Caloimetry and Molar Enthalpy

To measure heat absorbed/ released by water, we need to know:

• Temperature change (° C)
• Amount of Water (g, kg, mL, L)
• Specific Heat Capacity (kJ/kg ° C)- the heat needed to change 1 degree C in 1 kg

Example:
Calculate the amount of heat required to warm 400g of water from 20°C to 50°C.

ΔH = mCΔT
ΔH = (0,400kg)(4.19kJ/kg x T)(30°C)
ΔH = 50 kJ

Molar Enthalpy
- Heat absorbed / released by one mole

Example:
When a candle (C25H25) is burnt, heat is released according to the following reaction:

C25H52 + 38O₂ → 25CO₂ + 26H₂O +1100kJ

If 1.0g of wax is burnt, how much energy is released?

1.0g (mol / 352g)
= 0.00284mol

0.00284mol × 1100kJ / 1molC25H52
= 31.2 kJ/mol

Therefore: 31.2 kJ/mol of energy are released when 1.0g of wax is burnt

HEAT AND ENTHALPY

HEAT AND ENTHALPY

Ø Reactions that release heat are exothermic

Ø Reactions that absorb heat are endothermic

Ø Heat is a form of energy

Ø ENTHALPY - Stored energy

Ø Enthalpy of gasoline > Enthalpy of water

Ø Enthalpy symbol is H and change in enthalpy is ΔH

ENTHALPY GRAPHS:

EXAMPLES:
EXOTERMIC-
2C8H18 + 25O2 -----> 16 CO2 + 18H2O + 5076 kJ
2C8H18 + 25O2 -----> 16 CO2 + 18H2O ΔH= - 5076 kJ
ENDOTHERMIC-
3.2 C + 2H2 + 52.3 kJ -----> C2H4
3.2 C + 2H2 + -----> C2H4 ΔH= - 5076 kJ

MORE Practice:
State whether each of the following are exothermic or endothermic.
a.) H + Cl ---> HCl + 432 kJ EXOTHERMIC
b.) 12CO₂ + 11H₂O ---> C1₂H₂2O11 + 12 O1₂ ΔH= 5638kJ ENDOTHERMIC

jan 8 - types of chemical reactions

1) Synthesis

A+B ----> C (Two or more substances combine)
EX: H2 + Cl2 ----> 2HCl

6) Combustion


AB---->A+B (Breaking down into simpler substances)
EX: 2Ag2O---->4Ag + O2

3) Single Replacement

A + BX---->B + AX (Compunds must have a metal and a nonmetal-replacing one atom in a compound by anohter atom)
EX: Cl2 + 2KI ----> I2 + SKCl

4) Double Replacement

AB + XY ----> AY + XB (exchange of atoms between two different compounds)
EX: 2NaCl+ H2SO4 ----> 2 HCl + Na2SO4

5) Neutralization

(Products are water and an ionic salt/ Always between acids and bases)
EX: HCl + NaOH----> NaCL + H2O

6) Combustion

There are two types: Metallic (Can also be known as a synthesis reaction/includes oxygen) and hydro-carbon (includes carbon and oxygen). Also, the productes are always CO2 and H2O

EX: C5H12 + 8 O2 ----> 5CO2 + 6 H2O

Jan 6

Jan 6 - First class back from christmas break

Balancing with C, H, & O

1.) CH4 + O2 ---> CO2 + H2O
CH4 + O2 ---> CO2 + 2H2O
2.) CHH6 + O2 ---> CO2 + H2O
2CHH6 + 7O2 ---> 4CO2 + 6H2O
3.) C8H18 + O2 ----> CO2 + H2O
2C8H18 + 25O2 ----> 16CO2 + 18H2O

ALCOHOLS:
- Octane = most important chemical used to dilate fuel gas for cars
- OH means alcohol (Ex. C2H5OH = ethane)

Examples:
1.) CH3OH + O2 ---> CO2 + H2O
2CH3OH + 3O2 ---> 2CO2 + 4H2O
2.) C2H5OH + O2 ---> CO2 +H20
C2H5OH + 3O2 ---> 2CO2 +3H20

WORDS TO BALANCED EQUATIONS:

Example 1:
Aluminum chloride is mixed with potassium carvbonate. Aluminum carbonate and potassium chloride are formed.
Write the equation and balance:
AlCl3 + K2CO3 -----> Al2(CO3)3 + KCl
2 AlCl3 + 3 K2CO3 -----> 1 Al2(CO3)3 + 6 KCl

Example 2:
Aluminum metal reacts violently with bromine to produce aluminum bromide.
Al + Br2 -----> AlBr3
2 Al + 3 Br2 -----> 2 AlBr3

Example 3:
Magnesium sulphate hepta hydratee decomposes to form water and magnesium sulphate.
MgSO4 - 7 H2O -----> MgSO4 + 7 H2O


Acids - NEED TO KNOW:
HCl - Hydrochloric Acid
HNO3 - Nitric Acid
H2SO4 - Sulphuric Acid
H3PO4 - Phosphoric Acid
CH3COOH - Acetic Acid

