain volume in question 4.

(1.) Moles He (at 298 K, 760 mmHg)

= volume He / molar volume

= 49.0 L / 24.5 L mol-1 = 2.00 mol. 

Mass He = moles He x molar mass He

= 2.00 x 4.003 = 8.00 g.

(2.) Mass of H2O = moles H2O x molar mass H2O

= 3.00 x 18.02 = 54.1 g.

(3.) Mass of CH3CH2OH = volume CH3CH2OH x density CH3CH2OH 


= 100 cm3 x 0.785 g cm-3 = 78.5 g.


(4.) Moles N2 (at 298 K, 760 mmHg)

= volume N2 / molar volume 
= 25.0 L / 24.5 L mol-1 = 1.02 mol. 

The mass of N2 = moles N2 x molar mass N2

= 1.02 x (2 x 14.01) = 28.6 g.

Mass of Cr = moles Cr x molar mass Cr

= 1 mol x 52.00 g mol-1 = 52 g. 

The price of Cr = mass Cr x unit price

= 52 g x $1.00 g-1 = $52.

(1) The molecular formula of carbon tetrachloride is CCl4
Molar mass CCl4 =

molar mass C + (4 x molar mass Cl)

= 12.01 + (35.45 x 4) = 153.81. 


(2) Molar volume = volume of one mole

= molar mass / density
= 153.81 /1.584 = 97.10 g cm-3.

From the ideal gas approximation the volume of the gases can be obtained:

(1.) Volume H2 (at 298 K, 101 kPa)

= moles H2 x molar volume

= 1.00 mol x 24.5 L mol-1

= 24.5 L.

 

(2.) Volume He (at 298 K, 101 kPa)

= moles He x molar volume

= 1.00 mol x 24.5 L mol-1

= 24.5 L.

For solids and liquids, volume calculations involve the particular density of the substance:

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

= 1.00 mol x 26.98 g mol-1 = 26.98 g. 

Volume Al = mass Al / density Al

= 26.98 g / 2.7 g cm-3 = 10 cm3.

(4.) Mass H2O = moles H2O x molar mass H2O

= 1.00 mol x 18.02 g mol-1 = 18.0 g. 

Volume H2O = mass H2O / density H2O

= 18.0 g / 1.00 g cm-3 = 18.0 cm3.

Moles C2H4 = mass C2H4 / molar mass C2H4

= 112 g / 28.05 g mol-1 = 3.99 mol. 

Each molecule of C2H4 contains 2 C atoms, therefore 3.99 moles of C2H4 will contain 
2 x 3.99 moles of C = 7.98 moles C.

10 

The helium has the largest mass:
Mass of lead = 3.00 g (given).
Using the ideal gas approximation, 1 mole of gas occupies 22.4L at 273K, 101 kPa.
Moles helium = volume He / molar volume

= 22.4 L / 22.4 L mol-1 = 1.00 mol. 

The mass of He = moles He x molar mass He

= 1.00 x 4.003 = 4.00 g.

Moles  lithium = no. atoms Li / NA

= 3.01 x 1023 / 6.022 x 1023 = 0.500 mol. 

The mass of Li = moles Li x molar mass Li

= 0.500 x 6.941 = 3.47 g.

11

The fluorine has the largest mass:


(1.) Mass of lead = 63.0 g (given).


(2.) Moles fluorine, F2 , in 24.5 L at 298 K and 1.00 atm is found from
volume F2 / molar volume = 24.5 L / 24.5 L mol-1

= 1.00 mol. 

At a pressure of 2.00 atm, there will be twice as many molecules (assuming the volume and temperature are the same), ie 2.00 mol F2
Mass F2 = moles F2 x molar mass F2

= 2.00 mol x (2 x 19.00 g mol-1) = 76 g. 


(An equation that can be used to calculate this directly is covered later in the course.)


(3.) Moles caesium, Cs,  = no. atoms Cs / NA

= 3.01 x 1023 / 6.022 x 1023 = 0.500 mol. 


Mass Cs = moles Cs x molar mass Cs

= 0.500 x 132.9 = 66.5 g. 

12

Moles Al = mass Al / molar mass Al

= 2.7 / 26.98 = 0.10 mol. 


The stoichiometric ratio of Al to O is 1 to 1.5, therefore for each mole of Al, 1.5 moles O is required. No. mol O required = no. mol Al (0.10) x 1.5 = 0.15 mol O.  Mass of O atoms is 0.15 mol x 16.0 g mol-1 = 2.4 g.
Note that the question specifies oxygen atoms, not molecules, so the calculations are based on monatomic O, rather than the gaseous diatomic O2.

