Chemical equations
This presentation is the property of its rightful owner.
Sponsored Links
1 / 20

Chemical Equations PowerPoint PPT Presentation


  • 268 Views
  • Uploaded on
  • Presentation posted in: General

Chemical Equations. p.13 Q1 Write balanced equations including state symbols for the following: (a)zinc + oxygen → zinc oxide 2 Zn (s) + O 2 (g) → 2 ZnO (s) (b)potassium + water → potassium hydroxide + hydrogen 2 K (s) + 2 H 2 O (l) → 2 KOH (aq) + H 2 (g)

Download Presentation

Chemical Equations

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Chemical equations

Chemical Equations

p.13 Q1 Write balanced equations including state symbols for the following:

(a)zinc + oxygen → zinc oxide

2 Zn (s) + O2 (g) → 2 ZnO (s)

(b)potassium + water → potassium hydroxide + hydrogen

2 K (s) + 2 H2O (l) → 2 KOH (aq) + H2 (g)

(c)calcium carbonate → calcium oxide + carbon dioxide

CaCO3 (s) → CaO (s) + CO2 (g)


Ionic equations

Ionic Equations

  • These show what happens which ions actually react.

    • spectator ions do not change and are omitted. They just maintain electrical neutrality.

  • Start with the balanced full formula equation.

  • Covalent substances are written normally in the ionic equation.

  • Ionic substances which are in solution or molten have their ions written out separately.

  • Cancel out any ions appearing on both sides of the equation unchanged


Ionic equations1

Ionic Equations

2 K (s) + 2 H2O (l) → 2 KOH (aq) + H2 (g)

2 K (s) + 2 H2O (l) → 2 K+ (aq) + 2 OH- (aq) + H2 (g)

Fe2(SO4)3 (aq) + 3 Zn (s) → 3 ZnSO4 (aq) + 2 Fe (s)

2 Fe3+ (aq) + 3 SO42- (aq) + 3 Zn (s) →

3 Zn2+ (aq) + 3 SO42- (aq) + 2 Fe (s)

2 Fe3+ (aq) + 3 Zn (s) → 3 Zn2+ (aq) + 2 Fe (s)

Fe2(SO4)3 (aq) + 6 KOH (aq) → 2 Fe(OH)3 (s) + 3 K2SO4 (aq)

2 Fe3+ (aq) + 3 SO42- (aq) + 6 K+ (aq) + 6 OH- (aq) →

2 Fe(OH)3 (s) + 6 K+ (aq) + 3 SO42- (aq)

Fe3+ (aq) + 3 OH- (aq) → Fe(OH)3 (s)

3


Ionic equations for precipitation reactions

Ionic Equations for precipitation reactions

  • Start by writing the formula of the precipitate with its (s) state symbol on the product (right hand) side.

  • Write the 2 reactant ions needed to form this product on the left side of the equations with their (aq) state symbols.


P 23 q 2

p. 23 Q. 2

Al2O3 (s) + 6 HI (g) → 2 AlI3 (aq) + 3 H2O (l)

What masses of aluminium oxide and hydrogen iodide would be required to produce 102 g aluminium iodide?

R.F.M. of AlI3 = 27 + (3 × 126.9) = 407.7

Moles of AlI3 = 102 / 407.7 = 0.250

Moles Al2O3 = 0.250 / 2 = 0.125

R.F.M. of Al2O3 = (2 × 27) + (3 × 16) =102

Mass of Al2O3 = 0.125 × 102= 12.8 g

Moles of HI = 0.250 × 3 = 0.750

R.F.M of HI = 127.9

Mass of HI = 0.750 × 127.9 = 95.9 g


P 23 q 3

p. 23 Q.3

The substance ATP is important for transporting energy in living cells. a sample of 1.6270 g of ATP was analysed and found to contain 0.3853 g of carbon, 0.05178 g of hydrogen, 0.2247 g nitrogen and 0.2981 g of phosphorus. The rest was oxygen. The relative molecular mass of ATP is 507. What are its empirical and molecular formulae?

Oxygen mass= 1.6270 - 0.3853 – 0.05178 – 0.2247 – 0.2981 = 0.6671 g

C : H : N : P : O = (0.3853/12):(0.05178/1):(0.2247/14):(0.2981/31):(0.6671/16)

= 0.03211 : 0.05178 : 0.01605 : 0.009616 : 0.04169

0.009616 0.009616 0.009616 0.009616 0.009616

= 3.339 : 5.385 : 1.669 : 1.000 : 4.335

=10 : 16 : 5 : 3 : 13

empirical formula = C10H16N5P3O13 empirical formula mass = 507

molecular formula mass = 1 × empirical formula mass so molecular formula =

C10H16N5P3O13


P 27 q 1

p.27 Q. 1

A stock solution contains 10 mol dm-3 of hydrochloric acid. What volume of this solution should be dissolved in water to make 6 dm3 of acid of concentration 0.1 mol dm-3?

Moles of HCl = 6 dm3× 0.1 mol dm-3 = 0.6

Volume = 0.6 mole / 10 mol dm-3 = 0.06 dm3 = 60 cm3.


P 27 q 2

p.27 Q. 2

  • When solutions of potassium chloride and silver nitrate are mixed together, a precipitate of silver chloride forms.

  • Write a balanced molecular equation for this reaction.

