Moles P1

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Atomic WeightsYou must be able to…· Describe the mole as the SI unit for amount of substance

· Relate amount of substance to relative atomic mass

· Describe relationship between the mole and Avogadro’s

number

· Conceptualise the magnitude of Avogadro’s number

· Describe the relationship between molar mass and

relative molecular mass

· Calculate the molar mass of a substance given its

formula





THE NEUTRAL ATOM

• The atom consists of a ……………. containing protons

and neutrons surrounded by a cloud of ………………….

• Atomic Number (Z)

Number of ……………. in the Nucleus (= number of

electrons in a neutral atom.)

• Mass number (A) = Number of protons + neutrons.

Notation

Z

A

X………….. Number

(smaller) ………….

…………. Number

(bigger) ………….symbol

Neutrons =

………………………





THE NEUTRAL ATOM• The atom consists of a nucleus containing protons and

neutrons surrounded by a cloud of electrons.

• Atomic Number Z - Number of protons in the Nucleus =

number of electrons in a neutral atom.

• Mass number A = Number of protons + neutrons.

Notation

Z

A

XAtomic Number

(smaller) = PROTONS

Mass Number

(bigger) = P + N

symbol

Neutrons = Mass number –

Atomic Number

= A - Z





Relative Mass Atomic • Certain products, such as paper for example, are sold by the ream.

A ream is 500 sheets. Since it is impractical to actually count out 500 sheets, the weight (mass) of 500 sheets is determined; then each ream is packaged according to this mass.

• Atoms are even smaller than paper, so it is not possible to actually count them. However, it is possible to know the mass of an atom in respect to the mass of another atom.

• The Relative mass of an object is expressed by comparing it mathematically to the mass of another object. So the relative mass of an orange in relation to a grapefruit is .6. The relative mass of the grapefruit in relation to a grapefruit is 1.0.

• Atoms are compared to the lightest atom (hydrogen) which is 12 times lighter (1/12 of the mass of) one carbon atom.

• THE RELATIVE ATOMIC MASS IS THE NUMBER OF TIMES AN ATOM IS HEAVIER THAN 1/12 OF A C12 ATOM.





The MoleThe mole is defined as, “the amount of ………….. with the same number of

……………………… particles as ….. grams of carbon 12”. (n used as symbol for moles)

602 300 000 000 000 000 000 000Six hundred and two thousand, three hundred, billion billion !

6.023x1023 particles

12.00 g

CSymbol (….)

Number of particles = no of moles x no. particles in a mole

Particles = ……………..





The MoleThe mole is defined as, “the amount of matter with the same number of

elementary particles as 12 grams of carbon 12”. (n used as symbol for moles)

602 300 000 000 000 000 000 000Six hundred and two thousand, three hundred, billion billion !

6.023x1023 particles

12.00 g

CSymbol (L)

Number of particles = no of moles x no. particles in a mole

Particles = n x L





The Mole and MassThe mole is defined in such a way that the MASS NUMBER (A) of an element

is equal to the relative atomic mass mass of one mole of the substance. (in

grams) - THE MOLAR MASS

• Eg Na = 23g/mol, water(H2O)=18g/mol

Z

A

XAtomic Number

(smaller)

Mass Number

(bigger)

protons + neutrons

Periodic Table Symbol

Relative atomic

mass

or

mass(g) of one

mole





Relative Masses

• Relative atomic(Ar) - The mass of the atom relative to 1/12 of the mass of a C12 atom. (Number of times heavier than…)

O - 16 one atom of oxygen is 16 times heavier than 1/12 of the mass of a C12 atom, Na - 23 one atom of sodium… , H - 1 etc.

• Formula mass (Mr) - The sum of all the atomic masses of the atoms in a molecule.

Water H2O one molecule of water has a relative mass of (2x(1)+16) = 18 - that is the molecular or formula mass of water.

Mr(H2O) = 18 (Times heavier than…)





Atomic Structure

Symbol

NotationName Protons Neutrons

Mass

NumberElectrons

Calcium

3115P

14 30

13 6

Mg





ISOTOPES

Symbol PROTONS ELECTRONS NEUTRONS

Carbon 1212

6C

Carbon 1313

6C

Boron 1010

5B

Boron 1111

5B

Hydrogen 1

Hydrogen 2

Chlorine 35

Chlorine 37





Relative Masses - examples

Calculate the Formula (Molecular) masses of:

• O2 (oxygen gas) Mr (O2) = 2x16 = 32

• Cl2 (chlorine gas) Mr (Cl2) = 2x35. 5 = 71.0

• NaCl (sodium chloride - table salt)

Mr (NaCl) = 23+35.5 = 58.5

• CaCO3 (calcium carbonate)

Mr (CaCO3) = 40.1+12+(3x16) = 100.1

• (NH4)2Cr2O7 (ammonium dichromate)

Mr ((NH4)2Cr2O7 ) = 2(14+4)+2(52)+7(16) = 252





IsotopesChlorine has two isotopes 37

17Cl & 3517Cl

Cl(35) has 35-17=18neutrons Cl(37) has 20 neutrons!

