Chapter 3 Atoms, molecules and stoichiometry

Question 1

Atomic Mass Unit

Answer 1

• The mass of a single atom is so small that it is impossible to weigh it directly
• Atomic masses are therefore defined in terms of a standard atom which is called the unified atomic mass unit
• This unified atomic mass is defined as one-twelfth of the mass of a carbon-12 isotope
• The symbol for the unified atomic mass is u (often Da, Dalton, is used as well)
• 1 u = 1.66 x 10-27 kg

Question 2

relative atomic mass of an element x

Answer 2

=average mass of one atom of x/ (1/12) of the mass of one carbon-12 atom

Question 3

Relative atomic mass, Ar

Answer 3

• The relative atomic mass (Ar) of an element is the ratio of the average mass of the atoms of an element to the unified atomic mass unit
• The relative atomic mass is determined by using the average mass of the isotopes of a particular element
• The Ar has no units as it is a ratio and the units cancel each other out

Question 4

Relative atomic mass, =

Answer 4

weighted average mass of atoms in a given sample of an element / unified atomic mass unit

Question 5

Relative isotopic mass

Answer 5

The relative isotopic mass is the mass of a particular atom of an isotope compared to the value of the unified atomic mass unit

Question 6

To calculate the average atomic mass of an element the percentage abundance is taken into account

Answer 6

• Multiply the atomic mass by the percentage abundance for each isotope and add them all together
• Divide by 100 to get average relative atomic mass
• This is known as the weighted average of the masses of the isotopes

Question 7

, Mr

Answer 7

• The relative molecular mass (Mr) is the ratio of weighted average mass of a molecule of a molecular compound to the unified atomic mass unit
• The Mr has no units

Question 8

Mr can be found by adding up the

Answer 8

• relative atomic masses of all atoms present in one molecule
• When calculating the Mr the simplest formula for the compound is used, also known as the formula unit
• —Eg. silicon dioxide has a giant covalent structure, however the simplest formula (the formula unit) is SiO2

Question 9

Relative molecular mass=

Answer 9

weigted average mass of molecules in a given sample of a molecular compound / unified atomic masss unit

Question 10

Relative formula mass, Mr

Answer 10

-The relative formula mass (Mr) is used for compounds containing ions
It has the same units and is calculated in the same way as the relative molecular mass

Question 11

The Avogadro constant

Answer 11

(Na or L) is the number of particles equivalent to the relative atomic mass or molecular mass of a substance
The Avogadro constant applies to atoms, molecules, ions and electrons
The value of Na is 6.02 x 1023

Question 12

mole (mol)

Answer 12

The mass of a substance with this number of particles

Question 13

The mass of a substance containing the same number of fundamental units as there are atoms in exactly

Answer 13

12.00 g of 12C

Question 14

One mole of any element is equal to the relative atomic mass of that element in grams

Answer 14

• One mole of carbon, that is if you had 6.02 x 1023 atoms of carbon in your hand, would have a mass of 12 g
• One mole of water would have a mass of (2 x 1 + 16) = 18 g

Question 15

Ionic compounds

Answer 15

are formed from a metal and a nonmetal bonded together

-Ionic compounds are electrically neutral; the positive charges equal the negative charges

Question 16

Charges on positive ions

Answer 16

• All metals form positive ions
• –There are some non-metal positive ions such as ammonium, NH4+, and hydrogen, H+
• The metals in Group 1, Group 2 and Group 13 have a charge of 1+ and 2+ and 3+ respectively
• The charge on the ions of the transition elements can vary which is why Roman numerals are often used to indicate their charge
• Roman numerals are used in some compounds formed from transition elements to show the charge (or oxidation state) of metal ions

Question 17

Non-metal ions

Answer 17

• The non-metals in group 15 to 17 have a negative charge and have the suffix ‘ide’
• –Eg. nitride, chloride, bromide, iodide
• Elements in group 17 gain 1 electron so have a 1- charge, eg. Br–
• Elements in group 16 gain 2 electrons so have a 2- charge, eg. O2-
• Elements in group 15 gain 3 electrons so have a 3- charge, eg. N3-

-There are also more complex negative ions, which are negative ions made up of more than one type of atom