EXTRA Practice:
1.) NH3 + O2 ---> NO + H2O
4NH3 + 5O2 ---> 4NO + 6H2O
2.) (NH4)2C2O4 + AlCl3 ---> Al2(C2O4)2 + NH4Cl
3(NH4)2C2O4 + 2AlCl3 ---> Al2(C2O4)2 + 6NH4Cl

3.) aluminum metal reacts with bromine to form aluminum bromide
2Al + 3Br ---> 2AlBr3

NOTE:
***************
In this post, we have included all the missing posts from the past month. The dates are not here, but all the information and class notes should be in this post. If we are missing anything, let us know and we will include it in the next post***********
density, molar volume & molar volume procedure lab, percent composition, empirical formulas, concentration, dilution
- Chem11403 Team (Joanne H, Roxanne S, Christian B)
______________________________________________

DENSITY

Density= Mass/Volume Mass= Volume*Density Volume= Mass/Density



Finding the density of gases at STP is easier then finding it at solids and liquids because we know that the volume of the gas will be 22.4L. We can also find the mass by checking our periodic tables. After you just plug in the numbers and solve for the density of the gas.

Molar Mass g / 22.4L/mol = ?g/L

density of oxygen =
Density = Mass / Volume

32g / 22.4L/mol = 1.43g/L at STP

Finding the density of solids and liquids are much harder you do not have all the information you need such as the volume.

use the formula g / molar mass of element. 6.02 x 10 to the power of 23 subscripts
Density ---> Mass ----> Moles ---> Molecules ----> Atoms

Example: the density of Boron (solid) is 2.34g/mL how many molecules are in a 60ml piece?
2.34 g/ml * 60.0 mL = 140.4g
140.4 g * 1mol / 10.8 g = 13mol
13mol * 6.02 * 10 to the power of 23 / 1mol = 7.84 * 10 to the power of 24

_________________________________________________
MOLAR VOLUME

D=M*V 22.4L/1mol subscripts
Density Volume (STP) Atom
-Molar Mass = g/mol
- Molar Volume = L/mol

Example
-A sample of unknown gas contains 0.635 mol and occupies a volume of 482mL. Determine the molar volume.
482mL x 1L = 0.482L = 0.759 L/mol
1000mL 0.635mol


MOLAR VOLUME PROCEDURE LAB

1. Fill the sink with 3\4 of water and place the lighter in so no air remains.

2. Dry the lighter and weigh it

3. Place the gradulated cylinder into the sink and once again make sure there is no air.

4. Place the lighter in the water under the gradulated cylinder and press the button. Add 10 milileters of gas.

5. Record the volume of the gas and the mass of the lighter.





PERCENT COMPOSITION

Perecent Composition
-means the % mass of each element in a compound.

Examples
Find the % of composition of K2Cr2O7

2K-78.2
2Cr-104
2O-11.2
Total mass: 294.2

Formula: Mass of Element/Total Mass

2K-78.2/294.2=26.6%
1C-104/294.2=35.4%
3O-11.2/294.2=38%
100%

EMPIRICAL FORMULA
Find the total mass of carbon in a 3.0kg sample of Ethanol (C2H6O)

2C --> 24 52.1%
6H --> 6 13%
1O --> 16 34.7%
46g/mol
(0.521)(3.0kg) = 1.57kg

Empirical Formulas = gives the whole number ratios of elements in a given compound. Molecular Formula gives the actual numbers.

Emprical Formulas:

Molecular Emprical
P4O10 P205
C10H22 C5H11
C6H18O3 C2H6O
C5H120 C5H12O
N2O4 NO2

A sample of an unknown compound is analyzed and found to contain 8.4g of C, 2.1 of H, and 5.6g of O

elements mass(g) atomic mass moles Moles/smallest mole
C 8.4 12 0.7mol 0.7/0.35=2
H 2.1 1 2.1mol 2.1/0.35=6
O 5.6 16 0.35mol 0.35/0.35=1

RATIO:
0.5 2
0.33 or 0.66 3
0.25 or 0.75 4
0.2, 0.4, 0.6, 0.8 5



MOLAR CONCENTRATION
Concentration -- solution-- a homogeneous mixture

Concentration: Amount of solute
Amount of Solvent

Some units for Concentration: g/ml ; g/l ; mg/l ; mg/ml ; ug/l {{Not very useful to us}}
**The most common(and useful) units are mol/l = molarity <-- molar concentration M= mol/L mol= M times L L= Mol/M all of these formula are only for aqueous solutions, not gases Example: Stephanie dissolves 40.0g of NaOH in enough water to make 200ml of the solution. What is the concentration? Concentration= 40.0g = 0.200g/ml 200ml 40g times 1 mol =1.0mol 40.0g M= 1.0mol = 5.0mol/L 0.200L Example: Kira wants to evaporate some 3.0M NaCl to obtain 26.325g of NaCl, What volume should she evaporate? 26.325g times 1mol times 1L = 0.15L <--- = 150ml 58.5g 3mol

