2 - Amount of substance Flashcards by amelia b

Define relative atomic mass (Ar)

The average mass of an atom of an element relative to 1/12 of the mass of a carbon-12 atom

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Define relative molecular mass (Mr)

The average mass of a molecule relative to 1/12 of the mass of a carbon-12 atom

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What is a ‘mole’?

The amount of a substance that contains the same number of particles as the number of atoms in 12 grams of carbon-12

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1 mole of any substance contains 6.02x1023…

atoms / molecules / electrons / particles

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What equation do we use to calculate the number of particles?

number of particles = number of moles x Avogadro’s number

no of particles = n x L

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Example question

How many particles make up 0.67 moles of ammonia (NH3)?

Number of particles = 6.02x10²³ x 0.67

number of particles = 4.03x10²³

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Example question

How many atoms are in 57.0g of NH₃?

Mr of NH₃ = 14.0 + (3 x 1.0) = 17.0 g mol^-1

n = m / Mr
n = 57.0 / 17.0 = 3.35 mol

no of particles = n x L
3.35 x 6.022 x 10²³ = 2.02 x 10²⁴

There are 4 atoms present in a molecule of NH₃ so therefore we times 2.02 x 10²⁴ by 4 which equals 8.08 x 10²⁴

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Example question

What is the mass of 1 atom of Carbon?

Ar = 12.0

1 mole of C = 12g
6.022 x 10²³ atoms of C = 12.0g

Divide both sides by Avogadro’s number

1 atom of C = 1.9926 x 10^-23 g
3 sig fig = 1.99 x 10^-23 g

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How can we calculate the number of moles from the mass and Mr?

Number of moles = mass (g) / Mr or Ar

n = m / Mr or Ar

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Example question

Calculate the number of moles of 23g of gold. Give your answer to 2 significant figures

Number of moles = mass / Ar

Number of moles = 23 / 197

Number of moles = 0.12mol (2 s.f)

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Example question

Which contains the greatest number of moles?
10.8g of water or 4.8g of helium

H₂O
n = m / Mr
n = 10.8g / 18.0 = 0.6 moles of H₂O
There are 3 atoms present in a H₂O molecule so we therefore times the number of moles by the number of atoms
0.6 x 3 = 1.8 moles

He
n = m / Ar
n = 4.8 / 4 = 1.2 moles

Water contains the greatest number of atoms

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How can we calculate the number of moles in solutions?

Number of moles = concentration (moldm^-3) x volume (dm³)

n = c x v

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Example question - moles in solution

Calculate the number of moles of 200cm³ of 0.35moldm^-3 HCl

Number of moles = concentration x volume

200cm³ must be in dm³: 200 / 1000 = 0.2 dm³

Number of moles = 0.35 x 0.2 = 0.07moldm^-3

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What is the ideal gas equation?

pV = nRT

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What are the units of the ideal gas equation?

p = pressure (Pa)
V = volume (m³)
n = number of moles (mol)
R = gas constant (given in exam)
T = temperature (K)

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What are the standard conditions of temperature and pressure?

Temperature = 298K
Pressure = 100kPa

Example questions - ideal gas equation

Calculate the volume in cm³ of 0.36 moles of a gas at 100kPa and 298K

pV = nRT rearrange to give V = nRT / p

100kPa to 100,000 Pa

V = 0.36 x 8.31 x 298 / 100000

V = 8.91x10^-3 m³

V = 8.91x10^-3 x 1,000,000 = 8910cm³

Define ‘empirical formula’

The simplest whole number ratio of each element in a compound

How do we work out the empirical formula?
(Example) A compound contains 23.3% Magnesium, 30.7% Sulfur and 46.0% Oxygen. What is the empirical formula of this compound?

