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Chemistry A > CGP C3 > Flashcards

CGP C3 Flashcards

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1
Q

What is a relative formula mass (Mr)

A

. The relative atomic masses of all atoms in the molecular formula added together

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2
Q

What is the relative atomic mass

A

. It’s the same as an elements mass number
e.g Ar of Mg = 24

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3
Q

Relative formula mass worked example

A

Find the relative formula mass of MgCl2
Ar of Mg = 24 Ar of Cl = 35.5
Mg + Cl (2) = 24 x (2 x 35.5) = 95
Mr of MgCl2 = 95

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4
Q

How do you calculate the % Mass of an Element in a Compound

A

Percentage mass of an element in a compound = Ar x number of atoms of that element / Mr of the compound x 100

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5
Q

% Mass of an Element in a Compound worked example

A

Find the percentage mass of sodium in sodium bicarbonate : Na2CO3
Ar of Sodium = 23, Ar of Carbon = 12 Ar of Oxygen = 16
Mr of Na2CO3 = (2 x 23) + 12 + (3 x 16) = 106

(23 x 2 / 106) x 100 = 43%

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6
Q

Percentage mass complicated worked example
“A mixture contains 20% iron ions by mass. 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
    Mixture contains 20% iron by mass, in 50g (50x0.2) there would be 10g of iron
  2. Calculate percentage mass of iron in iron chloride

56 / 56 + (2 x 35.5) x 100 = 44.09%

  1. Calculate the mass of iron chloride that contains 10g of iron

Iron chloride contains 44.09% iron by mass, so there will be 10g of iron in 10 / (44.09 / 100) = 23g

So you need 23g of iron chloride to provide the iron in 50g of the mixture

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7
Q

Mole

A

. “The Mole” is simply the name given to an amount of a substance.
. Just like a hundred is 100 or a thousand is 1000, The Avogadro constant is 6.02 x 10^23 that’s all, it’s just a number.

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8
Q

One mole of a substance

A

. One mole of any substance is just an amount of that substance that contains 6.02 x 10^23 of particles. The particles could be atoms, molecules, ions or electrons

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9
Q

Why is the Avogadro constant that specific number

A

.The mass of that number of atoms or molecules of any substance os exactly the same number of grams as the Ar or Mr of the element or compound
In other words:
. One mole of atoms or molecules of any substance will have a mass in grams equal to the Ar or Mr for that substance e.g.
. Carbon’s Ar = 12, so one mole of carbon = 12g

. Nitrogen gas’ Mr = 28, so one mole of Nitrogen gas = 28g

This means that 12g of C, 28g of N2 or 44g of C02 contain one mole of atoms

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10
Q

What is the formula to find the number of moles in a given mass

A

. Number of moles = Mass in g / Mr

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11
Q

Finding number of moles worked example

A

How many moles are there in 66g of Carbon Dioxide (CO2)

  1. Calculate Mr of Carbon dioxide = 12 + (16x2) = 44
  2. Divide mass (66g) by the Mr (44) = 1.5 mol
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12
Q

Conservation of mass

A

. During a chemical reaction, no atoms are destroyed and no atoms are created
. This means there are the same number and types of atoms on each side of a reaction equation

. If the mass seems to change, there’s usually a gas involved

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13
Q

Moles in equations

A

. The big numbers in-front of the chemical formulas of the reactants and products tell you how many moles of each substance takes part or is formed, for example :
Mg + 2HCl = MgCl2 + H2
. In this reaction, 1 mole of magnesium and 2 moles of hydrochloric acid react together to form 1 mole of magnesium chloride and 1 mole of hydrogen gas

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14
Q

Balancing equations using reacting masses

A

If you know the masses of the reactants and products that took part in a reaction, you can work out the balanced symbol equation for the reaction.

