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Chemistry Module 2 - Own Flashcards > Chapter 3 - Amount of substance > Flashcards

Chapter 3 - Amount of substance Flashcards

1
Q

Why are chemicals usually measured by mass or volume?

A

Reactions occur on the atomic scale, so chemists need a way to convert measured mass or volume into the number of particles involved in the reaction.

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2
Q

What is the ‘amount of substance’? What is it represented by? Why is it measured in?

A

It is a measure used to count the number of particles in a substance, represented by n, and measured in moles (mol).

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3
Q

What is a mole?

A

A mole is the amount of substance that contains 6.02 × 10²³ particles. This number is known as the Avogadro constant.

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4
Q

What is the Avogadro constant?

A

6.02 × 10²³ particles.

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5
Q

Why is the Avogadro constant set at
6.02 × 10^23?

A

It is directly linked to the mass of carbon-12.
Example: 12 grams of carbon-12 contains 6.02 × 10^23 atoms.

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6
Q

How can you find the mass of one mole of atoms?

A

The mass of 1 mole of an element is its relative atomic mass in grams.

Example:
1 mole of carbon = 12.0g
1 mole of magnesium = 24.3g

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7
Q

Why is the mole an important concept in chemistry?

A

It provides an easy way to count atoms or particles by simply measuring the mass.

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8
Q

Does the mole only refer to atoms?

A

No, it can refer to anything, such as atoms, molecules, or ions. It’s important to specify the formula or name of the substance being measured.

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9
Q

What is molar mass?

A

Molar mass is the mass per mole of a substance, measured in g/mol

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10
Q

What is the molar mass of Carbon?

A

12.0 g/mol

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11
Q

What is the molar mass of NO2?

A

14.0 + (16.0 x 2) = 46.0g/mol

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12
Q

What is the formula linking the amount of substance, mass, and molar mass? Give it as the symbol equation as well.

A

Amount of substance = Mass ÷ Molar Mass

n = m ÷ M

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13
Q

What is the unit used to measure the Amount of substance?

A

mol

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14
Q

What is the unit used to measure mass?

A

g

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15
Q

What is the unit used to measure Molar Mass?

A

g/mol

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16
Q

PRACTICE QUESTION:
Calculate the amount of substance, in moles, in 96.0g of carbon, C.

A

n = m ÷ M

n = 96.0 ÷ 12.0

n = 8.0 mol

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17
Q

PRACTICE QUESTION:
Calculate the mass, in g, of 0.050mol of NO2.

A

Rearrange: n = m ÷ M

Rearranged formula: m = n x M

m = 0.050 x 46.0

m = 2.3g

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18
Q

PRACTICE QUESTION:
Calculate the molar mass when 2.65g contains 0.025mol of a substance.

A

Rearrange: n = m ÷ M

Rearranged formula: M = m ÷ n

M = 2.65 ÷ 0.025

M = 106.0 gmol-1

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19
Q

What is a molecular formula?

A

The molecular formula is the exact number of atoms of each element in a molecule.

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20
Q

What is an empirical formula?

A

The empirical formula is the simplest whole-number ratio of atoms of each element in a compound.

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21
Q

Why is the empirical formula important?

A

It is important for substances that do not exist as molecules, such as metals, some non-metals (e.g., carbon, silicon), and ionic compounds (e.g., sodium chloride).

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22
Q

Why can’t the actual number of atoms or ions in giant crystalline structures be used?

A

The actual numbers would run into billions and vary depending on the crystal size, making them impractical.

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23
Q

How is the empirical formula determined for giant crystalline structures?

A

It represents the ratio of atoms or ions in the structure, which is always the same.

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24
Q

How do you convert between molecular and empirical formulas?

A

Divide the molecular formula by the greatest common factor of the subscripts.

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25
Q

If the molecular formula is N2O4, what is the empirical formula?

A

NO2

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26
Q

If the molecular formula is H2O, what is the empirical formula?

A

H2O

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27
Q

If the molecular formula is C2H6, what is the empirical formula?

A

CH3

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28
Q

If the molecular formula is P4O6, what is the empirical formula?

A

P2O3

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29
Q

If the molecular formula is C9H12O3, what is the empirical formula?

A

C3H4O

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30
Q

Why are two terms needed for relative mass?

A
  1. Relative molecular mass (Mr): For simple molecules like water and carbon dioxide.
  2. Relative formula mass (Mr): For giant crystalline structures like ionic compounds.
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31
Q

What is relative molecular mass (Mr)?

A

It compares the mass of a molecule with the mass of an atom of carbon-12.

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32
Q

How is relative molecular mass calculated?

