new presentation in handin folder ‘chemical calculations” download and get ready please

• New presentation in handin folder• ‘CHEMICAL CALCULATIONS”

• DOWNLOAD AND GET READY PLEASE

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

Chemical Change - ReactionsObjectives: At the end of this section

you should be able to:-• Explain the difference between chemical and physical

change.• Use the periodic table to determine valency.• Explain the significance of the law of constant

composition for writing chemical reactions.*** (comeback to)

• Write and balance chemical formulae.• Calculate reacting masses for given reactions.• Explain energy changes in reactions.• Explain Avogadro’s law.• Work out reacting volumes of gases.

Chemical Change - ReactionsObjectives: At the end of this section

you should be able to:-• Distinguish between synthesis and decomposition

reactions• Discuss how energy released by chemical reactions is

essential for life• Describe how metals react with oxygen• Explain the reverse process in which metals are

separated from their oxides• Describe the reactions of non-metals with oxygen• Write balanced equations for reactions involving non-

metals.

Reacting Masses

1. 2Na + Cl2 --> 2NaCl

m(NaCl) = 2(23+35.5) = 117.0 g

2. C + Cl2 --> CCl4

m(

3. 2ZnS + 3O2 --> 2ZnO + 2SO2

4. FeS + 2HCl --> H2S + FeCl2

5. SO2 + 2H2S --> 3S + 2H2O

Calculate the mass of each underlined compound either produced or required. (Balance the reactions first)

Reacting Masses

1. 2Na + Cl2 --> 2NaCl

m(NaCl) = 2(23+35.5) = 117.0 g

2. C + Cl2 --> CCl4

m(

3. ZnS + O2 --> ZnO + SO2

4. FeS + HCl --> H2S + FeCl2

5. SO2 + H2S --> S + H2O

Calculate the mass of each underlined compound either produced or required. (Balance the reactions first)

Balancing Reactions1. 2Na + Cl2 --> 2NaCl

2NaCl: 2(23+35.5) = 2(58.5)=117g

2. C + 2Cl2 --> CCl4

CCl4: 12 + 4(35.5) = 154g

3. 2ZnS + 3O2 --> 2ZnO + 2SO2

SO2: 2(32 + 2(16)) = 128g

4. FeS + 2HCl --> H2S + FeCl2

FeS: 56 + 32 = 88g

5. SO2 + 2H2S --> 3S + 2H2O

H2S: 2(2 + 32) = 68g

Balancing More Reactions

1. Na + H2O --> NaOH + H2

2. H2 + O2 --> H2O

3. CaCO3 --> CaO + CO2

4. CaCl2 + Na2SO4 --> CaSO4 + NaCl

5. Al(NO3)3 + K2CO3 --> Al2(CO3)3 + KNO3

6. Na3PO4 + MgI2 --> Mg3(PO4)2 + NaI

Balancing More Reactions

1. 2Na + 2H2O --> 2NaOH + H2

2. 2H2 + O2 --> 2H2O

3. CaCO3 --> CaO + CO2

4. CaCl2 + Na2SO4 --> CaSO4 + 2NaCl

5. 2Al(NO3)3 + 3K2CO3 --> Al2(CO3)3 + 6KNO3

6. 2Na3PO4 + 3MgI2 --> Mg3(PO4)2 + 6NaI

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) or (NA)

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

Particles = n x L

The Carbon Standard• Carbon-12 is the standard upon which the relative mass of other atoms

is determined. It wasn’t always this way. • At first hydrogen was used and it was assigned the atomic mass of

one. If you have equal numbers of nitrogen atoms and hydrogen atoms, the nitrogen atoms are 14 times heavier than the hydrogen atoms. Therefore, nitrogen was assigned the atomic number of 14.

• Later oxygen was used as the standard with an atomic mass of 16. • However, carbon-12 proved to be more convenient to capture and

measure in pure form, so it became the standard. • However, now even carbon-12 is slowly losing its position as standard,

as sophisticated equipment makes it possible to give even more accurate measures of atomic mass. For this reason you will notice that on the periodic table the AMUs are not expressed as exact relative units to carbon-12.

Dozen & Particles

... ... ... ... ... ... ... ... ... ... ... ...particles

1 doz 1 doz 1 doz dozen

12

x36

12 12 12

?3

Moles & Particles

..........................................................................

..........................................................................

..........................................................................

..............................................

..........................................................................

..........................................................................

..........................................................................

..............................................

..........................................................................

..........................................................................

..........................................................................