  1. Divide the mass of each substance by its relative formula mas to find no. of moles
  2. Divide number of moles of each substance by the smallest number of moles in the reaction
  3. If any of the numbers aren’t whole numbers, multiply all the numbers by the same amount so that they become whole numbers
  4. Write the balanced symbol equation by putting these numbers in front of the chemical formula
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15
Q

Balancing equations using reactant masses worked example

“8.1g of zinc oxide (ZnO) reacts completely with 0.6g of carbon to form 2.2g of carbon dioxde and 6.5g of zinc. Write a balanced symbol equation for this reaction.
Ar(C) =12 Ar(O)=16 Ar(Zn) =65

A
  1. Work out the Mr for each substance
    ZnO = 65+16 = 81
    C = 12
    CO2 = 12 + (16x2) = 44
    Zn = 65
  2. Divide the mass of each substance by its Mr to work out how many moles are produced
    ZnO = 8.1/81 = 0.1 mol
    CO2 = 2.2/44 = 0.05 mol
    C = 0.6/12 = 0.05 mol
    Zn = 6.5/65 = 0.1mol
  3. Divide each number by the smallest amount of moles, in this case is 0.05
    ZnO = 0.1/0.05 = 2
    CO2 = 0.05/0.05 = 1
    C = 0.05/0.05 = 1
    Zn = 0.1/0.05 = 2
  4. Now that all the numbers are whole numbers, you can substitute them into the equation
    (2)ZnO + (1)C = (1)CO2 + (2)Zn
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16
Q

Limiting reactant definition

A

Reactions don’t go on forever, - you need stuff in the reaction flask that can react. If one reactant gets completely used up in a reaction before the rest, then the reaction will stop. That reactant’s called limiting

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17
Q

Limiting reactants

A

When some Magnesium Chloride (MgCO3) is placed into a beaker of hydrochloric acid, you can tell a reaction is taking place because you see lots of bubbles of gas being given off.
After a while the fizzing slows down and the reaction eventually stops.

  1. The reaction stops because one of the reactants have been used up completely, all the other reactants are in excess. They’re usually added in excess to make sure that the other reactant is used up
  2. The reactant that i used up in a reaction is called the limiting reactant (because it limits the amount of product formed)
  3. The amount of product formed is directly proportional to the amount of limiting reactant. For example, if you halve the amount of limiting reactant, the amount of product will also halve. If you double it, the amount of product formed will also double.
  4. This is because if you add more reactant, there will be more reactant particles to take part in the reaction meaning more product particles
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18
Q

How do we calculate the mass of product formed in a reaction

A

You can calculate the mass of a product formed in a reaction by using the mass of the limiting reactant and the balancing equation reaction:

  1. Write out the balanced equation
  2. Work out the Mr of the reactant and product you want
  3. Find out how many moles there are of the substance you know the mass of
  4. Use the balanced equation to work out how many moles there’ll be of the other substance. In this case, that’s how many moles of product will be made of this many moles of reactant
  5. Use the number of moles to calculate the mass
19
Q

Calculating the mass of a product worked example
“Calculate the mass of aluminium oxide formed when 135g of aluminium is burned in the air”

A
  1. Write out the balanced equation
    4Al + 3O2 = 2Al2O3
  2. Calculate the Mr’s (relative formula masses)
    Al:27 Al2O3=(2x27) + (3x16) = 102 (you don’t need to find the Mr of oxygen as it’s in excess)
  3. Calculate the number of moles of aluminium in 135g
    137/27=5
  4. Look at the ratio of moles in the equation
    4 moles of Al react to produce 2 moles of Al2O3 - half the number of moles are produced - so 5 moles of Al will react to produce 2.5 moles of Al2O3
  5. Calculate the mass of 2.5 moles of Al2O3
    2.5 x 102 = 255g
20
Q

Moles of gas

A

One mole of any gas occupies 24dm3 at 20°C
. At the same temperature and pressure, equal numbers of moles of any gas will occupy the same volume
. At room temp and pressure (20°C and 1 atm) one mole of any gas occupies 24dm3

You can use this formula to find the volume of a known mass of any gas at r.t.p
Volume of gas (dm3) = Mass of gas (g) / Mr of gas x 24

21
Q

Calculating the volume of a known mass of any gas at r.t.p worked example:
What’s the volume of 319.5g of chlorine at r.t.p?

