version 1.1

Version 1.1

2013

Chemistry NCEA L22.1 Quantitative Analysis

GZ Science resources 2013

The number of particles in a substance is measured in moles

1 mole of particles = 6.02214 x 1023 particles for any substance! called Avogadro’s number

Molar mass is calculated by adding up the atomic mass (or mass number ) or each individual atom.E.g. H2O Molar mass M = 1 + 1 + 16 = 18

The mole and molar mass


236.02x10

particles ofnumber

23

24

10x02.6

10x12

If the actual number of particles is known (atoms, ions or molecules) then the amount, n, is easily calculated since

For example 12 x 1024 sodium ions is equivalent to

How many moles of water molecules are there in 3 x 1022 molecules of H2O?

n=

mol = 20 mol of sodium ions.

Converting number of particles to amount, n


Sometimes information is provided in terms of the relative atomic mass, Ar, of an element rather than its molar mass. This has exactly the same numerical value but has no units e.g. Ar of oxygen is 16 and the M of oxygen is 16 g mol-1.

Note: The relative atomic mass is, in fact, the mass of an atom relative to the mass of an atom of the isotope carbon -12 which has a mass of 12. This means an atom of oxygen-16 is times heavier than an atom of carbon-12.

1216

The relative molecular mass, Mr, is the mass of a molecule relative to carbon-12. e.g. Mr of CuSO4 is 159.5, the same numerical value as the molar mass but having no units, (M = 159.5 g mol-1).

Relative atomic and molecular mass Ar and M


1. Identify how many of each atom

2. Write a sum using the molar mass for each atom

3. Calculate the total

e.g. H2SO4

2 H 1 S 4 O

2 x M(H) + M(S) + 4 x M(O)

(2 x 1) + 32 + ( 4 x 16)

M(H2SO4)= 98 gmol-1

e.g. CuSO4.5H2O

1 Cu 1 S 9 O 10 H

M(Cu) + M(S) + 9 x M(O) + 10 x M(H)

63.5 + 32 + (9 x 16) + (10 x 1)

M(CuSO4.5H2O)=249.5 gmol-1

Calculating Molar Mass of a compound


a) Select one of the above equations according to what information is in the question and rearrange if necessary. (i.e. mass and molar mass)

b) Write down the equation

n(CH4) = m/Mc) Write in the values (with units) underneath and calculate

n(CH4) =12.5g/16.0g mol-1

n(CH4) = 0.781mol

n = m / M n = c . v

EXAMPLECalculate the amount (in moles) of methane, CH4, in 12.5 g of the gas?

M(CH4) = 16.0 g mol1

One step calculations


2. How many moles? Given v or c

n = c . v

n = 0.100molL-1 x 0.010L

n = 1 x 10-3 moles

1. Mass of x moles? Given m or M

n = m / M m = n . M m = 0.241mol x 142.1gmol-1

m = 34.2g

n = m / M

n = c . v

n = moles (mol)

m = mass (grams)

M = molar mass (gmol-1)

n = moles (mol)

c = concentration (molL-1)

v= volume (L)



volume

mass

litres)(in volume

moles)(in amount

V

n

Concentration is a quantitative measure of how much dissolved substance there is in a given volume of solvent. The concentration can be expressed or measured in 3 different ways:

1) grams per litre, g L-1 If the mass is in grams and the volume in litres then concentration =

2) % w/v - or percentage weight per volume, is the mass of solute in 100mL of solvent.

3) moles per litre, mol L-1 This is the most common concentration unit used in chemistry. It is calculated using concentration = =

Concentration


The mass (m) of a substance

Symbol = m unit = grams

The amount n is related to mass m using relationship

1. Identify what is asked for, what is given and calculate the molar mass of the substance

2. Write an equation 3. Calculate the quantity

e.g how much SO2 is present in 1.8kg of SO2?

