FINAL Chapter 1/2/3 Flashcards
(14 cards)
CHAPTER 3
Find the molecular formula from combustion analysis (usually burned in excess oxygen).
You are given
- Mass of sample and mass of compounds, and you have to figure out the molecular formula of the sample
0.2524 g of their sample was combusted in excess oxygen to produce 0.6107 g of CO2, 0.1458 g of
H2O, and 0.1064 g of NO2. Find molecular formula of product.
1) Find the mass of carbon, hydrogen and nitrogen.
Find the moles of each compound then multiply by the molar mass of desired element and number of elements in the compound.
With hydrogen you would do
0.1458/18.02 x (2x1.008)
2) Find the mass of oxygen
mo = (total mass of sample) - (mass of carbon + mass of hydrogen + mass of nitrogen)
3) Find the moles of each substance, divide by smallest and multiply till whole!
4) Now find the molar mass of the empirical formula
Molar mass of substance/empirical formula, to figure out what factor to multiply each of the varibles by
CHAPTER 3
Is pouring a 25ml solution of volumetric flasks into a beaker a good way to make a precise solution?
- It is not a quantiative transfer, as you cannot guarantee that all the liquid leaves the flask. Use a pipette instead to transfer the solutions and then add distilled water until the solution reaches the meniscus at exactly 100 mL
CHAPTER 3
For any two-component aqueous solution (i.e. one solute in water),
there exists a mathematical relationship between molality (๐) and mole fraction (๐). Starting
from the expressions for molality and mole fraction of a hypothetical solute A in water, derive an
expression for mole fraction as a function of only molality.
Pretty easy question actually!
x = nA/nA + nW
now to rewrite moles of water, you can recognize that if you assume 1kg of water, then you can use m/M
nW = 1/0.018015kg/mol (make sure kg and not g) which equals 55.5
then for nA recognize taht based on the molality expression
m = nA/kgW, it is simply nA = m x kgW aka just m since kw = 1
Thus the expression would equal
m/m+55.5
CHAPTER 3
How do you find limiting reagent?
Moles/stoich coefficent
basically trying to figure out how many moles in 1
CHAPTER 3
Formula for density
mass/volume
kg/m^3 (same as kg/L) or g/mL
CHAPTER 3
Define strong and weak electrolytes
Strong is like strong acids and bases that completely dissociate into ions in solution and weak is like weak acids and bases that partially dissociate
Electrolytes conduct electricity (when in aq or molten (l) form) because the dissociated ions carry charges
Given molality and density, find molairity, mole fraction and %w/w
**even without asking to find %w/w this method is still useful
1) MOLAIRITY
- Calcualte moles of solute using the molality
Moles = molality x mass of solvent (assume 1kg)
= mol/kg x kg (donโt have to factor into g)
- Multiply moles of solute (same as molality) by the molar mass of solute to get mass of solute
- Combine with 1000g to get the total mass of solution
- divide this value by the density to get the volume
- you have the original moles of solution from the molality divided by this found volume
2) %W/W
- Mass of solute was determined in bullet 2 and then total mass of solution was also determined in step 3 so you can use these values
3)
What is
1) %w/w
2) %w/v
3) %v/v
1) g solute/100g solution
2) g solute/100mL solution
3) mL of solute/100mL solution
NOTICE that itโs always solute to solution
Given %w/w and density, find molairity, molality and mole fraction
1) MOLAIRITY:
- 98%w/w is simply 98g solute in 100g of solution. Multiply by density to cancel out solution
- Units are in g/mL, multiply by 1000 to make g/L
- Divide by molar mass of solute to cancel out the g for a final unit of mol/L
2) MOLALITY
- If 98% solute then 2% is solvent. 98% = 98g and 2% = 2g.
- You can find the moles of solute by dividing the 98/molar mass and the mass of solvent by taking the 2g and turning to kg
3) MOLE FRACTION
- If 98% solute then 2% is solvent. 98% = 98g and 2% = 2g. Set up as 98/2 and find the moles of each using the molar mass
- Divide desired mole value by total moles (total moles can be greater than 1 but the mole fractions SHOULD equal 1) to find the mole
Given molarity and molality, find mole fraction, density and %w/w
1) MOLE FRACTION:
- Use molality since it is in mol mol solute/kg solvent
- Find moles of solvent by converting to g and then using molar mass to convert to mol
- Add moles together to find total moles
- Divide mol of solute by total moles to find mole fraction
2) DENSITY:
- Calcualte mass of solute (moles is the molairity value given multiplied by molar mass of solute)
- Calcualte mass of solvent is moles divided by molality (mol x kg/mol) and turn into g
- Add both masses and assume volume of solution is 1L
- Density = total mass/volume
3) WEIGHT BY WEIGHT (use super easy strategy given in another q)
- Calculate mass of solute (use molairity for moles and molar mass)
- Calculate mass of solute (use moles of solute/molality which is mol * kg/mol and convert to g
- Find the total mass of the solution
- Divide desired mass by total mass and multiply by 100
**mole fraction assumed 1kg of water and density assumed 1L of water so this is why you used the moles of each
Given density and mole fraction, find, molairity and molality, %w/w
1) %W/W (FIND THIS ONE FIRST)
Suuuper easy!!
Letโs say mole fraction is 0.510 for acid. Then moles of water is 1-0.510 (assuming 1 mole total) which is 4.90. Now multiply both the by their respective molar masses to find the mass of each.
mass A/mass A + mass W * 100%
2) & 3) Just follow same method as above
When is molality equal to molarity?
When density is 1g/mL so volume is 1L and weights 1kg
CHAPTER 1
Significant figures for adding/sbstracting, multiplying/dividing and logs
+/- least number of decimal places
x/รท least number of sig figs
logs sig figs for inside log expression is number of decimals
-log(0.0250) is 1.979
3 sig figs = 3 deicmals
Given molarity and density find %w/w
3.10M and 1.31g/mL
1) You want to find mass of solute by multiplying molarity by its molar mass (assumes 1L of solvent)
2) Then you want to find mass of solution which is mass = density x volume and there are 1000mL involved
3) Divide the two!
