Any time you calculate molecular weight it should always have the units of g/mol.

Example:

H2O= 18.02g/mol

H= 1.01 x 2 = 2.02

O= 16.00 x 1= 16.00

= 18.02

Mole- is the amount of a substance that contains 6.02 x 10^23 representative particles of that substance.

Solution:

Solute- is what is being dissloved

Solvent- is what is doing the dissloving

Molarity(M)= moles of solute/ liters of solution

1 Liter = 1000 mililiters

A) A solution has a volume of 2.0 L and contains 36.0 g of glucose( C6H12O6). If the molar mass of glucose is 180 g/mol, what is the molarity of the solution?

Known:

M=mol/L

M=?

mol=?

L= 2.0

g=36.0

molar mass=180 g/mol

Step One:

180 g/mol x 1/36.0 g = 5 moles

Step two:

M= 5moles/2.0 L

M = 2.5

B) A solution has a volume of 250 mL and contains 0.70 mol NaCl. What is its molarity?

First Step:

250mL x 1L/ 1000 mL= 0.25 L

Second Step:

M= 0.70 mol/ 0.25 L

M= 2.8

C) How many moles of ammonium nitrate are in 335mL of 0.425 M NH4 NO3?

Step one:

335mL x 1L/1000 mL = 0.335 L

Step Two:

M=mol/ L

ML=mol

0.425 x 0.335L = 0.142 mol

D)

2) How many grams of sodium sulfate must be used to prepare 112 milliliters of a 0.23 M solution?

Dilutions- is when you are reducing the amount of moles of solute per unit volume, but the total number of moles of solute in solution does not change.

Moles of solute = M1 x V1 = M2 x V2

Example:

How many mL of 3.4 M hydrobromic acid would be needed to prepare 112 milliliters of 0.87 M hydrobromic acid solution?

Known/Given:

M1=3.4

M2=0.87

V1= ?

V2=112

Step 1 (convert):

112mL x 1L/1000mL =0.112L

Step 2

3.4M x V1 = 0.87M x 0.112L

Step 3

V1= 0.87M x 0.112L/ 3.4M

V1=0.0286L

Step 4

0.0286L x 1000mL/1L= 28.6mL

Concentation in percent( Volume/Volume) - if both of the solute and solvent are liquids

Percent by volume(%(v/v)) = (volume of solute/ volume of solution) x 100%

Example:

1)What is the percent by volume of ethanol in the final solution when 85mL of ethanol is diluted to a volume of 250mL with water.

(85mL/250mL) x 100% = 34 %

2) A bottle of the anitseptic hydrogen peroxide is labeled 3.0%. How many mL of hydrogen peroxide are in a 400.0mL bottle of this solution

3.0% = (Volume of solute/ Volume of solution) x 100%

3.0%/ 100% = Volume of solute/ 400mL

0.03 x 400mL = Volume of solute

12mL = Volume of solute

Concentration in Percent (Mass/ Mass) - Another way to express the concentration of a solution is as a percent (mass/ mass), which is the number of grams of solute in 100 grams of solution.

Percent by mass (%(m/m)) = (mass of solute/ mass of solution ) x 100%

1) How many grams of Potassium Sulfate would you need to prepare 1500g of 5.0% Potassium (m/m) solution?

5.0% = ( mass of solute/ 1500g ) x 100%

5.0%/100% = mass of solute/ 1500g

0.05 x 1500g = mass of solute

75 g = mass of solute

Vapor pressure- is the pressure that is exerted by a vapor that is in dynamic equilibruim with its liquid in a closed system.

Endothermic Reaction- is when the system absorbs energy form its surroundings in the form of heat

Exothermic Reaction- is when the system releases energy in the form of heat

There are three important colligative properties of solutions and those are vapor-pressure lowering, boiling-point elevation, and freezing-point depression.

Vapor pressure-is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system

Boiling-point- is the temperature at which the vapor pressure of the liquid equals the pressure surrounding the liquid and the liquid changes into vapor

Freezing-point- the temperature at which the liquid and solid phases of a substance of specified composition are in equilibrium at atmospheric pressure

Mole fraction- is the ratio of the moles of the solute to the total number of moles of solvent and solute

XA= nA/nA + nB  XB= nB/nA +nB 

Example:

Ethylene glycol (C2H6O2) is added to automobile cooling systems to protect against cold weather. What is the mole fraction of each component in a solution containing 1.25 mol of ethylene glycol and 4.00 mol of water?

XEG = nEG/ nEG + nH2O

The magnitudes of the freezing-point depression and the boiling-point elevation of a solution are directly proportional to the molal concentration, when the solute is molecular,not ionic

Ionic compound is a chemical compound in which ions are held together by ionic bonds(between metals and nonmetals)

Molecular compound is a compound where the atoms share electrons through covalent bonds( between nonmetals and nonmetals but can happen between nonmetals and metals as well but not common)

The unit molality(m) is the number of moles of solute dissolved in 1 kilogram of solvent. Molality refers to moles of solute per kilogram of solvent rather than moles of solute per liter of solution.

Molality= moles of solute/ Kilograms of solvent

Example:

How many grams of potassium iodide must be dissolved in 500.0 g of water to produce a 0.060 molal KI solution?

Knowns:

mass of water= 500.g= 0.5000kg

solution concentration=0.060m

molar mass KI= 166.0g/mol

According to te definition of molal, the final solution must contain 0.060 mol KI per 1000 g of H2O. Use the molality as a conversion factor convert from mass of water to moles of the solute(KI). Then use the molar mass of KI to convert from mol KI to g KI.

