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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)








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








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 = ……………..








Dozen & Particles

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

1 doz 1 doz 1 doz dozen

12

x36

12 12 12

?3








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








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








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!








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








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)








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








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 =








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)








Molar Volumes

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








ASKEDGIVEN

Mole Calculations

MOLES MOLES

MASS MASS

VOLUME VOLUME

CONCENTRATION CONCENTRATION

MOLARRATIO








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??








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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