Semester 2 Final Exam Flashcards
(35 cards)
Empirical Formula
Simplest whole # ratio of atoms in a compound
Molecular Formula
Actual formula (which is a whole multiple of the empirical formula)
Molecular: H2O
Empirical: HO
Molecular: C6H12O6
Empirical: CH2O
Molecular: C6H6
Empirical: CH
How to find the Empirical Formula
1) Find the moles of each element
2) Determine the mole ratio (Largest/Smallest)
Calculate the Empirical Formula of a compound containing 13.43g of Al and 53.18g Cl
O.497 mol Al
1.50 mol Cl
1.50/0.479= 3/1
AlCl3
Calculate the Empirical Formula of a compound containing 52.14% C, 13.12% H, and 34.73% O.
- 35 mol C (2)
- 2 mol H (6)
- 17 mol O (1)
C2H6O
How to find the Molecular Formula from the Empirical
1) Given the molecular mass, divide by the empirical mass
Empirical Formula: CH2O
(empirical mass=molar mass=30)
Molecular mass=180
Ratio: 180/30 = 6
Molecular Formula: 6(CH2O) -> C6H12O6
How to find % Yield
Lab (actual)
——————————— X 100%
Theoretical (Calculated)
Variables that limit a gas (4)
1) Temperature
2) Volume
3) Moles
4) Pressure
What is KMT
Kinetic Molecular Theory
The movement of Molecules/Particles
What does KMT involve?
Particle Size
**Volume occupied by a gas is mostly empty space
Particle Motion
**Gas particles move in a straight line in all directions
**CONSTANT RANDOM MOTION
Particle Energy
**No energy is lost by collision of gas particle w/ walls of container
NO IDEAL GAS EXISTS
They approach ideal behavior under certain conditions of temperature & pressure
What would be the partial (dry) pressure of N2 gas collected over water at 20 degrees C and 720. torr?
P atm= P N2 + P H20
- torr= P N2 = 17.5 torr
703 torr= P N2
What kind of relationship is Pressure vs. Volume?
Inverse
What kind of relationship is Pressure vs. Temperature?
Direct
What kind of relationship is Volume vs. Temperature?
Direct
When do Mole Ratios=Volume Ratios?
When Pressure and Temperature is the same
The equation for Heat
Heat= mass X change in temp X Specific Head
Specific Heat
The amount of heat energy required to raise 1 g of a substance 1 degree C
Specific heat for water
4.18 J/g degree C
A 110.0 g sample of Iron @ 55.5 degrees C raises the temp of 150.0 mL of water from 223.0 degrees C to 25.5 degrees C. Determine the specific heat of the iron.
Heat Gain (H20)= Heat loss (Metal)
(gained by H2O)
Heat =150.0 g X 25 X 4.18
Heat =1600 J
(gained = lost)
1600 J= 110.0g X 30.0 X Cp
Cp= 0.48