A

By adding the relative atomic masses of the elements in the molecule.

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33
Q

What is the relative molecular mass of: H2O?

A

Mr(H2O) = (1.0x2) + 16.0 = 18.0

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34
Q

What is the relative molecular mass of: CH4?

A

Mr(CH4) = 12.0 + (1.0x4) = 16.0

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35
Q

What is the relative molecular mass of: C6H12O6?

A

Mr(C6H12O6) = (12.0x6) + (1.0x12) + (16.0x6) = 180.0

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36
Q

What is relative formula mass?

A

It compares the mass of the formula unit with the mass of an atom of carbon-12.

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37
Q

How is relative formula mass calculated?

A

By adding the relative atomic masses of the elements in the empirical formula.

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38
Q

What is the relative formula mass of: NaCl?

A

Mr(NaCl) = 23.0 + 35.5 = 58.5

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39
Q

What is the relative formula mass of: Ca(NO3)2?

A

Mr(Ca(NO3)2) = 40.1 + ((14.0+(16.0x3))x2) = 164.1

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40
Q

How can the formula of an ionic compound be predicted?

A

From the known ions present in the compound.

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41
Q

How can you determine the formula of an ionic compound if the ions in a compound are unknown?

A

The formula can be determined through experiments analyzing the chemical composition of the substance (a process known as analysis).

42
Q

PRACTICE QUESTION:
In an experiment, 1.203g of calcium combines with 2.13g of chlorine to form a compound. Calculate the empirical formula.
[Ar: Ca=40.1; Cl=35.5]

A

Step 1: Convert the mass into moles of atoms using n = m ÷ M
n(Ca) = 1.203 ÷ 40.1 = 0.030 mol
n(Cl) = 2.13 ÷ 35.5 = 0.060 mol

Step 2: To find the smallest whole-number ratio, divide by the smallest whole number.
na(Ca):n(Cl) = 0.030÷0.030 : 0.060÷0.030 = 1:2

Step 3: Write the empirical formula: CaCl2

43
Q

PRACTICE QUESTION:
Chemical analysis of a compound gave the percentage composition by mass: C=40.00%; H=6.67%; O=53.33%. The relative molecular mass of the compound is 180.0
Calculate the molecular formula.
[Ar: C=12.0; H=1.0; O=16.0]

A

Step 1: Convert % by mass into moles of atoms using n = m ÷ M
n(C) = 40.00 ÷ 12.0 = 3.33 mol
n(H) = 6.67 ÷ 1.0 = 6.67 mol
n(O) = 53.33 ÷ 16.0 = 3.33 mol

Step 2: Find smallest whole-number ratio and empirical formula.
n(C):n(H):n(O) = 3.33÷3.33 : 6.67÷3.33 : 3.33÷3.33 = 1:2:1
Empirical formula = CH2O

Step 3: Find the relative mass of the empirical formula CH2O: 12.0 + (1.0x2) + 16.0 = 30.0

Step 4: Find the number of CH2O units in one molecule: 180 ÷ 30.0 = 6

Step 5: Write the molecule formula: CH2O x 6 = C6H12O6

44
Q

What are hydrated salts?

A

Hydrated salts are crystalline compounds that contain water molecules as part of their crystalline structure.

45
Q

What is the water in hydrated salts called?

A

It is called water of crystallization.

46
Q

What happens when hydrated copper(II) sulfate is heated?

A

The bonds holding the water molecules are broken, and the water is driven off, leaving behind anhydrous copper(II) sulfate (a white powder).

47
Q

What is the chemical equation for the removal of water from hydrated copper(II) sulfate?

A

CuSO4*5H2O(s) (hydrated) –> CuSO4(s) (anhydrous)+ 5H2O(l)

48
Q

What happens to the crystalline structure of copper(II) sulfate when the water is removed?

A

The crystalline structure is lost, and the compound becomes a white powder.

49
Q

Why is it difficult to remove all traces of water from hydrated copper(II) sulfate?

A

It is difficult because some water may remain trapped inside the crystals, and the anhydrous copper(II) sulfate retains a very pale blue color.

50
Q

How can you ensure that all the water has been removed from hydrated salts?

A

Heat the sample until it reaches a constant mass, meaning the mass of the residue no longer changes with further heating.

51
Q

What are two main assumptions made in experiments to determine the formula of hydrated salts?

A
  1. All the water has been lost.
  2. No further decomposition occurs.
52
Q

Why can it be challenging to determine when all the water has been removed?

A

You can only see the surface of the crystals, so some water may remain trapped inside.

53
Q

Why does color play a role in determining when water is removed?

A

If the hydrated and anhydrous forms have different colors, it is easier to judge when all the water has been removed.

