Recap – Chemical Equations

Formula equation – gives overview, balance for numbers and types of atoms

eg Mg + 2HCl MgCl2 + H2

Molecular equation – include states of reactants and productseg C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

Net ionic equation – focus on just what is involved and balance charges

eg Ca2+(aq) + CO32-(aq) CaCO3(s)

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Masses of Atoms

1H atom 1p 1 nucleon

4He atom 2p + 2n 4 nucleons 4 times the mass of H

7Li atom 3p + 4n 7 nucleons 7 times the mass of H

238U atom 92p + 146n

238 nucleons

238 times the mass of H

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Isotopes and Average Mass

1H atom

2H atom

1p

1p + 1n

99.99%

0.01%1.008

6Li atom

7Li atom

3p + 3n

3p + 4n

7.5%

92.5%(6 x 7.5%) + (7 x 92.5%) = 6.9

12C atom

13C atom

6p + 6n

6p + 7n

98.9%

1.1%(12 x 98.9%) + (13 x 1.1%) = 12.01

35Cl atom

37Cl atom

17p + 18n

17p + 20n

75.8%

24.2% (35 x 75.8%) + (37 x 24.2%)= 35.5

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Avogadro Number, NA

• Actual mass of H is 1.67 x 10-24 g • Use a scaling factor: the Avogadro

number, NA

NA = 6.022 1023

Definition: The Avogadro number is the number of 12C atoms present in exactly 12 g of 12C.

• One ‘mole’ of anything contains the Avogadro number, NA, of items

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Avogadro Number, NA

 Atomic mass

mass of NA atoms

‘Molar mass’

Hydrogen 1.008 amu 1.008 g

Lithium 6.94 amu 6.94 g

12C 12.00 amu 12.00 g

Carbon 12.01 amu 12.01 g

Chlorine 35.45 amu 35.45 g

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Molar Masses

Relative masses of molecules is simply the sum of the atomic masses of the component atoms.

eg H2 2 x 1.01 = 2.02 g mol-1

H2O (2 x 1.01) + 16.0 = 18.02 g mol-1

C6H12O6 (6 x 12.01) + (12 x 1.01) + (6 x 16.0) = 180.18 g mol-1

 For ionic solids, no molecules, use the empirical formula.

eg NaCl 23.0 + 35.5= 58.5 g mol-1

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The Mole

So now we can relate the number of grams of a substance we weigh out in the lab to the number of moles, and thus the number of particles of the substance that we have:

no. of moles (n) = mass (m) / molar mass (M)

no. of moles (n) = no. of items / 6.022 x 1023

or

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The Mole

Q: How many atoms present in 1.0 g of silver?

Moles = mass / molar mass = 1.0 / 107.9= 0.0093 mol

No. atoms= moles NA = 0.0093 6.022 1023 = 5.6 1021 atoms

 

n = no. of items / NA

n = mass / molar mass

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The Mole

Q: Calculate the no of moles of water in 1000 g water. Molar mass = 18.02.

Moles = mass / molar mass= 1000 / 18.02 = 55.49 moles

 

Q: How many water molecules would be present?

 

No. molecules = moles NA

= 55.49 6.022 1023 = 3.342 1025

molecules 

n = no. of items / NA

n = mass / molar mass

• By the end of this lecture, you should:− recognise the mass reported is a weighted average

of the mass of the individual isotopes of an element− be able to calculate an average mass from isotope

data− understand that NA is a scaling factor

− understand the difference between atomic mass and molar mass

− be able to calculate the molar mass of a compound− be able to convert between mass, moles and

numbers of atoms/ions/molecules

− be able to complete the worksheet (if you haven’t already done so…)

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Learning Outcomes:

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Questions to complete for next lecture:

1. Give the definition of a mole of substance.

2. How many atoms are there in 0.25 mole of silver?

3. Silver contains two isotopes: 107Ag (51.8%) and 109Ag(48.2%). What is the average atomic mass of silver?

4. Table sugar (sucrose) has the formula C12H22O11.

a) What is the molar mass of sucrose?

b) How many moles of sucrose are present in a sugar lump with mass 5.0 g?

c) How many molecules of sucrose are present in the sugar lump?