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GCSE Chemistry > C3 - Quantitative Chemistry > Flashcards

C3 - Quantitative Chemistry Flashcards

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1
Q

What is an isotope?

A

A variation of an element which occurs with different numbers of neutrons

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2
Q

Find the RFM of MgCl2.

Ar of Mg = 24

Ar of Cl = 35.5

A

1) Mg= 24
2) Cl2= 2 x 35.5 = 71
3) 24 + 71 = 95
4) RFM = 95

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3
Q

What’s the formula for the Mr of a substance?

A

RAM = (sum of: (isotopic mass x % abundance)) / 100

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4
Q

How do you find the percentage mass of an element in a compound?

A

(Ar/Mr) x 100

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5
Q

A mixture’s mass is 20% iron. What mass of iron chloride (FeCl2) would you need to provide the iron ions in 50g of the mixture?

Ar of Fe = 56

Ar of Cl = 35.5

A

1) Find the mass of iron in the mixture:

20% of 50g = 10g of iron needed

2) Calculate mass of Fe in FeCl2

56 / (56 + 2 x 35.5) = 56/127 (decimal)

3) Find the mass of FeCl2 which contains 10g of iron.

Mass of FeCl2 is 56/127 iron, and there is 10g of iron.

10g / (56/127) = 22.68g of iron chloride needed

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6
Q

What is a mole?

A

One mole is the amount of a substance which contains Avogadro’s number of particles.

The size or weight of particles does not change this value

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7
Q

What is Avogadro’s number?

A

6.02 x 1023

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8
Q

Why is Avogadro’s number used?

A

One mole (6.02 x 1023 particles) of any substance has a mass, in grams, equal to the Ar or Mr of that substance.

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9
Q

What is relative atomic mass?

A

The number of grams which one mole of an element weighs.

E.g. Mr of Carbon = 12, so 1 mole of carbon weighs 12g.

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10
Q

1) Carbon has an Ar of 12, so one mole of carbon weighs __g.

2) Oxygen has an Ar of 16, so CO2 has an Mr of __. Therefore, one mole of CO2 weighs __g

A

1) 12g
2) 44
3) 44g

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11
Q

The number of moles in a substance =

A

mass (g) ÷ Mr

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12
Q

Mr of a substance =

A

Mass ÷ moles

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13
Q

The mass of a substance =

A

Moles x Mr

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14
Q

1) What is the rule of conservation of mass in a chemical reaction?
2) What does this mean about the atoms on each side of the equation?

A

1) No atoms are created or destroyed- mass is conserved.

2) The same number of atoms, of each element, should be on each side of the equation.

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15
Q

Why could the total mass of the products of a reaction seem to differ from that of the reactants?

A

There would be a gas involved, and the reaction would have been in an unsealed vessel.

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16
Q

If the measured mass of the products of a reaction is smaller than that of the reactants, it will be because one of the ___ is a ___ which has ___ ___ ___ ___, whose mass can’t accounted for because it isn’t___ in the ___ ___

A

1) Products
2) Gas
3) Diffused into the air
4) Contained
5) Reaction vessel

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17
Q

If the measured mass of the products of a reaction is larger than that of the reactants, it will be because one of the ___ is a ___ in the ___, whose mass was not initially accounted for because it wasn’t ___ in the ___ ___.

A

1) Reactants
2) Gas
3) Air
4) Contained
5) Reaction vessel

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18
Q

Mg + 2HCl → MgCl2 + H2

What is the mole ratio of this reaction?

A

1 : 2 : 1 : 1

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19
Q

Mg + 2HCl → MgCl2 + H2

For every _ ___of magnesium and _ ___ of hydrochloric acid which react together, _ ___ of magnesium chloride and _ ___ of hydrogen are produced.

A

1 mole
2 moles
1 mole
1 mole

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20
Q

When given the masses of the reactants and products in a reaction, how would you work out the balanced symbol equation?

A

1) Find no. of moles of each substance: = mass/Mr
2) Simplify the numbers of moles if possible- e.g. 2:4:2:6 → 1:2:1:3
3) Write out the equation, putting the number of moles in front of the chemical formulae

21
Q

8.1g of ZnO reacts with 0.6g of C to form 2.2g of CO2 and 6.5g of Zn. Write a balanced symbol equation for this reaction.

Ar of Zn =65

Ar of O = 16

Ar of C = 12

A

1) Find Mr of ZnO = 65 + 16 = 81
2) Find Mr of CO2 = 12 + 2(16) = 44
3) Find no. of moles of each substance: = mass/Mr ZnO = 8.1 / 81 = 0.1 mol C = 0.6 / 12 = 0.05 mol CO2 = 2.2 / 44 = 0.05 mol Zn = 6.5 / 65 = 0.1 mol
4) Simplify the numbers of moles: 0.1 : 0.05 : 0.05 : 0.1 → 2:1:1:2
5) Write out the equation: 2ZnO + C → CO2 + 2Zn

22
Q

1) The reactant which is used up first in a reaction is referred to as what?
2) What are the other reactants classed as?

A

1) The limiting reactant.

2) Excess reactants.

23
Q

What is the relationship between the amount of product formed and the amount of limiting reactant?

A

The amount of product formed is directly proportional to the amount of limiting reactant.

24
Q

How would you calculate the mass of a product formed in a reaction, when given the mass of the limiting reactant?

A

1) Write out the balanced equation if not done for you.
2) Work out the Mr of the limiting reactant and the specified product.
3) Find out the no. moles of the reactant whose mass is given.
4) Work out how many moles there’ll be of the substance whose mass you need to find, using the balanced equation.
5) Use the no. moles to calculate the mass.

