Chemistry MDCAT UHS-1A,Lec#4

Topics

1. Percentage composition

2. Molarity 

3. Mole fraction

Phase

Every sample of matter with uniform properties and a fixed composition is called a phase

 Examples 

i)   Sample of pure water is a phase 

ii)  Homogeneous mixture of glucose in water is a phase.

 SOLUTION 

A mixture of two or more kinds of different molecular or ionic substances is called solution.

 Components of a solution

 Solute 

i) In the case of solution a gas or solid dissolved in a liquid, the gas or solid is the solute.

ii) Generally, the component of a solution, which is in smaller amount, is called solute.

 Solvent

i) In the case of solution of a gas or solid in a liquid,the liquid is the sovent.

ii) Generally the component of solution which is in larger amount is called solvent.

Example 

When sodium chloride is dissolved in water, sodium chloride is the solute and water is the solvent.

CONCENTRATION UNITS OF SOLUTIONS

1. PERCENTAGE COMPOSITION

Solute in percent of solvent.

Four ways of expressions of percentage composition.

i - w/w %

it is weight of solute dissolved per 100 grams of solution.

ii - v/w %

 it is volume of solute dissolved per 100 gram of solution.

iii - w/v %

 it is a weight of solute in grams dissolved per 100 cm3 of solution.

iv - v/v %

 it is volume of solute in cm3 dissolved per 100 cm3 of solution.

 All above expressed by percentage (%)

2.Molarity

The number of moles of solute dissolved per dm3(1000cm3) or 1 litre of a solution is called molarity.

 symbol

 it is denoted by 'M'.

Examples

 0.1 M NaOH solution and 0.02 M KMnO4 solution.

 formulae

i) Molarity(M)= Moles of solute/volume of solution in dm3 = n/v

ii) Molarity(M)= Mass of solute/ molar mass of solute × 1/volume of solution in dm3 = W2/M2 × 1000/V(cm3)

3.MOLE FRACTION

(Especially suitable for solution having more than two components)

The ratio of the number of moles of a component to the total number of moles of all the components of solution is called mole fraction.

Symbol

it is denoted by 'X'.

Explanation 

Let us having components A, B and C making a solution and their number of moles are nA,nB and nC respectively. XA ,XB and XC are their mole fractions and are given below:-

         XA = nA / nA + nB + nC

         XB = nB / nA + nB + nC

         XC = nC / nA + nB + nC

 •  The sum of the mole fractions of all the components of solution is unity (one)

•   To get mole percent, mole fraction is multiplied by 100.

•   There are no formal units of mole fraction because it is a ratio of same quantity.

•   Mole fraction of a component in a solution is always less than 1.

A comparison of various concentration units

• Molarity is temperature dependent and represent in its units as moles/ dm3 

• Molality is temperature independent and represent in its units as moles/kg

• Mole fraction is temperature independent and has no units.

• Parts per million(PPM) is temperature independent and represent in its units as PPM.

Multiple Choice Questions (MCQs)

1. 5.0 gram NaOH is dissolved in 1000 cm3 water. The molarity of NaOH is

A. 0.125 M

B. 0.250 M

C. 0.500 M

D. 1.00 M

2. 250 cm3 of 0.2 molar Potassium Sulphate solution is mixed with 250 cm3 of 0.2 molar potassium chloride solution.The molar concentration of potassium ions is

A. 0.2 Molar

B. 0.3 Molar

C. 0.25 Molar

D. 0.35 Molar

3. The number of molecules of glucose in 500 cm3 of its 1 molar solution are

A. 1.8 × 10^23

B. 1.2 × 10^24

C. 3.1 × 10^23

D. 6.02 × 10^23

4. 9 gram glucose is dissolved in 45 gram water. The mole fraction of glucose in solution is