0.5000kgH20 x 0.060molKI/1.000kg H2O x 166.0gKI/1mol KI = 5.0g KI

∆Tf = Kf x m - is the equation for solving the Freezing-Point Depression

∆Tb = Kb x m- is the equation for solving the boiling-point elevation

k is a constant

Example:

Antifreeze protects a car from freezing. It also protects it from over heating. Calculate the freezing-point depression and the freezing point of a solution containing 100 g of ethylene glycol antifreeze in 0.500 kg of water.

Knows

- mass of solute = 100 g ethylene glycol

-mass of solvent= 0.500 kg H2O

-Kf for H2O = 1.86C/m

MW=62.0 g/mol EG

MW=18.02g/mol H2O

Calculate the number of moles of solute and the molality. Then calculate the freezing-point depression and freezing point.

100g EG x 1mol/62.00g = 1.61mol

m= mole solute/kg solvent = 1.61 mol/ 0.500kg = 3.22m

∆Tf = Kf x m = 1.86C/m x 3.22m = 5.99C

The freezing point of the solution is 0.00C - 5.99C = -5.99

The energy stored in the chemical bonds of a substance is called chemical potenital energy.

Heat is represented by the letter q and it is the energy that is transferred from one object to another because of a temperature difference between them.

Heat always flow from a warmer object to a cooler object.

The law of conservation of energy states that in any chemical or physical process energy is neither created nor destroyed.

In thermochemical calculation, the direction of the heat flow is given from the view point of the system.

An endothermic process is one that absorbs heat from the surroundings( the system gains heat as the surroundings cool down)

An exothermic process is onne that releases heat to its surroundings( the system loses heat as the surroundings heats up)

The two most common units for measuring heat flow is calories and joules.

1 Calorie= 1 Kilocalorie= 1000 calories

1 J =0.2390cal 4.184J= 1 cal4

Specific heat- is the amount of heat it takes to raise the temperature of 1 g of the substance 1 degree C.

Example:

The temperature of 48.4 g piece of copper increases from 27.0 C to 63.0 C when the copper absorbs 472J of heat. What is the specific heat of copper?

Knowns:

q= 472J

m= 48.4g

T(f)= 63.0C

T(i)= 27.0C

C= q/ m x ∆T

C =472J/ 48.4g x (63-27)

C=472J/ 48.4g x(36)

C= 472J/ 1742.4

C= 3.691

Example:

How much heat is required to raise the temperature of 250.0g of mercury 52 degrees C.

Known:

m=250.0g

∆T=52

C= 0.14 J/g x C

C= q/m x ∆T

(m x ∆T)C= q

( 250.0g x 52)(0.14 J/g x C) =q

13000 x 0.14 J/g x C= q

1820J=q

In calorimetry, the heat released by the system is equal to the heat absorbed by its surroundings. Conversely, the heat absorbed by a system is equal to the heat released by its surroundings.

The device used to measure the absorption or release of heat in chemical or physical processes is called a calorimeter.

Enthalpy- is the heat released or absorbed by a reation at constant pressure.( measured in Jolues)

q(sys)=deltaH=-q(surr)= -m x C x deltaT

Example:

When 25.0 mL of water containing 0.025 mol HCl at 25.0 C is added to 25.0 mL of water containing 0.025 mol NaOH at 25.0 C in a foam cup calorimeter, a reaction occurs. Calculate the enthalpy change (in kj) during this reaction if the highest temperature observed is 32.0 C. Assume the densities of the solution are 1.00g/mL.

Knowns:

C(water)=4.18 J/ g x C

V(final)= V(HCl)+V(NaOH)=50.00mL

T(i)= 25C

T(f)= 32C

Denisty=1.00 g/mL

Unknown= DeltaH= kj

50.0mL x 1.00g/mL= 50.0g

DeltaT= 32-25=7 C

DeltaH= - m x C x deltaT

DeltaH= -50.0g x 4.18 (J/ g x C )x 7 C

DeltaH= -1463 J

DeltaH= -1463 J x 1kj/1000J = -1.5kJ

Thermochemical Equations- a chemical equation that includes the enthalpy.

In a chemical equation, the enthalpy change for the reaction can be written as either a reactant(endothermic) or a product(exothermic).

The heat of reaction is the enthalpy change for the chemical equation exactly as it is written( basically on the left or right hand side of a chemical equation). You will usually see heats of reaction reported as delta H, which is equal to the heat flow at constant pressure.

H2O(l) ->H2(g) + ½ O2(g)   Delta H= 285.8 KJ

H2O(g) -> H2(g) + ½ O2(g)   Delta H= 241.8 KJ

Difference= 44.0kJ

Example:

Calculate the amount of heat (in kJ) required to decompose 2.24 mol NaHCO3(s)

2NaHCO3(s) + 129kJ ----> Na2CO3(s) + H2O(g) + CO2(g)

Knowns:

2.24mol NaHCO3(s) decomposes

Delta H= 129 kJ (for 2 mol NaHCO3)

Unknowns:

delta H= ?kj

Problem:

Delta H= 2.24 mol NaHCO3(s) x 129kj/ 2mol NaHCO3(s)

Delta H= 144kj

The heat of combustion- is the heat of reaction for the complete burning of one mole of a substance

Whenever a change of state occurs by a gain or loss of heat, the temperature of the substance undergoing the change remains constant.

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