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Topic 1: Stoichiometric relationships Flashcards

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1
Q

Solids

A

The particles in solids are held in fixed positions. Therefore they have a fixed shape and volume.

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2
Q

Liquid

A

The particles are able to move more freely than in a solid. Therefore liquids have a fixed volume but no fixed shape.

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3
Q

Gas

A

In a gas, the particles move freely with high speeds, with very weak forces of attraction between the particles. For these reasons, gases have neither a fixed volume or a fixed shape.

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4
Q

What are these changes of state called?

	Solid to liquid
	Liquid to gas
	Solid to gas
	Gas to liquid
	Liquid to solid
	Gas to solid
A 
Melting
Boiling/evaporation
Sublimation
Condensation
Freezing
Deposition
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5
Q

What are the state symbols for the following states?

Solid
Liquid
Gas
Aqueous
A

(s)
(l)
(g)
(aq)

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6
Q

Physical changes

A

In a physical change, no new substances are produced.
The melting of ice is an example of a physical change.
No new substances are produced.

Evaporation and sublimation are other examples of physical changes

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7
Q

Chemical changes

A

A chemical change results in the formation of new chemical substances.
In a chemical reaction, the atoms in the reactants are rearranged to form new products.

e.g

CH₄(g) + 2O₂(g) -> CO₂(g) + 2H₂O(l)

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8
Q

Elements

A

An element is a substance that cannot be broken down into a simpler substance by chemical means.

All elements can be found on the periodic table.

Elements are arranged in order of increasing atomic number.

Some elements exist as diatomic molecules.

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9
Q

Molecules

What are the diatomic molecules?

A

A molecule is an electrically neutral group of two or more atoms bonded together.

Have no fear of ice cold beer.

Hydrogen, Nitrogen, Fluorine, Oxygen, Iodine, Chlorine, Bromine.

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10
Q

Compounds

A

A compound is formed from two or more different elements chemically joined in a fixed ratio.

Compounds have different properties from the elements that they are made from.

e.g

When Na and Cl react, they form NaCl. The compound formed has very different properties than Na or Cl.

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11
Q

Mixtures

A

Mixtures contain more than one element and/or compound that are not chemically bonded together and so retain their individual properties.

Mixtures can be homogeneous or heterogenous

Homogeneous mixtures have a constant composition throughout.

Heterogenous mixtures have visible different substances or phases (e.g a mixture of oil and water)

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12
Q

Atom economy equation

A

% atom economy = (molar mass of desired product / molar mass of all reactants) x 100

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13
Q

What is a mole

A

One mole contains exactly 6.02 x 10^23 mol^-1 (this is known as the Avogadro constant)

One mole of atoms contains 6.02 x 10^23 atoms
One mole of molecules contains 6.02 x 10^23 molecules
One mole of ions contains 6.02 x 10^23 ions

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14
Q

Determine the number of chlorine molecules and chlorine atoms in 1.00 mol of chlorine gas

A

Chlorine is a diatomic molecule with a formula of Cl2. One molecule of chlorine gas consists of two chlorine atoms bonded together by a single covalent bond.

In terms of molecules, there are 6.02 x 10^23 molecules in one mole of chlorine gas.
In terms of atoms, there are (6.02 x 10^23) x 2 atoms in one mole of chlorine gas.

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15
Q

Determine the number of hydrogen atoms and oxygen atoms in 0.500 mol of water, H2O.

A

H2O consists of two hydrogen atoms and one oxygen atom.

To calculate H atoms
0.5 x 6.02 x 10^23 x 2
= 6.02 x 10^23

To calculate O atoms
0.5 6.02 x 10^23 x 1
= 3.01 x 10^23

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16
Q

Determine the number of carbon atoms, hydrogen atoms, and oxygen atoms in 0.250 mol of ethanol, C2H5OH.