DILUTION
-when you add water con’c decreases
-if the volume is doubled con’c is halved
- Volume | Con’c | moles
6.0 | 2.0 | 6L x 2 = 12 mol
12.0 | 1.0 | 12 x 1 = 12 mol
48.0 | .25 | 48 x .25 = 12mol
-N1 = N2; C1V1=C2V2
-Karol adds 150.0mL of water to 50.0mL of .60M HCl. Find [HCl].
V1= 50mL C1= .60M V2=200.0mL C2=?
C1V1=C2V2
C2=.15M
-Jesse adds water to 100.0mL of .35M to a final volume of 400.0mL. Find the [HF].
C1=.25M V1=100mL C2=? V2=400mL
C1V1=C2V2
C2=.0625M
-Cheyenne dilutes 60.0mL of 0.40M HNO3 to 0.15M. What is the volume? How much water did she add?
C1=.40M V1=60mL C2=.15M V2=?
C1V1=C2V2
V2=160mL
160-60=100mL

Here is an interesting video I found on youtube:


DILLUTION PT. 2
(Making directions for experiment procedures)
**First step is to find the amount of mass you will need.
Apply proper unit conversions to achieve this.

Example:
Jeremy is asked to make a 0.55M solution of K2SO4.If he needs 250mL what procedure should he use?
2K-78.2
1S-32.1
4O-64.0
174.3g/mol
250mL x 1L = .25L
1000mL
.25L x .55mol = .1375 mol x 174.3g =24.0g

STEPS:
1. Measure 250mL of water
2. Weigh 424.0g of K2So4
3. Add K2SO4 to water
-Give directions to make 2.00L of 6.0M NaOH
1Na-23.0
1O- 16.0
1H- 1.0
40.0g/mol
2L x 6mol = 12mol x 40g = 480g
ANSWER:
1. Measure 2.00L of water
2. Weigh 480g of NaOH
3. Add NaOH to water and stir until dissolved

OCTOBER 26, 2009 [JOANNE]

GASES AND MOLES:

The volume of a balloon occupied by a certain gas depends on the temperature & pressure.

Standard Ambient Temperature & Pressure (SATP)

• 24.8L / 1mol
• 25°C and 100kPa

*We will focusing much on SATP yet

STP:

The molar volume of any gas at STP is 22.4L

• 22.4L / 1 mol
• 1 mol / 22.4L

EXAMPLES:

Ø Find the volume occupied by 0.060 mol of c02 at STP
0.060mole x 22.4L/1mole = 1.3L

Ø Find the mass of a 200.0 mL sample of NO2 at SATP
STEPS:

o Note that the 200.0mL given is not in L

o Before we can convert it by SATP, we must change the mL to L.

o We divide 200.0mL by 1000 for our conversion factor from mL to L.

o Then we use SATP (22.4L / 1 mol )

o To finish the question in grams, we multiply by 46.0g because N = 14.0g + !6.0(2) O = 46.0g

o The Equation:

§ 200.0mL x 1L/1000mL x 1mol/22.4L x 46 g/ 1 mole = 9200/22400 =.41g

Ø Find the volume occupied by 22.0g of CO2(g)

o 22.0g x 1mol / 44.0g x 22.4L / 1mol = 11.2 L

Oct. 21st /2009 [THE Christian Bondoc]

Herro.

Atomic Mass

• The mass of 1 mole of atoms in an element
• The mass of 1.0 mol of 'C' atoms is 12.0g
• The mass of 1.0mol of 'Ca' atoms 40.1g.

Molecular Mass

• The mass of 1.0 mole of molecules of an element or compound

N2, O2, H2, Br2, Cl2, F2, I2, P4, S8

• Assume the all the rest are monoatomic

(I apologize for the size, when i made it large, part of it got cut off)

Finding the molar mass of compounds

H2O

2 H = 2(1.0) = 2.0

1 O = 1(16.0) = 16.0

18.0g/mol

Find the molar of ammonium phosphate

NH4+

PO4 3-

(NH4)3PO4

3 N = 3(14.0) = 42

12H = 12(1.0) = 12

1P = 1(31.0) = 31

4O = 4(16.0) = 64

149g/mol

Converting mass <----> Moles

Conversion factor g/mol or mol/g

Find the mass of 2.5mols of water

H20 -> 18.0g/mol

(1mol/18.0g) x (1/2.5mol)

(1 mol/18.0g) x (1/2.5mol)

= 1/45g

= 45g

Find the number of moles in a 391g samples of nitrogen dioxide

NO2

1N = 14(1.0) = 14

2O = 2(16.0) = 32

45g/mol

(391g / mol ) (1 / g) =

(391g / mol) (1/ g) =

(391/mol) (1/46) = 8.5mol

1) Find the mass of 2.5 moles of water.

2H: (2)(1)

1O: (1)(16)

= 18g/mol

2.5 mol • 18.0g /mol= 45g

2) Find the number of moles in a 391g sample of nitrogen dioxide.

1N: (1)(14)

3O: (3)(16)

=45g/mol

391g • 1mol/45g = 8.5 mol

3) 3.6kg of sulphur trioxide = ? mol

11S: (1)(32.1)

3O: (3)(16.0)

= 80.1 g/mol

3.6kg • 1000kg/1g = 3600 g

3600g • 1mol/80.1g = 45mol