Solutions

Chemical calculations1- molecular weight (Mwt)

the sum of atomic weights (Awt) of all elements in the formula.

Examples:

NaCl: Mwt = 23 + 35.5 = 58.5 g/mol

CaBr2 : Mwt = 40 + (2x80) = 200 g/mol

Fe2O3: Mwt = (56x2) + (16x3) = 160 g/mol

glucose C6H12O6:Mwt =(12x6) + (12x1) + (16x6) = 180 g/mol

.2-Mole-gram-atom calculations The mole of any objects (ions, atoms, molecules) = 6 x 1023 Of these objects. This number is called Avogadro number (given the symbol NA).Why did chemists choose this large number for chemical calculations?1- The weight of a mole of any element = Atomic weight of this element, i.e. 1 mole of an element = AwtExample: 1 mole of oxygen = 16g. 1 mole of sodium = 23g2- The weight of a mole of any compound= Molecular weight of this element, i.e. 1 mole of a compound = MwtExample: 1 mole of H2O= 18g. 1 mole of H2SO4 = 98g

Calculation of moles (n) in a given weight:

→ m = n x Awt

→ m = n x Mwt

Examples:1- Calculate the mass of 2 moles calcium. (Ca = 40) → m = n x Awt = 2 x 40 = 80 g.2- Calculate the mass of 0.25 moles of silver. (Ag = 108) → m = n x Awt = 0.2 5x 108 = 27 g.3- Calculate the mass of 0.374 moles of KCN. (KCN = 65) → m = n x Mwt = 0.374 x 65 = 24.31 g.4- Calculate the mass of 100 moles of Al2O3. (Al2O3 = 102) → m = n x Mwt = 100 x 102 = 10200 g.5- Calculate the mass of 0.08 moles of PCl3. (PCl3 = 137.5 ) → m = n x Mwt = 0.08 x 137.5 = 11 g

3- Percentage weight calculationsThe percentage by weight of an element (%wt) in a given formula is the total weight of this element in the formula divided by the molecular weight of this compound.

Example:1- Calculate the %wt by weight of all elements in glucose C6H12O6 (Mwt = 180). → For carbon: %wt C = 6 x 12 / 180 = 0.400 x 100 % = 40 % For hydrogen: %wt H = 12 x 1 / 180 = 0.0667 x 100 % = 6.67 % For carbon: %wt C = 16 x 6 / 180 = 0.5334 x 100 % = 53.34 %

2- Calculate the %wt of all elements in NH4NO3 (Mwt = 80). → For nitrogen: %wt N = 2 x 14 / 80 = 0.35 x 100 % = 35 % For hydrogen: %wt H = 4 x 1 / 80 = 0.05 x 100 % = 5 % For oxygen: %wt C = 16 x 3 / 80 = 0.60 x 100 % = 60 %

The percentage by weight is useful to calculate the mass of each element in a given mass of a compound. The % of the element is multiplied by the given mass of the compound.

m(element) = %wt x m(compound) Example:

1- Calculate the mass of sulfur in 5 g of SO3. (S = 32, 0 = 16) → Mass of sulfur = (32/80) x 5 = 2 g.

2- Calculate the mass of all elements in 8 g of Na3PO4. (Na = 23, P= 31, 0 = 16) → Mwt = (23 x 3) + 31 + (16x4) = 164 g/mol Mass of sodium = (23x3/164) x 8 = 3.36585 g. Mass of phosphorus = = (31/164) x 8 = 1.5122 g Mass of oxygen = = (16x4/164) x 8 = 3.1219 g Note: the sum of all masses should be equal to 8.0 exactly.

4-Chemical reactions calculationsThe chemical reactions occur according to molar ratios in the balanced chemical equation.Steps in calculations:

1- Convert mass into moles using molecular weight .

