Stoichiometry Flashcards
Chromatography
The Process where a mobile phase (Which Contains the mixture to be analysed) moves over a stationary phase. As this happens, a process of adsorption and desorption leads to some components of the mixture moving faster over the stationary phase than others
Adsorption
The process by which molecules stick to the surface by other particles. It commonly involves liquid or gas molecules sticking to the surface of solid particles.
HPLC
High performance liquid chromatography is a method for measuring the concentration of organic compounds in a sample. It allows organic molecules at low concentrations to be separated from a mixture. These components can then be quantitatively.
Paper and Thin layer Chromatography
Submerging the edge of a paper into a solvent and watching the sample rise up the paper and undergo a continual process of adsorption and desorption. It aims to seperate out the components in the mixture and identify the presence of a component
Identifying sample components in thin layer chromatrography.
Rf (Retardation factor). This value describes how far the component has travelled up the stationery phase.
rf= 0 Means that component remaind at stationery phase
rf = 1 Means that component is at solvent front and showed no affinity for the stationery phase
rf = inbetween means it showed affinity for both stationary and mobile phase.
Equation to find length (Paper Chromo)
Distance travelled divided by total length from origion to solvent end.
A Chromatogram provides information about
The identity each component of a sample via the rentention time (Qualitative Anaysis).
The Concentration of each component of a sample via peak area (Quantitiatve Analysis)
Order of Molecules From most to least Polar
Most Polar
- Carboxylic acid (Hydrogen, dispersion , dipole, 3 Polar bonds)
- Alcohol (Hyrdogen, Dipole, Dispersion, 2 Polar Bonds)
- Ester (Dipole, Dispersion)
- Alkane (Weak Dispersion)
Concentration of Cl in 250ml (V) of solution 52.2g (Mass) of Aluminium chloride (AlCl3) is
4.76M
n(AlCl3) = 52.9 divided by 133.5 = 0.396 n(Cl) = 3 x 0.396 = 1.18 mol C = 1.18 divided by 0.25 = 4.76M
Source of Error Titration (Rinsing water left in Burette)
Burette (Underestimated) Titration Flask (Overestimated)
Source of Error Titration (Rinsing Water left in Pipette)
Burette (Overestimated) Titration Flask (Underestimated)
Source of Error Titration ( Indicator Chosen Changes Colour too Soon)
Burette (Overestimated) Titration Flask (Underestimated)
Source of Error Titration (Water in Titration Flask)
Burette (No effect) Titration Flask (No effect)
Source of Error Gravimetric (Insufficient Precipitating agent was added)
Underestimated
Source of Error Gravimetric (Precipitate not fully dried before weighing)
Overestimated
Source of Error Gravimetric (Precipitate not washed prior to drying and weighing)
Overestimated
Source of Error Gravimetric (Other Ions were present in the sample that reacted with the precipitating agent)
Overestimated
Source of Error Gravimetric (Using too much water for the initial dissolving)
No effect
Source of Error Gravimetric (Forming a precip thats too soluble)
Underestimated
Limestone contains 95% CaCO3 by mass. The reaction is
CaCO3 (s) + 2HCL (aq) -> CaCl2 (aq) + CO2 (g) + H2O (g)
0.750 of Co2 was evolved. The amount in mol of HCL acid reacting was
0.750 divided by 44 = 0.017 x 2
= 3.41 x 10 to the power of negative 2
Limestone contains 95% CaCO3 by mass. The reaction is
CaCO3 (s) + 2HCL (aq) -> CaCl2 (aq) + CO2 (g) + H2O (g)
The mass of lime stone added to the acid was
- 17x 100.1 (Caco3) = 1.1
- 7 x 100 Divided by 95 =
- 80g.
What will least likely alter the retention time of a substance passing through the column in a HPLC analysis
Diluting the sample being analysed (Will change peak but not retention time)
Things that will alter retention time in a HPLC analysis
- Replacing one solvent with another
- Increasing the temp of the column
- Decreasing the rate of flow of the mobile phase
How can Chromium be selectively analysed without interference from other metals
Because AAS has a hollow cathode lamp which is made from elements under investigation
How can a particular wave length used for analysis is chosen in UV- visible spectrometry
We will choose length that is highly absorbed in the component. We will also choose length that is exclusively absorbed. This assures accurate results.
The amount of NaCl in water was determined by Gravimetric Analysis.
A 25.0ml of water was mixed with 100ml of distilled water, stirred and mixture filtered. The Volume of filtrate was made up to 250ml.