n = ? m = 1800gM(SO2) = 64gmol

n = m/M n = 1800/64n = 28.1 moles

e.g. calculate mass of 0.5 moles of CuSO4.5H2O

m = ? n =0.5gM(CuSO4.5H2O) = 249.5gmol

m = n x M m = 0.5 x 249.5m = 125g

e.g. calculate the amount of Na+ ions in 25g of Na2CO3

n(Na+) = ? m = 25gFirst find n(Na2CO3)M(Na2CO3)=106gmol

n = m/M n=25/106n=0.236n(Na+)=0.472moles

-1

-1

-1

n = m/Mmol = mass

molar mass



a) Select one of the above equations depending on what 2 values you have (i.e you have mass – 2.3g, and M – 342gmol-1 but only volume – 12ml)

b) Write down the equation chosen and calculate

n(C12H22O11) = m/Mc) Use the calculated moles to insert in the other equation and

calculate

c(C12H22O11) = n / v information given in question (convert to

litres)

n = m / M n = c . v

EXAMPLEIf 2.3 g of sucrose (C12H22O11) is dissolved in 12mL of distilled water, what will the concentration be?M(C12H22O11) = 342 g mol-1

Two step calculations

a) Calculate n of each solution (n = c.v) and add together to get n (total)

b) Add volumes together to get v (total)

c) Calculate concentration of final solution

concentration (final) = n (total) / v (total)

EXAMPLE500 mL of 0.253 mol L1 NaHCO3 solution is mixed with 800 mL of 0.824 mol L1 NaHCO3 solution.What is the concentration of the final solution? c = n / v n = c . v



3. What is the concentration? ( M and m given)

c = n / v but n = m / M a) n = 2.12g /142.1gmol-1

n = 0.015mol

b) c = 0.015mol / 0.250L

c = 0.059molL-1



4. How much mass of x to make up solution? Given v + c

n = c . V conc = 0.200molL-1 vol = 250ml

x = CuSO4 M(CuSO4) = 249.5gmol-1

a) n = 0.200molL-1 x 0.250L

n = 5.00 x 10-3 mol

b) m = M . n

m = 249.5gmol-1 x 5.00 x 10-3 mol

m = 1.248g

n = c . v



a) Place each element from the formula into a column and write the above equation under each and calculate

C H S

m(C) = n x M m(H) = n x M m(S) = n x M

b) Calculate the mass of the compound by adding all the masses together

c) Calculate % of element in the compound

i.e. % S = mass (S) ÷ Mass (compound) x 100

d) All of the percentages should add up to 100%

EXAMPLEAllicin is the compound responsible for the characteristic smell of garlic. Its formula is C6H10S2O. What is the % of sulfur (S) in this compound?

m = n x M

C6 H10 S20 C=12 H=1 S= 32.1n (moles) is found from compound M(molar mass) is found in periodic table

Percentage composition

5a. What is the Percentage Composition of x in a compound?

Na2B4O7 x = B

Na B O

m = n . M m = n . M m = n . M

m = 2 x 23.0 m = 4 x 10.8 m = 7 x 16.0

m(Na) = 46g m(B) = 43.2g m(O) = 112.0g

m(Na) + m(B) + m(O) = 201.2g

%B = 43.2g / 201.2g x 100 %B = 21.5

Shortcut Note – you can calculate the Molar mass of the element in question (B) and divide it by the Molar mass of the compound but you lose the opportunity to gain partial marks if you make a mistake.



The mass of the different elements within a compound can be expressed as a percentage of the total mass – this can be used to identify a compound

1. From the chemical formula find the amount of each element present (in moles)

2. Use the molar masses of each element to find the mass of each element present.

3. Express the amount of each element as a percentage

4. Check to see that adding the percentages gives 100%

e.g. 1 mole of C2H6 contains 2 mol of C and 6 mol of H

The mass of C present is 2 x 12.0 = 24.0g

The mass of H present is 6 x 1.0 = 6.0g

The mass of 1 mole of C2H6 is 30.0g, so the percentage composition of C isC= 24.0/30.0 x 100=80.0%H=6.0/30.0 x 100=20.0%

80.0% + 20.0%=100%

checked



a) Place each element from the formula into a column and write the above equation under each and calculate mass.

Fe O

m(Fe) = n x M m(O) = n x M

b) Calculate the mass of the compound by adding all the masses from the elements together i.e. m (Fe2O3) = m(Fe) + m(O)

c) Calculate % of element in the compound

i.e. % Fe = mass (Fe) ÷ Mass (compound) x 100

d) Find the % Fe in the mass given (i.e 1.34g)

(mass given ÷100) x %

EXAMPLEA precipitate of 1.34 g of Fe2O3 was obtained by treating a vitamin supplement dissolved in water with NaOH and then heating. What is the mass of iron in the tablet?

m = n x M

Make sure your calculated mass is less than the given mass (unless 100% then it will be the same.