2- Compare moles of reactants and products

3- Obtain moles of unknown substance

4- Convert moles to mass using molecular weight

- Consider the balanced equation: C + O2 CO2 If 10 g carbon are reacted completely with oxygen, what mass of CO2 is produced? → moles C = 10/12 = 0.8334 mol 1 mol C 1 mol CO2

0.8334 mol C ? mol CO2

Hence moles CO2 = mol C = 0.8334 mol mass CO2 = 0.8334 X 44 = 36.66 g- Consider the balanced equation: Cl2 + H2 2HCl If 1.5 g of Cl are reacted completely, what mass of HCl is produced? → moles Cl2 = 1.5/35.5 = 0.0423 mol 1 mol Cl2 2 mol HCl

0.0423 mol Cl2 ? mol HCl Hence moles HCl= 2 mol Cl2 = 0.0846 mol mass CO2 = 0.0846 X 36.5 = 3.088 g

Consider the balanced equation: NaCl + AgNO3 AgCl + NaNO3 1- What mass of AgNO3 is required to reacts completely with 1.00 g of NaCl ? → moles NaCl = 1 / 58.5 = 0.0171 mol 1 mol NaCl 1 mol AgNO3

0.0171 mol NaCl ? mol AgNO3

Hence moles AgNO3 = mol NaCl = 0.0171 mol mass AgNO3 = 0.0171 x 169 = 2.888 g 2- What mass of AgCl will be produced from this reaction?→ moles AgCl = moles NaCl = 0.0171 mol mass AgCl = 0.0171 x 143.5 = 2.454 g

Chemical reactions do not go normally 100 % and the calculated yield is seldom achieved in the lab practically unless in ionic reactions which are very fast and complete.The following definitions are important:1- Theoretical yield (YTH): is the yield calculated theoretically depending on the moles.2- Practical yield (YPR): the actual yield obtained in the lab. It is always less than or equal to 3- Percentage yield (Y%): the ratio of practical yield to the theoretical yield multiplied by 100 %

Examples:1- in a particular reaction the actual yield was 5 g. the reaction was expected to give 7.9 g based on calculations. What is the percentage yield of this reaction?→ Y% = YPR/YTH X 100 % = [5/7.9] X 100 % = 63.29 %2- Consider P4 + 5O2 P4O10 If 10 g phosphorus were allowed to burn in enough oxygen, only 18 g of P4O10 were obtained. What is the percentage yield of P4O10 ? → Mwt P4 = 31 x 4 = 124 g / mol = (131 x 4) + (10 x 16) = 124 + 160 = 284 g/mol moles P4 = 10/124 = 0.08065 mol From the equation: 1 mol P4 1 mol P4O10 0.08065 mol P4 ? Mol P4O10 hence moles P4 = moles P4O10 = 0.08065 mol mass P4O10 = 0.08065 x 284 = 22.90 gPercentage yield of P4O10 ? Y% = YPR/YTH X 100 % = [18/22.9] X 100 % = 78.6 %

ACTIVITY 1- Consider the balanced equation: 2Na + Cl2 2NaCl calculate the mass of NaCl resulting from the reaction of 0.85 g Na with enough chlorine gas.2- In the reaction: 4Al + 3C Al4C3

Calculate the mass Al required to react completely with 8.88 g C. Also calculate the mass of the resulting product.3- In the reaction: 3NaOH + AlCl3 Al(OH)3 + 3NaCl a- what mass of NaOH will react completely with 2.4 g AlCl3? b- what mass of Al(OH)3 will form from the complete reaction of 11 g AlCl3 with enough NaOH?4- A reaction gave practically 4.4 g while expected to give 5.00 g theoretically. Calculate the percentage yield.

Pure water is not really pure. The purest water contains some hydronium ions and hydroxide ions. These two are formed by the self-ionization of two water molecules. This happens rarely. The process is an equilibrium where the reactants, intact water molecules, dominate the mixture. At equilibrium the molarities for the hydronium ion and hydroxide ion are equal. [H3O1+] = [OH1-]

self-ionization of water, Kw

The equation is H2O + H2O H3O1+ + OH1-

The equilibrium expression is the normal products over reactants.