1) Write out the elements involved
2) Write the percentages as masses
3) Divide these by their relative atomic masses to get the number of moles
4) Divide all the numbers by the smallest number of moles

Mg = 23.3g / 24.3 (Ar) = 1

S = 30.7g / 32.1 (Ar) = 1

O = 46.0g / 16.0 (Ar) = 3

Empirical formula = MgSO₃

Define ‘molecular formula’

The molecular formula is the actual number of atoms of each element in the compound

How do we work out the molecular formula from the empirical formula?
(Example) The empirical formula of a molecule is CH₂O. It has a relative molecular mass (Mr) of 180. What is its molecular formula?

1) Firstly work out the empirical mass of the molecule
2) Divide the molecular mass by the empirical mass
3) Multiply the empirical formula by the number of empirical units

Empirical mass = 12 + (1 x 2) + 16 = 30
Mr = 180

Mr / empirical mass = 180 / 30 = 6 (empirical units)

CH₂O x 6 = C₆H₁₂O₆

What is the purpose of a ionic equation?

Ionic equations show the ions that are formed in solution and show which particles are reacting

What form ions in solutions?

Acids, bases and salts

How can we use equations to work out masses?
(Example) How much CaO can be made when 34g of Ca is burnt completely in oxygen?

1) Write out the equation and balance it
2) Work out the Mr / Ar of species involved then write these as mass in grams
3) Divide the Ca side by 80 to find 1g then multiply by 34 to get 34g and do the same for the CaO side

2Ca _(g) + O₂ _(g) → 2CaO _(s)

Ar of Ca = 40
40 x 2 = 80
80g / 80 = 1g
1g x 34 = 34g

Mr of CaO = 56
56 x 2 = 112
112g / 80 = 1.4g
1.4g x 34 = 47.6g of CaO (Theoretical mass)

How can we use equations to work out masses of gases? (Example) What volume of H₂ is produced when 12g of potassium reacts with water at 100kPa of pressure and 298K? Gas constant is 8.31JK^-1mol^-1.

1) Write out the equation and balance it 2) Work out the number of moles of potassium 3) Use the equation to find out the molar ratio of K:H₂ 4) Use ideal gas equation to work out the volume 2K _(s) + 2H₂O _(l) → 2KOH _(aq) + H₂ _(g) Moles = mass / Ar moles = 12g / 39 = 0.31 moles of K 2 moles of K react to produce 1 mole of H₂ moles of H2 is 0.31 / 2 = 0.155 moles of H₂ pV = nRT but rearrange to get V = nRT / p V = 0.155 x 8.31 x 298 / 100,000 V = 3.84 x 10^-3 m³

What is the theoretical yield?

The amount of a product produced assuming no products are lost and all reactants react fully

How do we calculate percentage yield?

Percentage yield = (actual yield / theoretical yield) x 100

# Example question In a reaction involving the complete combustion of Calcium 32.6g of Calcium Oxide was produced. The theoretical mass is 47.6g. Calculate the percentage yield of this reaction.

Percentage yield = (actual / theoretical) x 100 Percentage yield = (32.6 / 47.6) x 100 = 68.5%

Why is a high percentage yield important?

Efficient conversion of reactants to products

What is atom economy?

How efficient a reaction is

How do we calculate the percentage atom economy?

% Atom economy = (Mr of desired product / sum of Mr of all reactants) x 100

# Example question Iron oxide (Fe₂O₃) can be reduced using carbon dioxide to make pure iron and carbon dioxide. Calculate the atom economy in the extraction of iron.

Fe₂O₃ + 1.5C → 2Fe + 1.5CO₂ Atomic mass of desired product = 2 x 55.8 = 111.6 Sum of molecular masses of all reactants = (2 x 55.8) + (3 x 16.0) + (1.5 x 12.0) = 177.6 % atom economy = 111.6 / 177.6 x 100 = 62.8%

Why is a high atom economy important?

- High atom economies produce less waste and will therefore benefit the environment - High atom economies mean that raw materials are used more efficiently which is more sustainable - Higher atom economy means less by-products so less time and money will be spent separating these from the desirable product - Companies will try to use reactions that tend towards 100% atom economy