A

C atoms
2 x 0.250 x 6.02 x 10^23
= 3.01 x 10^23

H atoms
6 x 0.250 x 6.02 x 10^23
= 9.03 x 10^23

O atoms
0.250 x 6.02 x 10^23
= 1.51 x 10^23

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17
Q

RAM (Ar)

A

The average mass of an atom of an element compared to 1/12th of carbon-12

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18
Q

How to calculate the RAM of an element

A

e.g

isotope abundance
24Mg 78.99%
25Mg 10.00%
26Mg 11.01%

Ar= (24x78.99)+(25x10.00)+(26x11.01)
————————————————
100

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19
Q

Relative molecular mass (Mr)

A

Relative molecular mass is the average mass of a molecule, compared to 1/12th of carbon 12.

The Mr is the sum of the Ar of the atoms in the molecule

e.g

Molecule		Atoms			Relative molecular mass
H2				2x(1.01)		2.02
H20			2x(1.01)		18.02
				1x(16.00)
C2H6			2x(12.01)		30.08
				6x(1.01)
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20
Q

Relative formula mass (Mr)

e.g question

Calculate Mr of NaCl

A

Relative formula mass is mostly used for compounds that do not form molecules, such as ionic compounds.

e.g

RAM (Na) = 22.99
RAM (Cl) = 35.45
Therefore Mr of NaCl = 58.44

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21
Q

Molar mass

What symbol is used to represent it? What are its units?

A

The mass of one mole of any substance is known as the molar mass.

It has the symbol M

It uses the units: g mol^-1

e.g

Hydrogen atoms have 1/12th of the mass of carbon-12 atoms. Therefore a mole of H atoms contains 6.02 x10^23 atoms and has a mass of 1.01 grams

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22
Q

e.g question

Determine the molar mass of ethanol C2H5OH

2 carbon atoms Ar= 12.01
1 oxygen atom Ar= 16.00
6 hydrogen atoms Ar= 1.01

A

First, calculate the Mr.

Mr = 2(12.01) + 1(16.00) + 6(1.01) = 46.08

Convert to M by multiplying by Mu (approx. 1 g mol^-1)

M = 46.08 g mol^-1

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23
Q

Calculating amount of substance

A

There is an equation triangle that represents the relationship between:

  • moles (n)
  • mass (m)
  • molar mass (Mr)

It looks like this:

	m

n Mr

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24
Q

Molar ratios

A

The coefficients in a balanced chemical equation tell us the mole ratios of reactants and products.

In the majority of chemical reactions, we have a reactant that is limiting and a reactant in excess.

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25
Q

e.g question

CaCO3(s) + 2HCl(aq) —> CaCl2(aq) + H2O(l) + CO2(g)
1 mol 2 mol 1 mol 1 mol 1 mol

Assuming that HCl is in excess, what amount of H2O can be produced from 0.667 mol of CaCO3

Assuming that CaCO3 is in excess, what amount of CO2 can be produced from 1.15 mol of HCl

A

From the equation we can see that the ratio of CaCO3 : H2O is 1:1

Therefore if we have 0.667 mol of CaCO3, we can make 0.667 mol of H2O

——————————————————————————————————

The ratio of HCl to CO2 is 2:1

Therefore 1.15 mol of HCl will produce 0.575 mol of CO2

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26
Q

Calculating % composition by mass

e.g question

Calculate the percentage composition by mass of Na, S, and O in Na2SO4

A

To calculate % composition by mass, use the formula:

% composition by mass = ( total mass of element / total mass of compound ) x 100

2(22.99) + 1(32.07) + 4(16.00) = 142.05 g mol^1

% Na by mass = (2x22.99 / 142.05) x 100 = 32.37%
% S by mass = (32.07 / 142.05) x 100 = 22.58%
% O by mass = (4x16.00 / 142.05) x 100 = 45.05%

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27
Q

Empirical formula vs molecular formula

e.g

Butane contains 4 carbons and 10 hydrogens. What is its molecular formula? What is its empirical formula?

A

The molecular formula is the actual number of atoms in a compound.

The empirical formula is the lowest whole number ratio of atoms in a compound.

e. g (butane)

Molecular formula of butane is C4H10

Empirical formula of butane is C2H5

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28
Q

The molecular formula of a compound can be determined from its empirical formula and its Mr (relative formula mass) or M (molar mass)

e.g question

A compound has the empirical formula CH2O and a Mr of 180.18. Determine its molecular formula.