Use units, especially in your final answer.

Mass and Percentage composition

5b. Find Mass of element in compound given mass of compound

% of O in BaSO4

a) Find Percentage Composition of each atom in a given compound

Ba S O

m(Ba) = n . M m(S) = n.M m(O) = n. M m(Ba) = 1mol x 137.3gmol-1 m(S) = 1mol x 32.0gmol-1 m(O) = 4mol x 16.0

m(Ba) = 137.3g m(S) = 32.0g m(O) = 64g

b) Calculate m of compound - m(BaSO4) = m(Ba) + m(S) + m(O) = 233.5g

m(BaSO4) = 233.5g

c) Calculate Percentage % O = 64g/233.5g x 100 %O = 27.4%

d) Calculate percentage of O in given mass of compound mass = 0.192g

m(O) = 27.4% of 0.192g m(O) = 0.0526g

Mass and Percentage composition


a) All % should add to 100%, change each into grams so they all total 100g (i.e. 28.9% = 28.9g)

b) Calculate the number of moles of each element (place each element in a column and write formula underneath, then calculate n)

c) Divide each n value by the smallest value – You may have to multiply all of the numbers up until you get the lowest common whole number of each

i.e n(C) = 4.13mol n(H) = 5.20mol n(N) = 2.06mol n(O) = 1.03mol --

1.03mol 1.03mol 1.03mol 1.03mol

d) Write the empirical formula underneath with the element followed by the number of moles i.e C4H5N2O

EXAMPLECaffeine is a stimulant which has a mass composition of 49.5% carbon, 5.20% hydrogen, 28.9% nitrogen and 16.5% oxygen.Calculate the empirical formula of caffeine. n = m / M

Empirical formula

6. Empirical Formula ( Find ratio of each atom in a compound)

a) Convert % of each atom to Mass (grams) - total g = 100g

C = 40% H = 6.67% O = 55.33%

b) Calculate n of each

C H O

n(C) = m/ M n(H) = m /M n(O) = m/ M n(C) = 40g/12.0gmol-1 n(H) = 6.67g/1.0gmol-1 n(O) 55.33g/16.0gmol-1

n(C) = 3.33mol n(H) = 6.67mol n(O) = 3.33mol

c) Divide each n by the smallest n value to obtain ratio

3.33/3.33 : 6.67/3.33 : 3.33/3.33

1 : 2 : 1

d) Convert to formula C H2O

Empirical formula


An empirical can be calculated from the percentage composition. It is the simplest whole number ratio of the atoms in a compound.

1. Find the mass of each element present in 100g of the compound

2. Change these masses to amounts using n = m/M

3. Divide each result by the smallest number of moles

4. Make small approximations or multiply all results by a constant to get a simple whole number ratio

5. This is the empirical formula

e.g. Analysis of a compound showed that it contained 14.29% H and 85.71% Cin 100g of compound there is14.29g of H85.71g of C

In 100g of compound there is14.29/1=14.29mol H since M(H)=1gmol

85.71/12 =7.14mol CSince M(C)=12gmol

Moles of H14.29/7.14 = 2.001

Moles of C7.14/7.14=1

Moles of H2.001≈ 2

Moles of C1

The empirical formula of the compound is CH2

-1

-1

Empirical formula


a) Calculate the empirical formula so you have a whole number formula - i.e porphryn = C10H5N2

molar mass of carbon

b) Calculate the molar mass of the empirical formula i.e C = (10 x 12) + …

c) Divide the molar mass of the molecular formula (given in question) by the molar mass of the empirical formula to calculate the multiplier i.e (molecular)306gmol-1 / (empirical) 153gmol-1 = 2

d) If the multiplier value is 1 then the empirical formula is the same as the molecular formula. Otherwise multiply each element by the value

C10H5N2 x 2 = C20H10N4

EXAMPLEThe porphryn molecule forms an important part of the haemoglobin structure. This porphryn molecule is 78.5% carbon, 3.2% hydrogen and 18.3% nitrogen.(i) Calculate the empirical formula of porphryn.(ii) If the molar mass of porphryn is 306 g mol1, calculate the molecular formula.

