Jan 18 Class - Lab
Heat of Combustion of Candle (paraffin) Wax

Materials:

• candle
• matches
• weigh scale
• lab stand
• wire mesh or triangle
• thermometer
• water

Procedure:

1. Set up the lab stand and wire mesh (low enough for fire to heat up)
   
2. Measure 150 mL of cold water and pour it into the metal can, using a graduated cylinder.
3. Record the temperature of the watter.
   
4. Record the mass of the candle before lighting it.
5. Place the candle under the can and carefully light it
6. Let the candle burn for aprx. 5 minutes. Record the new temperature of the water. Record the mass of the candle after the experiment.
7. Clean up your lab station.

Observations:
These are my groups observations...
Volume of water used: 150 mL
Initial Temperature of Water: 21°C
Initial Mass of Candle: 10.46g
Final Temperature of Water: 26°C
Final Mass of Candle: 9.8g

The accepted value for the molar heat of combustion of Paraffin Wax is aprx. 14500. We concluded that it was 1652. We were 88% wrong. Some possible sources of error were when the water and wax spilled, when we did not leave the thermometer in long enough, and when we did not record as precisely.

Calorimetry - Jan14 Class

Calorimetry
- To measure heat absorbed/released by water we need to know 3 things:
1) Temperature change °C
2) Amount of water (g, kg, ml, L)
3) Specific heat capacity (kJ/kg °C)

∆H = mC∆T
m = mass of water
C = specific heat capacity 4.19kJ/kg °C
∆T = change in temperature

Examples: Calculate the heat required to warm 400g of water from 20°C to 50°C
∆H = mC∆T
∆H = (0.400kg)(4.19 kJ/kg °C)(50° - 30°)
∆H = 50 kJ

75.0 kJ are added to 850 ml of water initially at 25.0°C. What will the final temperature be?

∆H = mC∆T

∆H = mC (Tf - Ti)

Molar Enthalpy: Heat absorbed/released by one mole

When a candle (C25H52) is burnt, heat is released according to the reaction:
C25H52 + 38 O2 -> 25 CO2 + 26 H2O + 11 000 kJ
If 1.0g of wax is burnt, how much energy is released?
1.0g x 1.0mol/352g = 0.00284mol

11 000kJ / 1mol C25H52 x 0.00284 mol C25H52 = 31.3 kJ

CALORIMETRY OF A BOMB!

Heat & Enthalpy - Jan. 12

In today's lesson we learned about Heat, Enthalpy, and the reactions.

-Reactions that release heat are exothermic. (Hot)
-Reactions that absorb heat are endothermic. (Cold)
-Heat is a form of energy
-All chemicals have energy stored in them
-Stored chemical energy is called enthalpy

Enthalpy of Gasoline > Enthalpy of Water







- Enthalpy has a symbol H
- Change in enthalpy is:






Enthalpy Diagrams (Exothermic & Endothermic)
Exothermic
- H is negative
- Products are lower than reactants
- Heat is released

Endothermic
- H is positive
- Reactants are lower than products
- Heat is absorbed




Two ways to write enthalpy in reactions













- Coefficients can stand for moles or molecules








1 mole of N2 produces 46.3 kJ
3 mole of H2 produces 46.3 kJ
2 mole of NH3 produces 46.3 kJ

This gives conversion factors:
46.3 kJ/ 1 mol N2
46.3 kJ/ 3 mol H2
46.3 kJ/ 2 mol NH3


Ex. Find the amount of heat released if 7.0 mol of N2 are consumed.
46.3 kJ/ 1 mol N2 x 7.0 mol of N2 = 324.1 kJ

Jan 6 Class Balancing Chemical Equations

Balancing with C, H & O (combustion)
To make combustion questions easier to balance, you can do them in particular steps.
1. carbons
2. hydrogens
3. oxygens

2C8H18 + 25O2 -> 16CO2 + 18H2O
C8H18 + 25/2O2 -> 8CO2 + 9H2O
Both are the same.

Alcohols
C2H5OH + 3O2 -> 2CO2 + 3H2O

Words -> Balanced Equations
EX: Sodium Chloride is mixed with Potassium Carbonate. Sodium Carbonate and Potassium Chloride are formed. Write a balance equation.

2NaCl + K2CO3 -> Na2CO3 + 2KCl


ACIDS
Name Formula
Hydrochloric acid HCl
Nitric Acid HNO3
Sulphuric Acid H2SO4
Phosphoric Acid H3PO4
Acetic Acid CH3COOH

Hydrogen makes it acid. -> (almost all)

EX: Nitric Acid is neutralized by barium hydroxide.
2HNO3 + Ba(OH)2 -> Ba(NO3)2 + 2HOH

December 4, 2009 (molecular formulas)

Finding Molecular Formulas
It is easy to do if you know the empirical formula and the Molar Mass.

If the empirical formula for a compound is CH2O and the molar mass is 90.0g/mol. Find the molecular formula.
Stock Solutions

• Are concentrated solutions from suppliers

12 M HCl
15 M HNO3
18M H2SO4

Makiki wants to make 200mL of 0.40M HCl. What volume of 10.0M HCl should she use?
V1= ? V2= 0.200L C2V2 = C1V1
C1= 10.0M C2= 0.40M (0.40M)(0.200L) = (10.0M) V1
V1 = 0.008L/ 8mL

Giving Directions - Dec2 Class

In today's class, we learned about which procedures we should use when it comes to concentrations. We also did a pre-lab about how much water we would need to dissolve Copper Chloride completely. We will be doing the lab this friday, Dec4.

Outline for experimental procedures:
- Find the mass you need

Johnny is asked to make 0.95M solution of K2S04. If he needs 300mL, what procedure should he use?
Con'c -> moles -> mass

Give directions to make 3.5 L of 4.5M NaOH

DILUTIONS OF SOLUTIONS
- When you add water, con'c decreases. If the volume is doubled, con'c is halved.

N1 = N2
C1V1=C2V2
Lauren adds 215.0 mL of water to 25.0 mL of 0.8M HCl. Find [HCl]
Janice adds water to 75.0 mL of 0.15M HF to a final volume of 310mL. Find the [HF]
Nicola dilutes 75.0 mL of 0.30M HNO3 to 0.10M. What is the final volume?

How much water did she add?
225mL-75mL = 150mL

Concentration- Nov. 30

In Tuesday's class we learned about concentration and its units.

Vocabulary:
Solution- A homogeneous mixture
Solute- The one present in smaller amount
Solvent- The one present in greater amount
Concentration- Amount of solute/Amount of solvent

Some units for Concentration:
g/mL, g/L, mg/L, mg/mL, ug/L

The most common (and useful) units are:
mol/L = Molarity = Concentration

Examples:
Stefano dissolves 108g of NaBr in enough water to make 300mL of the solution. What is the concentration?
Con'c = 108g/300 mL = 0.36g/mol [NaBr] = ? 0.300L = V

108g x 1 mol/108g M = 1.0mol/0.300L = 3.33mol/ L = 3.33M

Jayson wants to make 600.0mL of 0.60M CaCl2. What mass of solid CaCl2 is required?
V = 0.600L con'c ---> mol ---> mass
mol = M x L = 0.6m/L x 600L = 0.36mol x 111.1g/1mol = 39.996 or 40g