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stoichiometry


Quick Notes

  • Mole: The number of atoms in exactly 12 g of carbon-12
    • Number of moles = n
  • Avogadro's number: The number of atoms/molecules in one mole of a substance = 6.02 x 1023
  • Molar mass: The mass of one mole of a substance (element or compound)
    • Molar mass = Mr
    • Mr = addition of the mass of every element in the compound
    • Unit = g/moles
  • Law of definite proportions: A compound always contains exactly the same proportion of elements by mass.
  • Percentage yield: Measure the effectiveness of experiment
  • Empirical formula: simplest form of ratio
  • Molar volume: The volume occupied by 1 mole of any gas at r.t.p. 
  • Limiting reactant: A reactant is completely consumed in a chemical reaction
  • Excess reactant: The reactants left behind in a chemical reaction

Relative Atomic Mass

Comparing Atomic Masses with the Carbon Atom
  • To compare to a carbon atom, a carbon-12 atom is used.
  • The mass of the isotope is 12 times greater than hydrogen atom so of carbon-12 atoms is equivalent to the mass of one hydrogen atom.
  • Relative Atomic Mass: the average mass of one atom of the element (averaging isotopes) when compared with mass of a carbon-12 atom.
  • The Relative Atomic Masses are already stated on the periodic table above each chemical formula.

Relative Molecular Mass and Relative Formula Mass

  • Using Ar, we calculate Relative Masses of molecules and ionic compounds
Relative Molecular Mass
  • Molecules containes atoms joined together, e.g. Cl2
  • Average mass (molecular mass) of Cl2 = add relative masses of both atoms.
  • Relative Molecular Mass:
    • the average mass of one molecule of substance (averaging isotopes) when compared with mass of a carbon-12 atom.
    • same as relative molecular mass but for ions only
    • total Ar of all atoms in formula of ionic compound

Eg. Relative formula mass of MgSO4?

Mr = 24 + 32 + 4(16) = 120

Percentage Composition

Eg. Determine which oxides of iron of Fe2O3 or Fe3O4 has more iron.

Solution

Mr(Fe2O3) = 2(56) + 3(16) = 160

Percentage of Fe in Fe2O3
=
     x 100 %
= (2 x 56)/160 x 100%
= 70%

Mr(Fe3O4) = 3(56) + 4(16) = 232
Percentage of Fe in Fe2O3 = x 100 %
= x 100%
= 72%

Therefore, Fe3O4 has more iron composition than that of Fe2O3.

Calculating the Mass of an Element in a Compound


Use the example of Fe2O3 in the example above.
The percentage mass of iron in iron(III) oxide is 70%.
Therefore to calculate mass of iron in a 200g compound of iron(III) oxide is (0.7 x 200)g = 140g

Eg. Determine the mass of iron in 200g of Fe2O3.

Mr(Fe2O3)= 2(56) + 3(16) = 160
Mass of Fe in Fe2O3
=
    x 200g
= (2 x 56)/160 x 200g
= 140g

Calculating the Mass of Water in a Compound


Compound with water mass is ‘hydrated’ and has H2O in their formula.

Eg. Calculate water mass in 12.5g hydrated copper sulfate, CuSO4.5H2O


Mass of 5H2O in CuSO4.5H2O
=
x mass of sample
= (5 x 18)/250 x 12.5g
= 4.5g

Mole

Counting Particles
  • Unit for particles = mole
  • Symbol = mol
  • 1 mol = 6 x 1023 atoms

Moles of Particles - Calculating the Number of Moles

Eg 1: How many molecules in 6 x 1024 molecules of water, H2O?
= 5 mol

Eg 2: Calculate the number of molecules in 0.25 mole of CO2. Hence, how many atoms are present?