54
Q

What happens if a salt decomposes further during heating?

A

Some salts, like copper(II) sulfate, can decompose further (e.g., into black copper(II) oxide) if heated too strongly.

55
Q

Why is further decomposition hard to detect?

A

If there is no significant color change, it can be difficult to judge whether decomposition has occurred.

56
Q

Why are liquids and gases measured by volume in chemistry?

A

Liquids and gases are measured by volume because it allows the amount of substance in moles to be calculated, which provides a way to count particles present.

57
Q

What are the common units for measuring volume in chemistry?

A

Cubic centimeter (cm³) or millimeter (ml): 1cm³ = 1ml
Cubic deceimeter (dm³) or litre (L): 1dm³ = 1000cm³ = 1000ml = 1L

58
Q

Which units should you use when recording volume in practical chemistry work?

A

Use centimeters cubed (cm³) and decimeters cubed (dm³).

59
Q

What is the concentration of a solution?

A

The concentration of a solution is the amount of solute (in moles) dissolved in 1 cubic decimeter (dm³) of solution. It is measured in
moldm^−3.

60
Q

What does a concentration of 1 mol dm⁻³ mean?

A

It means 1 mole of solute is dissolved in 1 dm³ (or 1000 cm³) of solution.

61
Q

What is the equation that links moles, concentration, and volume for a solution?

A

n = c x V
n = amount of substance (in moles)
c = concentration (in mol dm⁻³)
V = volume (in dm³)

62
Q

How do you adjust the n = c x V, for volumes measured in cm³?

A

Divide the volume in cm³ by 1000 to convert it to dm³, giving:
n = c x (V(cm³)÷1000)

63
Q

Why do you convert between

A

Concentration is measure in mol dm⁻³, so the volume must also be in dm³ for calculations.

64
Q

What is a standard solution?

A

A standard solution is a solution with a known concentration.

64
Q

PRACTICE QUESTION: Calculate the amount of NaCl, in mol, in 30.0cm³ of a 2.00

A

n(NaCl) = c x (V(cm³)÷1000)

n(NaCl) = 2.00 x (30.0÷1000)

n(NaCl) = 0.0600 mol

64
Q

PRACTICE QUESTION:
Calculate the volume of a 0.160 mol dm⁻³ solution that contains 2.35 x 10^⁻³ mol of NaCl.

A

n(NaCl) = c x (V(cm³)÷1000)

so, V = (1000xn)÷c

V = (1000 x 3.25 x 10^⁻³) ÷ 0.160

V = 20.3cm³

65
Q

How can you calculate the mass required to prepare a standard solution?

A

Use your understanding of the mole equation to calculate the mass needed:
m = n x M, where:
m = mass of solute (g)
n = amount of solute (mol)
M = Molar mass of the solute (g mol⁻¹)

65
Q

How are standard solutions labeled in practical work?

A

Standard solutions are often labeled with their concentration, such as 1 mol dm⁻³.

65
Q

How is a standard solution prepared?

A
  1. An exact mass of solute is dissolved in a solvent.
  2. The solution is then made up to an exact volume using a volumetric flask.
66
Q

What are the units used for mass concentrations?

A

Mass concentrations are shown in g dm⁻³.

66
Q

PRACTICE QUESTION:
Calculate the mass of Na2CO3, required to prepare 100cm³ of a 0.250mol dm⁻³ standard solution.

A

Step 1: First work out the amount in moles required.
n(Na2CO3) = c x (V(cm³)÷1000)
n(Na2CO3) = 0.250 x (100÷1000)
n(Na2CO3) = 0.0250 mol

Step 2: Then work out the molar mass of Na2CO3.
M(Na2CO3) = (23.0 x 2) + 12.0 + (16.0 x 3) = 106.0g mol⁻¹

Step 3: Rearrange n = m ÷ M, to calculate the mass of Na2CO3 required.
m = n x M
m = 0.0250 x 106.0
m = 2.65g

67
Q

How do you calculate mass concentration from molar concentration?

A

Use the formula:
m = n x M, where:
m = mass of solute (g)
n = molar concentration (mol dm⁻³)
M = Molar mass (g mol⁻¹)

67
Q

Why is gas volume easier to measure than gas mass?

A

Measuring gas mass is difficult, but gas volumes can be easily measured.

68
Q

What is the relationship between gas volume and the number of gas molecules?

A

At the same temperature and pressure, equal volumes of different gases contain the same number of molecules.

69
Q

What is molar volume?

A

Molar volume is the volume per mole of gas molecules at a stated temperature and pressure.

70
Q

What is the molar volume at room temperature and pressure (RTP)?

