Section 3 - Equations, Calculations and Electrolysis

Question 1

Mr (relative formula mass)

Answer 1

all the relative atomic masses added together (Ar)

Question 2

mole value

Answer 2

6.02 x10²³

Question 3

mass of a mole

Answer 3

one mole of atoms/molecules of any substance will have a mass in grams equal to the relative particle mass (Ar or Mr ) of that substance

Question 4

moles formula

Answer 4

moles = mass in g/mr

Question 5

how to calculate amount of product from mass of reactant

Answer 5

1) write out balanced equation
2) work out Mr of substance you know the mass of
3) find out how many moles are in reactant
4) use ratio from equation to calculate how many moles in product
5) use moles to calculate mass using formula

Question 7

percentage yield

Answer 7

( actual yield/
theoretical yield) x100

Question 8

what is emperical formula

Answer 8

smallest whole number ratio of atoms in a compound

Question 9

how to get emperical formula

Answer 9

1) list all elements in compound
2) write experimental masses
3) find moles of each element
4) turn moles into a ratio for these by dividing by smallest number of moles
5) get ratio in simplest whole number form. thats the emperical formula

Question 10

emperical vs molecular formula

Answer 10

emperical = simples formula ratio for each element in a compound
molecular = actual n of atoms of each element in a single molecule

Question 11

molecular formula

Answer 11

1) find mass of emperical formula
2) divide molecular mass by this
3) tuimes original by emperical units

Question 12

finding emperical formula using combustion

Answer 12

1) get crucible and heat until red hot (so its clean without traces of oil or water)
2) leave to cool and weigh with lid
3) add magnesium ribbon to crucible and reweigh, use this to find mass of magnesium
4) heat with lid (to stop magnesium from escaping)
5) heat for 10 min or until magnesium turns white
6) allow crucible to cool and then reweigh with lids and contents, use this to find mass of magnesium oxide

Question 13

finding emperical formula using reduction

Answer 13

1) place rubber bung into test tube with a small hole in end and weigh them using a mass balance
2) take bung out of test tube and spread out a small amount of copper(II) oxide in middle of tube
3) re-insert bung and re-weigh
4) expel air from best tube by gently turning on gass and after 5 seconds light by holding burning splint up
5) use bunsen burner to heat for 10 min or until copperii oxide turns brownish pink
6) leave to cool
7) weigh with bung and contents

Question 14

salts

Answer 14

can be anhydrous or hydrated
- consist of lattice of positive and negative ions
- in some salts water molecyles are incorporated (water of cystalisation)
- a solid salt containing water of crystallisation is hydrated
- if it doesnt contain water of crystalisation its anhyfrous

Question 15

lattice in a hydrated salt

Answer 15

water molecules have a small +ve charge on hydrogen atoms and small -ve charge on the oxygen atoms, therefore they are attracted to ions in lattice and held in place

Question 16

how much water of cystallisation a salt contains

Answer 16

• one molecule of hydrated salt has a particular number of moles of water crystalisation - its formula shows how many
• many hydrated salts loose their water of cystallisation when heated to become anhydrous. you can work out its formula if you know mass when anhydrous and hydrated

Question 17

working out mass of salt knowing hydrated and anhydrous values

Answer 17

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