ATOMS

ATOMS, MOLECULES AND STOICHIOMETRY

Definitions

1.	The MOLE is defined as the amount of a substance which contains Avogadro's number of particles of the substance.
2.	Avogadro's Constant is defined as the number of atoms in exactly 12 g of carbon-12. It is equal to 6.02 x 10²³ mol^-1. K Solve this #1: How many atoms are there in 1 g of ¹²C? (ans)
3.	MOLAR VOLUME is the volume occupied by 1 mole of any substance in the gaseous state at standard temperature and pressure (s.t.p.) and this is equal to 22.4 dm³. s.t.p. = 0° C, 1 atm or 273 K, 101 kPa
4.	The MOLAR MASS of any substance is the mass of one mole of that substance. It is the RAM, RMM or RFM of that substance expressed in g/mol.
5.	Amount of Substance (mol) x Molar mass (g mol^-1) = Mass (g) K Solve this #2: 8 g of O² molecules contains 0.25 mol. What volume is this? (ans) K Solve this #3: 0.5 mol of Na₂CO₃.10H₂O = 143 g. Mass of Na⁺ and CO₃^2- ions? (ans)

Concentration of solutions

1.	Chemical reactions are often carried out between substances in solution.
2.	The CONCENTRATION of a solution is measured in terms of the number of moles of solute contained in a cubic decimetre of solution (mol dm^-3).
3.	Concentration or Strength of a solution is usually expressed in: a) mol dm^-3 or b) g dm^-3
4.	A MOLAR SOLUTION of a compound is one which contains one mole or the molar mass of the compound in one dm³ of the solution. K Solve this #4: What is the concentration in a) g dm^-3 and b) mol dm^-3 of a solution containing 0.5 g KOH in 200 cm³ solution? (ans) K Solve this #5: What mass of HCl is required to prepare 1.5 dm³ of 0.5 mol dm^-3 HCl? (ans)

Calculating Empirical & Molecular Formulae

1.	The EMPIRICAL FORMULA of a compound is the simplest form of the ratio of the atoms of different elements in it.
2.	The MOLECULAR FORMULA tells the actual number of each kind of atom in a molecule of the substance.
3.	Calculating E.F. from the masses of constituents. To determine the E.F. of a compound, we must first calculate the amount of each substance present in a sample and then calculate the simplest whole-number ratio of the amounts.
	Example : A 18.3 g sample of a hydrated compound contained 4.0 g of calcium, 7.1 g of chlorine and 7.2 g of water only. Calculate its E.F.

Element/substance	Ca	Cl	H₂O
Mass/g	4.0	7.1	7.2
Molar mass /g mol^-1	42	35.5	18
Amount /mol	0.1	0.2	0.4
Relative amount	1	2	4
Simplest ratio	1 : 2 : 4 E.F. is CaCl₂.4H₂O

4.	Calculating E.F. from Percentage Composition by mass. Example: An organic compound was analysed and was found to have the following percentage composition by mass: 48.8% C, 13.5% H and 37.7% N. Calculate the E.F. of the compound.

	C	H	N
% mass	48.8	13.5	37.7
Molar mass	12	1	14
Amount /mol	4.1	13.5	2.7
Relative amount	1.5	5	1
Simplest Ratio	3 : 10 : 2 E.F. is C₃H₁₀N₂

The M.F. can be found from the E.F. of the compound if the molar mass is known. The molar mass is a multiple of the E.F. mass.

Example : A compound has the E.F. CH₂O and molar mass 180 g mol^-1. What is the M.F.?

Let M.F. be (CH₂O)n

molecular mass = n(30) = 180

n = 6 M.F. is (CH₂O)₆ = C₆H₁₂O₆ (glucose)

Reacting Masses of Solids

1.	Equations tell us not only what substances react together but also what amounts of substance react together. Example: 2 NaHCO₃ (s) ® Na₂CO₃ (s) + CO₂ (g) + H₂O (g) 2 moles of NaHCO₃ give 1 mole of Na₂CO₃ i.e. 168 g of NaHCO₃ gives 106 g of Na₂CO₃ The amounts of substances undergoing reaction, as given by the balanced chemical equation, are called the stoichiometric amounts.
2.	STOICHIOMETRY is the relationship between the amounts of reactants and products in a chemical equation.
3.	If one reactant is present in excess of the stoichiometric amount required for reaction with another of the reactants, then the excess of one reactant will be left unused at the end of the reaction. K Solve this #6: How many moles of iodine can be obtained from 0.5 mole of potassium iodate (V)? (ans) Eqⁿ : KIO₃ (aq) + 5 KI (aq) + 6 H⁺® 3 I₂ (aq) + 6 K⁺ (aq) + 3 H₂O (l) K Solve this #7: What mass of barium sulphate can be precipitated from a solution which contains 4.0 g of barium chloride? (ans)
4.	In a chemical reaction, the reactants are often added in amounts which are not stoichiometric. One or more of the reactants is in excess and is not completely used up in the reaction. The amount of product is determined by the amount of the reactant that is not in excess and is used up completely in the reaction. This is called the limiting reactant. Example: 5.00g of sulphur are heated together with 5.00g of iron to form iron (II) sulphide. Which reactant is present in excess? What mass of product is formed? Fe (s) + S (s) ® FeS (s) 1 mole 1 mole 1 mole 5.00g of Fe = 5/56 mol of Fe = 0.0893 mol of Fe 5.00g of S = 5/32 mol of S = 0.156 mol of S i.e. there is insufficient Fe to react with 0.156 mol of S Þ Fe is the limiting factor mass of FeS formed = 0.0893 x 88 g = 7.86 g

Reacting Volumes of Gases

A MOLE of GAS occupies 22.4 dm³ at s.t.p.

i.e. the gas MOLAR VOLUME is 22.4 dm³ at s.t.p.

An equation which shows how many moles of different gases react together also shows the ratio of the volumes of the different gases that react together.

K Solve this #8: What volume of hydrogen is obtained when 3.00g of zinc react with an excess of dilute sulphuric acid at s.t.p. ? (ans)

K Solve this #9: 25 cm³ of a mixture of methane and ethane were completely oxidised by 72.5 cm³ of oxygen, measure at the same temperature and pressure. What was the composition of the mixture? (ans)

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Answers:

#1	12 g ® 6.02 x 10²³ atoms 1 g ® 1/12 x 6.02 x 10²³ = 5.02 x 10²³ atoms of C
#2	vol = 0.25 x 22.4 = 5.6 dm³
#3	mass of Na⁺ ions = 0.5 x 2(23) = 23 g mass of CO₃^2- ions = 0.5 x [12 + 3(16)] = 30 g
#4	conc of KOH = 2.5 g dm^-3 = 0.0446 mol dm^-3
#5	Required mass of HCl = 27.4 g
#6	1 mol of KIO₃ …………………. 3 mol of I₂ 0.5 mol of KIO₃ ………………… 3 x 0.5 = 1.5 mol of I₂
#7	4.5 g
#8	Volume = 1.03 dm³
#9	Volume of methane = 10 cm³ Volume of ethane = 15 cm³