Number of particles = 0.25 mol x 6 x 1023
= 1.5 x 1023 molecules

Number of atoms = total number of atoms in CO2 x number of particles
= 3 x 1.5 x 1023
= 4.5 x 1023 atoms

Molar mass

  • Molar mass – the mass of one mole of any substances
  • For substances consisting of atoms
    • It is the Ar of the element in grams. Eg. Ar(C) = 12, molar mass = 12g
  • For substances consisting of molecules
    • It is the Ar of the substance in grams. Eg. Ar(H2O) = 18, molar mass = 18g
  • For substances consisting of ions
    • It is the Ar of substance in grams. Eg. Ar(NaCl)= 58.5, molar mass= 58.5g
Eg. Find the mass of 0.4 mol of iron atom.

n = m / Mr
m = n x Mr
m = 0.4 x 56 = 22.4 g

Eg. Argon Fluorohydride gas, HArF, first known noble gas compound, has molar mass of 60g. Find the number of moles Argon atom in 6.66g of HArF.

n (HArF) = 6.66/60
= 0.111 mol

n (Ar) = 0.111 mol x 1 Ar atom in HArF
= 0.111 mol

Different Kinds of Chemical Formulae

  • Molecular Formula – shows the actual formula and kinds of atoms present, e.g. C2H6
  • Empirical Formula – shows the simplest whole number ratio of the atoms present, e.g. C2H6, ratio 1:3, therefore C1H3, simply CH3
  • Structural Formula – shows how atoms are joined in the molecule. It can be represented by ball-and-stick model or diagrammatically.

Calculating the Empirical Formula of a Compound

Find the empirical formula of an oxide of magnesium consisting of 0.32g of oxygen and 0.96g of magnesium.

Step 1: find the number of moles of the 2 elements.

n(Mg) =0.96 / 24
= 0.04 mol

 n(O) = 0.32 / 16
= 0.02 mol

Step 2: Divide the moles by the smallest number.

Mg = 0.04 / 0.02
= 2

O = 0.02 / 0.02
= 1

Therefore, the empirical formula is Mg2O

Calculating the Empirical Formula from Percentage Composition

An oxide of sulphur consists of 40% sulphur and 60% oxygen.
Take the total 100% to be 100g.

Step 1: find the number of moles of the 2 elements.

n(S) = 40 / 32
= 1.25 mol

n(O) = 60 / 16
= 3.75 mol

Step 2: Divide the moles by the smallest number.
S = 1.25 / 1.25
= 1

O = 3.75 / 1.25
= 3

Therefore, the empirical formula is SO3

Changing From Empirical formula to Molecular Formula

Find the molecular formula of propene, CH2, having molecular mass of 42.
Molecular formula will be CnH2n

Relative molecular mass = 12n(from carbon Ar) + 2n(2 x hydrogen Ar) = 14n

14 n = 42
n = 42 / 14 = 3

Therefore --> C3H6

Molar Volume of Gases

The Avogadro’s Law: 

Equal volume of gases at same temperature and volume contain equal number of particles or molecules.

Molar Volume of Gas: 

  • volume occupied by one mole of gas
  • All gases at room temperature and pressure (r.t.p.) = 24dm3
  • 1dm3 = 1000cm3
Formulae:
Eg. What is the number of moles of 240cm3 of Cl2 at r.t.p.?

        = 0.01 mol

Molar Volume and Molar Mass

Gases have same volume but not necessarily same mass
Example: Hydrogen -> 2g, Carbon Dioxide -> 44g

Eg. Find the volume of 7g of N2 at r.t.p.

Step 1: Find the number of moles from the mass of nitrogen
n = 7 / 28
= 0.25 mol

Step 2: Find the volume of nitrogen, now with formula of gas
0.25 mol = volume of gas / 24

Volume of gas = 0.25 mol x 24
= 6 dm3 (or 6000cm3)

Concentration of Solutions

Concentration of solution tells the number of solute in a volume of solution

Calculating the Amount of Solute

Moles of solute (n) = Concentration (mol/dm3 ) x Volume of solution (dm3)

Eg. What is the mass of solute in 600cm3 of 1.5 NaOH solution?

Volume of solution in dm3 = 0.60 dm3
n = 1.5 x 0.60
= 0.9 mol

Number of moles of NaOH = m / Mr
0.9 = m / 40

m = 0.9 x 40
= 36g

Calculations using Chemical Equations

Constructing Chemical Equations

Eg. 1: Reaction Between Hydrogen and Oxygen

Word Equation: Oxygen + Hydrogen --> Water

To write the chemical equation, we use symbols of atoms/molecules:
O2 + H2 --> H2O

Balance the equation!
O2 + 2H2 --> 2H2O

Calculations from Equations

Reacting Masses

  • In every equation, each atom is rational to each other.
  • Suppose we want to find moles of X atoms that reacted to form 0.25 mole of Y atoms.
  • We always put the atom we want to find as numerator and the denominator being the atom we know.

