3 - Amount of Substance Flashcards

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1
Q

What does a mole measure?

A

The amount of an substance. (The number of particles)

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2
Q

What is a mole?

A

The amount of any substance contains as many elementary particles as there are carbon atoms in exactly 12g of the carbon-12 isotope (6.02 x 10^23)

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3
Q

What is the equation to find the number of moles from the number of atoms or molecules?

A

Number of Moles = Number of particles you have / Number of particles in a mole (Avogadro Constant 6.02 x 10^23)

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4
Q

I have 1.50 x 10^24 carbon atoms. How many moles of carbon is this?

A

Number of Moles = 1.50 x 10^24 / 6.02 x 10^23
= 2.49 moles

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5
Q

What is the molar mass, M?

A

The mass per mole of a substance, in units of g mol^-1. Mass of ONE mole of a substance

Same as Relative Molecular Mass.

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6
Q

Find the Molar Mass of CaCO3?

A

M = (40.1) + (12.0) + (3 x 16.0) = 100.1 g mol-1

I.e 1 mole of CaCO3 weighs 100.1 grams.

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7
Q

What is the equation with moles, Mass and Molar Mass?

A

Number of Moles = Mass of Substance / Molar Mass.

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8
Q

How many atoms are in 8.5 g of H2S?

A

Molar Mass = 1.0 + 1.0 + 32.1 = 34.1 g mol-1

Number moles = 8.5 / 34.1 = 0.249 mol

Number of Molecules of H2S = 0.249 x (6.02 x 10^23) = 1.50 x 10^23

3 atoms in one molecule of H2S so, total no. atoms = (1.50 x 10^23) x 3 = 4.5 x 10^23

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9
Q

What is the molar gas volume?

A

The volume per mole of gas molecules at a stated temperature and pressure.
Units of dm^3mol^-1

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10
Q

At room temperature and pressure what is the Molar Gas Volume?

A

24 dm3mol-1 (298 K and 101 kPa)

N = V / 24

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11
Q

If the temperature and pressure stays the same for one mole of any gas what is the effect on volume?

A

One mole of gas always has the same volume, if temperature and pressure stay the same.

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12
Q

What is the formula for working out the number of moles in a volume of gas?

A

Number moles = Volume in dm^3 / Molar Gas Volume (24 at r.t.p)

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13
Q

How many moles are there in 6.0 dm^3 of oxygen gas at r.t.p?

A

Number of moles = 6.0 / 24 = 0.25 moles of oxygen molecules

(Molecules not atoms as gaseous oxygens only exist as O2 not O).

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14
Q

What does the ideal gas equation let you find?

A

The number of moles in a certain volume at any temperature and pressure.

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15
Q

What is the ideal gas equation, and it’s UNITS for each?

A

pV = nRT

Where:
p = pressure (Pa)
V = volume (m^3)
n = number of moles
R = 8.314 J K^-1 mol^-1 (gas constant)
T = temperature (Kelvin (°C + 273))

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16
Q

How many dm in cm?

A

1dm in 10cm

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17
Q

How many dm in m

A

1m is 10dm

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18
Q

At a temperature of 60.0°C and a pressure of 250 kPa, a gas occupied a volume of 1100 cm^3 and a mass of 1.60 g. Find the relative molecular mass.

A

n = pV / RT = (250000) x (1.1 x 10^-3) / 8.314 x (273+60) = 0.0993 mol

Mr = M/n = 1.60 / 0.0993 = 16.1

So 1 mole is 16.1 g and the Mr is 16.1.

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19
Q

What is the empirical formula?

A

A formula that shows the simplest whole-number ratio of atoms of each element present in a compound.

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20
Q

What is the Molecular Formula?

A

A formula that shows the actual number of atoms of each type of element in a molecule.

21
Q

A molecule has an empirical formula of C4H3O2, and a molecular mass of 166 g mol-1. Work out its molecular formula.

