Amount of substance

Question 1

Avogadro’s constant

Answer 1

6x10²³ particles

Question 2

Molar mass

Answer 2

The mass of one mole of something; relative molecular/ formula mass

Question 3

Moles formula

Answer 3

number of moles= mass of substance/molar mass

Question 4

Concentration formula

Answer 4

concentration is measured in mol dm^-3

number of moles = (concentration x volume cm³)/1000

number of moles = concentration x volume dm³

Question 5

Average room temperature and pressure

Answer 5

25^oC (298 K) and 100kPa

gives moles = volume in dm³/24

Question 6

Ideal gas equation

Answer 6

pV=nRT

p= pressure Pa

V= volume m³

n= number of moles

R= 8.31 JK^-1mol^-1

T= temperature K

Question 7

Balance C₂H₆ + O₂ —> C0₂ + H₂0

Answer 7

C2H6 + 3.5O2 —> 2C02 + 3H20

Question 8

Define an ionic equation

Answer 8

An equation in which only the reacting particles are included

Question 9

What are the steps to balance an ionic equation?

Answer 9

Remember Only Have Chips

Reacting particles

Oxygen by adding H₂O

Hydrogen by adding H⁺

Charges by adding electrons

Question 10

Use a balanced equation to work out masses

E.g. mass of iron oxide if 28g of iron is burnt in air

2Fe + 1.5O₂ —> Fe₂O₃

Answer 10

M of Fe = 56g mol^-1

moles of Fe = 28/56 = 0.5 moles

0.5mol of Fe makes 0.25mol of Fe₂O₃

M of Fe₂O₃= (2x56)+(3x16) = 160g mol^-1

Mass of Fe₂O₃= 0.25x160 = 40g

Question 11

Write and indentify each of the state symbols

Answer 11

s = solid

l = liquid

g = gas

aq = aqueous (solution in water)

gr = graphite (carbon only, not always used)

Question 12

Use a balanced equation to calculate gas volume

2Na_(s) + 2H₂O_(l) —> 2NaOH_(aq) + H_2(g)

Answer 12

M of Na = 23g mol^-1

moles of Na = 15/23 = 0.65mol

Ratio 2:1

0.65/2 = 0.326mol H₂

Volume = 0.326 x 24 = 7.8dm³

Question 13

Titration equipment

Answer 13

Question 14

Titration Concentration Calculation

25cm³ of 0.5M HCl was used to neutralise 35cm³ of NaOH. Calculate concentration of NaOH solution in mol dm^-3

Answer 14

HCl + NaOH —> NaCl = H₂O

25cm³ 35cm³

0.5M ?

mol of HCl = (0.5x25)/1000 = 0.0125mol

Ratio 1:1

Conc. o NaOH = (0.0125x1000)/35 = 0.36mol dm^-3

Question 15

Titration Volume Calculation

20.4cm^₃ of a 0.5M solution of sodium carbonate reacts with 1.5M of nitric acid. Calculate the volume of nitric acid required to neutralise the sodium carbonate.

Answer 15

Write a balanced equation

Na₂CO₃ + 2HNO₃ —> 2NaNO₃ + H₂O + CO₂

20. 4cm3 ?
21. 5M 1.5M

Moles of Na₂CO₃ = (0.5x20.4)/1000 = 0.0102mol

Ratio 1:2 so 0.0204mol HNO₃

Volume of HNO₃ = (0.0204x1000)/1.5 = 13.6cm³

Question 16

Define empirical formula

Answer 16

The smallest whole number ratio of atoms in a compound

Question 17

Calculate the molecular formula of C₄H₃O₂ with a molecular mass of 166g

Answer 17

Relative formula mass = (4x12)+(3x1)+(2x16) = 83g

166/83 = 2 empirical units

C₈H₆O₄

Question 18

Empirical formula from % by mass

56. 5% Potassium
57. 7% Carbon
58. 8% Oxygen

Answer 18

Assume you have 100g so your % can be transferred straight into mass, then work out moles as normal.

n=mass/M

56. 9/100=1.449mol of K
57. 7/12=0.725mol of C
58. 8/16=2.175mol of O

divide each by the smallest number

1. 449/0.725=2
2. 735/0.725=1
3. 175/0.725=3

Empirical formula= K₂CO₃

Question 19

Percentage yield formula

Answer 19

Percentage yield = 100(actual/theoretical)

Question 20

Atom economy equation

Answer 20

% atom econoomy = 100(mass of desired product/total mass of reactants)