..............................................particles

1 mol 1 mol 1 mol moles

(n)

L6.023 x 1023

x18.069 x 1023

6.023 x 1023 6.023 x 1023 6.023 x 1023

?3

Avogadro’s No Egs• If you have 6g of Hydrogen gas how

many...• A) Molecules• B) atoms• C) electrons do you have?

Avogadro’s No Egs• If you have 6g of Hydrogen gas how many...

• A) Molecules

n(H2) = m/Mr. = 6/(2) = 3 mol of H2 molecules

.: no of molecules = L x n = (6.023 x 1023)(3) = 1.8069 x 1024 • B) atoms

no atoms = molecules x 2 = 2 x 1.8069 x 1024 = 3.6138 x 1024

• C) electrons do you have?

no electrons = atoms = 3.6138 x 1024

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

AXAtomic Number

(smaller)

Mass Number (bigger)

protons + neutrons

Periodic Table Symbol

Relative atomic mass

or

mass(g) of one mole

The Mole and MassThe amount is defined in such a way that the relative atomic mass (MASS NUMBER) of an element is equal to the mass of one mole of the substance. (in grams) -

THE MOLAR MASS

• Eg Na = 23g/mol, water(H2O)=18g/mol

Z

AXAtomic Number

(smaller)


protons + neutrons


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…)

Relative Atomic Mass

Z

AXAtomic Number

(smaller)


protons + neutrons


or

mass(g) of one mole


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

Relative Atomic MassCalculate: The mass in grams –

1. of one mole of copper chloride (CuCl2)

2. one mole of carbon dioxide (CO2)

3. 1.5 mole of oxygen (O2)

4. TWO moles of methane (CH4)

5. Four moles of water.

Relative Atomic MassCalculate: The mass in grams -1. of one mole of copper chloride (CuCl2)

Mr(CuCl2) = 63.5 +2(35.5) = 134.5g

2. one mole of carbon dioxide (CO2)Mr(CO2) = 12 +2(16) = 44g

3. 1.5 mole of oxygen m(O2) = nxMr = 1.5 x 32 = 48g

4. TWO moles of methane (CH4) m(CH4) = 2x(12 +4(1)) = 2(16) = 32g

5. Four moles of water. m(H2O) = n x Mr = 4 (18) = 72g

Relative Masses - examples

Calculate the Relative Atomic Mass of:• O2 (oxygen gas) • Cl2 (chlorine gas)• NaCl (sodium chloride - table salt) • H2SO4 Sulphuric acid• CaCO3 (calcium carbonate)• (NH4)2Cr2O7 (ammonium dichromate)

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

Isotopes• Isotopes - Atoms of the same element which have different

numbers of neutrons. Eg: 613C & 6

12C

Relative atomic mass is (actually) the average mass (of all the isotopes in a random sample) of the atoms of an element relative to 1/12 of the mass of a carbon-twelve atom.

613C

• 6 protons

• 6 electrons

• 13-6 = 7 neutrons

612C

• 6 protons

• 6 electrons

• 12-6 = 6 neutrons

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)

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

Or4 atoms – 3 are 35 and one is 37!

Av Ar(Cl) = (37x1)+(35x3) = 35.504

ISOTOPESSymbol PROTONS ELECTRONS NEUTRONS

Carbon 1212 6C

Carbon 1313 6C

Boron 1010

5B

Boron 1111

5B

Hydrogen 1

Hydrogen 2

Chlorine 35

Chlorine 37

ISOTOPESSymbol PROTONS ELECTRONS NEUTRONS

Carbon 1212 6C 6 6 6

Carbon 1313 6C 6 6 7

Boron 1010

5B 5 5 5

Boron 1111

5B 5 5 6

Hydrogen 1

Hydrogen 2

Chlorine 35

Chlorine 37

IsotopesBoron’s two isotopes 5

10B & 511B

510B (19.8%) & 5

11B (80.2%) - exist in the ratio 1:4Calculate the average mass of a Boron atom. (Two methods)

In 100 atoms – 19.8 have a mass of 10 and 80.2 have mass 11!

Average Ar(B)= total mass = (10x19.8)+(11x 80.2) = 10.8 no of atoms 100

Or5 atoms – 4 are 10 and 1 is 11!

Av Ar(B) = (10x1)+(11x4) = 10.85

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 --> Moles n = m/Mr

• Eg calculate the number of moles of water that would have a mass of 100g.

• How many moles of iron (II) chloride are in 50 g of iron (II) chloride?

• Calculate the number of moles needed to get1kg of calcium carbonate.

• How many moles of CuSO4.5H2O would give you 0.1g of water?

The Mole & Mass --> Relative Mass Mr = m/n = 5.56 mol

• Eg Calculate the relative mass of a compound for which 0.001 moles have a mass of 0,0056 g.