• 37Cl (25%) & 35Cl (75%) - exist in the ratio 1:3

Calculate the average mass of a Cl atom. (Two methods)

In 100 atoms – 25 have a mass of 37 and 75 have mass 35!

Average Ar(Cl)= total mass = (37x25)+(35x75) = 35.50

no of atoms 100

Or

4 atoms – 3 are 35 and one is 37!

Av Ar(Cl) = (37x1)+(35x3) = 35.50

4





Relative Atomic Mass

Z

A

XAtomic Number

(smaller)

Mass Number

(bigger)

protons + neutrons

Relative atomic

mass

or

mass(g) of one

mole

Periodic Table Symbol

Calculate: The mass in grams -

1. of one mole of copper chloride (CuCl2)

2. one mole of carbon dioxide (CO2)

3. One and a half moles of oxygen (O2)

4. TWO moles of methane (CH4)

5. Four moles of water.

m = n x Mr

mass of substance = number of moles x mass of 1 mole





The Mole - moles --> Massm = n x Mr

Calculate the mass of

• 2 moles of copper oxide (CuO)

• 0.5 moles of copper (II) sulphate (CuSO4)

• 0.01 moles of calcium carbonate

• 5 moles of ammonium carbonate

mass = moles x relative mass





The Mole - moles --> Massm = n x Mr

Calculate the mass of

• 2 moles of copper oxide

(CuO) m = nxMr = 2x(63.5+16) = 159 g

• 0.5 moles of copper (II) sulphate

m (CuSO4) = n x Mr = 0.5 x ( (63.5) + 32.1 + 4(16) ) = 79.8 g

• 0.01 moles of calcium carbonate

Mr (CaCO3) = n x Mr = 0.01 x ( 40 + 12 + 3(16) ) = 1 g

• 5 moles of ammonium carbonate

m(NH4)2CO3 = n x Mr = 5 x ( 2(14+4)+12+3(16) ) = 5 x (96) = 480g





The Mole - mass calculationsC + O2 CO2

Carbon reacts with oxygen to form carbon dioxide as shown.

If 0.12g of carbon are reacted with excess oxygen what mass of carbon dioxide would be formed?

1. Balance the reaction

2. Work out moles of reactant(mass given).

3. Go through the equation to find out the number of moles being formed

4. 4. Work out quantity asked for.





The Mole - mass calculationsC + O2 CO2

Carbon reacts with oxygen to form carbon dioxide as shown.

If 0.12g of carbon are reacted with excess oxygen what mass of carbon dioxide would be formed?

1. Balance the reaction

2. Work out moles of reactant(mass given).

n(C) = m/Ar = 0.12/12 = 0.01 mol

3. Go through the equation to find out the number of moles being formed

- the molar ratio: C:CO2 1:1

- => n(CO2) = 0.01 mol

4. Work out quantity asked for.

m(CO2) = nxMr = 0.01 x (12+2(16)) = 0.01 x 44 = 0.44 g





Limiting reagent exampleAmmonia gas is made by reacting ammonium chloride with calcium hydroxide

according to:

NH4Cl + Ca(OH)2 NH3 + CaCl2 + H2O

If 32.1 g of ammonium chloride reacts with 7.5 g calcium hydroxide in solution,

Show by calculation; which is the limiting reagent and what mass of ammonia

is produced.

1. Balance the reaction.

2. Calculate moles given (both).

3. Work through molar ratio to decide which is limiting reagent.

4. Use limiting reagent to calculate quantity asked as before.





Limiting reagent exampleAmmonia gas is made by reacting ammonium chloride with calcium hydroxide

according to:

NH4Cl + Ca(OH)2 NH3 + CaCl2 + H2O

If 32.1 g of ammonium chloride reacts with 7.5 g calcium hydroxide in solution,

Show by calculation; which is the limiting reagent and what mass of ammonia

is produced.





Percentage CompositionAnalysis of a compound by mass makes it

possible to work out the % mass of each element.

eg Table salt: NaCl mass analysis:One mole of NaCl would have a mass of

23 + 35.5 = 58.5g

• The % composition can be found using the formula:

Mass element X x100

Total Mass Compound

• %Na = […../ (…..) ]x100 = …………..% (by mass)

• %Cl = (…../ (…….) )x100 = …………%

% Mass Element X =





Percentage Composition from mass.

Eg2 Calculate the % of oxygen in water.

Mr (H2O) = (m(O)/Mr(H2O))x100

= (16/18)x100

= 88.9%





Empirical and Molecular Formula.A compound consists of carbon, hydrogen and oxygen only. The

% by mass are Carbon 40.0% and 6.7% hydrogen. Calculate the

empirical and molecular formula of the compound if Mr = 60g·mol-1

%(O) = 100 – (40+6.7) = 53.3

C H O

In 100g: 40.0g 6.7g 53.3g

n=m/Mr:40/12

6.7/153.3/16

3.33 6.7 3.33

3.33 3.33 3.33

Simplest: 1 2.01 1

Empirical Formulae:

CH2O (12+2+16 = 30)

Molecular Formula: 2(CH2O)

C2H4O2 (Mr = 2x30)





Concentration - MolarityThe concentration of a solution is defined as the ………………. of

……………………… per ………………. (dm3) of ………………….

solute

solute

Final volume of

……………..