A

At RTP (20°C and 101 kPa or 1 atm)

71
Q

What is 1mol of gas equal to, in volume, at RTP?

A

1molofgas = 24.0dm³ = 24,000cm³.

72
Q

What is the equation to convert gas volume into moles?

A

Amount (n) is equal to the volume (V) divided by the molar gas volume.

73
Q

What is the ideal gas equation?

A

The ideal gas equation is:
pV=nRT

74
Q

What does each variable in the ideal gas equation represent?

A

p: Pressure, measured in Pascals (Pa).
V: Volume, measured in meters cubed (m³).
n: Amount of gas molecules, measured in moles (mol).
R: Ideal gas constant, always 8.314Jmol⁻¹K⁻¹.
T: Temperature, measured in Kelvin (K)

75
Q

What are the necessary unit conversions for the ideal gas equation?

A

cm³ → m³ = x10^-6
dm³ → m³ = x10^-3
°C → K = +273
kPa → Pa = x10^3

76
Q

QUICK CHECK:
Convert 25°C to Kelvin

A

T = 25 + 273 = 298K

77
Q

Why might room temperature and pressure (RTP) not always be accurate?

A
  1. RTP (20°C and 101 kPa) is approximate and convenient for experiments.
  2. At standard temperature (25°C) and atmospheric pressure, the molar gas volume is 23.5 dm³ mol⁻¹.
78
Q

What assumptions are made about an ideal gas?

A
  1. Molecules move in random motion.
  2. Collisions are elastic (no energy loss).
  3. Molecules have negligible size compared to the container.
  4. There are no intermolecular forces between molecules.
79
Q

Under what conditions do the assumptions of the ideal gas equation break down?

A
  1. High pressures: Gas molecules are close together, and their volume becomes significant compared to the container.
  2. Low temperatures: Molecules move slower, and intermolecular forces may become significant.
80
Q

What conditions favor the assumptions of the ideal gas equation?

A
  1. Low pressure: Molecules are far apart.
  2. High temperature: Molecules move quickly, reducing the effect of intermolecular forces.
81
Q

What does the term stoichiometry refer to in a balanced equation?

A

It refers to the ratio of the amounts (in moles) of each substance involved in the reaction.

82
Q
A
83
Q

In the equation 2H2(g) + O2(g) –> 2H2O(l), what is the stoichiometric ratio?

A

2 mol of H2: 1 mol of O2: 2 mol of H2O

84
Q

How do chemists use balanced equations in practice?

A
  1. To calculate the quantities of reactants required to produce a desired quantity of product.
  2. To determine the quantities of products expected from a given amount of reactants.
85
Q

What are the three steps for solving stoichiometry problems?

A
  1. Calculate the amount (in moles) of a known substance.
  2. Use the stoichiometric ratio to find the moles of the unknown substance.
  3. Calculate the required unknown information (mass, volume, etc.).
86
Q

What is theoretical yield?

A

The maximum possible amount of product from a reaction if all reactants were converted into the desired product.

87
Q

Why is the actual yield often lower than the theoretical yield?

A
  1. The reaction may not go to completion.
  2. Side reactions may occur.
  3. Loss of product during purification.
88
Q

How is percentage yield calculated?

A

Percentage yield = (Actual yield ÷ Theoretical yield) x 100%

89
Q

What is the limiting reagent in a reaction?

A

The reactant that is completely used up first and stops the reaction.

90
Q

How do you determine the limiting reagent?

A

Calculate the moles of each reactant and compare them using the stoichiometric ratio from the balanced equation.

91
Q

In the reaction 2H2(g) + O2(g) –> 2H2O(l), what is the limiting reagent if equal amounts of H2 and O2 are used?

A

H2 is the limiting reagent because 2 moles of H2 are required for every mole of O2. Half of the O2 will remain unreacted.

92
Q

What does atom economy measure?

A

Atom economy measures how efficiently reactants are converted into the desired product.

93
Q

What are the benefits of reactions with high atom economies?

A
  1. They produce a large proportion of desired products with minimal waste.
  2. They are more sustainable by making better use of natural resources.
94
Q

How is atom economy calculated?

A

Atom Economy = (Sum of molar masses of desired products ÷ Sum of molar masses of all products) x 100%

95
Q

What factors influence the sustainability of a chemical process?

A
  1. Atom economy: How efficiently raw materials are used.
  2. Percentage yield: How much of the theoretical yield is achieved.
  3. Environmental concerns: E.g., waste products like CO2 contributing to global warming.
  4. Costs: Availability and cost of starting materials and energy required for the process.

Chemistry Module 2 - Own Flashcards (10 decks)

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