A

(4 x 12.0) + (3 x 1.0) + (2 x 16.0) = 83.0 g mol-1

Mr= 166 g mol-1

166 / 83.0 = 2 empirical units in the molecule.

Empirical formula x 2 = C8H6O4

22
Q

When a hydrocarbon is burnt in excess oxygen, 4.4 g of carbon dioxide and 1.8 g of water are made. What is the empirical formula of the hydrocarbon?

A

No. of moles of CO2 = 4.4 / (12.0 + (2 x 16.0)) = 0.10 moles
1 mole of CO2 contains one mole of carbon atoms, so 0.10 moles of carbon.

No. of moles of H2O = 1.8 / ((2 x 1.0) + 16.0) = 0.10 moles
1 mole of H2O contain two moles of hydrogen atoms (H), so you must started with 0.20 moles of hydrogen atoms.

Ratio C:H = 0.10:0.20
0.10 / 0.10 = 1
0.02 / 0.10 = 2
So empirical formula must be CH2.

23
Q

A compound is found to have percentage composition 56.5% potassium, 8.70% carbon and 34.8% oxygen by mass. Calculate its empirical formula.

A

56.5 / 39.1 = 1.45 moles of K
8.70 / 12.0 = 0.725 moles of C
34.8 / 16.0 = 2.18 moles of O

Divide each number of moles by the smallest number- this case 0.725

K - 1.45 / 0.725 = 2.00
C - 0.725 / 0.725 = 1.00
O - 2.18 / 0.725 = 3.01

Empirical Formula = K2CO3

24
Q

When 4.6 g of an alcohol, with molar mass 46 g mol-1, is burnt in excess oxygen, it produces 8.8 g of carbon dioxide and 5.4 g of water. Calculate the empirical formula for the alcohol and then its molecular formula.

Remember Alcohols contain C,H and O.

A

No. of moles of CO2 = 8.8 / 44 = 0.2 moles
1 mole of CO2 contains 1 mole of C, so 0.2 moles of CO2 contain 0.2 moles of Carbon.

No. of moles of H2O = 5.4 / 18 = 0.3 moles
1 mole of H2O contains 2 moles of H, so 0.3 moles of H2O contain 0.6 moles of hydrogen.

Mass of C = 0.2 x 12 = 2.4 g
Mass of H = 0.6 x 1.0 = 0.6 g

Mass O = 4.6 - (2.4 + 0.6) = 1.6 g
Number of moles O = 1.6 / 16 = 0.1 moles

Moles ratio = 2:6:1 so empirical formula is C2H6O
Mr of C2H6O = 46 so empirical formula is also the molecular formula.

25
Q

What does hydrated mean?

A

A crystalline compound containing water molecules

26
Q

What is water crystallisation?

A

Water molecules that are bonded into a crystalline structure of a compound.

27
Q

How does water crystallisation work in salts?

A

Salt are ionic compounds so consist of a lattice of positive and negative ions. In some salts water molecules are incorporated in the lattice too.
Water molecules are polar so they are held in place in the lattice because they attracted to the oppositely charged ions. (Hydrogen to negative ions, oxygen to positive ions).
The water in the lattice is called the water of crystallisation.

28
Q

What does anhydrous mean?

A

Containing no water molecules.

29
Q

A solid salt containing water of crystallisation is …

A

Hydrated

30
Q

A solid salt containing no water of crystallisation is …

A

Anhydrous salt

31
Q

Heating 3.21 g of hydrated magnesium sulfate, MgSO4.XH2O, forms 1.567 g of anhydrous magnesium sulfate. Find the value of X and write the formula of the hydrated salt?

A
  1. Find the number of moles of water lost.
    Mass of water lost = 3.21 - 1.567 = 1.643
    No. of moles of water lost = 1.643 / 18 = 0.09127
  2. Find the number of moles of anhydrous salt.
    Molar Mass of MgSO4 = 24.3 + 32.1 + (4 x 16) = 120.4
    No. of moles = 1.567 / 120.4 = 0.01301 moles
  3. Ratio of moles.
    1 : (0.09127 / 0.01301) = 1 : 7.015
    So the formula is MgSO4.7H2O
32
Q

In a solution a concentrated is measured in what?