A. 1/5

B. 5.1

C. 1/55.5

D. 1/51

5. A solution has 46 gram ethanol, 48 gram methanol and 45 gram water. Mole fraction of ethanol in the solution is

A. 0.1

B. 0.2

C. 0.4

D. 0.5

 6. A solution of glucose is 5%.The volume in which 1g mole of it will be dissolved is 

A. 1.0 dm3

B. 1.6 dm3

C. 2.5 dm3

D. 3.2 dm3

7. When a liquid solute is dissolved in a liquid solvent then the best unit of concentration is

A. w/w %

B. w/v %

C. v/w % 

D. v/v %

8. How many grams of NaOH are present in 250 cm3 of its 0.2 M solution

A. 1.0 g

B. 2.0 g

C. 20 g

D. 40 g

9. When we dissolve 15.8 gram of KMnO4 in 500 cm3 of water. The solution is

A. 0.1 M

B. 0.2 M

C. 0.3 M

D. 1.0 M

10. Which one is temperature dependent

A. Molarity

B. w/ w %

C. Mole fraction

D. Molality

Answers

1. A

2. B

3. C

4. D

5. B

6. D

7. D

8. B

9. B

10. A

Chemistry 2017_18 Part _1 pairing scheme

Chemistry 2017_18 Part _1-Q 1 MCQs-1,4,5,6,8,9 two each
     2,3,7,10,11.  One each
Q-2.   1/3,2/2,3/3,8/4
Q-3.   4/4,6/4,7/2,9/2
Q-4.   5/4,10/3,11/2
Q-5.    1(N)+4
Q-6..   3+5
Q-7.    6+7
Q-8.     8+10
Q-9.      9(N)+11

Schedule for submission of admission forms for intermediate part I & II (Annual) examination 2018 for all BI & SES in Punjab

CHEMISTRY MDCAT UHS Topic 1A Fundamental Concepts, Lec#3,Combustion analysis

d) COMBUSTION ANALYSIS

     Combustion analysis is a method used in both organic chemistry and analytical chemistry to determine the elemental composition (more precisely empirical formula) of a pure organic compound by combusting the sample under conditions where the resulting combustion products can be quantitatively analyzed. Once the number of moles of each combustion product has been determined the empirical formula or a partial empirical formula of the original compound can be calculated.

Applications for combustion analysis involve only the elements of carbon (C), hydrogen (H), nitrogen (N), and sulfur (S) as combustion of materials containing them convert these elements to their oxidized form (CO2, H2O, NO or NO2, and SO2) under high temperature high oxygen conditions. Notable interests for these elements involve measuring total nitrogen in food or feed to determine protein percentage, measuring sulfur in petroleum products, or measuring total organic carbon (TOC) in water.

Types of elemental analysis

   Quantitative analysis is the determination of the mass of each element or compound present.

   Qualitative analysis is to qualitatively determine which elements exist in a sample.

   The detection and identification of elements in a compound by their mass ratio is called combustion analysis.

The sequence of combustion analysis is shown in the following diagram

By combustion analysis only those organic compounds can be analysed which simply contain carbon,hydrogen and oxygen.

From the masses the percentages is calculated by using following formula

 i) 

 % of carbon = mass of carbon dioxide obtained in experiment / mass of organic compound × 12/44 × 100

ii)

% of hydrogen = mass of water obtained in experiment / mass of organic compound × 2.0/18 × 100

iii)

 % of oxygen = 100 - (% of carbon + % of hydrogen)

example:

A sample of liquid consisting of carbon,hydrogen and oxygen was subjected to combustion analysis 0.5439 grams of the compound gave 1.039 gram of CO2 0.6369 gram of H2O.Determine the empirical formula of the compound.

Solution:

Element: C

% : 1.039g/0.5439g × 12/44 ×100=52.108

No.of gram atoms: 52.108 /12 = 4.34

Atomic ratio: 4.34/2.17 = 2

Element: H

%  : 0.6369g/0.5439g×2.016 /18×100=13.115

No.of gram atoms:  13.115/1008=13.01   

Atomic ratio: 13.01/2.17 = 6

Element:  O

%  :  100 - (52.108 + 13.115)= 34.77     

No.of gram atoms: 34.77/16.00 = 2.17      

Atomic ratio: 2.17/ 2.17 = 1
Empirical Formula: C2H6O

Difference between empirical and molecular formula

 Empirical formula
 •   A formula which represents the simplest whole number ratio of atoms of elements in a compound is called empirical formula.
•   It is obtained from percentage composition of elements that is chemical analysis.
•    This term is used for both molecular and ionic compounds examples NaCl, CH2O and CH are empirical formula of sodium chloride ,glucose and benzene respectively.
Molecular formula
•   A formula which represents actual number of atoms of each elements in a molecular compound is called molecular formula.
•  It is obtained by multiplying 'n' with empirical formula i.e; from empirical formula.
•  this term is used only for Molecular compounds examples C6H12O6 and C6H6 are molecular formula of glucose and benzene respectively.
 Note:
The term empirical formula is used for ionic compounds and Giant covalent structures (sand SiO2 ,graphite and diamond C). It is also used for covalent compounds as CH2O for glucose and acetic acid.