• What is the relative mass of a compound for which 0.01 mols has a mass of 0.18g

• Identify the element for which 0.05 moles has a mass of 0.16 g ?

Mr (X) = m/n= 0.0056/0.001 = 5.6 g/mol

The Mole & Mass --> Relative Mass Mr = m/n = 5.56 mol

• Eg Calculate the relative mass of a compound for which 0.001 moles have a mass of 0,0056 g.

• Mr = m/n = 0.0056/0.001 = 5.6 g.mol-1

• What is the relative mass of a compound for which 0.01 mols has a mass of 0.18g

• Mr = m/n = 0.18/0.01 = 18 g.mol-1 .: H2O

• Identify the element for which 0.005 moles has a mass of 0.16 g ? Mr (X) = m/n

= 0.16/0.005 = 32 g/mol .: X is sulphur!

The Mole - ReactionsSodium reacts with water to form hydrogen and sodium hydroxide

according to the equation.

Na + H2O H2 + NaOHIf 46g of sodium are reacted with excess water what mass of hydrogen

would be formed?1. Balance the reaction

2. Work out moles of substance GIVEN.

3. Go through the equation to find out the number of moles reacting and being formed. (Molar ratio).

4. Work out quantity asked for.

Steps1. Balance equation2. Calculate moles given3. Use Molar Ratio to find moles asked4. Calculate quantity asked.

The Mole - Reactions

GIVEN ASKED

2. Moles given (m/mr)

1 & 3 MOLAR RATIO

4. Moles asked(nxMr/v)

2H2 + O2 2H2O

4g of O2? g H2O

n(O2 ) = m/Mr

M:R O2 :H2O 1:2 .: n(H2O) = 2xn(O2)

m(H2O) = nxMr

Amount GIVEN

Amount ASKED

The Mole - equationsSodium reacts with water to form hydrogen and sodium hydroxide

according to the equation.

Na + H2O H2 + NaOHIf 46g of sodium are reacted with excess water what mass of

hydrogen would be formed?1. Balance the reaction

2Na + 2H2O H2 + 2NaOHMOLAR RATIO:2 : 2 : 1 : 22 Work out moles of reactant (given).

n(Na)=m/Ar=46/23=2mol3 Go through the equation to find out the number of moles

reacting and being formed - the molar ratio:

Na : H2 2:1 => 1 mole H2 formed4 Work out quantity asked for.

m(H2) = nxMr = 1 x 2 = 2 g

GIVEN ASKED

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 reaction2. 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 reaction2. Work out moles of reactant(mass given).

n(C) = m/Ar = 0.12/12 = 0.01 mol3. Go through the equation to find out the number of moles

being formed - the molar ratio: C:CO2 1:1 - => n(CO2) = 0.01 mol4. Work out quantity asked for.

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

Mole examples - B & J p119 21 & p120 22

1. Na + Cl2 NaClCalculate the mass of salt formed if 2.3g of

sodium is reacted with XS chlorine.2. Zn + HCl ZnCl2 + H2

What mass of HCl is needed to produce 100g of hydrogen?

3. KClO3 KCl + O2

What mass of oxygen is produced from 1kg of potassium chlorate?

4. Fe2O3 + H2 Fe + H2OWhat mass of iron is produced if 3g of rust (Fe2O3)

is reacted with XS(100g )of hydrogen?


1. 2Na + Cl2 2NaClCalculate the mass of salt formed if 2.3g of

sodium is reacted with XS chlorine.

2. Balanced

3. n(Na) = m/Ar =(2.3)/23 = 0.1 mol

4. M:R 1:1 n(NaCl) = 0.1 mol

5. m(NaCl) = nxMr =(0.1)*(23+35.5) = 5.85 g


1. 2Na + Cl2 2NaClCalculate the mass of salt formed if 2.3g of

sodium is reacted with XS chlorine.

2. n(Na) = m/Ar = 2.3/23 = 0.1mol3. Molar Ratio

Na:NaCl 2:2 ie 1:1 => n(NaCl) = 0.1mol

3. m(NaCl) = nxMr = 0.1x(23+35.5) = 5.85g

Mole examples - B & J p119 21 & p120 22 2. Zn + 2HCl ZnCl2 + H2

What mass of HCL is needed to produce 100g of hydrogen?

2.1 n(H2) = m/Mr = 100/2 = 50mol

2.2 Ratio HCl:H2 2:1

.;n(HCl) = 2n(H2)= 2*50 = 100mol

2.3 m(HCl) =nxMr = (100)*(1+35.5) = 3650 g

Mole examples - B & J p119 21 & p120 22 3. 2KClO3 2KCl + 3O2

What mass of oxygen is produced from 1kg of potassium chlorate?