A

First, find the mass of its empirical formula. CH2O = 12.01 + 2(1.01) + 16.00
= 30.03

Next, divide the relative formula mass by the mass of the empirical formula.
180.18 / 30.03
= 6

Finally, multiply the empirical formula by 6. Therefore:
CH2O —> C6H12O6

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29
Q

How to calculate an empirical formula from a % composition.

e.g question

An organic compound contains 62.0% carbon, 13.9% hydrogen, and 24.1% nitrogen by mass. Determine its empirical formula.

A

First, we must assume that we have 100g of the compound. Therefore the percentages can be converted to masses (g).

In 100g of this compound we will have:

62. 0g of carbon
13. 9g of hydrogen
24. 1g of nitrogen

The next step is to convert the masses into moles. We can do this by diving the masses of the elements by their respective molar masses.

n(C) = 62.0 / 12.01 = 5.16
n(H) = 13.9 / 1.01 = 13.8
n(N) = 24.1 / 14.01 = 1.72

The next step is to divide each amount by the smallest value
C 5.16/1.72 = 3
H 13.8/1.72 = 8
N 1.72/1.72 = 1

This gives us the empirical formula:
C3H8N

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30
Q

Water of crystallization

A

The water of crystallization is the fixed number of water molecules present in one formula unit of a salt.

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31
Q

What happens when you heat a hydrated salt?

Give an example of an equation showing this.

A

Upon heating, the hydrated salt decomposes and formes a white anhydrous salt and water vapor.

Care must be taken as over-heating of the anhydrous salt can cause further decomposition to take place.

e.g equation

CuSO4 • 5H2O(s) —> CuSO4(s) + 5H2O(g)

32
Q

Experimental procedure of determining the water of crystallization

A
  1. Measure the mass of an empty crucible and lid.
  2. Add a known mass of the sample and record the mass.
  3. Heat the crucible for five minutes, holding the lid at an angle so the gas can escape.
  4. After cooling, reweigh the crucible, lid and contents.
  5. Repeat steps 3 and 4 until the mass remains constant (known as heating to constant mass).
33
Q

How to determine the water of crystallization for hydrated salts from experimental data.

e.g BaCl2 • xH2O

(we will be finding the value of x)

Experimental Data:
Mass of crucible and lid (g) 21.50
Mass of crucible, lid, and sample (g) 26.50
Mass of crucible, lid, and sample after 1st heating (g) 25.76
Mass of crucible, lid, and sample after 2nd heating (g) 25.76

A

First we need to determine the mass of the hydrated sample. To do this, we subtract the mass of the crucible and lid from the mass of the crucible, lid, and sample.

26.50-21.50= 5.00g

Next we determine the mass of water driven off when the sample was heated. To do this, we subtract the mass of crucible, lid, and sample after heating from the mass of crucible, lid, and sample.

26.50-25.76= 0.74g

Finally we can determine the mass of the anhydrous ‘BaCl2’. To do this, we subtract the mass of water driven off when the sample was heated from the mass of the hydrated sample.

5.00-0.74= 4.26g

Now we need to convert this mass in grams to amount in moles. We do this by diving the mass of the sample by its molar mass

n(BaCl2) = 4.26 / 208.23 = 0.02 mol

Now we need to convert the amount of water driven off in moles. Divide the mass of water by its molar mass.

n(H2O) = 0.74 / 18.02 = 0.04 mol

Divide each of these values by the smallest value to give us our smallest whole number ratio.

BaCl2 0.02/0.02 = 1
H20 0.04/0.02 = 2

This gives us the formula of the hydrated salt which is:

BaCl2 • 2H2O

34
Q

What is the name of this hydrated salt?

BaCl2 • 2H2O

A

Barium chloride Dihydrate

Remember that the number in front of the H2O determines the name.

1		mono
2		di
3		tri
4		tetra
5		penta
6		hexa
7		hepta
8		octa
9		nona
10		deca
35
Q

What assumptions do we make in a water of crystallization experiment?