Relationship between empirical and molecular formula

Molecular formula = n × empirical formula

STOICHIOMETRY

    Stoichiometry is a branch of chemistry which tells us a quantitative relationship between reactants and products in a balanced chemical equation.
Chemical equation
Chemical equation in the statement that describes chemical reaction in terms of symbols and chemical formula.
Limitation of Balanced chemical equations
 They do not tell about the
i) Conditions (temperature and pressure)
ii) Rate of reaction
iii) Physical state of reactants and products
 iv) mechanism of reaction
 v) We sometime write a chemical equation that never happens
Conditions for stoichiometric calculations 
stoichiometric calculations are based on the following conditions
i) All the reactants must be completely converted into the products.
ii) The side ration must not occur
iii) The law of conservation of mass and the law of Definite proportions must be obeyed while doing the calculations
The following types of relationship can be studied with the help of a balanced chemical equation at STP
2H2     +      O2       -------> 2H2O
2 moles       1 mole           2 moles
4g                32g.                 36g
44.828dm3 22.414dm3    44.828dm3

Mass-mass relationship
If we are given the mass of one substance we can calculate the mass of the other substance.
Mass-mole relationship or mole-mass relationship
 If we are given the mass of one substance we can calculate the moles of other substance and vice versa.
 Mass-volume relationship
if we are given the mass of one substance we can calculate the volume of the other substances and vice versa.
Mole-mole relationship
If we are given the mole of one substance we can calculate the mole of the other substances and vice versa.

Limiting reactant

  Limiting reactant is a reactant that controls the amount of the product formed in a chemical reaction due to being less than the required amount.It can also be defined as follows:-
 i)  It is a reactant that produces least number of moles of product.
ii) It is consumed earlier in the reaction.
 Identification of limiting reactant 
To identify a limiting reactant, the following three steps are performed.
i) Calculate the number of moles from the given amount of the reactants.
ii)Calculate the number of moles of product formed from the given moles of each reactant.
iii) identify the reactant as limiting reactant which produces least moles of the product.

Yield

 The amount of the product obtained as a result of the chemical reaction is called yield.
Types of yield
a) Theoretical Yield
 The amount of the product calculated from the balanced chemical equation is called theoretical yield.
b) Actual Yield
The amount of the product obtained in a chemical reaction experimentally is called actual yield.
c) Percentage Yield
% Yield=Actual yield/Theoretical yield×100
Point to be remembered:-
 Actual yield is always less than theoretical yield.
Reasons:-
i) A practially inexperienced worker has many shortcomings and cannot get the expected yield.
ii) The processes like filtration, crystallization etc if not properly carried out,decrease the theoretical yield.
iii) Some of the reactants might take part in a competing side reaction and reduce the amount of the desired product.

Multiple Choice Questions:

1. Amount of KOH used in combustion analysis
A) 50%
B) 60%
C) 70%
D) 80%

2. Amount of oxygen in combustion analysis is determined by method of
A) Product
B) Difference
C) Addition
D) Division

3. The H2O absorber used in combustion analysis is

A. Magnesium hyochlorite

B. Magnesium chlorite

C. Magnesium chlorate

D. Magnesium perchlorate

4. It is determined by combustion analysis that there is 0.1 g hydrogen is present in 1g of organic sample compound.The percentage of hydrogen in the sample is

A. 1%

B. 9%

C. 10%

D. 11%
5.The study of quantitative relationship between reactant s and products in a balanced chemical equation is called
A. Limiting Reactant
B. Yield
C. Stoichiometry
D. Mass spectrometry
6.All are assumptions of stoichiometry except
A. Balanced chemical equation
B. No side products formed
C. Physical states of reactants and products
D. Completion of reaction
7. The best stoichiometric relationship is
A. Mole - mole
B. Mole - mass
C. Mole - volume
D. Mole - volume
8. If we knows mass of reactant then we can calculate_______ of product formed.
A. Mole
B. Mass
C. Volume
D. All of above
9. The maximum yield which can be formed according to balanced chemical equation is called
A. Actual yield
B. Theoretical yield
C. Percentage yield
D. Average yield
10. The efficiency of chemicap reaction can be measure by its
A. Actual yield
B. Theoretical yield
C. Percentage yield
D. Average yield

Answers 

1. A

2. B

3. D

4. B

5. C

6. C

7. A

8. D

9. B

10. C

Explanation

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