500cm3

=+

Concentration =Amount of ……… (……….)

Volume of ………………

30g of

NaCl

C =n

v





Volume Conversions

1 dm = ….. cm

1 dm3 = ………… cm3

1 m3 = …………….. dm3 = ………………….cm3 (10….)

1cm3

1 dm3 (1 litre)

10 cm3

10 cm3

10 cm3





Decimal Conversions

King Henry Died a miserable death called measles

Kilo Hecta Decca m(unit) deci centi milli

1000 100 10 1 1/101/100

1/1000

1m

?km

0. 0 0 1km

1 m = 0.001 km

1 km = 1000 m





Conversions

1 cm = 0.1 dm

1 cm2 = (0.1)2 dm2

1 cm2 = 0.01 dm2

1 cm3 = (0.1)3 dm3

1 cm3 = 0.001 dm3

1 dm3= 1000 cm3

25 cm3 = 0.025 dm3





Molar VolumesOne mole of an ideal (ANY) gas occupies a volume of 22,4dm3 at

standard temperature and pressure. (STP)

STP: T= 0ºC, 273K P =1 atmosphere (101,3kPa)

Fe2O3 + 3H2 2Fe + 3H2OWhat volume of hydrogen reacts with 50g of Fe2O3

Fe2O3 : H2

1 : 3

n(H2) =3n(Fe2O3) = 3(0.3125) = 0.9375

v(H2) = nxMv = 0.9375x22.4 = 21dm3

n(Fe2O3) = m/Mr = 50/(2(56)+3(16)) = 0.3125mol

moles = volume/molar volume ==> n = v/Mv22.4 dm3





ASKEDGIVEN

Mole Calculations

MOLES MOLES

MASS MASS

VOLUMEVOLUME

CONCENTRATIONCONCENTRATION

MOLAR

RATIO

Number

Of

particles

Number

Of

particles





Volume - Volume Calculations

H2 + N2 --> NH3

If 3.00 dm3 of nitrogen are reacted to produce ammonia, what

volume of hydrogen will be required? (At STP)

1. 3H2 + N2 --> 2NH3

2. n(N2) = v/Mv = 3/22.4 = 0.134mol

3. N2 : H2 1:3 n(H2) = 3(N2)

4. n(H2) = 3(0.13) = 0.401mol

5. v(H2) = n(H2)Mv = 0.401(22.4) = 8.98dm3





Mass Volume Calculations

1. Balance the equation - 3H2 + N2 --> 2NH3

2. Calculate the moles of the substance given.

n(NH3) = v/Mv = 46/22.4 = 2.05 mol

3. Work through the molar ratio to find out the moles of the substance

asked.

N2 : NH3 as 1 : 2

n(N2) = 1/2(n(NH3)) = 1/2(2.05) = 1.03 mol

4. Calculate the quantity asked for.

m(N2) = n(N2)Mr = 1.03(28) = 28.84 g

2. H2 + N2 --> NH3

What mass of nitrogen (in dm3) would be needed to produce 46dm3 of ammonia?





Standard Solution

A standard solution is one

for which the concentration is precisely known.

Since

c = n(solute)/v(solvent)

= m/Mr V

• The number of moles of solute (Mass)

• The volume of solution.

These values must be accurately determined.

2.45g

Mass is determined

accurately using an

electronic balance.

• Possible accuracies of

0.1 - 0.0001g

KMnO4

Volume is measured

using a volumetric

flask.

• 250 cm3

• 100 cm3, 200 cm3,





Weighing Technique

Procedure - Weighing by difference.

1. Zero scales and clean the pan.

2. Weigh the weighing container.

3. Add (approximately) the

required amount of salt. Take

care not to drop any salt onto

the pan.

4. Transfer the salt to a clean

beaker.

5. Reweigh the weighing

container.

6. Subtract the final mass of the

container from the mass of salt

and container to give the mass

of salt transferred to the beaker.

Mass is determined accurately

using an balance (electronic

or triple beam).

• Possible accuracies of 0.1 -

0.0001g

2.45g

KMnO4

Results:

Mass salt + container: …………

Final Mass container: …………

Mass salt transferred:





Volumetric FlaskMaking a standard solution.

1. Rinse a clean & dry 100 cm3 beaker with a little distilled water.

2. Transfer the correctly weighed amount of salt to the beaker. Ensure NO SALT IS LOST.

3. Add 50 - 80 cm3 water the salt and stir gently with a glass rod until all salt is dissolved. DO NOT REMOVE THE ROD FROM THE SOLUTION NOR ALLOW ANY DROPS OF SOLUTION TO ESCAPE.

4. Add ALL the solution to volumetric flask via funnel. Ensure glass rod and beaker are thoroughly rinsed. (Include rinsings.)

5. Add enough solvent to bring the level up to the mark.





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Moles P1

Education

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