A
  • All mass lost is due to the loss of water
  • All water of crystallization is driven off
  • The crucible does not absorb water
  • The anhydrous salt does not decompose further
36
Q

Conducting experiments to calculate empirical formula

e.g using magnesium ribbon

A

We start the experiment by weighing 10 cm of Mg ribbon.

Next, we form the Mg ribbon into a coil (this can be done by wrapping it around a pencil).

To calculate the empirical formula, we must calculate the mass of the crucible and lid, and the mass of the Mg, crucible, and lid.

Mass of crucible and lid 40.19g
Mass of Mg, crucible, and lid 40.30g

Next, the crucible is heated with a bunsen burner. Here we can see the reaction starts to take place, with the Mg glowing as it reacts with oxygen. The lid should be opened periodically to let oxygen in for the reaction. Care must be taken not to lose too much white powder.

Once all the Mg has reacted, we are left with a white powder- which is MgO. When the crucible is cooled, the mass of the contents, crucible, and lid is recorded

Mass of contents, crucible, and lid 40.37g

37
Q

How to calculate empirical formula from experimental data

e.g using Mg ribbon. Calculate the empirical formula of MgO

Data from experiment:

Mass of crucible and lid 40.19
Mass of Mg, crucible, and lid 40.30
Mass of contents, crucible, and lid 40.37
(after heating)

A 
Data from experiment:
Mass of crucible and lid				40.19
Mass of Mg, crucible, and lid			40.30
Mass of contents, crucible, and lid	40.37
(after heating)

First, we muss calculate the mass of Mg. We can do this by subtracting the mass of crucible and lid from mass of Mg, crucible, and lid.

40.30-40.19= 0.11g

Then we must calculate the mass of O. We do this by subtracting the mass of Mg, crucible, and lid from the mass of contents, crucible, and lid.

40.37-40.30= 0.07g

Now we must convert these values from mass in grams, to amount of moles.

n(Mg) = 0.11 / 24.31 = 4.52 x 10^-3 mol
n(O)= 0.07 / 16.00 = 4.38 x 10^-3 mol

Now we divide these values by the smallest value.

Mg 4.52 x 10^-3 mol / 4.38 x 10^-3 mol
= 1.03

O 4.38 x 10^-3 mol / 4.38 x 10^-3 mol
= 1

This gives us our empirical formula:
MgO

38
Q

What errors can be made when calculating empirical formula from experimental data?

(Example using Mg ribbon to calculate the empirical formula of MgO)

A
  • The magnesium was not pure.
  • The product was not only magnesium oxide.
  • The product was lost when the lid was removed.
39
Q

Calculating empirical formula from combustion analysis

e.g question (menthol)

Menthol is an organic compound composed of C, H, and O atoms.

The complete combustion of 0.1005g of menthol produces 0.2829g of CO2 and 0.1159g of H2O.

Calculate the empirical formula of menthol.

A

First, we must calculate the mass of C in 0.2829g of CO2

To do this, we divide the molar mass of C in the molecule by the total molar mass of the CO2 molecule. Then we multiple it by the mass (g) of CO2 produced in the reaction.

( 12.01 / 44.01 ) x 0.2829 = 0.07720g of C

Next we convert this to moles.

n(C) = 0.07720 / 12.01
= 6.428 x 10^-3 mol

Next we will calculate the mass of H in 0.1159g of H2O

(2.02 / 18.02) x 0.1159 = 0.01299g of H

Convert this to moles.

n(H) = 0.01299 / 10.01
= 0.01286 mol

Next we will calculate the mass of oxygen in 0.1005g of menthol. To do this, we subtract the mass of carbon and hydrogen from the original mass of menthol.

0.1005-0.07720-0.01299 = 0.01031g of O

Convert this to moles

n(O) = 0.01031 / 16.00
= 6.444 x 10^-4 mol

Divide these values by the smallest value to get our lowest whole number ratio.

This will give us a ratio of 10:20:1, meaning that the empirical formula is:

C10H20O

40
Q

Limiting and excess reactants

A

The limiting reactant is the reactant that limits the amount of product that can be made.

