Moles
Combustion Analysis
(C,H, and O) + O2 (g) ---> CO2 (g) + H20 (l) C,H,and O are the only factors that will change
Dilution Formula
(Initial Concentration)(Initial Volume)=(Final Concentration)(Final Volume) M1V1=M2V2
Limiting Reactant
1) Balance equation 2) convert given moles of one element to find # of moles present of the other 3) Determine Limiting Reactant (How many moles needed of one element to use all of the moles of the other) 4) Determine greatest amount of product that can be formed limiting reactant in calculated # moles (# of moles of product/moles of limiting reactant)
Combustion Steps
1) Find FW of CO2 and H2O 2) Convert CO2 and H2O to GRAMS of C and H Ex. (1 mol/FW CO2)(1 mol C/ 1 mol CO2)(FW C/1 mole C) & (1 mol H20/FW H20)(2 mol H/1 mol H2O)(FW H/1 mol H) 3) Find Oxygen Given total mass - (mass of C + mass H) 4) CONVERT TO MOLES. Divide each element by its atomic weight Ex. C= mass/atomic weight (12.01) H= mass/atomic weight (1.00797) O=mass/atomic weight (15.9994) 5) Divide by smallest number of moles 6) Determine empirical formula
net ionic equations
1) Write out equation and find charges 2) Balance Equation 3) breakup equation, leaving solid alone because it cannot be dissolved 4) Find and cross out spectator ions (ions on both sides of equation)
1 mole
6.022 x 10EE23 atoms (Avogadro's number)
Mole
A mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12 (i.e., 6.023 X 1023)
Percent Yield
Actual yield/theoretical yield x 100
Percent Composition
Mass of element/mass of compound x100
Heat absorbed
Q=M (C)(Delta T) Q=Heat absorbed by water m=mass of water c=specific heat of water (1 cal/1 g/C) delta t= change in temp of water SPECIFIC HEAT = 4.186 J/g Celsius Heat Absorbed is always in Joules
Specific Heat
Q=smt specific heat is measured in j/g degrees celsius
What is the mass in g of 4.30 moles of Aluminum
Step 1: Find the atomic mass of Aluminum Al=26.98 u 26.98 u = 1 mole Step 2: Multiply 4.3 (26.98) = 116 g
Empirical Formula
a) Assume every sample is 100g b) Convert % of each element to grams Ex. 72.7% Mg = 72.7 g c) Convert elements to moles (g--->moles) -Find Formula Weight of Mg (24.31g) d) Divide element by the smallest # of moles e) Make sure moles are a whole number Ex. If result is 1.5 moles, multiply by 2 to get 3
Molecular Formula
a) Molecular formula mass/empirical formula mass b) round up or down to get a whole number c) Multiply the empirical formula by that number Ex. Empirical formula = CH20 Number = 3 Molecular formula = (CH2O)3 = C3H6O3
Change in enthalpy
change= q + w
Mole Conversion
grams -->moles-->molecules (atoms)
Molarity
moles of solute/liters of solution