Molecular formula


7. Molecular Formula ( Find moles of each atom in a compound)

Given the Molecular M = 180.0gmol-1

a) Find M of the empirical formula (add up all M x each atom)

b) Calculate multiplier = Molecular M = 180.0gmol-1

Empirical M 30gmol-1

= 6

c) Multiply empirical formula by multiplier

CH2O x 6 = C6H12O6

Molecular formula


The Molecular Formula can be determined from the empirical formula and molar mass It gives the actual number of atoms present

1.Find empirical formulaand Calculate molar mass

2.Given molar mass of the molecular formula calculate units present (x)

3.Multiply empirical formula by unit (x)

4. Write molecular formula

Example.The empirical formula of a compound was found to be CH2Omolar mass is 60.0gmol

X= molar mass

Molar mass of empirical formula

X= 60.0 30.0

X= 2

=(CH2O)x=(CH2O)2

The molecular formula is C2H4O2

Molecular formula


This is an experimental procedure involving the measurement of masses of substances to determine the composition of unknown substances.

The technique is: Highly accurate (since masses can be measured accurately) Useful for substances involved in chemical reactions that go to completion.

Calculations should be carried out to three significant figures.

If the masses of the elements combined to make a compound are known accurately, than the empirical formula can be calculated.

Masses of compounds can be determined from precipitation.

Gravimetric Analysis


The crystals of some salts contain molecules of water. These molecules fit into the crystal lattice. The ratio of water to salt is always a whole number The water present in hydrated salts is known as water of crystallisation When a hydrated salt is heated, the water is driven off and the salt is said to

become anhydrous

Water of Crystalisation


a) Write down two columns : Anhydrous salt and Water

b) Calculate the mass of each from the data given.

> Anhydrous salt = (mass of crucible, lid and MgSO4 after second heating) – (mass of crucible and lid)

> Water = (mass of hydrated salt) – (mass of salt calculated)

c) Calculate n of each (n=m/M)

d) Divide each by smallest value and write down formula -

MgSO4 . (number of moles) H2O

EXAMPLEHydrated magnesium sulfate is heated in a crucible. The following data is collected.mass of crucible and lid = 26.49 gmass of hydrated magnesium sulfate = 2.12 gmass crucible, lid and magnesium sulfate after first heating = 27.55 gmass of crucible, lid and magnesium sulfate after second heating = 27.52 gUse these results to determine the formula of hydrated magnesium sulfate.

M(MgSO4) = 120 g mol1 M(H2O) = 18.0 g mol1


Anhydrous salt Water

mass = m(heated) – m(crucible) mass = m(hydrated) – m(heated)

m(CaCO3) = 1.56g m(H2O) = 1.52g

n= m/M n = m/Mn=1.56g/90.08gmol-1 n= 1.52g/18.0gmol-1

n=0.017mol n = 0.080mol

n(CaCO3) : n(H2O)

0.017mol : 0.080mol

0.017mol : 0.017mol

1 : 5

CaCO3.5H2O

Divide each by the smallest value –this will be n (anhydrous salt)

The dot (.) means + water

Data from experiment or given in question



EXAMPLEWine makers convert sugars, such as glucose, to ethanol and carbon dioxide.

C6H12O6 2C2H5OH + 2CO2

Starting with 540g of glucose what is the maximum amount of ethanol, in moles and also in grams that could be produced.M (C2H5OH) = 46 g mol-1 M (C6H12O6) = 180g mol-1

C6H12O6 2C2H5OH + 2CO2

M = 180gmol-1 M = 46gmol -1

- m = 540g m = ?

n = m/M

n = 540g / 180gmol-1

n = 3.0mol

1 mole of C6H12O6 forms 2 moles of C2H5OH

So n (C2H5OH) = n(C6H12O6) x U/K = 3.0mol x 2/1 = 6.0mol

m = n x M

m = 6.0 mol x 46gmol-1

m = 276g

Write down the equation with the known information underneath

1

2

31. Calculate moles of known. 2. convert mole ratios. 3. calculate mass of unknown.

Equations and mole ratios n = m/M


9. Calculate mass of x given equation and mass of a compound y x = NaCl y = Cl2 m(Cl2) = 12.8g

Cl2 + 2NaI 2NaCl + I2

a) Calculate moles of compound y

n = m/M n(Cl2) = 12.8g/71.0gmol-1

n(Cl2)=0.180moles

b) Convert moles y to x

1mol Cl2 forms 2mol NaCl so n(NaCl) = 2 x n(Cl2) = 0.361moles

c) Calculate mass (or concentration) of NaCl

m = n . M c = n/v

m(NaCl) = 0.361mol x 58.5gmol-1 c(NaCl) = 0.361mol x 0.100L m(NaCl) = 21.12g c(NaCl) = 3.61molL -1


