Chemistry Bridging Course Flashcards
Atomic Number
Number of protons in the nucleus
Mass Number
The total number of protons and neutrons in the nucleus
Cation
Positive ion which has less electrons than protons
Anion
Negative ion which has more electrons than protons
Isotopes
Atoms of the same element with the same number of protons but different number of neutrons
Do isotopes of the same element have the same chemical properties
Yes, they have the same electron configuration (same number of electrons)
What physical properties of isotopes of the same element might be different ?
The mass, density, boiling point, diffusion rate
Relative Atomic Mass , Ar
The weighted mean mass of an atom compared with 1/12th mass of an atom of carbon-12
Relative Isotopic Mass
The mass of an isotope compared with 1/12th mass of an atom of carbon-12
Ar Value
Mass of an element (mass number)
Relative Formula Mass (Mr)
Mass of each atom multiples by how many of each atom there is
Ar values of each atom
e.g O2 (oxygen) = 16 x 2
Mr = 32
Water of crystallisation
eg CuSO4.5H20
The full stop is the water of crystallisation
You multiply the Mr of water (1+1+16=18) by the number in front of H2O in this case 5
so 18x5
Diatomic Substances
O2 Oxygen
H2 Hydrogen
N2 Nitrogen
F2 Fluorine
Br2 Bromine
I2 Iodine
Cl2 Chlorine
Polyatomic Substances
H2O Water - Molecule
NaOH Sodium Hydroxide - Lattice
Molecular Formula
Actual number of atoms in each element in a molecule
e.g Glucose = C6H12O6
6 Carbons
12 Hydrogens
6 Oxygen
Empirical Formula
Simplest whole number ratio of atoms of each element in a compound
e.g Glucose = C6H12O6 = CH2O
Calculating Empirical Formula
Step 1: Mass or Percentage/Ar for each element
Step 2: Divide all the answers by the smallest answer from step 1
Step 3: Find the simplest whole number ratio
1.49 or 1.5 can’t be rounded up to 2.
Multiply all rations till they are whole numbers
Calculating The Molecular Formula From The Empirical Formula (with Mr known)
Step 1: Calculate Empirical Formula
Step 2: Work out empirical mass e.g CH2 = 12+2 = 14
Step 3: Divide Mr by empirical mass which is the multiple e.g 56/14 = 4
Step 4: Multiple empirical formula by the multiple = CH2 x 4 = C4H8
Calculating The ‘Water of Crystallisation’
Example 1
CuSO4.xH2O
x = water of crystallisation
CuSO4.5H2O
5H2O = water of crystallisation
. = 18 x number in front if H2O
Step 1: Calculate the mass of the anhydrous salt (mass of crucible&resiude - mass of crucible)
Step 2: Calculate the mass of water lost
(mass of crucible&hydrated salt - mass of crucible&residue)
Step 3: Ratio of Anhydrous (without water) salt : water
Use empirical formula
Step 4: Find the simplest whole number ratio of anhydrous salt : water
Step 5: The number for water is the number that goes in front of H2O
Calculating the ‘Water of Crystallisation’
Example 2
Step 1: Calculate the mass of water lost:
(Mass of hydrated salt - mass of anhydrous salt) which equals the mass of water
Step 2: Ratio of anhydrous salt : water using empirical formula
Step 3: Find the simplest whole number ratio
Endings
‘ide’ - element negative ion (anion) with a negative charge
‘ate’ - compound negative ion (anion) with oxygen
Polyatomic Ions
Ammonium- NH4(+)
Hydroxide- OH(-)
Nitrate (v) NO3(-)
Carbonate - CO3(2-)
Sulfate (VI) - SO4(2-)
Phosphate - PO4(3-)
Combustion of Fuels
Burning - Reacting with Oxygen
Balance those equations in alphabetical order
1-Carbon 2-Hydrogen 3-Oxygen