Question 18

Balancing Equations: Ionic equations

Answer 18

• In aqueous solutions ionic compounds dissociate into their ions
• Many chemical reactions in aqueous solutions involve ionic compounds, however only some of the ions in solution take part in the reactions
• The ions that do not take part in the reaction are called spectator ions
• An ionic equation shows only the ions or other particles taking part in a reaction, without showing the spectator ions

Question 19

Empirical & Molecular Formulae

Answer 19

• The molecular formula is the formula that shows the number and type of each atom in a molecule\
• The empirical formula is the simplest whole number ratio of the elements present in one molecule or formula unit of the compound

Question 20

Empirical & Molecular Formulae: organic molecules, simple inorganic molecules and ionic compounds

Answer 20

• Organic molecules often have different empirical and molecular formulae
• Simple inorganic molecules however have often similar empirical and molecular formulae
• Ionic compounds always have similar empirical and molecular formulae

Question 21

Calculating molecular formula

Answer 21

1. calculate relative formula mass of empirical formula
2. divide relative fromula by mass of X by relative formula mass of empirical formula
3. multiply each number of elements by empirical number

Question 22

Empirical formula from mass

Answer 22

1. note the mass of each element
2. divide the masses by atomic masses
3. divide by the lowest figure to obtain the ration

Question 23

Empirical formula from %

Answer 23

1. note the % by mass of each atom
2. divide the % by atomic masses
3. Divide by the lowest figure to obtain the ration

Question 24

Water of crystallisation

Answer 24

is when some compounds can form crystals which have water as part of their structure

Question 25

hydrated compound

Answer 25

A compound that contains water of crystallisation

Question 26

The water of crystallisation is separated from the main formula by a

Answer 26

dot when writing the chemical formula of hydrated compounds

Eg. hydrated copper(II) sulfate is CuSO4∙5H2O

Question 27

anhydrous compound

Answer 27

A compound which doesn’t contain water of crystallisation

Question 28

A compound can be hydrated to different degrees

Answer 28

Eg. cobalt(II) chloride can be hydrated by six or two water molecules
CoCl2 ∙6H2O or CoCl2 ∙2H2O

Question 29

The conversion of anhydrous compounds to hydrated compounds is

Answer 29

reversible by heating the hydrated salt
Anhydrous to hydrated salt:
CuSO4 + 5H2O → CuSO4∙5H2O

Hydrated to anhydrous salt (by heating):
CuSO4∙5H2O → CuSO4 + 5H2O

Question 30

mole calculation

Answer 30

no. of moles = mass of substance in grams / molar mass (g mol^-1)

Question 31

calculating Reacting masses

Answer 31

• the chemical equation is required
• This equation shows the ratio of moles of all the reactants and products, also called the stoichiometry, of the equation
• To find the mass of products formed in a reaction the following pieces of information is needed:
• —The mass of the reactants
• —The molar mass of the reactants
• —The balanced equation

Question 32

why are reacting mass calculations needed

Answer 32

The masses of reactants are useful to determine how much of the reactants exactly react with each other to prevent waste

Question 33

Percentage yield

Answer 33

• In a lot of reactions, not all reactants react to form products which can be due to several factors:
• —Other reactions take place simultaneously
• —The reaction does not go to completion
• —Reactants or products are lost to the atmosphere

Question 34

percentage yield calculation

Answer 34

=actual yield/ predicted yield (theoretical yield)

x100

Question 35

Excess & limiting reagents

Answer 35

excess=of one or more of the reactants (excess reagent)

limiting reagent= is not in excess

Question 36

determining which reactant is limiting

Answer 36

• The number of moles of the reactants should be calculated

- The ratio of the reactants shown in the equation should be taken into account

Question 37

Volumes of gases calculation

Answer 37

volume of gas (dm3) = amount of gas (mol) x 24

-At room temperature (20 degrees Celsius) and pressure (1 atm) one mole of any gas has a volume of 24.0 dm3

Question 38

concentrations of solutions

Answer 38

-The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 of solution

• A concentrated solution is a solution that has a high concentration of solute
• A dilute solution is a solution with a low concentration of solute

Question 39

concentration of solutions calculation

Answer 39

number of moles (mol) = concentration (mol dm-3) x volume (dm3)

• When carrying out calculations involve concentrations in mol dm-3 the following points need to be considered:
• –Change mass in grams to moles
• –Change cm3 to dm3