9. Calculate Mass of x given Formula and equation and mass of y m(CH4) = 6800g m(CH3OH) = ?

3CH4 + CO2 + 2H2O 4CH3OH

a) Calculate using mass ratios – write under formulas

3CH4 + CO2 + 2H2O 4CH3OH

Mass= 6800g = x

M = 48gmol-1 =128.0gmol-1

m(CH3OH) = m(CH4) x M(CH3OH)

M(CH4)

m(CH3OH) = 6800g x 128.0gmol-1

48.0gmol-1

m(CH3OH) = 18133.3g or 18.133kg



EXAMPLE20.0 mL of an aqueous solution of 0.105 mol L1 sodium carbonate is titrated with a 0.250 mol L1 hydrochloric acid solution.What volume of hydrochloric acid is required to reach the equivalence point of the titration.2 HCl (aq) + Na2CO3(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)

Na2CO3 + 2HCl → 2NaCl + CO2 + H2Oc = 0.105molL-1 c = 0.250molL-1

v = 0.020L v = ?

n = c x v

n = 0.105 x 0.020 n = 2.10 x 10-2mol

1 mole of Na2CO3 forms 2 moles of HCl

So n (HCl) = 2 x n(Na2CO3) = 2 x (2.10 x 10-2)mol = 4.20 x 10-2mol

v = n / c

v = 4.20 x 10-2 mol / 0.250

v = 0.0168L or 16.8mL


1

2

31. Calculate moles of known. 2. convert mole ratios. 3. calculate volume of unknown.



13. Calculate concentration of x given conc. of y and formula

Na2CO3 + 2HCl 2NaCl + H2O + CO2

10.0ml of 0.106molL-1 sodium carbonate reacted with 21.56ml HCl

a) Calculate mols of y (Na2CO3)

n(Na2CO3) = c x v

= 0.106molL-1 x 0.0100L

= 0.00106mol

b) Convert mols y to x

1mol Na2CO3 = 2molHCl

n(HCl) = 2 x n(Na2CO3)

=0.00212mol

c) Calculate concentration

c = n/v

c(HCl) = 0.00212mol / 0.02156L = 0.0983molL-1


EXAMPLEA chemist found that 4.69 g of sulfur combined with fluorine to produce 15.81 g of gas. Determine the empirical formula of the compound.

?S + ?F SF?

m = 4.69g m = m(SF?) – m(S) = 11.12g m = 15.81g

M = 32.1gmol-1 M = 19.0gmol-1 n(S) = m/M n(F) = m/M

n(S) = 4.69/32.1 n(F) = 11.12/19.0

n(S) = 0.146mol n(F) = 0.585mol n(S) = 0.146 n(F) = 0.585mol

0.146 0.146mol

= 1 : 4

S F4


Calculate number of moles of each element making up compound

1

2

3Divide each number of moles by the smallest value to get the smallest common whole number

Use those numbers to write formula of compound

4

Empirical formula and mole ratios n = m/M


10. Calculate compound Formula given mass and equation m(product) = 43.5g m(Fe) =15.0g

Fe + Cl2 Fe Clxa) Calculate moles of reactant (for which given mass)

n(Fe) = m/M n(Fe) = 15.0g/55.9gmol-1

n(Fe)=0.268moles

b) Convert moles reactant to product

1mol Fe forms 1mol FeClx

c) Calculate mass of each possible molecular formula of FeClx

m = n . M m = n . Mm(FeCl2) =0.268mol x 126.9gmol-1 m(FeCl3)= 0.268mol x 162.3gmol-1

M(FeCl2) = 34.1g m(FeCl3) = 43.5g

d) Match m(product) with calculated mass = FeCl3

Empirical formula and mole ratios n = m/M

concentration (new) = concentration (original) x volume (original)

volume (new)

EXAMPLE20.0 mL of a 0.00500 mol L–1 FeCl3.6H2O. solution was pipetted into a volumetric flask and diluted to 100 mL. What is the concentration of the diluted solution?

