C3 - Quantitative chemistry Flashcards

1
Q

What is the Law of Conservation of mass?

A
  • The Law of Conservation of Mass states that no matter is lost or gained during a chemical reaction. Mass is always conserved, therefore the total mass of the reactants is equal to the total mass of the products, which is why all chemical equations must be balanced

The sum of the relative atomic/molecular masses of the reactants will be the same as the sum of the relative atomic/molecular masses of the products

  • If the reaction flask is closed and no other substance can enter or leave the system, then the total mass of the reaction flask will remain constant otherwise If the reaction flask is open and a gaseous product is allowed to escape, then the total mass of the reaction flask will change as product mass is lost when the gas leaves the system.
  • If carried out in a closed system then the mass before and after the reaction will be the same
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2
Q

Example 1Balance the following equation:

aluminium + copper(II)oxide ⟶ aluminium oxide + copper

Unbalanced symbol equation:

Al + CuO ⟶ Al2O3 + Cu

A
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3
Q

Example 2Balance the following equation:

magnesium oxide + nitric acid ⟶ magnesium nitrate + water

Unbalanced symbol equation:

MgO + HNO3 ⟶ Mg(NO3)2 + H2O

A
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4
Q

What is Relative Atomic Mass and Relative Formula Mass?

A
  • This is calculated from the mass number and relative abundances of all the isotopes of a particular element
  • The symbol for the relative formula mass is Mr and it refers to the total mass of the molecule whereas The symbol for the relative atomic mass is Ar.
  • To calculate the Mr of a substance, you have to add up the relative atomic masses of all the atoms present in the formula
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5
Q

How do you calculate % mass of an element in a compound?

A
  • The percentage by mass of an element in a compound can be calculated using the following equation:
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6
Q

Why might mass of the products decrease if the products are gases?

A
  • Some chemical reactions may appear to involve a change in mass due to the presence of a gaseous reactant or product
  • If the reaction flask is open and a gaseous product is allowed to escape, then the total mass of the reaction flask will decrease as product mass is lost when the gas leaves the system
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7
Q

How do you explain Observed changes?

A
  • By analysis of the balanced chemical equations and the corresponding state symbols, you should be able to deduce the changes in mass for non-enclosed reaction systems
  • For example, the reaction between hydrochloric acid and calcium carbonate produces carbon dioxide gas:

2HCl (aq) + CaCO3 (s) → CaCl2 (aq) + H2O (l) + CO2 (g)

  • Mass will be lost from the reaction flask as carbon dioxide gas escapes to the atmosphere
    • So, the mass of the reaction mixture will decrease
  • If the mass of a reaction flask is found to increase then it is probably due to one of the reactants being a gas found in the air and all of the products are either solids or liquids
  • For example, the reaction of magnesium with oxygen produces magnesium oxide:

2Mg (s) + O2 (g) → 2MgO (s)

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8
Q

What is Uncertainty and Errors?

A

Uncertainty & Error

  • An error is the difference between a value or quantity obtained in an experiment and an accepted or literature value for an experiment
  • There are two types of errors in experiments, random errors and systematic errors
  • Uncertainties are the same as random errors
  • Uncertainties express the confidence to which the measurement can be taken
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9
Q

What are Random Errors?

A

Random Errors

  • When you are reading an instrument and estimate the final digit, there is an equal chance that you may read it slightly too high or slightly too low
    • This is a random error
  • Random errors are can be affected by:
    • How easily the instrument or scale is to read
    • The person reading the scale poorly
    • Changes in the environment, for example
      • fluctuations in the temperature of the lab
      • air currents in the room
  • Random errors will pull a result away from an accepted value in either direction (either too high or too low)
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10
Q

What are Systematic Errors?

A
  • Systematic errors are errors that occur as a result of a faulty or poorly designed experimental procedure
  • Systematic errors will always pull the result away from the accepted value in the same direction (always too high or always too low)
  • For example,
    • If you forget to zero an electronic balance (using the tare button) the mass weighings will always be higher than they should be
    • If you don’t read the volume in a burette at eye level, the volumes will always be smaller than they should be due to a parallax error
    • If you fail to keep a cap on a spirit burner in a calorimetry experiment, the alcohol will evaporate and give you a larger mass loss
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11
Q

How do calculate Uncertainty?

