## Wednesday, November 5, 2008

### chemical calculation - igcse practice

Introducing the connection between moles, mass and formula mass

The mole is most simply expressed as the 'formula mass in g' of the defined chemical 'species', and that is how it is used in most chemical calculations.

Every mole of any substance contains the same number of the defined species.

The actual particle number is known and is called the Avogadro Constant and is equal to 6.023 x 1023 'defined species' per mole. This means there are that many atoms in 12g of carbon (C = 12) or that many molecules of water in 18g* (H2O = 1+1+16 = 18, H = 1; O = 16) * this is about 18cm3, so picture this number of molecules in a nearly full 20cm3 measuring cylinder or a 100ml beaker less than 1/5th full!

However, the real importance of the mole is that it allows you to compare ratios of the relative amounts of reactants and products, or the element composition of a compound, at the atomic and molecular level. If you have a mole ratio for A:B of 1:3

it means 1 particle of A to 3 particles of B irrespective of the atomic or formula masses of A and B. (see also section 6. for reacting masses not using moles)

Important Note. Relative is just a number based on the carbon-12 relative atomic mass scale. Molar mass is a term used to describe the mass of one mole i.e. the relative atomic/formula/molecular mass in grams (g).

Examples:

Example 7.1.1: 1 mole of ammonia, NH3,

consists of 1 mole of nitrogen atoms combined with 3 moles of hydrogen atoms.

Or you could say 2 moles of ammonia is formed from 1 mole of nitrogen molecules (N2) and 3 moles of hydrogen molecules (H2).

Example 7.1.2: 1 mole of aluminium oxide,

Al2O3, consists of 2 moles of aluminium atoms combined with 3 moles of oxygen atoms

(or 1.5 moles of O2 molecules).

For calculation purposes learn the following formula for 'Z' and use a triangle if necessary.

(1) mole of Z = g of Z / atomic or formula mass of Z,

(2) or g of Z = mole of Z x atomic or formula mass of Z

(3) or atomic or formula mass of Z = g of Z / mole of Z

where Z represents atoms, molecules or formula of the particular element or compound defined in the question.

Example 7.2.1: How many moles of potassium ions and bromide ions in 0.25 moles of potassium bromide?

1 mole of KBr contains 1 mole of potassium ions (K+) and 1 mole of bromide ions (Br-).

So there will be 0.25 moles of each ion.

Example 7.2.2: How many moles of calcium ions and chloride ions in 2.5 moles of calcium chloride?

1 mole of CaCl2 consists of 1 mole of calcium ions (Ca2+) and 2 moles of chloride ion (Cl-).

So there will be 2.5 x 1 = 2.5 moles of calcium ions and 2.5 x 2 = 5 moles chloride ions.

Example 7.2.3: How many moles of lead and oxygen atoms are needed to make 5 moles of lead dioxide?

1 mole of PbO2 contains 1 mole of lead combined with 2 moles of oxygen atoms (or 1 mole of oxygen molecules O2).

So 1 x 5 = 5 mol of lead atoms and 2 x 5 = 10 mol of oxygen atoms (or 5 mol oxygen molecules) are needed.

Example 7.2.4: How many moles of aluminium ions and sulphate ions in 2 moles of aluminium sulphate?

1 mole of Al2(SO4)3 contains 2 moles of aluminium ions (Al3+) and 3 moles of sulphate ion (SO42-).

So there will be 2 x 2 = 4 mol aluminium ions and 2 x 3 = 6 mol of sulphate ion.

Example 7.2.5: How many moles of chlorine gas in 6.5g? Ar(Cl) = 35.5)

chlorine consists of Cl2 molecules, so Mr = 2 x 35.5 = 71

moles chlorine = mass / Mr = 6.5 / 71 = 0.0944 mol

Example 7.2.6: How many moles of iron in 20g? (Fe = 56)

iron consists of Fe atoms, so moles iron = mass/Ar = 20/56 = 0.357 mol Fe

Example 7.2.7: How many grams of propane C3H8 are there in 0.21 moles of it? (C = 12, H = 1)

Mr of propane = (3 x 12) + (1 x 8) = 44, so g propane = moles x Mr = 0.21 x 44 = 9.24g

Example 7.2.8: 0.25 moles of molecule X was found to have a mass of 28g. Calculate its molecular mass.

