Formulae, Stoichiometry And The Mole Concept

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Chemistry (5070)

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Formulae, Stoichiometry And The Mole Concept

Formulae, Stoichiometry And The Mole Concept

Formulae, Stoichiometry And The Mole Concept

The Modern Symbols: usually taken for the first letter of the element but not the case always. 

The first twenty elements with their symbols and atomic numbers are:

Hydrogen ( H ) 1
Carbon ( C ) 6
Sodium ( Na ) 11
Sulfur ( S ) 16
Helium ( He ) 2
Nitrogen ( N ) 7
Magnesium ( Mg ) 12
Chlorine ( Cl ) 17
Lithium ( Li ) 3
Oxygen ( O ) 8
Aluminium ( Al ) 13
Argon ( Ar ) 18
Beryllium ( Be ) 4
Fluorine ( F ) 9
Silicon ( Si ) 14
Potassium ( K ) 19
Boron ( B ) 5
Neon ( Ne ) 10
Phosphorus ( P ) 15
Calcium ( Ca ) 20
Symbols equations are used with reactants on the left side of the arrow and products on the right. These equations must be balanced as matter isn’t created or destroyed during a chemical reaction and hence it must be made certain that the number of atoms of all elements at each side is the same. 

For Example: 

Mg + O2   à   MgO

As there are two atoms of oxygen, 2 must be added with Mg and MgO

The final equation would look like this: 

2Mg + O2 à 2MgO

The reaction between Zinc and Dilute Hydrochloric acid.

Zn + HCl à ZnCl2 + H2 

As there are two Cl and H in the product side, 2 must be added with HCl.

The final equation would look like this,

Zn + 2HCl à ZnCl2 + H2

The reaction between Sodium Carbonate and Dilute Sulfuric acid to produce sodium sulphate, carbon dioxide and water.

Na2Co3 + H2SO4 à Na2SO4 + CO2 + H2O

This equation is perfectly balanced and would require no further addition. 

The elements are present in four different states in these reactions. They are solid (s), liquid (l), gas (g) and aqueous (aq). An aqueous solution is such that a substance is dissolved in water. 

These states are also written with the elements in these chemical reaction for better understanding. 

PbNO3 (aq) + 2NaCl (aq) à PbI2 (s) + 2NaNO3 (aq) 

This shows that the lead iodide obtained in the product is a precipitate.

Ionic Equations:

These equations only include the only ions that take part in the chemical reaction.

HCl (aq) + NaOH (aq) à NaCl (aq) + H2O (l)

This equation will be broken down into ions.

H+ (aq) + Cl (aq) + Na+ (aq) + OH-  (aq) à Na+ (aq) + Cl-(aq) + H2O (l)

After removing spectator ions from the equation, the ionic equation would come up to:

H+(aq) + OH- (aq) à H2O (l)

Empirical Formulas:


Show the formula of a compound in its simplest form. 

Different compounds may have the same empirical formula.

Example: 


Ethanoic Acid has following composition by mass,

Carbon 40%
Hydrogen 6.67%
Oxygen 53.33 %

Solution:
Divide the composition percentage by relative atomic mass

Carbon (40/12) = 3.33

Hydrogen (6.66/1) = 6.66

Oxygen (53.33/16) = 3.33

3.33 : 6.66 : 3.33

Divide by smallest value à (1:2:1)

Empirical formula of ethanoic acid is CH2O.

The ionic compounds exist as structures of oppositely charged ions. For example, in HCl, this indicates that the H ions and Cl ions exist in the ratio of 1:1. For Na2SO4, the ions for Sodium and Sulphate exist in the lattice in ratio of 2:1. 

Molecular Formula 


Shows the number of atoms of different elements in a compound. 

Benzene has carbon and hydrogen in the composition 92.3% carbon and 7.7% hydrogen. The relative molecular mass of benzene is 78. 

By using this information, we can find both empirical and molecular formula of benzene. 

Relative atomic mass of carbon is 12 and hydrogen is 1. 

Divide the composition percentage by the relative atomic mass. 