3.1m(KClO3) = m/Mr

= 1000/(39+35.5+(3x16))

= 8.16mol

3.2Ratio KClO3:O2 2:3

n(O2) = (3/2)n(KClO3)=(3/2)(8.16) = 12.24mol

3.3m(O2) = nxMr = 12.24(32)=391.68g


4. Fe2O3 + H2 Fe + H2OWhat mass of iron is produced if 3g of rust (Fe2O3)

is reacted with XS(100g )of hydrogen?

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 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 = [23/ (58.5) ]x100 = 39.3 % (by mass)

• %Cl = (35.5/ (58.5) )x100 = 60.7%

% Mass Element X =

Percentage Composition from mass.Eg2 Calculate the % of oxygen in water.

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: …….g ……..g ….…g

n=m/Mr: …/… 6.7/….

53.3/……

…… …… ……..……. …… …….

Simplest: … …… ….

Empirical Formulae: ……. (12+2+16 = …..)

Molecular Formula: 2(CH2O) ……… (Mr = …. X 30)

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 6.7/1

53.3/16

3.33 6.7 3.333.33 3.33 3.33

Simplest: 1 2.01 1

Empirical Formulae: CH2O (12+2+16 = 30)

Molecular Formula: 2(CH2O) C2H4O2 (Mr = 2x30)

Empirical and Molecular Formula.

OH

O

CH3

C

MOLECULAR FORMULA: CH3COOH or C2H4O2

Actual formula

Mr: 2(12)+4(1)+2(16)=60g.mol-1

%C: (24/60)x100 = 40.0%

%H: (4/60)x100 = 6.7%

%O: (32/60)x100 = 53.3%

EMPIRICAL FORMULA: CH2O

(Simplest whole number ratio)

Empirical and Molecular Formulae.

Eg3. If a compound consisting of nitrogen and oxygen only - contains 30.4% by mass of nitrogen. What is the molecular formula of the compound? >>

Formula from Percentage Composition.Eg3. If a compound consisting of nitrogen and oxygen only - contains 30.4% by mass of nitrogen. What is the molecular formula of the compound?

1 Assume you have 100g of the compound. You would therefore have 30.4g of nitrogen.... and 100-30.4 = 69.6g of oxygen.

Mass Ratio- N : O 30.4 : 69.6

Mole Ratio n = 30.4/14= 2.2 : 69.6/16= 4.4 (n=m/Ar)

1(2.2/2.2) : 2 (4.4/2.2)

Compound Formula NO2 (More egs B & J p112/3 ex1-4)

pg 119 No 17

Molar VolumesOne mole of an ideal (ANY) gas occupies a volume of ………….3 at ………………………… temperature and pressure. (STP)

STP: T= ….ºC, ……K P =1 atmosphere (……...kPa)

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

Fe2O3 : H2

… : …..

n(H2) =…..n(Fe2O3) = ……………………

v(H2) = …………………………… dm3

n(Fe2O3) = m/Mr = ………………….= ………………mol

moles = volume/molar volume ==> n = …./…..

Molar Volumes

One mole of an ideal (ANY) gas occupies a volume of 22,4dm3 at standard temperature and pressure. (STP)

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 : H2

1 : 3

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

v(H2) = nxMv = 0.9375x22.4 = 21.0 dm3

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

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

REACTANTS PRODUCTS

MOLES MOLES

MASS MASS

VOLUME VOLUMEMOLARRATIO

ASKEDGIVEN

Mole Calculations

1. balance rxn. REACTANTS PRODUCTS

MOLES MOLES

MASS MASS

VOLUME VOLUMEMOLARRATIO

n=m/Mr m=nxMr2.

3.

4.

ASKEDGIVEN

Mole Calculations

MOLES MOLES

MASS MASS

VOLUME VOLUME

CONCENTRATION CONCENTRATION

MOLARRATIO

ASKEDGIVEN

Mole Calculations

MOLES MOLES

MASS MASS

VOLUME VOLUME

CONCENTRATIONCONCENTRATION

MOLARRATIO

Number Of

particles

Number Of

particles

Volume - Volume Calculations1. Balance the equation2. Calculate the moles of the substance given.3. Work through the molar ratio to find out the moles of the

substance asked.4. Calculate the quantity asked for. (Volume V = n x Mv)

Mv = 22.4dm3 At STP

EG: H2 + N2 --> NH3If 3.00 dm3 of nitrogen are reacted to produce ammonia, what

volume of hydrogen will be required? (At STP)

H2 + N2 --> NH3

Volume - Volume CalculationsH2 + N2 --> NH3

If 3.00 dm3 of nitrogen are reacted to produce ammonia, what volume of hydrogen will be required? (At STP)

H2 + N2 --> NH3

Volume - Volume CalculationsH2 + 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

Reactions – Limiting reagentThe reagent that runs out first and stops the

reaction is known as the LIMITING REAGENT.

