chemical calculations p
DESCRIPTION
A set of slides created to teach Chemicals Calculations P to grade 10 learners following the NSC Caps syllabus in Cape Town.TRANSCRIPT
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Chemical Calculations
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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)
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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
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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 = ……………..
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Dozen & Particles
... ... ... ... ... ... ... ... ... ... ... ...particles
1 doz 1 doz 1 doz dozen
12
x36
12 12 12
?3
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Relative Atomic Mass
Z
AXAtomic 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
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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
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ISOTOPESSymbol PROTONS ELECTRONS NEUTRONS
Carbon 1212 6C
Carbon 1313 6C
Boron 1010
5B
Boron 1111
5B
Hydrogen 1
Hydrogen 2
Chlorine 35
Chlorine 37
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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!
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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
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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
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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. (5.58 g)2. Zn + HCl ZnCl2 + H2
What mass of HCl is needed to produce 100g of hydrogen? (3650 g)
3. KClO3 KCl + O2
What mass of oxygen is produced from 1kg of potassium chlorate? (391.68g)
4. Fe2O3 + H2 Fe + H2OWhat mass of iron is produced if 3g of rust (Fe2O3)
is reacted with XS(100g )of hydrogen? (2.1 g)
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Mole examples - B & J p119 21 & p120 22
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 2:2 .; 1:1 n(NaCl) = n(Na) = 0.1 mol
5. m(NaCl) = nxMr
=(0.1)*(23+35.5)
= 5.85 g
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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 =
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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)
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Molar Volumes
One mole of an ideal (ANY) gas occupies a volume of 22,4dm3 at standard temperature and pressure. (STP)
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ASKEDGIVEN
Mole Calculations
MOLES MOLES
MASS MASS
VOLUME VOLUME
CONCENTRATION CONCENTRATION
MOLARRATIO
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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
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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
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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
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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??
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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)
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Keith Warne
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