A

Volume = mass of gas/mr of gas x 24 =
319.5/71 x 24 =108dm3

22
Q

Using the volume of one gas to to find the volume of another:
“How much carbon dioxide is formed when 30 dm3 of oxygen reacts with carbon monoxide 2CO + O2 = 2CO2

A

1 mol of O2 = 2 moles of CO2
1 volume of O2 = 2 volumes of CO2
30 dm3 of O2 = 60 dm3 of CO2

23
Q

What is concentration

A

Concentration is a measure of how crowded things are

24
Q

Concentration explained

A

. Lots of reactions in chemistry take place between substances that are dissolved in a solution. The amount of a substance (e.g the mass or the number of moles) in a certain volume of solution is called it’s concentration

. The more solute (the substance that’s dissolved) there is in a given volume, the more concentrated the solution

25
Q

How do we measure concentration

A

. One way to measure the concentration of a solution is by calculating the mass of a substance in a given volume of solution. (The units will be units of mass/units of volume)

26
Q

How do we measure the concentration of a solution

A

concentration (g/dm3) or (mol/dm3) = mass of solute(g) or (mol)/volume of solvent(dm3)

27
Q

Finding concentration of a solution worked example:
“What’s the concentration in g/dm3 of a solution of sodium chloride where 30g of sodium chloride is dissolved in 0.2dm3 of water”

A

30 / 0.2 = 150

28
Q

What is 1dm3 in cm3

A

1000cm3

29
Q

How do we convert cm3 into dm3

A

dividing by 1000

30
Q

What’s the concentration in mol/dm3 of a solution with 2 moles of salt in 500cm3

A

500cm3/1000 = 0.5dm3
2 / 0.5 = 4 mol/dm3

31
Q

What is a titration

A

Titrations are experiments that allow you to find the volume needed for two solutions to react together completely. If you know the concentration of one of the solutions, you can use the volumes from the titration experiment, as well as the reaction equation, to find the concentration of the solution

32
Q

A student started with 30.0cm3 of sulfuric acid H2SO4 of unknown concentration in a flash. She found that it took an average of 25.0cm3 of 0.100 mold/dm3 sodium hydroxide (NaOH) to neutralise the sulfuric acid. Find the concentration of the acid in mol/dm3. The balanced symbol equation for the reaction is:
2NaOH + H2SO4 = Na2SO4 + 2H2O

A
  1. Work out how many moles of the known substance you have using the formula (no. moles = conc. x volume)
    0.100 mol/dm3 x (25.0 / 1000) dm3 = 0.00250 moles of NaOH
  2. Use the reaction equation to work out how many moles of the “unknown” stuff you must have had
    Using the equation, you can see that two moles of sodium hydroxide reacts with one mole of sulfuric acid. So 0.00250 moles of NaOH must have reacted with 0.00250 / 2 = 0.00125 moles of H2SO4
  3. Work out the concentration of the unknown stuff
    Concentration = number of moles / volume
    = 0.00125 mol / (30.0 / 1000) dm3 = 0.41666… mol/dm3 = 0.0417 mol/dm3
33
Q

Converting mol/dm3 to g/dm3

A
  1. To find the concentration in g/dm3, start by finding the concentration in mol/dm3 using the steps above
  2. Then, convert the concentration in mol/dm3 to g/dm3 using the equation mass = moles x Mr
34
Q

Converting mol/dm3 to g/dm3 worked example:
“What’s the concentration in g/dm3, of the sulfuric acid solution in the example above?”

