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GCSE Chemistry > C4 - Chemical Calculations > Flashcards

C4 - Chemical Calculations Flashcards

1
Q

What are Chemical Equations?

A

Equations that show the reactants and products of a reaction.

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2
Q

What are reactants?

A

The substances you start with.

Always on the left-hand side of the equation.

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3
Q

What are products?

A

The new substances made from the reactants after the chemical reaction has taken place.

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4
Q

What is a word equation?

A

A word equation is an equation where the names of the elements involved are used to show what took place.

E.g - Hydrogen + Oxygen ———> Water

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5
Q

What is a symbol equation?

A

A symbol equation is an equation where the periodic symbols of the elements involved are used to show what took place.

E.g - CaCO3 ———> CaO + CO2

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6
Q

What is a balanced symbol equation?

A

A balanced symbol equation is when there is the same number of each type of atom on both sides of the equation.

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7
Q

Why is it so important to balance symbol equations?

A

Because atoms cannot be created or destroyed in a chemical reaction, according to the ‘Law of Conservation of Mass’.

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8
Q

What does the law of the conservation of mass state?

A

The total mass of the products formed in an equation, is always equal to the total mass of the reactants.

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9
Q

Balance the following equation:

H2 + O2 ———> H20

A

H2 + O2 ———> H20

Left = 2 Hydrogen’s and 2 Oxygen’s

Right = 2 Hydrogens but 1 Oxygen.

So to balance you would write:

2H2 + O2 ———> 2H2O

Now there are 4 Hydrogen’s on both sides and 2 Oxygen’s on both sides.

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10
Q

What are the state symbols for Solids, Liquids, Gases and Aqueous Solutions?

A

Solid (s)
Liquid (l)
Gas (g)
Aqueous solution (aq)

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11
Q

What is the relative atomic mass, Ar of an element?

A

The relative atomic mass of an element is the mass of an atom relative to Carbon-12 (standard reference point).

*also called the mass number (protons + neutrons)

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12
Q

Why do scientists work with relative atomic masses instead of actual masses?

A

The mass of a single atom is so tiny that it would be impractical to use exact values in experiment calculations.

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13
Q

Why is carbon-12 used as the standard reference point? What does it mean?

A

Carbon-12 has has 6 protons, 6 neutrons and 6 electrons.

Hydrogen has an Ar of 1 - this means that hydrogen’s atoms are 1/12 the mass of carbon-12’s atoms.

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14
Q

What is the symbol for relative atomic mass?

A

Ar

E.g Ar of Hydrogen = 1

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15
Q

What are relative formula masses?

A

Relative formula masses are the relative masses of all of the elements in a chemical compound added together.

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16
Q

What is the relative formula mass of H2SO4?

A

Ar of H = 1
Ar of S = 32
Ar lf O = 16

So H2SO4 = (1 x 2) + 32 + (16 x 4) = 98

Mr = 98

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17
Q

What is the symbol for relative formula mass?

A

Mr

E.g Mr of H2O = 18

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18
Q

What is a mole?

A

A unit of measure that describes the number of atoms within a substance.

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19
Q

What is one mole of anything in terms of its mass?

A

One mole is the relative atomic mass of ANY substance expressed in grams.

E.g 1 mole of hydrogen = 1 mole of carbon

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20
Q

What is Avagadro’s constant?

A

Avagadro’s constant is the number of atoms or molecules in one mole of a substance.

6.02 x 10^23

This means that there are 602000000000000000000000 atoms in 1 mol of every substance.

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21
Q

What is the equation used to calculate the number of moles?

A

Number of moles = mass of the substance (g) / relative atomic mass (Ar) or relative formula mass

Moles = mass (g) / Ar

or

Moles = mass (g) / Mr

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22
Q

If you rearrange that equation, how can you calculate the mass of a substance?

A

mass (g) = number of moles x Ar

OR

mass (g) = number of moles x Mr

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23
Q

How many moles of sulfuric acid molecules are there in 4.9g of sulfuric acid?

