MASS-VOLUME RELATIONSHIP

Week:                                                                   Date:                                                     Time:

Period:                                 Duration: 1 HR 20 MIN.                                                  Average age of learners: 16YEARS

Subject:                               CHEMISTRY                                                                         Class: SS TWO

Topic:                                    MASS- VOLUME RELATIONSHIP

Sub topic:

Reference materials:

(1) ESSENTIAL CHEMISTRY, TONALD PUBLISHERS, I. O ODESINA

(2) NEW SCHOOL CHEMISTRY, AFRICAN FIRST PUBLISHERS, OSEI YAW ABABIO

(3) INTERNET

Instructional materials: beaker, beam balance

Entry behavior: The students have been taught density.

Behavioural objective: At the end of the lesson the students should be able to:

1.       Define mole.

2.       State the relationship between relative densities and relative molar mass.

3.       Solve problems relate to mass- volume relationship

CONTENT

MASS AND VOLUME RELATIONSHIP

STOICHIOMETRY OF REACTIONS

QUANTITY

SYMBOL

UNITS

Mass

M

g or Kg

Molar Mass

M

g mol-1

 Volume

V

cm3 or dm3

Molar Volume

Vm

dm3 mol-1

Amount

N

mol

Molar Concentration

C

mol dm-3

Mass concentration

P

g mol-1

Avogadro’s Concentration

L

mol-1

Number

N

-

NB Molar Mass

M

g mol-1

 

RELATIONSHIP BETWEEN QUANTITIES

Molar Volume =  , M =      (b)  Molar Volume =  , M =     (c ) Avogadro’s Constant =  , L =

A mole is the amount of substance containing Avogadro’s number of formular units.

Mole = molar mass (weight in gm) = molar Gas Volume at STP 22.4dm3 = Formula units 6.02x1023 units (Avogadro’s Number).

The formula of a compound tells us the elements present in it. It also tells us the amounts of the different elements present in it. These amounts are usually expressed as a mole ratio of different elements in the compound.

Formula

                                      CaCO3

Elements present

Number of moles

Mole ratio

    Calcium                  Carbon                           Oxygen

         1                              1                                      3

         1            :                1              :                        3

                A balanced equation of a chemical reaction tells us the relationship of the amounts of the reactants to one another and to the products. This relationship between the amounts of the reactants and products is known as the stoichiometry of the reaction.

Mole ratios and mass relationships

A mole is the amount of substance which contains as many elementary particles as there are carbon atoms in 12g of carbon- 12.

The mole ratio in which reactants in which reactants combine and products are formed gives the stoichiometry of the reaction.

Example

                                        Zn (s)  +  2HCl(aq)                     ZnCl2(aq)  +  H2 (g)

Number of moles            1              2                                  1                    1

Mole ratio                           1     :       2            :                   1             :        1

Molar mass                         65.4g     36.5g                         136.4g         2g         

Number of moles            65.4g     (2 x 36.5g)             136.4g         2g  

Amount  x  Molar mass    =  Reacting mass

                Amount =   

EXAMPLE

Calculate the number of moles of magnesium chloride produced by reacting 168g of magnesium trioxocarbonate (iv) with excess hydrochloric acid. [ Mg =24, C =12, O =16, H =1, Cl =35.5].

Solution

Equation of reaction

MgCO3(s)  +2HCl(aq)                MgCl2(aq)   +   CO2(g)  + H2O(l)

From the equation,

1 mole of MgCO3(s)  produce 1 mol of MgCl2(aq)

Molar mass of MgCO3 = Mg + C + 3O

                                                = 24 + 12 + (3x16)

                                                = 36 + 48 = 84gmol-1

Number of moles of MgCO3 reacted =   =   = 2mol

2 mole of MgCO3(s)  produce 2 mol of MgCl2(aq)

 REACTIONS INVOLVING GAS VOLUMES

Equal volume of gases at a given temperature and pressure contain the same number of molecules. We can calculate the volumes of gases from a balanced chemical equation, provided that the gases are under the same conditions of temperature and pressure. The coefficients of gaseous reactants and products in a balanced equation give the mole relations as well as the volume relation among the gases.

4NH3(g) +  5O2(g)                    4NO(g)   +         6H2O(g)

4 vol      :  5 vol                               4 vol      :    6 vol

4 mol     :  5 mol                           4 mol      :    6 mol

4 volume of ammonia react with  5 volume of oxygen to produce 4 volume of nitrogen (ii) oxide and 6 volumes of water vapour.

One mole of any gas at s.t.p. molar volume occupies 22.4 dm3. Thus , we can calculate the volumes, masses, moles and number of molecules of gaseous reactants and products from a balanced chemical equation.

Example

What volume of steam is produced when 10g of propyne is burnt in excess oxygen at s.t.p.?

Solution

2C3H6(g)  +  9O2(g)               6H2O(g)  + 6CO2(g)

 1 mol    :  4.5 mol              3 mol    :   3 mol

1 mol of propyne  produce 3 vol of steam at s.t.p.

(3x12)+(1x6)]g propyne  produce 3x 22.4 dm3 of steam at s.t.p

42g of propyne produce 67.2 dm3 of steam at s.t.p

10g of propyne produce    x10g of steam at s.t.p

42g of propyne produce 16 dm3 of steam at s.t.p

PRESENTATION

i.The teacher explains mole and mole ratio.

ii. The students chorus the definition of mole and mole ratio

iii. The teacher guides the students in solving some problems on mass-volume relationship

EVALUATION

 The teacher evaluates the lessons by asking the following questions:-

1.       Define mole

2.       Calculate the relative molecular mass of 0.4g of oxygen occupied at s.t.p 280cm3

ASSIGNMENT

If 2g zinc react with excess hydrogen chloride acid. Calculate (a) the number of mole of hydrogen liberated (b) the volume of hydrogen liberated [Zn =63] (ans- (a) 0.032mol ; (b) 716cm3)

  

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