c(new) = 0.00500molL-1 x (0.020L / 0.100L)

c(new) = 0.00500molL-1 x 0.2

c(new) = 0.00100molL-1

Because this is a dilution the new concentration must be less than the original

Make sure you convert all of your volumes into litres first

Dilution calculations


11. Calculate new concentration given old conc. and volumes

e.g. 40ml of 2.0molL-1 diluted by adding 60ml water

c(new) = c(old) x old volume

new volume

c(new) = 2.0molL-1 x 40ml

(40ml + 60ml)

C(new) =0.80molL-1

Also can calculate new concentration through evaporation

Dilution calculations


n(solution 1)= c1 x v1 + n(solution 2)= c2 x v2 = total moles

Final concentration = total mols / total volume (v1 + v2)

EXAMPLE500 mL of 0.253 mol L1 NaHCO3 solution is mixed with 800 mL of 0.824 mol L1 NaHCO3 solution. What is the concentration of the final solution?

n(solution 1 ) = 0.253molL-1 x 0.50L =

c(new) = 0.00500molL-1 x 0.2

c(new) = 0.00100molL-1 Make sure you convert all of your volumes into litres first

Mixed solution calculations


Volumetric analysis involves a titration. >In a titration, a solution, of unknown concentration, is typically titrated (added) from a burette into a flask. >The flask typically contains a known amount of solution. A pipette is used to measure a precise set volume called an aliquot.>In an acid-base titration the end point of the titration is reached when there is a neutralisation. Indicator (usually phenolphthalein) is added into the flask and a colour change (lasting more than 10 seconds) will indicate a neutralisation.>Using c=n/v and a balanced equation, the concentration of the solutions can be calculated.

Volumetric Analysis


Pipette>calibrated to provide an exact volume.>clean and rinsed with solution to be pipetted.>reading from the bottom of the meniscus.

Titration equipment


Burette>burette must be cleaned and rinsed with small amount of solution that will go into it.>The volume of the solution in the burette is read from the bottom of the meniscus. Volume is recorded before and after the titration.>the volume of solution delivered by the burette is called a titre.

Titration equipment


Conical Flask>rinsed with distilled water before use.>aliquot of solution (unknown or known concentration) placed into the flask from the pipette.>a few drops of indicator are placed into the flask>the flask is swirled during the titration.

Titration equipment


Reading is to be taken from the bottom of the meniscus curve.

a) Read the ml

29ml

b) Read the 1/10 of ml

29.9ml

c) Divide the 1/10ml up into 0.00, 0.02, 0.04…0.08 and read to closest

29.98ml

Reading the burette


Make sure you label your beaker with the solutions they contain

Rinse all glassware appropriately

Titration equipment setup


Before starting the titration a suitable pH indicator must be chosen. The endpoint of the reaction, when all the products have reacted, will have a pH dependent on the relative strengths of the acids and bases. The pH of the endpoint can be roughly determined using the following rules:A strong acid reacts with a strong base to form a neutral (pH=7) solution. A strong acid reacts with a weak base to form an acidic (pH<7) solution. A weak acid reacts with a strong base to form a basic (pH>7) solution. When a weak acid reacts with a weak base, the endpoint solution will be basic if the base is stronger and acidic if the acid is stronger. If both are of equal strength, then the endpoint pH will be neutral.A suitable indicator should be chosen, that will experience a change in colour close to the end point of the reaction.

Indicator Acid Base pH range

Methyl Orange Red Yellow 3.1 - 4.4

Phenolphthalein Colourless Pink 8.3 - 10.0

Indicator selection


A standard solution is a solution whose concentration is known accurately. Its concentration is usually given in mol L–1. When making up a standard solution it is important that the correct mass of substance is accurately measured and all of this is successfully transferred to the volumetric flask used to make up the solution.

Making a 1molL-1 of NaCl solution

a) Calculate the mass of 1mole of NaCl

m (NaCL) = n x M

m (NaCl) = 58.5g

b) Determine the volume of the volumetric flask being used and mass of substance added.

1L flask = 1mol

100mL flask = 1/10mol

c) Weigh and add substance to the volumetric flask with wash bottle

d) Continue filling flask with distilled water to line – bottom of meniscus must touch line

Standard solutions


1. First, the burette should be rinsed with the standard solution, the pipette with the unknown solution, and the conical flask with distilled water.

