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Unit 2 Chem > Stoichiometry > Flashcards

Stoichiometry Flashcards

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1
Q

Chromatography

A

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

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2
Q

Adsorption

A

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.

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3
Q

HPLC

A

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.

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4
Q

Paper and Thin layer Chromatography

A

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

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5
Q

Identifying sample components in thin layer chromatrography.

A

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.

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6
Q

Equation to find length (Paper Chromo)

A

Distance travelled divided by total length from origion to solvent end.

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7
Q

A Chromatogram provides information about

A

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)

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8
Q

Order of Molecules From most to least Polar

A

Most Polar

  • Carboxylic acid (Hydrogen, dispersion , dipole, 3 Polar bonds)
  • Alcohol (Hyrdogen, Dipole, Dispersion, 2 Polar Bonds)
  • Ester (Dipole, Dispersion)
  • Alkane (Weak Dispersion)
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9
Q

Concentration of Cl in 250ml (V) of solution 52.2g (Mass) of Aluminium chloride (AlCl3) is

A

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
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10
Q

Source of Error Titration (Rinsing water left in Burette)

A 
Burette (Underestimated)
Titration Flask (Overestimated)
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11
Q

Source of Error Titration (Rinsing Water left in Pipette)

A 
Burette (Overestimated)
Titration Flask (Underestimated)
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12
Q

Source of Error Titration ( Indicator Chosen Changes Colour too Soon)

A 
Burette (Overestimated)
Titration Flask (Underestimated)
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13
Q

Source of Error Titration (Water in Titration Flask)

A 
Burette (No effect)
Titration Flask (No effect)
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14
Q

Source of Error Gravimetric (Insufficient Precipitating agent was added)

A

Underestimated

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15
Q

Source of Error Gravimetric (Precipitate not fully dried before weighing)

A

Overestimated

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16
Q

Source of Error Gravimetric (Precipitate not washed prior to drying and weighing)

A

Overestimated

17
Q

Source of Error Gravimetric (Other Ions were present in the sample that reacted with the precipitating agent)

A

Overestimated

18
Q

Source of Error Gravimetric (Using too much water for the initial dissolving)

A

No effect

19
Q

Source of Error Gravimetric (Forming a precip thats too soluble)

A

Underestimated

20
Q

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

A

0.750 divided by 44 = 0.017 x 2

= 3.41 x 10 to the power of negative 2

21
Q

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

A
  1. 17x 100.1 (Caco3) = 1.1
  2. 7 x 100 Divided by 95 =
  3. 80g.
22
Q

What will least likely alter the retention time of a substance passing through the column in a HPLC analysis

A

Diluting the sample being analysed (Will change peak but not retention time)

23
Q

Things that will alter retention time in a HPLC analysis

A
  • Replacing one solvent with another
  • Increasing the temp of the column
  • Decreasing the rate of flow of the mobile phase
24
Q

How can Chromium be selectively analysed without interference from other metals

A

Because AAS has a hollow cathode lamp which is made from elements under investigation

25
Q

How can a particular wave length used for analysis is chosen in UV- visible spectrometry

A

We will choose length that is highly absorbed in the component. We will also choose length that is exclusively absorbed. This assures accurate results.

26
Q

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

A

0.150 divided by 143.4 = 1.046x10^-3 mol

27
Q

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

A

1:1
n (Nacl) =n (AgCl) = 1.046x10^-3
250 divided by 20 x 1.046x 10^-3 =
0.0131 mol

28
Q

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

A

m (Nacl) = nx M (0.0131) x(58.5)
= 0.766g in 25ml of water

  1. 766 divided by 25 X 100 =
  2. 06%
29
Q

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

A
  • Low Solubility but must be stable when heated

- Have a Known formula

30
Q

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

A

n(Hcl) = Cv
(0.110) (0.0235)
= 2.59x 10^-3 mol

31
Q

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

A

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

32
Q

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

A

m (NH3) = nx M
= 3.23 x 10^-2 x 17
=0.546g

33
Q

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

A

0.549 Divided by 9.971 x 100 = 5.51 %

34
Q

During Titration the burette was incorrectly given a final rinse with distilled water. Would this lead to a higher or lower than expected result?

A

Higher than true value because the acid has been diluted so you will need more of it on the base

Unit 2 Chem (13 decks)

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