Sulfuric Acid = H2SO4

A

Number of moles = mass(g) / Mr

= 4.9 / 98

= 0.05mols

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24
Q

What is the mass of 7.5 x 10^-3 moles of aluminium sulfate?

Aluminium sulfate = Al2(SO4)3

A

Mass (g) = number of moles x Mr

= 0.0075 x 342

= 2.6g

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25
Q

How do balanced symbol equations tell us the number of moles of substances?

A

Because if there is a big number in front of an element or compound it shows the number of moles of that substance.

Example:

H2O = 1 mol 
2H2O = 2 mols
3H2O = 3 mols
26
Q

How can you calculate the masses of reactants and products in a chemical reaction?

A

We know that the mass (g) of a substance is the number of moles x Mr or Ar. Therefore if you have a balanced chemical equation such as:

CH4 + 2O2 ———> CO2 + 2H2O

Left-hand side:
CH4 = 16g
2O2 = 64g (32x2)
64 + 16 = 80g

Right-hand side:
CO2 = 44g
2H2O = 36g (18x2)
44 + 36 = 80g

That is how to calculate the masses of the reactants and products. Notice that they are both 80g due to the Law of Conservation of Mass.

27
Q

How do you work out the molar ratio of reactants and products?

A

LOOK at the equation.

Example:

Mg + 2HCl ———> MgCl2 + H2

The molar ratio can be worked out by looking at the front of each compound.

So the molar ratio of the reactants is 1:2

And the molar ratio of the products is 1:1

28
Q

Why is molar ratio important?

A

Molar ratio is important because when calculating reacting masses, it makes things a lot easier.

Example:

2Mg + O2

Let’s say we know that the number of moles in 12g of that Magnesium is 0.5 mols. Because the molar ratio of these 2 reactants is 2:1, we know that the number of mols of that particular oxygen molecule is 0.25.

29
Q

When calculating reacting masses what 4 steps need to be followed?

A
  1. Write a ✅ on the compound that is named with a mass, and a❓on the mass that you are asked to calculate.
  2. Calculate the number of moles of the compound with a ✅. Moles = mass / Mr
  3. Calculate the number of moles of the compound with a ❓. You use molar ratio here (look at the front of the numbers).
  4. Finally calculate the mass of the compound with a ❓. Mass = number of moles x Mr
30
Q

What mass of Oxygen reacts with 12g of Magnesium?

A
  1. Label balanced equation with ✅ and ❓

2Mg(✅) + O2(❓) ———> 2Mg O

  1. Moles of ✅ (2Mg) = 12g / 24 (Ar of 1 mol of Mg)
    =0.5 mols
  2. Moles of ❓(O2) = 0.25 mols (molar ratio = 2:1)
  3. Mass of ❓(O2) = (16 x 2) x 0.25 = 8g

8g of Oxygen is needed to react with 12g of Magnesium.

31
Q

What mass of calcium hydroxide is made from 14kg of calcium oxide?

A
  1. Label balanced equation with ✅ and ❓

CaO (✅) + H2O ———> Ca(OH)2 (❓)

  1. Moles of ✅ (CaO) = 14000g / 56 = 250mols
  2. Moles of ❓(Ca(OH)2) = 250 mols (molar ratio = 1:1:1)
  3. Mass of ❓(Ca(OH)2) = 74 x 250 = 18500g
  4. 5kg of Calcium Hydroxide is made from 14kg of Calcium Oxide.
32
Q

Show how the molar ratio of reactants and products be used to produce a balanced equation?

A

Example:

NaNO3 : NaNO2 : O2

= 0.1 : 0.1 : 0.05

After calaculating this using knowledge from the previous flashcards, if you divide by the smallest number given you will get a simpler ratio

divide all by 0.05 =:

2 : 2 : 1

Once you’ve calculated this, you can deduce a balanced equation for the original equation:

2NaNO3 ———> 2NaNO2 + O2

33
Q

What is the limiting reactant?