The excess reactant is the reactant that remains when the limiting reactant has been consumed.

41
Q

Determining the limiting and excess reactants

e. g question
50. 0g of N2H4 is reacted with 75.0g of N2O4. Determine the limiting and excess reactants.

2N2H4(l) + N2O4(l) —> 3N2(g) + 4H2O(g)

A
  1. We must calculate the amount (in mol) of each reactant.
  2. Divide the amount of each reactant by its coefficient in the balanced equation.
  3. The lowest value is the limiting reactant.

e.g question

First we must determine the amount in moles of N2H4 and N2O4

n(N2H4) = 50.0g / 32.06g mol^-1
= 1.56 mol

n(N2O4) = 75.0g / 92.02g mol^-1
= 0.815 mol

Now we need to divide the amounts in moles by the coefficients

N2H4 1.56 / 2 = 0.780
N2O4 0.815 / 1 = 0.815

0.780<0.815 therefore N2H4 is the limiting reactant and N2O4 is the excess reactant.

42
Q

How to determine the amount of excess reactant that remains at the end of the reaction

e.g question

2N2H4(l) + N2O4(l) —> 3N2(g) + 4H2O(g)

n(N2H4) = 1.56 mol

n(N2O4) = 0.815 mol

Limiting reactant = N2H4
Excess reactant = N2O4

A

First we must look at the molar ratio of N2H4 : N2O4

It is 2:1

We use this ratio to determine the amount of N2O4 left at the end of the reaction

1.56 mol N2H4 x 1/2 = 0.780 mol N2O4

This is the amount of N2O4 that reacts completely with 1.56 mol of the limiting reactant (N2H4)

The final step is to subtract this number from the original amount of moles (found in the previous step)

n(N2O4) remaining = 0.815 - 0.780
= 0.0350 mol

43
Q

What is the formula that links moles, concentration, and volume?

What are the units?

A

n

c V

n mol
c mol dm^-3
V dm^3

44
Q

Determining limiting and excess reactants

e. g question (using concentration and volume)
3. 00g of Zn is reacted with 50.0cm^3 of 1.00 mol dm^-3 HCl. Determine the limiting and excess reactants,

Zn(s) + 2HCl(aq) —> ZnCl2(aq) + H2(aq)

A

First we need to determine the amount in moles of zinc. We can do this by diving the mass (g) by its molar mass.

n(Zn) = 3.00 / 65.38
= 0.0459 mol

To determine the amount in moles of HCl, we need to use the equation n=cV.

n = (50/1000) dm^3 x 1.00 mol dm^-3
= 0.0500 mol

The next step is to divide the amounts in moles by their coefficients (from the equation).

Zn 0.0459/1 = 0.0459
HCl 0.0500/2 = 0.0250

HCl is a smaller value and is therefore the limiting reactant.
Zn is the excess reactant.

45
Q

How do you convert cm^3 to dm^3

How do you convert dm^3 to cm^3

A

cm^3 —> dm^3 = divide by 1000

dm^3 —> cm^3 = multiply by 1000

46
Q

Theoretical yield, experimental yield, and percentage yield

A

The theoretical yield is the maximum amount of product that can be formed in a chemical reaction

The theoretical yield is calculated based on the stoichiometry of the reaction and the amount of limiting reactant.

The experimental yield (or actual) is the actual amount of product that is formed in a chemical reaction. This will almost always be less than the theoretical yield.

Percent yield = (actual yield / theoretical yield) x 100

47
Q

e. g question - Theoretical yield, experimental yield, and percentage yield
100. 0g of iron(II) oxide is reacted with 100.0g of carbon. 46.73g of iron is produced. Calculate the % yield of Fe.

2Fe2O3(s) + 3C(s) —> 4Fe(s) + CO2(g)

A

The first step is to determine which reactant is the limiting reactant. To do this, we must convert from mass in g to amount in moles.

n(Fe2O3) = 100/159.70 = 0.6262 mol
n(C) = 100/12.01 = 8.326 mol

Next we divide these values by their respective coefficients in the equation.

