C4. Stoichiometry

Question 1

Define Relative atomic mass

Answer 1

average mass of isotopes of an element compared to 1/12th of the mass of an atom ¹²C

Question 2

Define the Molecular formula

Answer 2

the number and type of different atoms in one molecule

Question 3

Define relative molecular mass

Answer 3

sum of the relative atomic masses (Mᵣ)

Question 4

Relative formula mass

Answer 4

Mᵣ, is used for ionic compounds

Question 5

Define Mole

Answer 5

Symbol ‘Mole’ is the unit of amount of substances, 1 mole contains 6.02 x 10²³ particles,

e.g. Atoms, Ions, molecules.
This number is called the Avogadro constant

Question 6

Mole of atoms [2]

Answer 6

=6.02 x 10²³
this is called Avagadro’s number

Question 7

Steps to reacting masses [4]

Answer 7

n=m/Mr

1) identify 2 substances
2) n,m,Mᵣ
3) Calculate n. of moles + consider ratios (1:1)
4) calculate missing mass

Question 8

define Molar mass Mᵣ

Answer 8

Mass of 1 mole of a substance is a molar mass
- Relative formula mass in grams

Question 9

Describe what a limiting reactant is

Answer 9

The reactant that is used up the fastest

Question 10

Explain why particular reactants are added in excess

Question 11

Worked example of limiting reactant

Answer 11

2Ca + O₂ –> 2CaO
n 2.0 n 2.0
m 80g m 64g
M

Question 12

ratio of 3:2 is

Answer 12

ratio of 3:2 is
÷2, x3

Question 13

state the equations [2]

Answer 13

n = m/Mᵣ
n = V/Vm

Question 14

state conversion of cm³ and dm³

Answer 14

1000cm³ = dm³

cm to dm –>

Question 15

state conversion of cm³ and dm³

Answer 15

1000cm³ = dm³

cm³ to dm³ –> ÷ by 1000
dm³ to cm³ –> x by 1000

Question 16

define Molar mass volume

Answer 16

volume occupied by one mole of any gas, at room temperature and pressure

The molar volume is equal to 24 dm³ (24,000 cm 3) (24,000 cm 3)

Question 17

(Mole Equation)
Example: determine the number of moles in 95g of MnO₂

Answer 17

Example 1: Determine the number of moles in 95g of MnO₂.

Solution:
Mass of MnO2 = 95g

Mass of one mole MnO2 = 86.94g

Number of moles formula is
n = m/Mr

Number of moles = 95 / 86.94

Number of moles = 1.092 mol

Question 18

(Reacting masses simple ratios e.g. 1:2)
Example: 4.8g of magnesium ribbon reacts in a solution f hydrochloride acid containing 7.3g of HCL

Answer 18

Example: 4.8g of magnesium ribbon reacts in a solution f hydrochloride acid containing 7.3g of HCL
Mg (s) + 2HCL (g) ⟶ 2 MgCL₂ + H₂

Step 1. calculate the moles of magnesium used in the reaction
n = m/Mr

Mg 2HCL
n = 0.2 ml n = 0.2 m;
m = 4.8 m = 7.3
Mᵣ = 24 Mᵣ = 36.5

step 2. find the ratio of Mg to MgO using the molar ratio from the balanced equation
2Mg : 2MgO
1 : 2
0.2 : 0.4

Mg is in excess
HCL is the limiting reactant

Question 19

Metal ions are ____
most non-metals ions are

Answer 19

positve, negative

Question 20

ionic compounds are formed when _____ react with ______

Answer 20

ionic compounds are formed when metals react with non-metals