2 Amounts of Substance

Learning Objectives:2.1 Ar & Mr, Avogadro’s number, and the

mole2.2 Ideal Gas Law2.3 Empirical and Molecular Formula2.4 Moles in solutions2.5 Titrations and calculations2.6 Atom economy and yields

2.1 Relative Atomic Mass

• Relative atomic mass = mass of an atom relative to relative to 12C

• It is used to compare masses as the actual masses of atoms is too small

• Ar can be found from the periodic table

2.1 Relative Molecular Mass

• Relative molecular mass = relative mass of one molecule

• Mr = sum of Ar of all atoms in the molecule

• Mr is used to compare masses of molecules as the actual masses are too small

2.1 Avogadro’s Number

Avogadro’s constant = 6.02 x 1023 atoms/molecules in 1 mole

• This is true for ALL atoms/elements/molecules.• This allows us to compare them.

Mass of 1 mole = Mr in grams

2.1 Calculations with Mr

Mass of 1 mole = Mr in grams

• This can be used to convert from grams to moles.

• Moles can then be used to convert between different atoms/elements/molecules. (We will look at this in section 2.5).

How many moles are there in 0.53 g of Na2CO3?

Given: What do we need? 0.53 g

Mr = (2 x Ar Na) + (1 x Ar C) + (3 x Ar O)Mr = (2 x 23.0) + (12.0) + (3 x 16.0) = 106.0 g

So… 1 mole of Na2CO3 = 106.0 g

So… 0.53 g x = 0.0050 moles Na2CO31 mole106.0 g

We use this as a conversion factor and the units we don’t want cancel out.

Mr

2.2 Ideal Gas Law

• Ideal gas law relates pressure, volume, moles, and temperature.

2.2 Ideal Gas Law: Units

• P = Pascals (Pa)• V = meters cubes (m3)• n = moles• R = constant (8.31 J K-1 mol-1)• T = Kelvin (K)

How many moles of hydrogen are present in a100 cm3 at 20.0 oC and a pressure of 100 kPa?

PV = nRT

P = 100 kPa x = 100,000 Pa1 kPa1000 Pa

V = 100 cm3 x1 m3

106 cm3 = 0.0001 m3

n = ? R = 8.31 J K-1 mol-1 T = 20.0 oC = 293 K

100,000 x 0.0001 = n x 8.31 x 20n = (100,000 x 0.0001) / (8.31 x 20) = 0.00411 moles

2.3 Empirical Formula

Empirical formula = lowest whole number ratio of atoms in a molecule

• Molecular formula = C6H12

• Empirical formula = CH2

Calculating Empirical Formula

• If you have the masses of all elements in a compound you can calculate the empirical formula by converting to moles.

• Moles can be compared but grams cannot as all elements have different masses.

Calculating Empirical Formula

1. Find masses from experiment.2. Work out number of moles.3. Convert number of moles to whole number

ratio.

Calculating Empirical Formula10.1 g of a white solid contains 4.01 g calcium,

1.20 g of carbon, and 4.80 g of oxygen.

Ca C OMass = 4.01 g 1.20 g 4.80 gAr = 40.1 12.0 16.0Moles = 0.10 0.10 0.30/ smallest = 0.10 0.10 0.10Ratio = 1 1 3

Empirical Formula = CaCO3

Calculating Molecular Formula

1. Find the ratio between the Mr’s molecular formula and the empirical formula.

2. Multiply by the ratio.

What is the molecular formula of the compound with empirical formula C2H6O and an Mr of 92.

Empirical FormulaMr = (2 x 12.0) + (6 x 1) + (1 x 16) = 46.0

Molecular FormulaMr = 92

Ratio = 92/46 = 2 Multiply by 2 C4H12O2

2.4 Moles in Solutions

• Mr can be used to convert between grams, moles, and concentrations of different substances (see section 2.5).

• Concentration is measured in number of moles on a certain volume.

Concentration = = mol dm-3 molesdm3

What is the concentration of a solution made of1.17 g NaCl dissolved in 500 cm3 of water?

Given: mass NaCl = 1.17 g V = 500 cm3

Concentration = mol / dm3

1.17 g

Mr = 23.0 + 35.5 = 58.5

x 1 mol58.5 g = 0.0200 mol NaCl

0.0200 mol500 cm3 x 1000 cm3

1 dm3 = 0.0400 mol dm-3

2.5 Calculations

• Mr and balanced equations can be used to convert between grams, moles, and concentrations of different substances.

What mass of gas is produced by 0.12 g ofmagnesium and excess hydrochloric acid?

Mg + 2HCl H2 + MgCl2

Given: mass Mg = 0.12 g Mr = 24.3

Needed: mass of H2 Mr = 2 x 1 = 2

0.12 g 1 mol24.3 g

x = 0.0050 mol Mg

0.0050 mol Mg 1 mol H2

1 mol Mgx = 0.0050 mol H2

0.0050 mol H2 2 g

1 mol = 0.010 g H2x

Titrations

• Titrations are used to determine the amount of one of the substances.

We can use the acid base equation to determine the molar ratio and calculate the amount of substance.

HCl + NaOH --> NaCl + H2O

25.0 cm3 of a solution of sodium hydroxide,NaOH, of unknown concentration wasneutralised by 20.0 cm3 of a 0.100 mol dm-3

solution of HCl. What is the concentration of theNaOH?

HCl + NaOH --> NaCl + H2O

V = 25 cm3 NaOH ? = conc. NaOHV = 20.0 cm3 HCl reacted conc = 0.100 mol dm-3 HCl

HCl + NaOH --> NaCl + H2O

V = 25 cm3 NaOH ? = conc. NaOHV = 20.0 cm3 HCl reacted conc = 0.100 mol dm-3 HCl