Fe2O3 0.6262/2 = 0.3131
C 8.326/3= 2.775

Iron oxide is the limiting reactant and carbon is the excess reactant.

Next we need to calculate the theoretical yield of iron. To do this, we need to know the molar ratio of the limiting reactant (iron oxide) to the product in question (iron)

The molar ratio of Fe2O3 : Fe = 2:4

The next step is to multiple the amount in moles of Fe2O3 by 4/2

0.6262 x 4/2 = 1.252 mol

Next we must convert this amount in moles to mass in grams

m(Fe) = 1.252 x 55.85 = 69.92g

This value is the theoretical yield of the experiment.

Percent yield = (actual yield / theoretical yield) x 100

(46.73/69.92) x 100
= 66.83%

48
Q

Percentage purity

e.g question

A 150.0g sample of copper ore contains 87.3g of pure copper. Calculate the percentage purity.

A

Percentage purity is the percentage of a pure compound in an impure sample.

% purity = (mass of pure compound in sample / total mass of impure sample) x100

e. g question
(87. 3 / 150.0) x 100 = 58.2%

49
Q

Molar volume of a gas

A

The molar volume of a gas (Vm) is the volume occupied by one mole of an idea gas. The molar volume must be stated at specific conditions.

Under STP conditions, one mole of any gas occupies a volume of 0.0227m^3 or 22.7dm^3

50
Q

STP conditions

A

Temperature = 273 K

Pressure = 1.00 x 10^5 Pa

51
Q

e.g question 1 - molar volume of a gas

Calculate the volume in dm^3 occupied by 0.250 mol of N2 at STP

A

V = n x 22.7

V = 0.250 x 22.7 = 5.68 dm^3

Remember to check for units. Multiplying by 22.7 will always give an answer in dm^3

52
Q

e.g question 2 - molar volume of a gas

Calculate the volume in cm^3 occupied by 0.00619 mol of CO2 at STP.

A

V = n x 22.7

V = 0.00619 x 22.7 = 0.141 dm^3

This question asks for the answer in cm^3 so convert the value.

0.141 x1000 = 141cm^3

53
Q

e.g question 3 - molar volume of a gas

Calculate the amount in moles of N2 in a 0.742 dm^3 sample.

A

n = V/22.7

n = 0.742/22.7

n= 0.0327 mol

54
Q

e.g question 4 - molar volume of a gas

Determine the volume of H2 in cm^3 produced at STP when 2.00g of Mg is reacted with excess HCl(aq).

Mg(s) + 2HCl(aq) —> MgCl2(aq) + H2(g)

A

The first step is to determine the amount in moles of magnesium

n(Mg) = 2.00/24.31 = 0.0823 mol

Mg is the limiting reactant as the question tells us that HCl is the excess reactant.

Next we must determine the molar ratio of Mg:H = 1:1

Meaning that 0.0823 mol of Mg will produce 0.0824 mol of H2

n(H2) = 0.0823 mol

V = n x 22.7

V = 0.0823 x 22.7 = 1.87dm^3

The question wants the answer in cm^3 so convert this value accordingly.

1.87 x1000 = 1870cm^3

55
Q

Gas laws: Boyle’s law

A

The volume occupied by a gas is inversely proportional to its pressure

In equation form, this law can be represented as:

  1. PV = k
  2. P ∝ 1 / V
  3. P1 V1 = P2 V2
56
Q

Gas laws: Charles’s law

A

The volume occupied by a gas is directly proportional to its absolute temperature.

In equation form, this law can be represented as:

  1. V ∝ T
  2. V / T = k
  3. V1/T1 = V2/T2
57
Q

Gas laws: Gay-Lussac’s law

A

The pressure exerted by a gas is directly proportional to its absolute temperature.