Eg. X + 2Z --> 2Y


1. Find the ratio first:

2. Then multiply the ratio by no. of moles of Y to find the reacting mole of X.
1/2 x 0.25 = 0.125 mole

Therefore 0.125 mole of X reacted with 0.25 mole of Y.

3. To find the reacting mass of X, e.g. Y is given as 35g, we just multiply the mole by the mass of Y as they are always in ratio:

0.125 x 35 = 4.375 g

Reacting Masses and Volumes

First, find the ratio of moles and multiply the mole of the gas volume you want to find with the volume of gas at room temperature (24dm3)

Example

MgCl2 is formed by reacting Mg and HCl according to equation:
Mg (s) + 2HCl (aq) --> MgCl2 (s) + H2 (g)

Find the amount of hydrogen gas, in cm3, formed when 14.6g of HCl is reacted.

m(HCl) = 14.6/36.5
=0.4 mol

Multiply ratio by mole of HCl = 1/2 x 0.4 = 0.2 mol

Multiply mole by molar volume of gas at r.t.p. = 0.2 x 24 dm3= 4.8 dm3
1dm3 = 1000cm3

Therefore, 4.8dm3 x 1000 = 4800 cm3

4800cm3 of gas is formed

Example Questions

1 mole of CO2 is equal to

- 6.02 x 1023 molecules
- 2 moles of O
- 1 mole of C
- 44.0095 grams
- 22.4 L at r.t.p.
- 31.9988 grams of O
- 12.0107 grams of C



MCQ Questions

1. One mole of each of the following compounds is burnt in excess oxygen. Which compound will produce three moles of carbon dioxide and three moles of steam only?
a. C3H8
b. C3H7OH
c. C3H7CO2H
d. CH3CO2CH3

2. 20 cm3 of carbon monoxide are reacted with 10 cm3 of oxygen. The equation for the reaction is shown:
2CO + O2 --> 2CO2
Which volume of carbon dioxide will be produced?


a. 10 cm3
b. 20 cm3
c. 30 cm3
d. 40 cm3

3. What has a mass equal to that of one mole of water?
a. 24  dm3 of water
b. one mole of steam
c. one molecule of water
d. two moles of hydrogen molecules and one mole of hydrogen molecules

4. An 8 g sample of oxygen contains the same number of atoms as 16 g of element X. What is the relative atomic mass, Ar of X?
a. 4
b. 8
c. 16
d. 32

5. Which quantity is the same for one mole of ethanol and one mole of ethane?
a. mass
b. number of atoms
c. number of molecules
d. volume at r.t.p.

6. Which ion is present in the highest concentration in a 2 mol/dm3 aqueous solution of sodium sulphate?
a. hydrogen ions
b, hydroxide ions
c. sodium ion
d. sulphate ion

7. Magnesium reacts with hydrochloric acid. Which solution would give the fastest initial rate of reaction?
a. 40 g of HCl in 1000 cm3 of water
b. 20 g of HCl in 1000 cm3 of water
c. 10 g of HCl in 100 cm3 of water
d.  4 g of HCl in 50 cm3 of water

8. One mole of hydrated copper(II) sulphate, CuSO4.5H2O is dissolved in water. How many moles of ions does the solution contain?
a. 1
b. 2
c. 6
d. 7

9. What is the ratio of the volume of 2 g of hydrogen to the volume of 16 g of methane, both volumes at r.t.p.?
a. 1 to 1
b. 1 to 2
c. 1 to 8
d. 2 to 1

10. Which of the following contains the same number of molecules as 9 g of water?
a. 2 g of hydrogen gas
b. 14 g of nitrogen gas
c. 32 g of oxygen gas
d. 44 g of carbon dioxide gas

11. Calcium reacts with water as shown.
Ca (s) + 2H2O (l) --> Ca(OH)2 (aq) + H2 (g)

What is the total mass of the solution that remains when 40g of calcium reacts with 100 g of water?

a. 58 g
b. 74 g
c. 138 g
d. 140 g

12. 20 cm3 of oxygen are reacted with 20 cm3 of carbon monoxide. What are the volumes of the gases remaining, at the original temperature and pressure?

