Chapter 3- Amount of Substance

Question 1

Avagadro’s Constant

Answer 1

the number of atoms in one mole of carbon-12 isotope
- 6.02X10^23

Question 2

Equation for the number of moles of a substance

Answer 2

n=m/M
n=number of moles
m=mass (g)
M= Molar mass (g/mol)

Question 3

Define a mole

Answer 3

amount of any substance containing as many particles as there are particles in 12g of the carbon-12 isotope

Question 4

Molar mass

Answer 4

mass (g) of 1 molee of a substance in g/mol

Question 5

Equation for number of atoms using avagadro’s constant

Answer 5

no, of atoms=
(m/ M) x 6.02x10^23
m=mass (g)
M= molar mass (g/mol)

Question 6

Why must you be careful when the question says “amount of substance/moles”

Answer 6

Amount of substance and moles can refer to anything:
- 1 mol of H: 1 mole of hydrogen ATOMS
- 1 mol of H2: 1 mole of hydrogen MOLECULES

Question 7

Relative molecular mass

Answer 7

the number of atoms of each element in a molecule

Question 8

Relative formula mass

Answer 8

the weighted mean mass of the formula unit of a molecule compared to 1/12 the mass of a carbon-12 atom

Question 9

Empirical formula

Answer 9

the simplest whole-number ratio of atoms of each elements in a compound

Question 10

How to workout empirical formula from mass

Answer 10

1. Convert mass into moles using n=m/M
2. Find the smallest whole number ratio by dividing both sides by the smallest whole number
3. Write empirical formula

Question 11

How to work out the molecular formula

Answer 11

1. Convert mass into moles using n=m/M
2. Find the smallest whole number ratio and then the empirical formula
3. Write relative mass (M) of empirical formula
4. Find no of empirical formula units in one molecule
   (M of molecule/ M of em.form)
5. Multiply empirical formula by result of step 4

Question 12

How to convert percentage composition by mass to moles

Answer 12

Think of whole compound as 100%.
e.g. 40% C; 6.67% H; 53.33% 0
n= %composition/ Molar mass
n(C)= 40%/ 12
=3.33 mol

Question 13

Components of a hydrated salt

Answer 13

water= water of crystallisation
solute= anhydrous salt

Question 14

How to work out the formula of a hydrated salt

Answer 14

1. calculate no. mole for the anhydrous salt using n=m/M
2. Calculate the no. mole of water using n=m/M
3. Find smallest whole ratio
   (tip: make anhydrous salt=1 so ratio is 1:n)
   e.g. 0.04:0.2= 1:5

Question 15

Problems to experimental formula of hyrated salts

Answer 15

Assumption 1: All water has been lost
Assumption 2: No further decomposition

Question 16

Equation for the no. of moles using conc & volume

Answer 16

n=cv aka (v=n/c or c=n/v)
n= moles
c= concentration (moldm^-3)
v= volume dm3

Question 17

Chemistry Conversions (liquid)

Answer 17

1cm3=1ml
1dm3=1000cm3
1000cm3=1000ml=1L

Question 18

Chemistry conversions
(ideal gas equation)

Answer 18

cm3 x(10^-6) = m3
dm3 x (10^-3) = m3
cm3 x (10^-3) = dm3

Celsius +273= Kelvin
kiloPascals x(10^3)= Pascals

Question 19

Molar gas volume (Vm)

Answer 19

the volume per mole of gas molecules at stated temperature and pressure

Question 20

Conditions for RTP

Answer 20

20 celsius/293 K
101kPA=1ATM

At RTP, 1 mole of gas molecules has the volume of 24dm3=24000cm3

Question 21

Equation for molar gas volume Vm

Answer 21

n=V/ Vm
volume (dm3 OR cm3)
Molar volume (24dm3 or 24000cm3)

Question 22

Ideal gas equation

Answer 22

pV=nRT
p=pressure (Pa)
V= Volume (m3)
n=moles
R= ideas gas constant (8.314 J/mol/K
T= Temperature (Kelvin)

Question 23

Assumptions on ideal gas equation

Answer 23

random motion
elastic colliisons
negligible size
no intermolecular forces

Question 24

Electron shell capacity

Answer 24

2.8.18.32.50.72