K = [H3O1+] [OH1-] / [H2O] [H2O] The concentration for the water is a constant at any specific temperature. This means the equation can be rewritten as:K[H2O] [H2O] = [H3O1+] [OH1-]

The quantity on the right hand side of the

Equation " K[H2O] [H2O] = Kw " is formally defined as Kw. The numerical value for Kw is different at different temperatures. At 25oC Kw = 1.0 x 10-14

Kw = K[H2O] [H2O] Kw = [H3O1+] [OH1-] = 1.0 x 10-14

Kw, pH and buffer calculations Acidic solutions are those having

[H+] > 1 x 10-7 or [OH-] < 1 x 10-7 Basic solution are those having [OH-] > 1 x 10-7 or [H+] < 1 x 10-7 A neutral solution must have [H+] = = 7.00The pH of solution is the negative logarithm of the hydronium ion concentration. That is:

pH = - log[H+]Hence if H+ concentration is 1 x 10-3 then pH = 3Acidic solutions are defined to have pH < 7Basic (alkaline) solutions are defined to have pH > 7Neutral solutions are those with pH = 7.00 exactly.Similarly we can write for the hydroxyl ion:

pOH = - log[OH-]

What is the relation between pH and pOH?It is found that in all aqueous solutions:

[H+][OH-] = 1 x 10-14 = KwWhere Kw is called the water dissociation constant. If we apply the negative logarithm to the above equation we get:

pH + pOH = pKw = 14For example an acidic solution with pH = 4 must have pOH = 10 and a basic solution with pOH = 2.5 must have pH = 14-2.5 = 11.5

Examples: Calculate pH of the following solutions: 1- A solution with [H+] = 3.5 x 10-5 M→ pH = - log [H+] = -log 3.5 x 10-5 = 4.4562- An HCl solution that has molar concentration = 0.05 M→ HCl H+ + Cl- HCl ionize completely, hence [H+] = [Cl-] = [HCl] = 0.05pH = - log [H+] = -log 0.05 = 1.303- A basic solution with [H+] = 2 x 10-10 M→ pH = - log [H+] = -log 2 x 10-10 = 9.694- A basic solution with [OH-] = 4.4 x 10-4 M→ pOH = - log [OH-] = -log 4.4 x 10-4 = 3.356pH = 14 – pOH = 14 – 3.356 = 10.6445- A solution of KOH with molarity of 1.6 x 10-3 → KOH ionize completely, hence [OH-] = [K+] = [KOH] = 1.6 x 10-3 M pOH = - log [OH-] = -log 1.6 x 10-3 = 2.79pH = 14 – pOH = 14 – 2.79 = 11.21

Definitions•A solution is a homogeneous mixture

•A solute is dissolved in a solvent.–solute is the substance being dissolved

–solvent is the liquid in which the solute is dissolved–an aqueous solution has water as solvent

•A saturated solution is one where the concentration is at a maximum - no more solute is able to dissolve.

–A saturated solution represents an equilibrium: the rate of dissolving is equal to the rate of crystallization. The salt continues to dissolve, but crystallizes at the same rate so that there “appears” to be nothing happening.

Dissolution of Solid SoluteWhat are the driving forces which cause solutes to dissolve to form solutions?1. Covalent Covalent solutes dissolve by H-bonding to water or by London

dispersion forces (LDF).

2. Ionic solutes dissolve by dissociation into their ions.

Solution and Concentration

the expressing concentration by:the expressing concentration by:

Percent Composition by Mass (% m/m)Volume Percent (% v/v)Mole Fraction (X) Molarity (M) Molality (m) Normality (N)

* Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution.

1. Percent Composition by Mass (%):

This is the mass of the solute divided by the mass of the solution (mass of solute plus mass of solvent), multiplied by 100.

mass of the solute / mass of the solution x100% =

e.g:Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt.Solution:20 g NaCl / 100 g solution x 100% = 20% NaCl solution

2. Volume Percent (% v/v):

Volume percent or volume/volume percent most often is used when preparing solutions of liquids. Volume percent is defined as:v/v % = [(volume of solute)/(volume of solution)] x 100%Note that volume percent is relative to volume of solution, not volume of solvent.