  oxygen/cm3 carbon monoxide/cm3 carbon dioxide/cm3
 a 0 020
 b 0 040
 c 10 0 20
 d 10 10 20

13. What is the mass of oxygen contained in 72 g of pure water?
a. 16 g
b. 32 g
c. 64 g
d. 70 g

14. A volume of ethane, C2H6, at r.t.p. has a mass of 20 g. What is the mass of an equal volume of propene, C3H6 at r.t.p.?
a. 20 g
b. 21 g
c. 28 g
d. 42 g

15. Sodium reacts with water according to the equation below.
2Na + 2H2O --> 2NaOH + H2
Which volume of hydrogen is produced at r.t.p. when 0.2 mol of sodium reacts?

a. 1.2 dm3
b. 2.4 dm3
c. 4.8 dm3
d. 9.6 dm3

16. What is the mass of aluminium in 204g of aluminium oxide, Al2O3?
a. 26g
b. 27g
c. 54g
d. 108g

17. The equation for the burning of hydrogen is: 2H2 (g) + O2 (g) ---> 2H2O (g)
One mole of hydrogen gas is made to react with one mole of oxygen gas. What will be present after the reaction?
a. 1 mol of steam only
b. 1 mol of steam + 0.5 mol of oxygen gas
c. 1 mol of steam + 1 mol of hydrogen gas
d. 2 mol of steam + 0.5 mol of oxygen gas

18. The compound SO2Cl2 reacts with water according to the equation:
SO2Cl2 + 2H2O ---> H2SO4 + 2HCl
How many moles of sodium hydroxide will neutralise the solution produced by one mole of SO2Cl2 and excess water?
a. 1
b. 2
c. 3
d. 4

19. How many atoms are there in (NH4)SO4?
a. 15
b. 14
c. 10
d. 7

20. 3Cu + 8HNO3 --> XCu(NO3)2 + 2NO + YH2O
What are the appropriate values of X and Y for the balanced reaction equation above?
a. X = 3, Y = 8
b. X = 4, Y = 3
c. X = 3, Y = 4
d. X = 4, Y = 8

21. What is the percentage by mass of calcium in calcium carbonate?
a. 4%
b. 12%
c. 40%
d. 44%

22. Which of the following compounds contain the same percentage by mass of nitrogen as ammonium cyanate, NH4CNO?
a. NH4NO3
b. NH4Cl2
c. N2H2
d. (NH2)2CO

23. A hydrocarbon contains 20% of carbon and 80% of hydrogen. What is its molecular formula if it has a relative molecular mass of 30?
a. CH3
b. CH4
c. C2H6
d. C2H4

24. 0.1 mol of metal X (Ar = 27) was burned in oxygen to give an oxide with mass of 5.1g. What is the formula of the metal oxide if it has a relative molecular mass of 30?
a. XO
b. XO2
c. X2O3
d. X2O

25. 100cm3 of gaseous hydrogen contains n molecules. How many molecules are there in 100cm3 of gaseous methane (CH4) under the same temperature and pressure?
a. n
b. n/5
c. 5n/2
d. 2n/5

26. Which of the following consists of the greatest number of atoms?
a. 4g of H2
b. 8g of O2
c. 71g of Cl2
d. 72dm3 of argon gas at rtp

27. Which of the following consists of the most number of molecules?
a. 32g of O2
b. 18g of water
c. 28g of N2
d. 4g of H2

28. Which of the following formulae for ionic compounds is not correct?
a. Mg(OH)2
b. KF
c. Al2O3
d. Ca2Cl

29. What is the maximum mass of chromium, Cr, that can be extracted from 76g of chromium(III) oxide?
a. 38g
b. 48g
c. 52g
d. 152g

30. How many oxygen atoms are present in 0.2 moles of N2O5?
a. 6.02 x 1023
b. 4.05 x 1023
c. 3.01 x 1022
d. 6.02 x 1021

MCQ Answers

1. d
2. b
3. b
4. d
5. c (Avogadro's law)
6. c
7. c
8. b
9. a
10. b
11. c (no. of moles of Ca = 40/40 = 1; no. of moles of water = 100/18. So Ca is the limiting reactant and H2O was present in excess. Mass of hydrogen gas liberated/lost = 2g. Mass of solution = 140 - 2 = 138 g)
12. c
13. c
14. c
15. b
16. d
17. b
18. d
19. a
20. c
21. c
22. d
23. c
24. d
25. a
26. a
27. d
28. d
29. c
30. a

Structured Question Worked Solutions

1. When iron is heated in a steam of dry chlorine, it produces a chloride that contains 34.5% by mass of iron.
a. calculate the empirical formula of this chloride

b. the relative molecular mass of this chloride (Mr) is 325.
i. what is the molecular formula of this chloride?
ii. hence construct an equation, including state symbols, for the reaction of iron with chlorine.