A
  • Treatment of uncertainties depends on the type of instrument used

Using analogue instruments

  • Any instruments that have an analogue scale, the uncertainty is taken as half the smallest division on the scale
  • For example,
    • A thermometer that reads to 1oC, the uncertainty would be _+_0.5 o C
    • A burette that reads to 0.10 mL, the uncertainty would be _+_0.05 mL

Using digital instruments

  • Any instruments that have a digital scale , the uncertainty is taken as the smallest division on the scale
  • For example,
    • An electronic balance that reads to 0.01 g, the uncertainty would be _+_0.01 g

Other uncertainties

  • Other sources of uncertainty can arise where the judgement of the experimenter is needed to determine a changing property
  • For example,
    • Judging the end point of a titration by looking at the colour of the indicator
    • Controlling a stopwatch in a rate of reaction experiment
    • Deciding when to extinguish the flame in an experiment
  • These uncertainties are very difficult to quantify, but they should be commented on as a source of error in an evaluation
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12
Q

What is the mole?

A

The Mole

  • Chemical amounts are measured in moles
  • The symbol for the unit mole is mol
  • One mole of a substance contains the same number of the stated particles, atoms, molecules, or ions as one mole of any other substance
  • The number of atoms, molecules or ions in a mole (1 mol) of a given substance is the Avogadro constant. The value of the Avogadro constant is 6.02 x 1023 per mole

For example:

  • One mole of sodium (Na) contains 6.02 x 1023atoms of sodium
  • One mole of hydrogen (H2) contains 6.02 x 1023molecules of hydrogen
  • One mole of sodium chloride (NaCl) contains 6.02 x 1023 formula units of sodium chloride
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13
Q

What is the relationship between moles and Relative Atomic mass?

A

Linking the Mole and the Atomic Mass

  • One mole of any element is equal to the relative atomic mass of that element in grams
  • This is called the molar mass
  • To find the mass of one mole of a compound, we add up the relative atomic masses
    • So one mole of water would have a mass of (2 x 1) + 16 = 18 g
    • So one carbon atom has the same mass as 12 hydrogen atoms
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14
Q

What is the equations for moles, mass and molar mass?

A
  • Although elements and chemicals react with each other in molar ratios, in the laboratory we use digital balances and grams to measure quantities of chemicals as it is impractical to try and measure out moles
  • Therefore we have to be able to convert between moles and grams
  • We can use the following formula to convert between moles, mass in grams and the molar mass:
  • Mass is in g
  • Moles is in mol
  • M is measure in gmol-1
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15
Q

How many moles are in 2.64 g of sucrose, C12H11O22 (Mr = 342.3)?

A

Answer:

  • The molar mass of sucrose is 342.3 g mol-1
  • The number of moles is found by mass ÷ molar mass
  • This comes to 2.64 g ÷ 342.3 g mol-1 = 7.71 x 10-3 mol
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16
Q

What is the mass of 0.250 moles of zinc?

A

Answer:

    • From the periodic table the relative atomic mass of Zn is 65.38
      • So, the molar mass is 65.38 g mol-1
      • The mass is calculated by moles x molar mass
      • This comes to 0.250 mol x 65.38 g mol-1 = 16.3 g
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17
Q

How do you calculate masses using balanced equations?

A
  • Chemical equations can be used to calculate the moles or masses of reactants and products
  • To do this, information given in the question is used to find the amount in moles of the substances being considered
  • Then, the ratio between the substances is identified using the balanced chemical equation
  • Once the moles have been determined they can then be converted into grams using the relative atomic or relative formula masses

Example 1Calculate the mass of magnesium oxide that can be made by completely burning 6.0 g of magnesium in oxygen in the following reaction:

2Mg (s) + O2 (g) ⟶ 2 MgO (s)

18
Q

How do you use moles to balance Equations?

A
  • Stoichiometry refers to the numbers in front of the reactants and products in an equation, which must be adjusted to make sure that the equation is balanced
  • These numbers are called coefficients (or multipliers) and if we know the masses of reactants and products, the balanced chemical equation for a given reaction can be found by determining the coefficients
  • First, convert the masses of each reactant and product in to moles by dividing by the molar masses using the periodic table
  • If the result yields uneven numbers, then multiply all of the numbers by the same number, to find the smallest whole number for the coefficient of each species
    • For example, if the resulting numbers initially were 1, 2 and 2.5, then you would multiply all of the numbers by 2, to give the whole numbers 2, 4 and 5

For example

64 g of methanol, CH3OH, reacts with 96 g of oxygen gas to produce 88 g of carbon dioxide and 72 g of water. Deduce the balanced equation for the reaction.(C = 12, H = 1, O = 16)

Answer

  • Calculate the molar masses of the substances in the equation

CH3OH = 32 g mol-1 O2 = 32 g mol-1

CO2 = 44 g mol-1 H2O = 18 g mol-1

  • Divide the masses present by the molar mass to obtain the number of moles

CH3OH = 64 g ÷ 32 g mol-1 = 2 mol

O2 = 96 g ÷ 32 g mol-1 = 3 mol

CO2 = 88 g ÷ 44 g mol-1 = 2 mol

H2O = 72 g ÷ 18 gmol-1 = 4 mol

  • The mole ratios are the same as the coefficients in the balanced equation

2CH3OH + 3O2 ⟶ 2CO2 + 4H2O

19
Q

What are Limiting Reactants?