A

The limiting reactant is the reactant that gets used up first as it is not in excess. The amount of product produced in a reaction depends on the limiting reactant.

34
Q

How do you show that a substance is the limiting reactant in a reaction?

A

1) Calculate the number of moles for each reactant using the masses they have given. Write as a ratio.
2) Compare to the ratio of the reactants in the balanced equation. Divide the calculated moles by the moles in the balanced equations.
3) See which reactant is in excess and which is limiting.

36
Q

If you have 4.8g of magnesium ribbon reacting in a solution of dilute hydrochloric scid containing 7.3g of HCl, which reactant is the limiting reactant?

A

Balanced: Mg + 2HCl ———> MgCl2 + H2

  1. Only interested in reactants. Moles = Mass / Mr

Moles of Mg = 4.8 / 24 = 0.2 mol
Moles of HCl = 7.3 / (1+35.5) = 0.2 mol

  1. From the balanced equation at the top, you can see that 1 mole of Mg will react with 2 moles of HCl.
    Therefore 0.2 moles of Mg will need 0.4 moles of HCl to react completely. In this case, we haven’t got 0.4 moles of HCl, so the dilute hydrochloric acid is the limiting reactant and the magnesium is in excess.
37
Q

What is the Yield of a Chemical Reaction?

A

The yield of a chemical reaction describes how much product is actually made.

38
Q

What is the Percentage Yield of a Chemical Reaction?

A

The amount of product actually made, out of the maximum theoretical amount of product that could be made - according to the Law of Conservation of mass.

It’s about PRODUCT!

39
Q

Why is a 100% yield not possible?

A

Although mass cannot be lost - certain practical factors during production will always reduce the yield.

40
Q

What are some factors that can reduce yield?

A
  • Product left behind in the apparatus.
  • Losses in separating the products from the reaction mixture.
  • Gases given off.
  • Reversible reactions not going to completion.
41
Q

What is the equation for Percentage Yield?

A

Percentage Yield = actual mass of product produced / maximum theoretical mass of product produced x 100

Y = (m/m) x 100

% = (g) / (g) x 100

42
Q

Calculate the Percentage Yield:

78.5g of Iron was formed

Theoretical mass = 88.2g

A

PY = (78.5 / 88.2) x 100%

= 89%

43
Q

What is atom economy?

A

The extent to which atoms in the reactants end up in the desired product.

44
Q

Why is atom economy more important than ever?

A

We are moving towards a ‘greener’ chemical industry - so minimalising the wastage of energy and raw materials is vital to the fight against climate change.

45
Q

What is the equation for percentage atom economy?

A

Percentage Atom Economy = Mr of desired product / Sum of Mr of ALL reactants from equation x 100

AE = Mr (desired product) / Mr (reactants) x 100

46
Q

Calculate the atom economy to form copper(II) oxide from copper (II) carbonate.

CuCO3 ———> CuO + CO2

A
  1. Desired Product = CuO

Mr of CuO = 63.5 + 16 = 79.5

  1. Mr of CuCO3 = (63.5 + 12 + (16 x 3) = 123.5
  2. (79.5 / 123.5) x 100% = 64.4%
47
Q

What is the concentration of a solution?

A

A measure of the relationship between the volume of the solvent, and the mass of the solute.

More solute in the same volume of solvent = higher concentration.

48
Q

What is the equation for concentration?

A

Concentration (g/dm^3) = mass of solute (g) / volume of solute (dm^3)

c = m / v

OR:

Concentration (g/dm^3) = mass of solute (g) / volume of solute (cm^3) X1000

-You x1000 because there are 1000cm^3 in 1dm^3 (decimetre cubed).

49
Q

50g of Sodium Hydroxide is dissolved in 200cm^3 of water. What is the concentration in g/dm^3?

A

Concentration (g/dm^3) = 50/200 x1000

= 0.25 x 1000 = 250g/dm^3

50
Q

Re-arrange the equation to make the amount of solute the subject.