Solution

1a. molar ratio of Fe : Cl = 34.5/56 : (100 - 34.5)/35.5
                                             =   0.616   : 1.845
                                             =        1       :   3

Hence empirical formula is FeCl3

1bi. let the molecular formula be (FeCl3)n

n x (56 + 3 x 35.5) = 325
--> n = 2

Hence molecular formula is Fe2Cl6

1bii. 2Fe (s) + 3Cl2 (g) --> Fe2Cl6 (s)

2. CFCs are compounds that contain only carbon, chlorine and fluorine. They are atmospheric pollutants and destroy ozone in the upper atmosphere.

a. 'CFC11' has the following composition by mass

C: 8.7%
F: 13.8%
Cl: 77.5%

Calculate the empirical formula of CFC11

b. 'CFC12' has the molecular formula CF2Cl2. It can be made by the reaction of hydrogen fluoride, HF, with tetrachloromethane, CCl4

CCl4 + 2HF --> CCl2F2 + 2HCl

What is the maximum mass of CFC12 that can be made from 10.0 g of hydrogen fluoride?


Solution

2a.
  C F Cl
 % mass
 8.7 13.8 77.5
 Ar 12 19 35.5
 molar ratio
 8.7/12
= 0.725
= 1
 13.8/19
= 0.726
= 1
 77.5/35.5
= 2.183
= 3

Empirical formula: CFCl3

2b. Mr of HF = 20
Mr of CF2Cl2 = 121
2 x 20 g of HF give 121 g of CF2Cl2

Therefore, 10 g of HF give (121/40) x 10 = 30.25 g of CF2Cl2

3. Potassium superoxide, KO2 is an ionic solid. It can be used in spacecraft to supply oxygen according to the following equation.
4
KO2 (s) + 2H2O (l) --> 4KOH (s) + 3O2 (g)

The potassium hydroxide formed removes carbon dioxide.

a. show that 1.0 g of potassium superoxide will supply about 0.25 dm3 of oxygen at room temperature and pressure.

bi. name the compound formed when carbon dioxide reacts with solid potassium hydroxide

bii. give the equation for the formation of this compound

c. Supplies of oxygen in hospitals are stored in cylinders.
i. state one other use for oxygen
ii. describe briefly how oxygen is obtained from air


Solution

3a. Mr of KO2 = 71
4 x 71 = 284 g of KO2 give 3 x 24 = 72 dm3 of O2 at r.t.p.

hence, 1.0 g of KO2 give (72/284) x 1 = 0.25 dm3 of O2 at r.t.p

3bi. potassium carbonate

3bii. CO2 (g) + 2KOH (s) --> K2CO3 (s) + H2O (l)

3ci. together with acetylene in welding

3cii. oxygen is obtained by fractional distillation of liquid air

4. Many cars are fitted with air-bags which inflate in an accident. Air-bags contain the solid sodium azide, NaN3, which decomposes rapidly to form sodium and nitrogen. The nitrogen fills the air-bag.

a. construct the equation, including state symbols, for the decomposition of sodium azide.

b. in a crash, an air-bag fills with 72 dm3 of nitrogen at room temperature and pressure. What mass of sodium azide is needed to provide the nitrogen?

c. Sodium azide,
NaN3 reacts with dilute hydrochloric acid to give sodium chloride and a compound A.
Compound A contains 2.33% hydrogen and 97.7% nitrogen by mass.

i. what is the empirical formula of compound A?

ii. construct the equation for the reaction between sodium azide and dilute hydrochloric acid.