Mr = mass X / moles of X = 28 / 0.25 = 112

Example 7.2.9: What mass and moles of magnesium chloride is formed when 5g of magnesium oxide is dissolved in excess hydrochloric acid?

reaction equation: MgO + 2HCl ==> MgCl2 + H2O

means 1 mole magnesium oxide forms 1 mole of magnesium chloride (1 : 1 molar ratio)

formula mass MgCl2 = 24+(2x35.5) = 95,

MgO = 24+16 = 40, 1 mole MgO = 40g, so 5g MgO = 5/40 = 0.125 mol

which means 0.125 mol MgO forms 0.125 mol MgCl2,

Mass = moles x formula mass = 0.125 x 95 = 11.9g MgCl2

Example 7.2.10: What mass and moles of sodium chloride is formed when 21.2g of sodium carbonate is reacted with excess dilute hydrochloric acid?

reaction equation: Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2

means 1 mole sodium carbonate gives 2 moles of sodium chloride (1:2 ratio in equation)

Formula mass of Na2CO3 = (2x23) + 12 + (3x16) = 106

Formula mass of NaCl = 23 + 35.5 = 58.5

moles Na2CO3 = 21.2/106 = 0.2 mole

therefore 2 x 0.2 = 0.4 mol of NaCl formed.

mass of NaCl formed = moles x formula mass = 0.4 x 58.5 = 23.4g NaCl

Using the Avogadro Constant, you can actually calculate the number of particles in known quantity of material.

Example 7.3.1: How many water molecules are there in 1g of water, H2O ?

formula mass of water = (2 x 1) + 16 = 18

every mole of a substance contains 6 x 1023 particles of 'it' (the Avogadro Constant).

moles water = 1 / 18 = 0.0556

molecules of water = 0.0556 x 6 x 1023 = 3.34 x 1022

Since water has a density of 1g/cm3, it means in every cm3 or ml there are

33 400 000 000 000 000 000 000 individual H2O molecules or particles.

Example 7.3.2: How many atoms of iron (Fe = 56) are there in an iron filing of mass 0.001g ?

0.001g of iron = 0.001 / 56 = 0.00001786 mol

atoms of iron in the nail = 0.00001786 x 6 x 1023 = 1.07 x 1019 actual Fe atoms

(10.7 million million million atoms!)

Example 7.3.2: (a) How many particles of 'Al2O3' in 51g of aluminium oxide?

Atomic masses: Al =27, O = 16, f. mass Al2O3 = (2x27) + (3x16) = 102

moles 'Al2O3' = 51/102 = 0.5 mol

Number of 'Al2O3' particles = 0.5 x 6 x 1023 = 3 x 1023

(b) Aluminium oxide is an ionic compound. Calculate the number of individual aluminium ions (Al3+) and oxide ions (O2-) in the same 51g of the substance.

For every Al2O3 there are two Al3+ and three O2- ions.

So in 51g of Al2O3 there are ...

0.5 x 2 x 6 x 1023 = 6 x 1023 Al3+ ions, and

0.5 x 3 x 6 x 1023 = 9 x 1023 O2- ions.

More advanced use of the mole and Avogadro Number concepts (for advanced level students only)

You can have a mole of whatever you want in terms of chemical species e.g.

In terms of electric charge, 1 Faraday = 96500 C (coulombs) = 6 x 1023 electrons

If you have 2.5 moles of the ionic aluminium oxide (Al2O3) you have ...

2 x 2.5 = 5 moles of aluminium ions (Al3+) and 3 x 2.5 = 7.5 mol of oxide ions (O2-)

When you write ANY balanced chemical equation, the balancing numbers, including the un-written 1, are the reacting molar ratio of reactants and products.

Extra Advanced Questions - more suitable for Advanced AS-A2 students which can be completely tackled after ALSO studying section 9 on the molar volume of gases and ANSWERS to QA7.1

QA7.1 This question involves using the mole concept and the Avogadro Constant in a variety of situations.

The Avogadro Constant = 6.02 x 1023 mol-1. The molar volume for gases is 24dm3 at 298K/101.3kPa.

Atomic masses: Al = 27, O = 16, H = 1, Cl = 35.5, Ne = 20, Na = 23, Mg = 24.3, C = 12

Where appropriate assume the temperature is 298K and the pressure 101.3kPa.

Calculate ....

(a) how many oxide ions in 2g of aluminium oxide?

(b) how many molecules in 3g of hydrogen?

(c) how many molecules in 1.2 cm3 of oxygen?

(d) how many molecules of chlorine in 3g?

(e) how many individual particles in 10g of neon?

(f) the volume of hydrogen formed when 0.2g of sodium reacts with water.

(g) the volume of hydrogen formed when 2g of magnesium reacts with excess acid.

(h) the volume of carbon dioxide formed when the following react with excess acid

(1) 0.76g of sodium carbonate and (2) 0.76g sodium hydrogencarbonate

(i) the volume of hydrogen formed when excess zinc is added to 50 cm3 of hydrochloric acid, concentration 0.2 mol dm-3.

(j) the volume of carbon dioxide formed when excess calcium carbonate is added to 75 cm3 of 0.05 mol dm-3 hydrochloric acid.