For example: Solution is about 12% v/v ethanol. This means there are 12 ml ethanol for every 100 ml of H2O. It is important to realize liquid and gas volumes are not necessarily additive. If you mix 12 ml of ethanol and 100 ml of H2O, you will get less than 112 ml of solution.

3. Mole Fraction (X)

This is the number of moles of a compound divided by the total number of moles of all chemical species in the solution. Mole Fraction (X) =

Keep in mind, the sum of all mole fractions in a solution always equals 1.

the number of moles

the total number of moles

Example:What are the mole fractions of the components of the solution formed when 92 g glycerol is mixed with 90 g water? (molecular weight water = 18; molecular weight of glycerol = 92)Solution:90 g water = 90 g x 1 mol / 18 g = 5 mol water92 g glycerol = 92 g x 1 mol / 92 g = 1 mol glyceroltotal mol = 5 + 1 = 6 molxwater = 5 mol / 6 mol = 0.833xglycerol = 1 mol / 6 mol = 0.167It's a good idea to check your math by making sure the mole fractions add up to 1:xwater + xglycerol = .833 + 0.167 = 1.000

4. Molarity (M)Molarity is probably the most commonly used unit of concentration. It is the number of moles of solute per liter of solution (not necessarily the same as the volume of solvent!).

Molar Concentration calculations

The molar concentration (Molarity) of solutions is the number of moles of solute per liter of solution. That is:

Since n = m/Mwt , substituting in the above formula:

and m = Mwt x M x V(L)

Example:What is the molarity of a solution made when water is added to 11 g CaCl2 to make 100 mL of solution?Solution:11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2100 mL x 1 L / 1000 mL = 0.10 Lmolarity = 0.10 mol / 0.10 Lmolarity = 1.0 M

Examples:1- Calculate the molar concentration of a solution made by: A- dissolving 3 g MgBr2 in 500 mL aq. solution. → M = m/Mwt x V(L) = 3/(184 x 0.5) = 0.0326 M B- dissolving 1.11 g FeSO4 in 75 mL aq. solution. FeSO4 = 56 + 32 + 64 = 154 g/mol → M = m/Mwt x V(L) = 1.11 / (154 x 0.075) = 0.0961 M 2- What is the molarity of a 23 mL aqueous solution of NH4CN containing 0.023 g solute? NH4CN = 14 + 4 + 12 + 14 = 44 g/mol → M = m/Mwt x V(L) = 0.023 / (44 x 0.023) = 0.0227 M3- if 15 g of K3PO4 were dissolved in water and the volume completed to 2.5 L, calculate the molar concentration. K3PO4 = (39 x 3) + 31 + (16 x 4) = 212g/mol → M = m/Mwt x V(L) = 15 / (212 x 2.5) = 0.0283 M

5. Molality (m)Molality is the number of moles of solute per kilogram of solvent.

Molality (m) = moles of solute / kilogram of solvent

Because the density of water at 25°C is about 1 kilogram per liter, molality is approximately equal to molarity for dilute aqueous solutions at this temperature.

Example:What is the molality of a solution of 10 g NaOH in 500 g water?Solution:10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol NaOH500 g water x 1 kg / 1000 g = 0.50 kg watermolality = 0.25 mol / 0.50 kgmolality = 0.05 M / kgmolality = 0.50 m

6. Normality (N)Normality is equal to the gram equivalent weight of a solute per liter of solution. A gram equivalent weight or equivalent is a measure of the reactive capcity of a given molecule. Normality is the only concentration unit that is reaction dependent. Normality (N)= gram equivalent weight of a solute/ liter of solution

Example:1 M sulfuric acid (H2SO4) is 2 N for acid-base reactions because each mole of sulfuric acid provides 2 moles of H+ ions. On the other hand, 1 M sulfuric acid is 1 N for sulfate precipitation, since 1 mole of sulfuric acid provides 1 mole of sulfate ions.

Dilution

When a solution is diluted, solvent is added to lower its concentration.