Carbon 92.3/ 12 à 7.7

Hydrogen 7.7/1 à 7.7

Divide these two numbers by the smallest one. (7.7/7.7) for both. 

The ratio for these two elements in the compound is 1:1. So the empirical formula is CH for benzene. The relative mass for CH is 13. Divide the relative molecular mass of benzene by the empirical mass. (78/13) à 6. 

6 x CH à C6H6 molecular formula for benzene. 

Percentage Composition


The formula for percentage composition by mass in an element is, 

=  relative atomic mass x number of atoms of the element in the compound x 100
                    Relative molecular or formula mass of the compound

For example, CH4 methane, relative molecular mass is 16 (12 + 4*1 )

Percentage of carbon = 12/16 * 100 = 75%

Percentage of Hydrogen = 4/16 * 100 = 25%

The Mole

One mole of a substance has a mass equal to its relative atomic mass, relative molecular mass or relative formula mass as appropriate, expressed in grams. For example, 

12 g of magnesium would have 0.5 mol as the relative atomic mass of magnesium is 24. (12g/24 à 0.5 mol)

2 g of H2O would have 0.11 mol as the relative atomic mass is 18. (2g/18 à 0.11 mol)

Reacting Mass And Volume


CH4(g) + 2O2(g) à CO2(g) + 2H2O(l)

1 molecule of methane and two molecules of oxygen react to form one molecule of carbon dioxide and 2 molecules of water. These molecules can also be referred as moles and hence in terms of grams as per the relative atomic mass. It would mean that 16g of methane react with 64g of oxygen to form 44g of carbon dioxide and 36g of water. This proportion would remain the same and would produce predictable results. 

Finding The Emperical Formula Of A Compound

Magnesium oxide exists in different formulas such as MgO, Mg2O and MgO2

0.30g of magnesium burns in air to form 0.50g of magnesium oxide. 

Ratio of the two is 1:1.67

One mole of magnesium is 24 and so by multiplying by the ratio (0.50/0.30 * 24) = 40g

24 g of magnesium in 40g Magnesium oxide, so remaining 16g for oxygen, therefore, one molecule of oxygen is there by relative atomic mass of oxygen (16) 

16 + 24 = 40. 

Predicting The Mass Of Products Of A Reaction

An equation for zinc nitrate is given that decomposes into zinc oxide, oxygen and nitrate on heating. 

2Zn(NO3)2 (s) à 2ZnO(s) + O2  (g) + 4NO2 (g)

5g of zinc nitrate is used in the reaction. 

As per the equation 2 moles of zinc nitrate will decompose to form 2 moles of zinc oxide. 

The molar mass of these two as per the molecular mass in grams is 379g for zinc nitrate and 163g for zinc oxide. 

163/379 * 5g = 2.15g of zinc oxide will form. 

Reactions Involving Gases


•   One Mole Of Any Gas At Room Temperature Which Is 250c Will Have A Volume Of 24 dm3 (Avagadro’s Law)

Zinc reacts with excess of dilute sulphuric acid to form zinc sulphate and 100cm3 hydrogen gas.

Zn + H2SO4 à ZnSO4 + H2

1 mole of zinc produces 1 mole of hydrogen gas. 

Zinc mass as per Mr is 65.5g à 1 mol

Hydrogen volume 24000cm3 à 1 mol

For hydrogen, 24000/24000 * 100 = 100cm3

For zinc, 65.5/24000 * 100 = 0.27g 

0.27g of zinc will produce 100cm3 of hydrogen gas. 

Reacting Gases Only 

For these reactions we will take into consideration the equation of oxidization of carbon monoxide to form carbon dioxide, 250 cm3 of carbon monoxide is oxidized at r.t.p. 

2CO(g) + O2(g) à 2CO2(g)

2 moles of carbon monoxide will oxidize with 1 mol of oxygen to form 2 moles of carbon dioxide. 2 moles is 250cm3. By ration of 2:1:2 with 2 representing 250cm3. 