Solution

4a. 2NaN3 (s) --> 2Na (s) + 3N2 (g)

4b. no. of moles of nitrogen = 72/24 = 3

no. of moles of NaN3 required = 2

mass of NaN3 = 2 x (23 +  3 x 14) = 130 g

4ci.
 element HN
 % 2.3397.7
 Ar 114
 No. of moles
 2.336.98
 Ratio 13

Empirical formula: N3H

4cii. NaN3 + HCl --> NaCl + N3H

5. Lead white, a white pigment used in old paintings, contains lead(II) carbonate. It darkens when exposed to air containing traces of hydrogen sulphide, due to the formation of black lead(II) sulphide, PbS. The white colour can be restored by treating the painting with aqueous hydrogen peroxide which converts lead(II) sulphide into lead(II) sulphate and water.
ai. Write the chemical equation of the reaction between lead(II) sulphide and hydrogen peoxide.
aii. Calculate the volume of 0.100 mol/dm3 hydrogen peroxide required to react with 0.25 g of lead(II) sulphide.

b. The element germanium (Ge) was once an important component of transistors. The flow chart below shows how germanium can be made from its ore germanite.
When 1.00 g of germanite was treated in this way, the germanium present was completely converted into 0.177 g of a chloride containing 33.9% by mass of germanium.
i. Determine the empirical formula of the chloride.
ii. Write down the valency of germanium in the chloride.
iii. Calculate the percentage of germanium in germanite

Solution

ai. PbS + 4H2O2 ---> PbSO4 + 4H2O
aii. no. of moles of PbS = Mass / Mr
                                    = 0.25 / (207 + 32)
                                    = 0.001046
1 mole of PbS reacts with 4 moles of H2O2
0.001046 moles of PbS react with 0.001046 / 1 x 4 = 0.004184 moles of H2O2
Volume of H2O2 required = no. of moles / concentration
                                     = 0.004184 / 0.100
                                     = 0.0418 dm3

bi. 
ElementsGermanite Chloride 
Mass in %33.966.1 
Ar7335.5
No. of moles33.9 / 73 = 0.46466.1 / 35.5 = 1.861
Divide by smallest number0.464 / 0.464 = 11.861 / 0.464 = 4

--> Empirical formula = GeCl4

bii. valency = 4
biii. Mass of germanium in germanium chloride = 33.9 / 100 x 0.177
                                                                     = 0.0600 g
% of germanium in germanite = 0.0600 / 1.00 x 100%
                                            = 6.00% (3 sf)

6. Calculate the number of moles of the following:
a. 62.1g of BaCl2
b. H2SO4 in 55cm3 of 0.25 mol/dm3 H2SO4  solution
c. 17dm3 of nitrogen gas N2

Solutions

6a. Mr of BaCl2 = 208
no. of mol of BaCl2 = 62.1/208 = 0.2985 mol --> 0.3mol (2 dp)

6b. no. of mol of H2SO= concentration x volume = 0.25 x 55/1000 = 0.01375 --> 0.01 mol (2 dp)

6c. Mr of N2 gas = 28
no. of mol of N2 gas = 17/28 = 0.6071 mol --> 0.61 mol (2 dp)

7. Calculate 
a. the concentration of 32g of KOH in 500cm3 of water
b. the mass of iron (Fe) that can be extracted from 14.9g of iron(II) oxide (Fe2O3)
c. the volume of 100g of CO2 at normal atmospheric pressure and temperature

Solutions

7a. Mr of KOH = 56
no. of mol of KOH = 32/56 = 0.5714mol --> 0.57mol (2 dp)
conc of KOH = no. of mol/(500/1000) = 1.14 mol/dm3

7b. Mr of Fe2O3 = 160
Mass of Fe extracted = (56 x 2)/160 x 14.9 = 10.43g

7c. Mr of CO2 = 44
no. of mol of CO2 = 100/44 = 2.272 mol --> 2.27 mol (2 dp)
vol of CO2 = 2.27mol x 24dm3 = 54.48dm3

8. A solution of 250cm3 of ethanol in water contained 22.3g of ethanol, C2H5OH. Calculate the concentration of ethanol in g/dm3 and mol/dm3.

Solutions

8. Conc of ethanol in g/dm3 = 22.3 / (250/1000) = 89.2 g/dm3
Mr of ethanol = 46
conc of ethanol in mol/dm3 = 89.2/46 = 1.94 mol/dm3
 
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