13 

Note that the question refers to the reaction of gaseous H2 and Cl2, so the diatomic nature of these gases must be taken into account in subsequent calculations. 
Balanced equation:
H2 + Cl2®2HCl
Therefore, in the presence of an excess of Cl2, 2.00 mol H2 will react completely to give 4.00 mol HCl. 
Mass HCl = moles HCl x molar mass HCl

= 4.00 mol x 36.46 g mol-1 = 146 g. 

14 

Empirical formula of nickel(II) sulfate-6-water: NiSO4.6H2O (molar mass 262.86)
Empirical formula of nickel(II) chloride-6-water: NiCl2.6H2O (molar mass 237.69)
Since both formulae contain one nickel ion, each mole of NiSO4.6H2O will produce one mole of NiCl2.6H2O. 
Moles NiSO4.6H2O = mass NiSO4.6H2O / molar mass NiSO4.6H2
= 2.63 g / 262.86 g mol-1

= 0.0100 mol, and so 0.0100 mol NiCl2.6H2O will be produced (theoretically, ie ignoring losses due to experimental technique). 
Mass NiCl2.6H2O = moles NiCl2.6H2O x molar mass NiCl2.6H2O
= 0.0100 mol x 237.69 g mol-1 = 2.38 g.

15

(1.) Balanced equation:
2NaOH + CO2 ® Na2CO3 + H2O
From the equation, each mole of Na2CO3 produced requires two moles of NaOH. 
Moles Na2CO3 = mass Na2CO3 / molar mass Na2CO3 = 53.0 g / 105.98 g mol-1 = 0.500 mol. Therefore moles NaOH required = 2 x 0.500 mol

= 1.00 mol. 
Mass NaOH = moles NaOH x molar mass NaOH

= 1.00 mol x 40.00 g mol-1 = 40.0 g.

(2.) Balanced equation:
Na2CO3 + 2HCl ® 2NaCl + H2O + CO2
Moles Na2CO3 = mass Na2CO3 / molar mass Na2CO3 = 53.0 g / 105.98 g mol-1 = 0.500 mol. 
With excess hydrochloric acid present, all the sodium carbonate will react, each mole of Na2CO3 producing one mole of CO2, therefore 0.500 mol CO2 is produced. 
At 273 K and 1.00 atm, 1 mol of gas occupies 22.4 L (ideal gas approximation), 
volume CO2 = moles CO2 / molar volume

= 0.500 mol x 22.4 L mol-1 = 11.2 L.

16 0.100 M implies 0.100 mol of substance per L of solution. 
Therefore 250 mL of solution will require 0.0250 mol Na2CO3
Mass Na2CO3 = moles Na2CO3 x molar mass Na2CO3 = 0.0250 mol x 105.98 g mol-1 = 2.65 g.
17

Empirical formula of nickel(II) nitrate-6-water: Ni(NO3)2.6H2O. 
Ni(NO3)2.6H2O will dissolve in water as shown below:
Ni(NO3)2.6H2O(s) ®Ni2+(aq) + 2NO3-(aq) + 6H2O
(Note that the water of crystallisation is now part of the water solution). Therefore each mole of Ni(NO3)2.6H2O dissolved will produce one mole of Ni2+, and two moles of NO3-
Moles Ni(NO3)2.6H2O

= mass Ni(NO3)2.6H2O / molar mass Ni(NO3)2.6H2O
= 29.1 g / 290.8 g mol-1

= 0.100 mol. 
This amount of Ni(NO3)2.6H2O is dissolved in 1.00 L of water, so in 100 mL there will be 
0.100 x 0.100 = 0.0100 moles of Ni(NO3)2.6H2O dissolved. 


(1) Moles Ni2+in 100 mL = moles Ni(NO3)2.6H2O dissolved as the formula contains only one Ni2+ ion

  = 0.0100 mol, 

(2) Moles NO3- in 100 mL

= 2 x moles Ni(NO3)2.6H2O dissolved

= 0.0100 x 2 = 0.0200 mol.

(3) No. of Ni2+ions = moles of Ni2+ions x  NA

= 0.0100 x 6.022 x 1023 = 6.02 x 1021 ions.

(4) If there are 6.02 x 1021 ions present in 100 mL, then on average, 
100 / 6.02 x 1021 mL would contain a single Ni2+ion.