A

mass of solute (g) = concentration (g/dm^3) x volume (dm^3)

m = c x v

51
Q

A solution of sodium chloride has a concentration of 200g/dm^3. What is the mass of the sodium chloride in 700cm^3 of the solution?

A

700cm^3 = 0.7dm^3

m = c x v

200g/dm^3 x 0.7dm^3 = 140g

52
Q

How can you increase the concentration of an aqueous solution? - 2 ways

A

1) Evaporate some of the solution.

2) Add more solute.

53
Q

What is a Titration?

A

A technique used to accurately measure the volume of acid or alkali needed to neutralise eachother and react completely.

54
Q

What is ‘the end point’ of a Titration?

A

The point at which a reaction between an acid and an alkali is complete.

55
Q

How can you see the ‘end point’ of a Titration?

A

Use an acid based indicator and wait until there is a colour change.

56
Q

What is the step-by-step protocol for carrying out a Titration?

A

1) In a clean, dry pipette, transfer 25cm^3 of an alkali into a conical flask.
2) Add a few drops of a suitable indicator to the alkali.
3) Fill a clean, dry burette with acid. Note the starting volume.
4) Gradually add acid into the conical flask and swirl.
5) When the colour changes, stop adding acid and note the new volume.
6) Calculate the titration and repeat until two concordant results are obtained. (Within 0.1cm^3).

57
Q

After carrying out a titration, you have established the volume of acid needed to neutralise a known volume of alkali.

What calculation is used to calculate the concentration of the alkali or the acid?

A

Concentration (mol/dm^3) = mass of solute (moles) / volume of solution (dm^3)

c = m / v

OR:

Concentration (mol/dm^3) = mass of solute (moles) / volume of solution (cm^3) X 1000

58
Q

What is the concentration of 10 moles dissolved in a 2dm^3 solution?

A

10 moles / 2 dm^3 = 5 mol/dm^3

59
Q

In a Titration, a 12.5cm^3 sample of nitric acid, reacts exactly with 10cm^3 of a 0.4 mol/dm^3 solution of sodium hydroxide?

Calculate the concentration of the nitric acid in moles/dm^3.

A

1) Write balanced equation:
KOH + HNO3 ———>KNO3 + H2O

2) Number of moles of potassium hydroxide (known substance).

Number of moles = volume (dm^3) x concentration (mol/dm^3)

= 10 cm^3 / 1000 = 0.01 dm^3

  1. 4 x 0.01 = 0.004 moles
    3) Ratio of potassium hydroxide to nitric acid is 1:1, so therefore there is 0.004 moles of nitric acid.
    4) Concentration of the nitric acid (mol/dm^3) = 0.004 moles / 12.5 cm^3 X1000 = 0.32 moles/dm^3
    5) Might be asked to convert mol/dm^3 to g/dm^3.

conc (g/dm^3) = conc (mol/dm^3) x Mr

= 0.32 x Mr of HNO3 (1 + 15 + 14x3) = 20.2 g/dm^3

60
Q

What is the rule about volumes of gases?

A

A certain volume of gas always contains the same number of gas molecules under the same conditions.

The volume of 1 mole of ANY GAS at room temperature is ALWAYS 24dm^3 (24000cm^3).

61
Q

What is the equation for the number of moles of a gas?

A

Number of moles of gas = volume of gas (dm^3) / 24 dm^3

N = v / 24

OR

Number of moles of gas = volume of gas (cm^3) / 24000 cm^3

N = v / 24000

62
Q

Re-arrange the equation to make the volume of gas the subject.

A

Volume of gas (dm^3) = n of moles x 24dm^3

63
Q

An air bag is inflated with 70g of Nitrogen, N2. What volume would the gas occupy at room temperature?

A

Volume = number of moles x 24dm^3

Moles = Mass/ Mr = 70/14x2 = 70/28 = 2.5 moles

Volume = 2.5 x 24 = 60dm^3

GCSE Chemistry (17 decks)

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