Chapter 2 & 3 Atoms, Ions, Moles Flashcards
What are isotopes
Atoms of the same element with different numbers of neutrons and different masses
what is relative isotopic mass
the mass of an atom of an isotope compared to 1/12 of the mass of a 12C atom.
what is relative atomic mass
the weighted mean mass of 1 mole of an element compared to the mass of 1 mole of 12C
atomic number =
amount of protons
A represents the […] number
Mass
Z represents the […]
Proton number
How come isotopes still react in the same way
The number of neutrons has no effect on the reactions of an element
Cations are
Positive ions (atoms w fewer electrons than protons)
Anions are
Negative ions (atoms with more electrons than protons)
Ions and atoms of an element have the same number of […] but a different number of […]
Protons, electons
mass to charge ratio =
relative mass of ion / relative charge on ion
In a mass spectrometer, heavier ions move […] than lighter ones and are more difficult to […]
Slower, deflect
What happens after a sample is placed in the mass spectrometer
The sample is vaporised and then ionised to form postive ions
-> then the ions are accelerated and move
relative isotopic mass =
( the abundance x the mass number ) / 100
Relative molecular mass is used for
Simple molecules
For compounds with giant structures, we use
Relative formula mass
What is a binary compound
Compound that only contains 2 elements
When constructing a balanced equation what do you do
- first get the ions and charges right and come up with the formulae
e.g Aluminium oxide will be Al2O3 due to Al being 3+ and O being 2-
- then get the atoms equal on boths sides so its
4Al (s) + 3O2 (g) -> 2Al2O3
What is amount of substance
The number of particles in a substance, measured in moles (mol)
What is the avogadro constant
6.02 x 10^23 mol^-1
-> the number of particles in each mole of carbon 12
What is the molar mass
Mass per mole of a substance
Amount of substance (n) =
Mass (m) / molar mass (M)
what is the molar gas volume (the volume occupied by any 1 mole of any has at rt and rtp)
24 dm^3 mol^-1
What is empirical formula
The simplest whole number ratio of atoms of each element present in a compound
What is molecular formula
The number and type of atoms of each element in a molecule
Empirical formula from mass worked example
1.203g of calcium combines with 2.13g of chlorine to form a compound
n(Ca) = 1.203/40.1 = 0.03 mol
n(Cl) = 2.13/35.5 = 0.06 mol
n(Ca):n(Cl) = 0.03/0.03 : 0.06/0.03
= 1:2
therefore it’s CaCl2
Molecular formula worked example
Chemical analysis of a compound gave the percentage composition by mass C: 40.00%; H: 6.67%:
O: 53.33% [Ar: C, 12.0; H, 1.0; 0, 16.0] The relative molecular mass of the compound is 180.0.
Step 1: Convert % by mass into moles of atoms using n=m/M:
C = 40.00/12 = 3.33 mol
H = 6.67 / 1 = 6.67 mol
O = 53.33 / 16 = 3.33 mol
Step 2: Find smallest whole-number ratio and empirical formula.
3.33/3.33 = 1 and 6.67/3.33 = 2
Ratio of C:H:O= 1:2:1 = CH2O
-> Write the relative mass of the empirical formula CH2O: 12.0 + 1.0 x 2 + 16.0 = 30.0
-> Find number of CH2O units in one molecule: 180 / 30= 6
->Write the molecular formula: CH2O x 6 = C6H12O6
Atom economy =
Sum of desired products / sum of all products
Benefits of high atom economy
- more efficient
- reduces waste
- preserves raw materials
Percentage yield =
(Actual yield / theoretical yield) x 100
why is it hard to achieve theoretical yield
- reaction may not go to completion
- side reactions may have taken place
- purification of produce may result in some loss of it
What is the limiting reagent
The reactant that is used up first
2H2 (g) + O2 (g) -> 2H2O (l)
Why is hydrogen the limiting reagent
2 mol of H are needed for every mol of O
-> if an equal amount of both were used, the H would get all used up ans only half of the O
1cm ^3 = ? Ml
1 ml
1 dm ^ 3 = ? cm ^3 and ? ml
1000 for both
amount [n] (mol) =
concentration [c] (mol dm^-3) x volume [V] dm^3
Worked example converting between solution volumes and moles (mol from solution)
Calculate the amount of NaCI, in mol, in 30.0 cm^3 of a 2.00 mol dm-3 solution.
n (NaCl) = c x V (cm^3) / 1000 = 2.00 x 30/1000 = 0.06 mol
Worked example converting between solution volumes and moles (solution volume from mol)
Calculate the volume of a 0.160 mol dm-3 solution that contains 3.25 x 10^-3 mol of NaCI.
n = c x V(cm^3) / 1000 so
V = 1000 x n / c = 1000 x 3.25 x 10^-3 / 0.16
= 20.3 cm^3
What’s a standard solution
A solution of known concentration
How are standard solutions prepared
By dissolving an exact mass of the solute in a solvent and then making the solution an exact volume
Worked Example:
Calculate the mass of Na2CO3 required to prepare 100cm^3 of a 0.250 mol dm^3 standard solution
- Work out the amout in moles required
n (Na2CO3) = c x [V (in cm^3)/1000] = 0.250 x 100/1000 = 0.0250 mol - Work out the molar mass of Na2CO3
-> its 106 g - Work out the mass using n = m/M
m = n x M = 0.0250 x 106 = 2.65g