125cm3 of oxygen will be required to form 250cm3 carbon dioxide.

Reactions In Solution

Concentration

A one molar concentration of a substance consists of one mole of the substance dissolved in 1dmof solution. 

Amount of substance (moles) = Volume of solution (dm3) x concentration (mol/dm3) 

Example, 
The concentration of potassium hydroxide solution is 0.10g/cm3. What is the concentration expressed in mol/dm3?

0.10g/cm3= 100g/dm3

The molar mass of potassium hydroxide is 56g. 56g à 1dm3

100g = 100/56 = 1.79 mole.

The concentration is 1.79mol/dm3.

Example,
What mass of potassium nitrate would be required to produce potassium nitrate of volume 50cm3 and concentration 0.5 mol/dm3

Volume of solution 0.050 dm3.

Number of moles of potassium nitrate = 0.050 x 0.5 = 0.025 mol

Mr of potassium nitrate = 101

Mass of potassium nitrate = 101 x 0.025 = 2.53g

Reactions Involving Solids And Solutions


Example,
Solid iron fillings and dilute hydrochloric acid will react to form iron (II)chloride. How much of dilute hydrochloric acid would be required of concentration 0.5mol/dm3 to react completely with 1g of iron fillings. 

Fe(s) + 2HCl(aq) à FeCl2(aq) + H2 (g) 

1g of iron fillings would be 0.018 moles. As molar mass is 56g and 56g is 1 mole. 

The equation shows the ratio in which HCl and Fe react which is 2:1 and so 0.036moles of HCl would be required to react. The concentration of HCL is 0.5mol/dm3. Which means in 1dm3 there will be 0.5 mol. So for 0.036 à (1000/0.5 * 0.036) = 72cm3

Reactions Involving Only Solutions

Potassium iodide solution reacts with aqueous lead(II) nitrate solution forming an insoluble precipitate of lead(II) nitrate and a solution of potassium nitrate. 

Pb(NO3)2 (aq) + 2KI(aq) à PbI2(s) + 2KNO3(aq)

What volume of potassium iodide solution of concentration 0.5mol/dm3 is needed to react exactly with 10cm3 of lead(II) nitrate solution of concentration 0.1mol/dm3, what mass of lead(II) iodide is formed?

1dmor 1000cm3 of lead(II) nitrate à 1 mol, 10cm3 à 0.01 mol.

1 mol of lead(II) nitrate reacts with 2 mol of potassium iodide and so 0.02 mol of potassium iodide is required for this reaction. 

1000cm3 of potassium iodide contains 0.5 mol and so a solution of 0.02 mol will be of 40 cm3.

1 mole of lead(II) nitrate produces 1 mole lead(II) iodide as per the ratio and 0.01 mol of lead(II) nitrate must be produced. 

The Mr of lead(II) iodide is 461g and so the precipitate would be 4.61g (461 x 0.01)

Percentage Yield And Percentage Purity


Percentage yield is the ratio of actual and theoretical yields in percentage. The formula for it is as follows:

Actual yield x 100
Theoretical yield

For example, 
The ionic equation for lead(II) nitrate with excess of potassium iodide to form lead(II) iodide.

3.31g of lead(II) nitrate was dissolved to obtained 3.89g of lead(II) iodide.

1 mol of lead(II) nitrate produces 1 mol of lead(II) iodide.

Pb2+(aq) + 2I-(aq) à PbI2(s)

1mol/dm3 solution of lead nitrate has a concentration of 2.0 mol/dm3 of nitrate ions. 

Lead(II) nitrate Mr in mass = 331g à produces 461g of Lead(II) iodide

3.31x331/331 = 3.31g of lead(II) nitrate produces 3.31x461/331 = 4.61g of lead(II) nitrate.

Theoretical yield is 4.61g
Actual yield is 3.89g 
Percentage yield = 3.89 x 100/ 4.61 = 84.4%

Purity

A sample of 5g of impure sodium chloride has 4.85g of pure sodium chloride. The purity of this 5g would be 

4.85/5 x 100 = 97%

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