If 46g of sodium are reacted with excess water what mass of hydrogen would be formed?

Na + H2O H2 + 2NaOH46g 2 moles XS

Na will run out first

Na is LIMITING REAGENT

What is the minimum amount of water needed to react completely with 46g of sodium??

Reactions – Limiting reagentThe reagent that runs out first and stops the reaction is known as the

LIMITING REAGENT.

If 46g of sodium are reacted with excess water what mass of hydrogen would be formed?

2Na + 2H2O H2 + 2NaOH46g 2 moles XS

Na will run out first

Na is LIMITING REAGENT

What is the minimum amount of water needed to react completely with 46g of sodium??

n (H2O) = n(Na) = 2 mol .: m(H2O) = nxMr = 2x(18)=36g

Therefore 36g of water would react COMPLETELY with 46g of sodium.

If you had less than 46 g of sodium – less water would be needed.

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.

Mass Volume Calculations1. KClO3 KCl + O2

What volume of oxygen is produced by the decomposition of 1kg of potassium chlorate?

2. H2 + N2 --> NH3

How much nitrogen (in dm3) would be needed to produce 46dm3 of ammonia?

3. S + O2 --> SO2

What volume of sulphur dioxide could be produced from 20.0dm3 of oxygen?

4. Zn + 2HCl ZnCl2 + H2

What mass of zinc is needed to produce 100dm3 of hydrogen?

5. Fe2O3 + H2 Fe + H2OIf 3.00kg of iron oxide is reacted with 0.256dm3 of

hydrogen, what mass of water would be produced?

Mass Volume Calculations1. KClO3 KCl + O2

What volume of oxygen is produced by the decomposition of 1kg of potassium chlorate?

2. Balance the equation - 2KClO3 2KCl + 3O2 (1)

3. Calculate the moles of the substance given.

n(KClO3) = m/Mr = 1000/(39+35.5+3(16)) = 8.16mol (1)

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

KClO3 : O2 2 : 3

n(O2) = 3/2n(KClO3) = 3/2(8.16) = 12.24 mol (1)

4. Calculate the quantity asked for. (Volume V = n x Mv)

Mv = 22.4dm3 At STP

v(O2) = n(O2)Mv = 12.24(22.4) = 275 dm3 (2)

Mass Volume Calculations

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


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


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?

Volume -Volume Calculations

1. Balance the equation - Done


n(O2) = v/Mv = 20/22.4 = 0.893 mol


O2 : SO2 as 1 : 1

n(SO2) = n(O2) = 0.893 mol

4. Calculate the quantity asked for. (Volume V = n x Mv)

Mv = 22.4dm3 At STP

v(SO2) = n(SO2)Mv = 0.893(22.4) = 20.0 dm3

3. S + O2 --> SO2

What volume of sulphur dioxide could be produced from 20.0dm3 of oxygen?

Mass Volume Calculations

1. Balance the equation - Done


n(H2) = v/Mv = 100/22.4 = 4.46 mol


Zn : H2 as 1 : 1

n(Zn) = n(H2) = 4.46 mol

4. Calculate the quantity asked for. (mass m = n x Mr) m(Zn) = n(Zn)Mr = 4.46(65.4) = 291.9g

4. Zn + 2HCl ZnCl2 + H2

What mass of zinc is needed to produce 100dm3 of hydrogen?

Volume - Mass Limiting Reagent

1. Balance the equation - Fe2O3 + 3H2 2Fe + 3H2O


n(Fe2O3) = m/Mr = 3000/(2(56)+3(16)) = 18.75 mol XS

n(H2) = v/Mv = 0.256/22.4 = 0.0114 mol LIMITING REAGENT


H2 : H2O as 1 : 1

n(H2O) = n(H2) = 0.0114 mol

4. Calculate the quantity asked for. (mass m = n x Mr)

m(H2O) = n x Mr = 0.0114(2+16) = 0.206g

5. Fe2O3 + H2 Fe + H2OIf 3.00kg of iron oxide is reacted with 0.256dm3 of hydrogen,

what mass of water would be produced?

new presentation in handin folder ‘chemical calculations” download and get ready please

Documents

molar mass

weight mass

relative atomic mass

relative molecular mass

chemical reactions

reactions of nonmetals

gc cl2

h2s fecl2so2 h2s