A
  1. Work out the relative formula mass for the acid
    Mr (H2SO4) = (2x1) + 32 + (4x16) = 98
  2. Convert the concentration in moles (that you’ve already worked out) into concentration in grams. So, in 1dm3
    Mass in grams = moles x relative formula mass = 0.041666 x 98 = 4.08333…g
    So the concentration in g/dm3 = 4.08 g/dm3
35
Q

Atom economy explained

A
  1. A lot of reactions make more than one product. Some of them will be useful, but others will be a waste.
  2. The atom economy (atom utilisation) of a reaction tells you how much mass of the mass of the reactants is wasted when manufacturing a chemical and how much ends up as useful products
  3. 100% atom economy means that all the atoms in the reactants have been turned into useful (desired products). The higher the atom economy, the ‘greener’ the process.
36
Q

Atom economy equation

A

Atom economy = (relative formula mass of desired products) / (relative formula mass of all reactants) x 100

37
Q

Atom economy worked example:
“Calculate the atom economy of the following reaction to produce hydrogen gas:
CH4(g) + H2O(g) = CO(g) + 3H2(g)

A
  1. Identify the desired product - hydrogen gas
  2. Work out the Mr of all the reactants:
    Mr (CH4) = 12 + (4x1) = 16
    Mr (H2O) = (2x1) + 16 = 18
    16 + 18 = 34
  3. Work out the Mr of the desired product
    3 x Mr(H2) = 3 x (2x1) = 6
  4. Use the formula to calculate atom economy:
    6/34 x 100 = 17.6%
    17.6% of starting materials are useful
    Over 80% of the starting materials are wasted
38
Q

High atom economy

A
  1. Reactions with low atom economy use up resources quickly and make lots of waste material that have to be disposed of somehow. That tends to make these reactions unsustainable - the raw materials will run out and the wate has to go somewhere
  2. Low atom economy reactions aren’t usually profitable. Raw materials are often expensive to buy, and waste products can be expensive to remove and dispose of responsibly
  3. The best way around the problem is to find a use for the waste products rather than throwing them away. There’s often more than one way to make the product you want, so the trick is to come up with a reaction that gives useful “by-products” rather than useless ones
  4. Reactions with the highest atom economy are the ones that only have one product. Those reactions have an atom economy of 100%. The more products there are, the lower the atom economy is likely to be
39
Q

What does percentage yield tell you

A

Percentage yield tells you about the overall success of an experiment. It compares what you think you should get (theoretical yield) with what you get in practise (actual yield).

40
Q

Percentage yield explanation

A

. The amount of product you get is known as the yield. The more reactants you start with, the higher the actual yield will be. But the percentage yield doesn’t depend on the amount of reactants you started with - it’s a percentage.

. Percentage yield is always somewhere between 0 and 100%. 100% yield means that you got all the product you expected to get. 0% yield means that no reactants were converted into product i.e no product at all was made

. Industrial processes should have as high a percentage yield as possible to reduce waste and reduce costs.

41
Q

Yield in real life

A

Yields are always less than 100%
. In real life, you never get a 100% yield. Some product reactant always gets lost along the way and that goes for big industrial processes as well as school lab experiments. How this happens depends on what sort of reaction it is and what apparatus is being used. Lots of things can go wrong, but three common problems are:
. Not all reactants react to make a product.
. There might be side reactions
. You lose some product when you separate it from reaction mixture

42
Q
  1. Not all reacts react to make a product
A

. In reversible reactions, the products can turn back into reactants, so the yield will never be 100%

For example in the Haber process, at the same time N2 + 3H2 = 2NH3 is taking place, the reverse reaction 2NH3 = N2 + 3H2 is also happening

This means the reaction N2 + 3H2 = 2NH3 never goes to completion (the reactants don’t all get used up)

43
Q
  1. There might be side reactions
A

The reactants sometimes react differently to how you expect. They might react with gases in the air, or impurities in the reaction mixture, so they end up forming extra products other than the ones you want

44
Q
  1. You lose some product when you separate it from the reaction mixture
A

When you filter a liquid to remove solid particles, you nearly always lose a bit of liquid or a bit of solid.
. If you want to keep the liquid, you’ll lose the bit that remains with the solid and filter paper (as they always stay a bit wet)
. If you want to keep the solid, some of it’ll get left behind when you scrape it off the filter paper.

You’ll also lose a bit of material when you transfer it from one container to another - even if you manage not to spill it. Some of it always gets left behind on the inside surface of the old container.

Chemistry A (25 decks)

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