2. Secondly, a known volume of the unknown concentration solution should be taken with the pipette and placed into the conical flask, along with a small amount of the indicator chosen. The burette should be filled to the top of its scale with the known solution with a funnel. Remove the funnel after the burette is filled. Record the starting volume.

3. The known solution should then be allowed out of the burette, into the conical flask. At this stage we want a rough estimate of the amount of this solution it took to neutralize the unknown solution. Let the solution out of the burette until the indicator changes colour and then record the value on the burette. This is the first titre and should be excluded from any calculations.

4. Perform at least three more titrations, this time more accurately, taking into account roughly of where the end point will occur. Record of each of the readings on the burette at the end point. Endpoint is reached when the indicator just changes colour for more than 10 seconds. The titres must be concordant to 0.2ml to gain Excellence ( 0.4ml Merit and Achieved)

Titration method

1. Read your instructions carefully. Understand clearly which solution goes into the conical flask and which goes into the burette. Make sure you convert all volumes into litres i.e. 10ml aliquots from a pipette is 0.010L

2. Record the given concentration on your sheet

Concentration of standard sodium hydroxide solution = _____________ mol L1

3. Rule up a table to record your titrations. Titre = final reading – start reading

initial titration

1st titration

2nd titration

3rd titration

4th titration

Final reading

Start reading

titre

Recording Titrations


1. At least 3 titrations must be within 0.2ml of each other to be concordant at Excellence level

2. Any titres not falling within this range must be discarded and not used in the average.

3. If more there are more than 3 concordant titres within this range they will be included in the average calculation

EXAMPLE

successive titrations – 24.50 mL, 23.25 mL, 23.35 mL and 23.28 mL.

23.35mL – 23.25mL = 0.10ml

This difference is less than 0.2ml allowed, and 23.28 falls within this range so all three are used. 24.50mL falls outside this range so is discarded.

v = (23.35mL + 23.28mL + 23.25mL) ÷ 3 = 23.29mL = 0.0233L

Concordant titres


EXAMPLEA standard solution of 0.180 mol L1 hydrochloric acid was titrated against 25.0 mL samples of a solution of sodium carbonate. The following volumes of hydrochloric acid solution were used in successive titrations – 24.50 mL, 23.25 mL, 23.35 mL and 23.28 mL.The equation for the reaction is

Na2CO3 + 2 HCl 2 NaCl + CO2 + H2OUse this information to determine the concentration of the sodium carbonate solution.

Give your answer to three significant figures.

Na2CO3 + 2HCl → 2NaCl + CO2 + H2Ov = 0.0250L v = 0.0233L (concordant titres)c = ? c = 0.180molL-1

n = c x v

n = 0.180molL-1x 0.0233L

n = 4.19 x 10-3mol

2 moles of HCl forms 1 mole of Na2CO3

So n (Na2CO3) = n(HCl) ÷ 2 = (4.19 x 10-3) ÷ 2 = 2.10 x 10-3mol

c = n / v

c =2.10 x 10-3 mol /0.0250L

c = 8.39 x 10-2molL-1


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31. Calculate moles of known. 2. convert mole ratios. 3. calculate concentration of unknown.

Titration calculations c = n/v


EXAMPLE20ml ethanoic acid solution (unknown) is titrated with a solution of 0.105 mol L1 sodium hydroxide (known) What is the concentration of hydrochloric acid.CH3COOH (aq) + NaOH(aq) → NaCH3COO(aq) + H2O(l)

NaOH + CH3COOH → NaCH3COO + H2Oc = 0.105molL-1 c = ?v = titration (eg 0.018L) v = 0.020L

n = c x v

n = 0.105 x 0.018 n = 1.89 x 10-3mol

So n (CH3COOH) = n(NaOH) x U/K (1/1) = 1.89 x 10-3mol

c = n / v

c = 1.89 x 10-3 mol / 0.020

c = 0.0945molL-1


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31. Calculate moles of known. 2. convert mole ratios. 3. calculate volume of unknown.

Titration calculations c = n/v


version 1.1

Documents

mass number

molar mass gmol

mass gramsm

mass of solute

h2o molar mass

relative atomic mass

relative molecular mass

avogadros number molar