A
  • A chemical reaction does not go on indefinitely and stops when one of the reagents is used up
  • The reagent that is used up first is the limiting reactant, as it limits the duration of the reaction and hence the amount of product that a reaction can produce
  • The one that is remaining is the excess reactant
  • The limiting reagent is the reactant which is not present in excess in a reaction
20
Q

How do you determine limiting reactants?

A
  • In order to determine which reactant is the limiting reagent in a reaction, we have to consider the amounts of each reactant used and the molar ratio of the balanced chemical equation
  • When performing reacting mass calculations, the limiting reagent is always the number that should be used, as it indicates the maximum possible amount of product that can form
    • Once all of a limiting reagent has been used up, the reaction cannot continue
  • The steps are:
  1. Convert the mass of each reactant into moles by dividing by the molar masses
  2. Write the balanced equation and determine the molar ratio
  3. Look at the equation and compare the moles
21
Q

In a reaction to produce sodium sulfide, Na2S, 9.2 g of sodium is reacted with 8.0 g of sulfur. Which reactant is in excess and which is limiting?

A
22
Q

What is Concentration and how do you calculate it?

A
  • A solid substance that dissolves in a liquid is called a solute, the liquid is called a solvent and the two when mixed together form a solution
  • Most chemical reactions occur between solutes which are dissolved in solvents, such as water or an organic solvent
  • Concentration simply refers to the amount of solute there is in a specific volume of the solvent
  • The greater the amount of solute in a given volume then the greater the concentration
  • A general formula for concentration is thus:
23
Q

How do you go from dm^3 and cm^3 and vice versa?

A
    • 1 decimetre cubed (dm3) is the same as 1 litre
  • You may be given data in a question which needs to be converted from cm3 to dm3 or the other way around
      • To go from cm3 to dm3 :
        * Divide by 1000
        • To go from dm3 to cm3 :
          • Multiply by 1000
24
Q

How do you calculate the concentration of a solution in terms of mass per unit volume

A
  • To calculate the concentration of a solution in terms of mass per unit volume, you need to
    • Identify the solute and solvent
    • Convert the volume units into decimetres cubed
    • Divide the mass of the solute by the volume of the solution in decimetres cubed

A student dissolved 10 g of sodium hydroxide, NaOH, in 2 dm3 of distilled water. Calculate the concentration of the solution.

25
Q

What is Yield?

Why do you never get 100% yield

A
  • Yield is the term used to describe the amount of product you get from a reaction
  • In practice, you never get 100% yield in a chemical process for several reasons
  • These include:
    • Some reactants may be left behind in the equipment
    • The reaction may be reversible and in these reactions a high yield is never possible as the products are continually turning back into the reactants
    • Some products may also be lost during separation and purification stages such as filtration or distillation
    • There may be side reactions occurring where a substance reacts with a gas in the air or an impurity in one of the reactants
    • Products can also be lost during transfer from one container to another
26
Q

What is Actual & Theoretical Yield?

A
  • The actual yield is the recorded amount of product obtained
  • The theoretical yield is the amount of product that would be obtained under perfect practical and chemical conditions
  • It is calculated from the balanced equation and the reacting masses
  • The percentage yield compares the actual yield to the theoretical yield
  • For economic reasons, the objective of every chemical producing company is to have as high a percentage yield as possible to increase profits and reduce costs and waste
27
Q

How do you calculate percentage yield?

A
  • The percentage yield is a good way of measuring how successful a chemical process is
  • There are often several methods of creating a compound and each method is called a reaction pathway
  • Reaction pathways consist of a sequence of reactions which must occur to produce the required product
  • Companies often investigate and try out different reaction pathways and these are then compared and evaluated so that a manufacturing process can be chosen
  • The percentage yield of each pathway is a significant factor in this decision making process
  • The equation to calculate the percentage yield is:
28
Q

How do you obtain theoretical masses of products?

A
  • We can obtain theoretical masses of products using a balanced equation and a given mass of reactant
  • The process is as follows:
  1. Write out the balanced equation for the reaction(if not already given in the question)
  2. Convert the given mass of reactant(s) into moles, by dividing the masses by the molar masses
  3. Use the coefficients(multipliers) in the equation to deduce the number of moles of the product(s)
  4. Convert the moles of product into mass by multiplying by the molar mass
29
Q

In an experiment to displace copper from copper(II)sulfate, 6.5g of zinc was added to a solution of copper(II)sulfate. The copper was filtered off, washed, dried and weighed. the final mass of copper obtained was 4.8 g. Calculate the percentage yield of copper.The balanced equation for the reaction is

Zn (s) + CuSO4 (aq) ⟶ Cu (s) + ZnSO4 (aq)

A
30
Q

What is Atom Economy and how do you calculate it?

