Chapter 2 & 3 Atoms, Ions, Moles

Question 1

What are isotopes

Answer 1

Atoms of the same element with different numbers of neutrons and different masses

Question 2

what is relative isotopic mass

Answer 2

the mass of an atom of an isotope compared to 1/12 of the mass of a 12C atom.

Question 3

what is relative atomic mass

Answer 3

the weighted mean mass of 1 mole of an element compared to the mass of 1 mole of 12C

Question 4

atomic number =

Answer 4

amount of protons

Question 5

A represents the […] number

Answer 5

Mass

Question 6

Z represents the […]

Answer 6

Proton number

Question 7

How come isotopes still react in the same way

Answer 7

The number of neutrons has no effect on the reactions of an element

Question 8

Cations are

Answer 8

Positive ions (atoms w fewer electrons than protons)

Question 9

Anions are

Answer 9

Negative ions (atoms with more electrons than protons)

Question 10

Ions and atoms of an element have the same number of […] but a different number of […]

Answer 10

Protons, electons

Question 11

mass to charge ratio =

Answer 11

relative mass of ion / relative charge on ion

Question 12

In a mass spectrometer, heavier ions move […] than lighter ones and are more difficult to […]

Answer 12

Slower, deflect

Question 13

What happens after a sample is placed in the mass spectrometer

Answer 13

The sample is vaporised and then ionised to form postive ions
-> then the ions are accelerated and move

Question 14

relative isotopic mass =

Answer 14

( the abundance x the mass number ) / 100

Question 15

Relative molecular mass is used for

Answer 15

Simple molecules

Question 16

For compounds with giant structures, we use

Answer 16

Relative formula mass

Question 17

What is a binary compound

Answer 17

Compound that only contains 2 elements

Question 18

When constructing a balanced equation what do you do

Answer 18

• 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

Question 19

What is amount of substance

Answer 19

The number of particles in a substance, measured in moles (mol)

Question 20

What is the avogadro constant

Answer 20

6.02 x 10^23 mol^-1

-> the number of particles in each mole of carbon 12

Question 21

What is the molar mass

Answer 21

Mass per mole of a substance

Question 22

Amount of substance (n) =

Answer 22

Mass (m) / molar mass (M)

Question 23

what is the molar gas volume (the volume occupied by any 1 mole of any has at rt and rtp)

Answer 23

24 dm^3 mol^-1

Question 24

What is empirical formula

Answer 24

The simplest whole number ratio of atoms of each element present in a compound

Question 25

What is molecular formula

Answer 25

The number and type of atoms of each element in a molecule

Question 26

Empirical formula from mass worked example

Answer 26

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

Question 27

Molecular formula worked example

Answer 27

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

Question 28

Atom economy =

Answer 28

Sum of desired products / sum of all products

Question 29

Benefits of high atom economy

Answer 29

• more efficient
• reduces waste
• preserves raw materials

Question 30

Percentage yield =

Answer 30

(Actual yield / theoretical yield) x 100

Question 31

why is it hard to achieve theoretical yield

Answer 31

• reaction may not go to completion
• side reactions may have taken place
• purification of produce may result in some loss of it

Question 32

What is the limiting reagent

Answer 32

The reactant that is used up first

Question 33

2H2 (g) + O2 (g) -> 2H2O (l)
Why is hydrogen the limiting reagent

Answer 33

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

Question 34

1cm ^3 = ? Ml

Answer 34

1 ml

Question 35

1 dm ^ 3 = ? cm ^3 and ? ml

Answer 35

1000 for both

Question 36

amount [n] (mol) =

Answer 36

concentration [c] (mol dm^-3) x volume [V] dm^3

Question 37

Worked example converting between solution volumes and moles (mol from solution)

Answer 37

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

Question 38

Worked example converting between solution volumes and moles (solution volume from mol)

Answer 38

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

Question 39

What’s a standard solution

Answer 39

A solution of known concentration

Question 40

How are standard solutions prepared

Answer 40

By dissolving an exact mass of the solute in a solvent and then making the solution an exact volume

Question 41

Worked Example:

Calculate the mass of Na2CO3 required to prepare 100cm^3 of a 0.250 mol dm^3 standard solution

Answer 41

1. Work out the amout in moles required
   n (Na2CO3) = c x [V (in cm^3)/1000] = 0.250 x 100/1000 = 0.0250 mol
2. Work out the molar mass of Na2CO3
   -> its 106 g
3. Work out the mass using n = m/M
   m = n x M = 0.0250 x 106 = 2.65g