A

mol dm-3

How many moles are dissolved per 1 dm3 of solution.

33
Q

What is the equation for number moles using concentration?

A

Number of moles = concentration (mol dm-3) x volume (dm3)

34
Q

What mass of solid sodium carbonate, Na2CO3, is needed to make 250 cm3 of 0.300 mol dm-3 sodium carbonate solution?

A

n = conc x volume
n = 0.300 x 0.25 = 0.075 mol

Mass = 0.075 x [(2 x 23.0) + 12.0 + (3 x 16.0)] = 7.95 g

35
Q

What is theoretical yield?

A

Theoretical yield is the mass of a product that should be made if no chemicals are lost in the process.

36
Q

How do you calculate the percentage yield?

A

Percentage Yield = Actual yield / theoretical yield x 100%

37
Q

What does percentage yield tell us?

A

how much of our reactants (in percent) successfully turned into a product

38
Q

Why might have a theoretical yield not been reached?

A
  • Reaction may not have gone to competition.
  • ‘Side’ reactions may have taken place.
  • Purification of the product may result in loss of some product.
  • Reaction May involve several steps with product loss.
39
Q

Ethanol can be oxidised to form ethanal: C2H5OH + [O] —> CH3CHO + H2O
9.2 g of ethanol was reacted with an oxidising agent in excess and 2.1 g of ethanal was produced. Calculate the theoretical yield and the percentage yield.

A

n (C2H5OH) = 9.2 / (24 + 5 + 16 + 1) = 0.2 mol

Mole ratio 1:1 so n (CH3CHO) = 0.2 mol

Theoretical Yield (mass of CH3CHO) = 0.2 x (24 + 4 + 16) = 8.8 g

Percentage Yield = 2.1 / 8.8 x 100 = 24%

40
Q

What does a percentage yield not tell us about a reaction itself?

A

How wasteful the reaction itself is. A reaction has an 100% yield could still be very wasteful if a lot of the reactants end up in by-products rather than the desired products.

41
Q

What does atom economy measure?

A

Measure of the proportion of reactant atoms that become part of a desired product. How well toms have been utilised.

42
Q

How do you calculate atom economy?

A

% Atom Economy = (Molecular Mass if Desired Product / Sum of molecular masses of all products) x 100

43
Q

What is the atom economy in an addition reaction?

A

The atom economy is always 100% as the reactants combine in an addition reaction to form one product.

44
Q

What is the atom economy in an substitution reaction?

A

Atoms are swapped from one reactant to another reactant, so there always at least two products, at least one by-product. So will never be 100%

45
Q

Find the atom economy for the reaction of bromoethane (CH3Br) with sodium hydroxide (NaOH) to make Methanol (CH3OH)?

A

CH3Br + NaOH —> CH3OH + NaBr

Atom Economy = (12 + (3 x 1) + 16 + 1) / (12 + (3 x1) + 16 + 1)+(23 + 79.9) x 100
= 32 / (32 + 102.9) x100
23.7%

46
Q

Why is it important to make reactions with high atom economies?

A

Important for sustainability as they make the best use of natural resources.

47
Q

What are the economic benefits from a high atom economy?

A

Reactant chemicals are usually costly. Waste of money if a high proportion of them end up in useless product.
A low atom economy produces lots of waste, it cost money to separate the desired product from the waste product and more money to dispose of the waste products safely so they don’t harm the environment.

48
Q

How can we make reactions more sustainable?

A
  • Many raw materials are limited so high atom economy are needed to use them efficiently so they last as long as possible.
  • Reaction conditions with low energy demands as these are cheaper and better for the environment.
  • Use renewable sources instead of non-renewable sources (crude oil).