The amount of solute remains constant before and after the dilution:

moles BEFORE = moles AFTER

C1V1 = C2V2

Suppose you have 0.500 M sucrose stock solution. How do you prepare 250 mL of 0.348 M sucrose solution ?

Concentration 0.500 M Sucrose

250 mL of 0.348 M sucrose

A bottle of 0.500 M standard sucrose stock solution is in the lab.

Give precise instructions to your assistant on how to use the stock solution to prepare 250.0 mL of a 0.348 M sucrose solution.

A bottle of 0.500 M standard sucrose stock solution is in the lab.

Give precise instructions to your assistant on how to use the stock solution to prepare 250.0 mL of a 0.348 M sucrose solution.

Example:How many millilieters of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?Solution:5.5 M x V1 = 1.2 M x 0.3 LV1 = 1.2 M x 0.3 L / 5.5 MV1 = 0.065 LV1 = 65 mLSo, to prepare the 1.2 M NaOH solution, you pour 65 mL of 5.5 M NaOH into your container and add water to get 300 mL final volume.

3 Stages of Solution Process• Separation of Solute– must overcome Intermolecular Forces (IMF)

or ion-ion attractions in solute– requires energy, ENDOTHERMIC ( + DH)

• Separation of Solvent– must overcome Intermolecular Forces (IMF)

of solvent particles– requires energy, ENDOTHERMIC (+ DH)

• Interaction of Solute & Solvent– attractive bonds form between solute

particles and solvent particles– “Solvation” or “Hydration” (where water =

solvent)– releases energy, EXOTHERMIC (- DH)

Factors Affecting Solubility

11. Nature of Solute / Solvent. Nature of Solute / Solvent. - Like dissolves like (IMF)

2. Temperature -2. Temperature -i) Solids/Liquids- Solubility increases with Temperature

Increase K.E. increases motion and collision between solute / solvent.

ii) gas - Solubility decreases with TemperatureIncrease K.E. result in gas escaping to atmosphere.

3. Pressure Factor -3. Pressure Factor -i) Solids/Liquids - Very little effect

Solids and Liquids are already close together, extra pressure will not increase solubility.

ii) gas - Solubility increases with Pressure.Increase pressure squeezes gas solute into solvent.

Definition of Chemical Equilibrium:

Chemical equilibrium applies to reactions that can occur in both directions. In a reaction such as:CH4(g) + H2O(g) CO(g) + 3H2(g)

The reaction can happen both ways. So after some of the products are created the products begin to react to form the reactants. At the beginning of the reaction, the rate that the reactants are changing into the products is higher than the rate that the products are changing into the reactants. Therefore, the net change is a higher number of products.

Even though the reactants are constantly forming products and vice-versa the amount of reactants and products does become steady. When the net change of the products and reactants is zero the reaction has reached equilibrium. The equilibrium is a dynamic equilibrium. The definition for a dynamic equilibrium is when the amount of products and reactants are constant. (They are not equal but constant. Also, both reactions are still occurring.)

Equilibrium Constant

To determine the amount of each compound that will be present at equilibrium you must know the equilibrium constant. To determine the equilibrium constant you must consider the following equation:

aA + bB cC + dD

Use the equation to determine the equilibrium constant (Kc).

For example: Determining the equilibrium constant of the following equation can be accomplished by using the Kc equation.Using the following equation, calculate the equilibrium constant.

N2(g) + 3H2(g) 2NH3(g)

N2(g) + 3H2(g) 2NH3(g)

A one-liter vessel contains 1.60 moles NH3, .800 moles N2, and 1.20 moles of H2. What is the equilibrium constant?

Le Chatelier's PrincipleLe Chatelier's principle states that when a system in chemical equilibrium is disturbed by a change of temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change of variable.

The three ways that Le Chatelier's principle says you can affect the outcome of the equilibrium are as follows:

1- Changing concentrations by adding or removing products or reactants to the reaction vessel.

2- Changing partial pressure of gaseous reactants and products.

3- Changing the temperature.

THE END