25
Q

Calculate the mass of aluminium oxide formed when 135g of aluminium is burned in air.

Ar Al = 27

Ar O = 16

A

1) Write out the balanced equation: 4Al + 3O2 →2Al2O3
2) Calculate the relative formula masses: Al = 27 Al2O3 = (2 x 27) + (3 x 16) = 102 (Don’t need to find Mr of oxygen because it’s the excess reactant.)
3) Find out the no. moles of aluminium in 135g. = mass/Mr = 135/27 = 5
4) Look at the ratio of moles in the equation to find no. moles of Al2O3. 4 moles of Al react to make 2 moles of Al2O3. So ratio = 2 : 1 = 5 : 2.5
5) Use the no. moles to calculate the mass. Mass = moles x Mr = 2.5 x 102 = 255g.

26
Q

At the same ___ and ___, equal numbers of ___ of any gas will occupy the same ___.

A

1) temperature
2) pressure
3) moles
4) volume

27
Q

What volume does one mole of any gas occupy at room temperature and pressure?

A

24dm3

28
Q

What formula can be used to find the volume of a known mass of any gas, at room temperature and pressure?

A

Volume = (mass of gas (g) / Mr of gas) x 24

1 mole of any gas occupies 24cm2 at room temp. & pressure. Moles = mass/Mr.

29
Q

What is the volume of 319.5g of chlorine at room temperature and pressure? Mr of Cl = 35.5

A

Volume = (mass of gas (g) / Mr of gas) x 24

= (319.5 / 35.5 x 2) x 24 = 108dm3

30
Q

What volume of carbon dioxide is formed when 30dm3 of oxygen reacts with carbon monoxide? The equation for this reaction is:

2CO + O2 →2CO2

A

Only need to focus on oxygen (because volume is given) and carbon dioxide (because you need to find its volume).

1) Mole ratio of oxygen : carbon dioxide = 1:2
2) Equal no. moles of gases occupy the same volume (assume same temperature and pressure)
3) So 1 volume O2 : 2 volumes CO2
4) So 30dm3 : 2 x 30dm3
5) 2 x 30dm3 = 60dm3 of CO2

31
Q

What is concentration a measure of?

A

How much of a substance there is within a certain volume

32
Q

When given mass and volume, how can concentration be calculated?

A

Concentration = mass of solute / volume of solvent

33
Q

When given moles and volume, how can concentration be calculated?

A

Concentration = no. moles of solute / volume of solvent

34
Q

What is concentration measured in?

A

g/dm3 OR mol/dm3

35
Q

What is volume of liquids measured in?

A

dm3

36
Q

How do you convert cm3 into dm3?

A

÷ 1000

37
Q

How do you convert dm3 into cm3?

A

x 1000

38
Q

How do you convert m3 into dm3?

A

x 1000

39
Q

What is the concentration, in mol/dm3, of a solution with 2 moles of salt in 500cm3?

A

Convert cm3 to dm3: 500 ÷ 1000 = 0.5dm3

Concentration = no. moles / volume = 2 ÷ 0.5 = 4 mol/dm3

40
Q

What are titrations?

A

Experiments which allow you to find the exact volumes of two solutions needed for them to react completely

41
Q

What’s the concentration, in g/dm3, of a 0.417 mol/dm3 solution of sulfuric acid (H2SO4)

A

1) Mr H2SO4 = (1 x 2) + 32 (16 x 4) = 98
2) In 1 dm3: mass = moles x Mr = 0.417 x 98 = 40.866g
3) so concentration = 40.87g/dm3

42
Q

Having done a titration, what would you be able to do if you knew the concentration of one of the reacting solutions?

A

You could use the volumes found in the titration and the reaction equation to find the concentration of the other reacting solution

43
Q

A student started with a 30cm3 solution of sulfuric acid (H2SO4), of unknown concentration.

She found by titration that it took an average of 25cm3 of 0.1mol/dm3 sodium hydroxide (NaOH) to neutralise the sulfuric acid.

Find the concentration of the sulfuric acid in mol/dm3. The balanced symbol equation for the reaction is:

2NaOH + H2SO4 → Na2SO4 + 2H2O

A

1) Find no. moles of known substance: = concentration x volume = 0.1 x (25 ÷ 1000) = 0.0025mol of NaOH
2) Find no. moles of unknown substance using the reaction equation: Mole ratio NaOH: H2SO4 = 2:1 = 0.0025 : 0.00125 mol H2SO4
3) Find concentration of unknown substance: = moles ÷ volume = 0.00125mol ÷ (30 ÷ 1000) = 0.416r mol/dm3 = 0.417mol/dm3

44
Q

What is the atom economy/utilisation of a reaction?

A

The percentage mass of the reactants which form useful products when manufacturing a chemical.

45
Q

What is the formula for atom economy?

A

atom economy = 100 x (Mr of desired product ÷ Mr of all reactants)

Like efficiency equation.

46
Q

What are the disadvantages of reactions with low atom economies?

A

1) Quickly use up resources
2) Make lots of waste
3) May not be profitable: raw materials are expensive and waste is expensive to dispose of

47
Q

What is the formula for percentage yield?

A

Percentage yield = 100 x (mass of product made ÷ max. theoretical mass of product made)

48
Q

Give the 3 main reasons why yields are always less than 100%.

A

1) In reversible reactions, products turn back into reactions- so it never goes to completion.
2) Side reactions- reactants could react with gases in the air or impurities in the reaction mixture.
3) Product is lost when separated from others- i.e. when filtering/transferring a liquid.

GCSE Chemistry (10 decks)

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