  • KCl (aq) + AgNO3 (aq) → AgCl (s) + KCl (aq)

  • (b) In an experiment, 24.0 cm3 of 0.05 M silver nitrate solution exactly reacts with 15.0 cm3 of potassium chloride solution. Calculate the molar concentration of the potassium chloride.

  • Moles of AgNO3 = 24.0/1000 dm3 × 0.05 mol dm-3 = 0.0012

  • 1:1 AgNO3:KCl so 0.0012 mol KCl

  • Concentration = 0.0012 mol / (15/1000) dm3 = 0.0800 mol dm-3

  • (c) How much KCl would be needed to produce 100 cm3 of this concentration?

  • Moles = 100/1000 dm3× 0.0800 mol dm-3 = 0.008

  • Mass = 0.008 mol × 74.5 g mol-1 = 0.596 g


P 31 question 1

p.31 Question 1

A chemist mixed 12.0 g of phosphorus with 35.5 g of chlorine gas to synthesise phophorus (III) chloride. The yield was 42.4 g of PCl3. the equation for the reaction is:

2 P (s) + 3 Cl2 (g) → 2 PCl3 (l)

Moles of P = 12 / 31 = 0.3871;Moles of Cl2 = 35.5 / 71 =0.5 Moles of PCl3 = 42.4 / 137.5 = 0.3084.

This amount of PCl3 required 3/2 × 0.3084 moles Cl2 which is 0.463 moles, and 0.3084 moles P.

Therefore P was used in excess and Cl2 was limiting.

Percentage yield in terms of Cl2 = (0.463 / 0.5)*100 = 92.6%


P 31 q2

p.31 Q2

2-bromo-2-methylpropane, C4H9Br can be used to produce methylpropene, C4H8. 2-bromo-2-methyl propane reacts with sodium ethoxide, NaOC2H5 to produce ethanol C2H5OH and sodium bromide, along with methylpropene. Work out the atom economy for the reaction.

Ratio C4H9Br : C2H5OH : C4H8 = 1 : 1 : 1

R.F.M. 137 46 56

Atom economy = (56/[137 + 46]) × 100

= 30.6%


P 29 q1

p.29 Q1

0.025 mol Pb(NO3)2 to 0.50 mol KI.

1:2 ratio

Pb(NO3)2 (aq) + 2 KI (aq) → PbI2 (aq) + 2 KNO3 (aq)


Isotopes

Isotopes

  • An isotope is a sample of an element in which the atoms have the same number of neutrons as well as the same number of neutrons.

  • Isotopes are samples of the same element in which the atoms have different numbers of neutrons whilst having the same number of protons.


Relative atomic mass

Relative atomic mass

  • R.A.M. = (weighted average mass of an atom) / (1/12 mass of an atom of 12C)

  • Relative isotopic mass, relative formula mass and relative molecular mass are defined on the same scale.

    • Relative isotopic mass is not necessarily an integer and identical to mass number

  • R.A.M., R.F.M. and R.M.M. have no units

    • ratios of masses

  • Molar mass has the same numerical value as R.F.M. but has units g mole-1 (‘the mass of 1 mole’)


Mass spectrometer

Mass spectrometer


Mass spectrometer1

Mass spectrometer

  • Gaseous sample needed (vaporisation by heating or laser dye absorption)

  • Vacuum so ions move freely without colliding

  • Electron gun bombards sample molecule with high energy electrons causing ionisation, e.g.

    CH4 (g) + e- → CH4+ (g) + 2 e-

  • Negative plates accelerate sample ions and select for uniform velocity

  • Variable electromagnet deflects the ions

  • Detector, connected to computer, generates a graph of ion abundance against mass / charge (a function of electromagnetic field strength)

  • Molecular ions can fragment by random breaking of any covalent bond within the molecule. Every species detected in the mass spectrum is a positive ion.


Calculating relative atomic mass

Calculating relative atomic mass

Bromine (50.5% 79Br, 49.5% 81Br)

RAM = 80.0

Silver (51.3% 107Ag, 48.7% 109Ag

RAM = {(51.3 × 107) + (48.7 × 109)} / 100 = 108.0

Chromium ( 4.3% 50Cr, 83.8% 52Cr, 9.6% 53Cr, 2.3% 54Cr)

RAM = {(4.3 × 50) + (83.8 × 52) + (9.6 × 53) + (2.3 × 54)} / 100

= 52.1

Tips: SUS, check answer is plausible.


P 55 q2

p. 55 Q2

Notice that relative isotopic mass is not identical to mass number (due to nuclear binding energy).

R.A.M. = {(0.12 × 77.92) + (2.0 × 79.92) + (12 × 81.91) + (12 × 82.91) + (57 × 83.91) + (17 × 85.91)} / 100

= 83.90

(SUS)


Some other uses of mass spectrometry

Some other uses of mass spectrometry

  • Determining 12C / 13C in plant derived material

    • hence photosynthetic pathway and type of plant

  • Dating rocks

  • 14C dating (age of materials of biological origin)

  • Isotopic composition of hormones in athletes (is the molecule made ‘recently’ from material of photosynthetic origin and is therefore ‘natural’ or was it made from petroleum derived material by an organic chemist?

  • Analysis of atmospheres on other planets by spacecraft.


  • Login