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Quantitative analysisC3.2

Love , Heba

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1 Mole = 6.023 x

One mole of atoms or a molecule of any substance will have the same mass in Grams as the Relative Mass ( Mr and Ar) for that substance

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MEASURING AMOUNTS ( MASS AND MOLES )

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Examples :Substance Relative Mass One molar mass

Carbon ( C ) 12 12 g

Nitrogen ( ) Mr = ( 14 x 2 ) 28

28 g

Carbon Dioxide () Mr = ( 12 x 2 ) 44

44g

Water () Mr = (1 x 2) + 16Mr = 18

18g

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Moles & Mass Formula

MASS

Moles x Mr

In GRAMS

Number of Moles given

Example 1) What is the mass in grams of 3.5 moles of magnesium (mg)

Step 1) Mr of Mg = 24Step 2) Identify which formula to use Step 3) Mass = Moles x Mr Step 4 ) Mass = 3.5 x 24Step 5) Mass= 84 g

Example 2) How many moles are there in 66g of Carbon dioxide?

Step 1) Mr of = 12+ ( 16 x 2 ) = 44 Step 2) Identify which formula to use Step 3) Moles = Mass ÷ Mr Step 4 ) Moles = 66 ÷ 44 = 1.5Step 5) Moles = 1.5 g

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SOLUTIONS AND CONCENTRATIONS

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Remember :

Mass Concentration =

Volume = ( if given in divide by 1000)

Mole concentration = or

Mass = g ( grams)

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How much has dissolved?You can find this out by evaporating the water…

If you have say, 500g of solution , you don’t need to use the whole lot, you can just use for example 10g…

1) Weigh a clean, dry evaporating basin 2) Weigh 10 g of the solution and put it in the basin3) Gently heat the basin to evaporate the water from the solution4) Check if the water seems to have evaporated5) Weigh the dry basin and remaining solid6) Reheat and reweigh until there's no further change in mass

STEP 6 IS TO ENSURE ALL WATER HAS EVAPORATED

Calculations on next page

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Example: Mass of clean Basin = 54.6gMass of Basin with Solid = 56.9g

0.3 g difference..

We used 10g of 500g 0.3g is how much that has dissolved in 10 g

500 g = 10 x 50

0.3 x 50 = 15 g

15 g of substance has dissolved in 500g of Water

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Concentration IMPORTANT FORMULAS

MASS CONCENTRATION

𝑔 /𝑑𝑚3

Mass Conc Volume

Mass

EXAMPLE…

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EXAMPLE WORKING OUT MASS CONCENTRATION

3.75g of Sodium chloride are dissolved in of water. What is the concentration in of the final solution ?

1)Know which formula your using2)Mass concentration = 3)Mass = 3.27 and Volume =

250/1000= 4)3.75 5)Mass concentration =

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Concentration IMPORTANT FORMULAS

MOLE CONCENTRATION

𝑚𝑜𝑙/𝑑𝑚3

Mass Conc Mr

Mole Conc

EXAMPLE…

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EXAMPLE 1 – converting Mass concentration to Mole conc.

Find the mole concentration of 196 solution of

1)Know which formula your using2)Mole concentration = 3)Mass conc. = 196 4)Mr = ( 1 x 2) + 32 + ( 16 x 4) = 985) 196 98 6)Mole concentration =

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EXAMPLE 2 – converting Mole concentration to Mass conc.

Find the mass concentration of solution of HCL

1)Know which formula your using2)Mass concentration = 3)Mole conc. = 4)Mr = 1 + 35.5 = 36.5 5)36.5 x 2.5 = 91.256)Mass concentration =

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Hard Water

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Hard water makes SCUM ( a nasty ppt ). This is because it doesn’t lather with soap.

Hard water contains calcium ions or / and

magnesium ions In some areas water flows over rocks and through soils containing Magnesium/ Calcium ions.

Magnesium Sulfate, dissolves in water, and so does Calcium Sulfate ,

Calcium Carbonate usually exists as chalk, limestone or marble. It can react with acid rain to create calcium hydrogencarbonate. This is soluble and dissolves in water , releasing Calcium ions.

Calcium hydrogencarbonate =

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Removing Temporary Hardness – BOILING.

Temporary hardness is caused by Calcium Hydrogencarbonate

Calcium hydrogencarbonate

. Calcium hydrogencarbonate decomposes.

The lime scale in your kettle is Calcium Carbonate – its insoluble

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Removing Permanent and Temporary Hardness – ion exchange resin

Permanent hardness is caused by

EXAMPLE: .

Resin is an insoluble solid polymer. Water is supplied trough an ion exchange resin. The resin contains a lot of Sodium ions ( or hydrogen) and exchanges them for the C and M ions when the water runs through them .

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Titrations

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Basic’s about TitrationsAcid-base titration is a neutralisation reaction

IONIC equation:

(aq) +

REMEMBER : ACID + ALKALI

Titrations help find out exactly how much of acid is required to neutralise a quantity of alkali ( or vice versa)

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How to do a titration - Step 1Using a pipette and pipette filler, add some

alkali ( About ) to a conical flask , along with two or three drops of indicator.

( use of indicator depends on strength) Type of Indicator Strength of Acid Strength of Alkali

Phenolphthalein Weak Strong

Methyl orange Strong Weak

Any Strong Strong

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How to do a titration - Step 2

-Fill a burette with acid.-Using the burette add the acid to the

alkali at a time giving it a swirl

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How to do a titration - Step 3The indicator changes colour when all the alkali has been neutralised.

( stop adding in alkali). Repeat this whole experiment 2/3 times

E.G Phenolphthalein turns colourless in acids and pink in Alkalis

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Moles

VolumeConc.

Calculations – Working out number of moles to find Conc.

Moles in Balanced Eq. 2 1

Moles in Experiment 0.025 x 0.1 = 0.0025 0.0025/2 = 0.00125

Volume 25 /1000 = 0.025 30/1000 = 0.03

Concentration 0.1 0.00125/0.03= 0.0416

it

Start off by writing the balanced equation then adding in the information given to you.

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ANSWER : 0.0416

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Moles

VolumeConc.

Calculations – Working out number of moles to find Conc.

Moles in Balanced Eq. 1 2

Moles in Experiment 0.025 x 0.5 = 0.0125 0.0025/2 = 0.00125

Volume 25/1000 = 0.025 32.5/1000 = 0.0325

Concentration 0.5 0.00125 / 0.0325 = 0.77

Start off by writing the balanced equation then adding in the information given to you.

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ANSWER : 0.77