A
  • Along with the percentage yield, atom economy is used to analyse the efficiency of reactions
  • Most reactions produce more than one product and very often some of them are not useful
  • Atom economy studies the amount of reactants that get turned into useful products
  • It illustrates what percentage of the mass of reactants become useful products
  • It is used extensively in the analysis of systems and procedures in industries, in an effort to obtain sustainable development
  • It is also a very important analysis for economic reasons as companies prefer to use processes with higher atom economies
  • The higher the atom economy of a process then the more sustainable that process is
  • The equation for calculating atom economy is:
31
Q

Hydrogen gas is obtained from methane in a process called steam-methane reforming. The reaction is as follows:

CH4 (g) + H2O (g) → CO (g) + 3H2 (g)

Calculate the atom economy of this reaction.

A
32
Q

How do you choose a reaction pathway?

A
  • Reactions that have low atom economies use up a lot of resources and produce a lot of waste material which then needs to be disposed of, a very expensive procedure
  • These reactions are thus unsustainable as they use up too much raw material to manufacture only a small amount of product
  • They are not economically attractive as raw materials tend to be expensive, as does waste disposal which requires chemicals, equipment, space and transport
  • Companies continually analyse reactions and processes and evaluate several factors in an effort to improve efficiency
  • Atom economy, percentage yield, rates of reaction and equilibrium position are important factors which need to be considered when choosing a reaction pathway
  • High percentage yields and fast reaction rates are desirable attributes in industrial chemical processes
  • In reversible reactions, the position of the equilibrium may need to be changed in favour of the products by altering reaction conditions
  • If the waste products can be sold or reused in some way that would improve the atom economy
  • Alternative methods of production could also be considered that may produce a more useful by-product
33
Q

How do you calculate the concentration of solution in moles?

A
  • It is more useful to a chemist to express concentration in terms of moles per unit volume rather than mass per unit volume
  • Concentration can therfore be expressed in moles per decimetre cubed
  • We can modify the concentration formula to include moles
    • The units in the answer can be written as mol dm-3 or mol / dm3:
34
Q

How do you convert g dm^-3 to mol d^-3?

How do you convert mol dm^-3 to g dm^-3?

A
  • You may have to convert from g dm-3 into mol dm-3 and vice versa depending on the question
    • To go from g dm-3 to mol dm-3:
      • Divide by the molar mass in grams
    • To go from mol dm-3 to g dm-3:
      • Multiply by the molar mass in grams
35
Q

Example 1 Calculate the amount of solute, in moles, present in 2.5 dm3 of a solution whose concentration is 0.2 mol dm-3.

A
36
Q

What is Avogadro’s law?

A

Avogadro’s Law

Volume(dm^3) = amount of gas(mol) * 24dm^3

  • Avogadro’s Law states that at the same conditions of temperature and pressure, equal amounts of gases occupy the same volume of space
  • At room temperature and pressure, the volume occupied by one mole of any gas was found to be 24 dm3 or 24,000 cm3
  • This is known as the molar gas volume at RTP
  • RTP stands for “room temperature and pressure” and the conditions are 20ºC and 1 atmosphere (atm)
  • From the molar gas volume the following formula triangle can be derived:

If the volume is given in cm3 instead of dm3, then divide by 24,000 instead of 24:

37
Q

How do you find the volume using Avogadro’s constant?

A

To find the volume

Volume = Moles x Molar Volume

Examples of Converting Moles into Volumes Table

38
Q

How do you find the Moles using Avogadro’s constant?

A

To find the moles

Moles = Volume ÷ Molar Volume

39
Q

Example 1What is the volume of 154 g of nitrogen gas at RTP?

A
  • A second style of gas calculation involves calculating the volumes of gaseous reactants and products from a balanced equation and a given volume of a gaesous reactant or product
  • These problems are straightforward as you are applying Avogadro’s Law, so the moles ( and cofficients) in equations are in the same ratio as the gas volumes
40
Q

Example 2The complete combustion of propane gives carbon dioxide and water vapour as the products

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)

Determine the volume of oxygen needed to react with 150 cm3 of propane and the total volume of the gaseous products

A

Answer

  • The balanced equation shows that 5 moles of oxygen are needed to completely react with 1 mole of propane
  • Therefore the volume of oxygen needed would be = 5 moles x 150 cm3 = 750 cm3
  • The total number of moles of gaseous products is = 3 + 4 = 7 moles
  • The total volume of gaseous products would be = 7 moles x 150 cm3 = 1050 cm3