A 20ml aliquot of solution was pipetted into a conical flask and an excess of AgNo3 solution was added.
The Precip weighed 0.150g
AgNo3+Nacl -> AgCl+ NaNO3
Calc amount in mol of precip
0.150 divided by 143.4 = 1.046x10^-3 mol
The amount of NaCl in water was determined by Gravimetric Analysis.
A 25.0ml of water was mixed with 100ml of distilled water, stirred and mixture filtered. The Volume of filtrate was made up to 250ml.
A 20ml aliquot of solution was pipetted into a conical flask and an excess of AgNo3 solution was added.
The Precip weighed 0.150g
AgNo3+Nacl -> AgCl+ NaNO3
Calc amount in mol of NaCl in the 250ml solution
1:1
n (Nacl) =n (AgCl) = 1.046x10^-3
250 divided by 20 x 1.046x 10^-3 =
0.0131 mol
The amount of NaCl in water was determined by Gravimetric Analysis.
A 25.0ml of water was mixed with 100ml of distilled water, stirred and mixture filtered. The Volume of filtrate was made up to 250ml.
A 20ml aliquot of solution was pipetted into a conical flask and an excess of AgNo3 solution was added.
The Precip weighed 0.150g
AgNo3+Nacl -> AgCl+ NaNO3
Calc Concentration in %m/v of sodium chloride in water
m (Nacl) = nx M (0.0131) x(58.5)
= 0.766g in 25ml of water
- 766 divided by 25 X 100 =
- 06%
The amount of NaCl in water was determined by Gravimetric Analysis.
A 25.0ml of water was mixed with 100ml of distilled water, stirred and mixture filtered. The Volume of filtrate was made up to 250ml.
A 20ml aliquot of solution was pipetted into a conical flask and an excess of AgNo3 solution was added.
The Precip weighed 0.150g
AgNo3 (aq) +Nacl (aq) -> AgCl (s) + NaNO3 (aq)
State one property that Agcl must have to be used in the gravimetric analysis
- Low Solubility but must be stable when heated
- Have a Known formula
Titration. (Determining Ammonia Content in Window Cleaner)
9.971g sample of cleaner was transformed to a 250ml standard flask and the volume made up to the mark with distilled water.
A 20 ml aliquot of diluted window cleaner was pipetted into a conical flask and titrated with 0.110M HCl solution.
The average Titre was 23.5 ml .
HCl (aq) + NH3 (aq) -> NH4Cl (aq)
Determine amount in mol of HCl used in the titration
n(Hcl) = Cv
(0.110) (0.0235)
= 2.59x 10^-3 mol
Titration. (Determining Ammonia Content in Window Cleaner)
9.971g sample of cleaner was transformed to a 250ml standard flask and the volume made up to the mark with distilled water.
A 20 ml aliquot of diluted window cleaner was pipetted into a conical flask and titrated with 0.110M HCl solution.
The average Titre was 23.5 ml .
HCl (aq) + NH3 (aq) -> NH4Cl (aq)
Determine amount in mol of Ammonia in the 9.971g sample
1:1
n(NH3) = n (HCl) = 2.59 x10 ^ -3
n (NH3) in 250ml flask= 250 divided by 20 x 2.59x 10 ^-3 = 3.23x 10^-2
Titration. (Determining Ammonia Content in Window Cleaner)
9.971g sample of cleaner was transformed to a 250ml standard flask and the volume made up to the mark with distilled water.
A 20 ml aliquot of diluted window cleaner was pipetted into a conical flask and titrated with 0.110M HCl solution.
The average Titre was 23.5 ml .
HCl (aq) + NH3 (aq) -> NH4Cl (aq)
Determine mass of Ammonia in the 9.971g sample
m (NH3) = nx M
= 3.23 x 10^-2 x 17
=0.546g
Titration. (Determining Ammonia Content in Window Cleaner)
9.971g sample of cleaner was transformed to a 250ml standard flask and the volume made up to the mark with distilled water.
A 20 ml aliquot of diluted window cleaner was pipetted into a conical flask and titrated with 0.110M HCl solution.
The average Titre was 23.5 ml .
HCl (aq) + NH3 (aq) -> NH4Cl (aq)
Determine Percentage by mass (%m/m) of ammonia in cleaner
0.549 Divided by 9.971 x 100 = 5.51 %
During Titration the burette was incorrectly given a final rinse with distilled water. Would this lead to a higher or lower than expected result?
Higher than true value because the acid has been diluted so you will need more of it on the base
