Chemical Calculations

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Reacting Masses

1. 2Na + Cl2 --> 2NaCl

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

1. C + Cl2 --> CCl4

m(

1. 2ZnS + 3O2 --> 2ZnO + 2SO2

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

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

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





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 dozdozen

12

x36

12 12 12

?3





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





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





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





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 + NaOH

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

1. Balance the reaction

2Na + 2H2O H2 + 2NaOH

MOLAR RATIO:2 : 2 : 1 : 2

2 Work out moles of reactant (given).

n(Na)=m/Ar=46/23=2mol

3 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 formed

4 Work out quantity asked for.

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

GIVEN ASKED





Mole examples - B & J p119 21 & p120 22

1. 2Na + Cl2 2NaCl

Calculate the mass of salt formed if 2.3g of sodium is reacted with XS chlorine.

1. Balanced

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

3. M:R 2:2 .; 1:1 n(NaCl) = n(Na) = 0.1 mol

4. 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)





ASKEDGIVEN

Mole Calculations

MOLES MOLES

MASS MASS

VOLUMEVOLUME

CONCENTRATIONCONCENTRATION

MOLAR

RATIO





Volume - Volume Calculations1. Balance the equation

2. 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 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





Reactions – Limiting reagent

The 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?

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

2. 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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