In equation form, this law can be represented as:

  1. P ∝ T
  2. P/T = k
  3. P1/T1 = P2/T2
58
Q

Gas laws: Avogadro’s law

A

The volume occupied by a gas is directly proportional to the amount (in mol) of gas

In equation form, this law can be represented as:

  1. V ∝ n
  2. V/n = k
  3. V1/n1 = V2/n2
59
Q

Gas laws: The combined gas law

A

The combined gas law combines Boyle’s law, Charles’s law and Gay-Lussac’s law

(P1 x V1) (P2 x V2) (P1 x V1)
——— = ——— ——— = k
T1 T2 T1

60
Q

Ideal gas equation

A

PV = nRT

P is pressure (Pa)
V is volume (m^3)
n is amount (mol)
R is the gas constant (8.31 J K-1 mol -1)
T is temperature (K)

The equation can be rearranged to solve for P, V, n, and T

61
Q

e.g question 1 - ideal gas equation

Calculate the volume in dm^3 occupied by 0.500 mol of gas at 1.50x10^5 Pa and 25.0 celsius.

A

We will be calculating for volume so the equation will be rearranged to

V = (nRT) / P

We need to make sure that all of the data we have for the question is in the right units. For example, the question has given us the temperature in celsius, but we need it in kelvin. It should be converted.

V = (0.500 x 8.31 x 298) /150000
=8.25 x 10^-3 m^3
= 8.25 dm^3

62
Q

How to convert a value from degrees celsius —> kelvin?

A

Add 273 to the value

63
Q

e.g question 2 - ideal gas equation

Calculate the pressure (in kPa) of 0.200 mol of gas that occupies a volume of 10.0 dm^3 at 20.0 celsius.

A

We are calculating for pressure so the equation will be rearranged to

P= (nRT) / V

P= (0.200 x 8.31 x 293) / 0.0100

P= 4.87 x 10^4 Pa

Convert this value to kPa by dividing by 1000

P= 48.7 kPa

64
Q

How to convert a value from dm³ —> m³

How to convert a value from m³ —> dm³

A

Divide the value by 1000

Multiply the value by 1000

65
Q

e.g question 3 - ideal gas equation

Calculate the amount (in mol) of gas that occupies a volume of 20.0dm^3 at 50.0 degrees celsius and 85.0 kPa

A

We are calculating for moles. Rearrange the formula to give

n= PV/RT

n= (85000 x 0.0200) / (8.31 x 323)

n= 0.633 mol

66
Q

Deviation from ideal gas behaviour

What is the name of a type of gas that deviates from ideal gas behaviour? What characterize them?

A

An ideal gas is a hypothetical gas that obeys the gas laws and the kinetic-molecular theory.

A real gas is a gas that deviates from ideal gas behaviour. They have finite, measurable volume. They also have intermolecular forces that act between particles. They exhibit nearly ideal behaviour at relatively high temperatures and low pressures. They deviate the most from ideal behaviour at low temperatures and high pressures.

67
Q

What is the kinetic-molecular theory?

A

The kinetic molecular theory states the following:

  • Particles of an ideal gas are in constant, random, straight-line motion
  • Collisions between particles of an ideal gas are elastic- meaning that total kinetic energy is conserved
  • The volume occupied by the particles of an ideal gas is negligible relative to the volume of the container
  • There are no intermolecular forces acting between particles of an ideal gas
  • The average kinetic energy of the particles of an ideal gas is directly proportional to the absolute temperature in kelvin
68
Q

Calculating molar mass of a gas

A

M = (mRT) / (PV)

69
Q

Determining the molar mass of a gas experimentally

A

Method:

To collect the gas, a measuring cylinder is inverted and filled with water, whilst also submerged in a water bath.
The measuring cylinder is filled with a known volume of gas from a cigarette lighter.
The mass of the lighter is recorded before and after the measuring cylinder is filled.

Calculations:

To calculate the molar mass of a gas (e.g butane, C4H10) we need to know the change in mass of the lighter, the volume of gas collected, and the pressure and temperature.

We will then use the equation:

M = mRT / PV

Change in mass of lighter = 0.230 grams

Volume of gas collected = 100.0cm^3 (1x10^-4 m^3)

THIS MUST BE IN METERS TO USE IT IN THE EQUATION
Temperature = 22.0C = 295K

TEMP. MUST BE IN KELVIN
Atmospheric pressure =100717 Pa

Vapour pressure of water at 22.0C = 2634 Pa

To find the pressure of the gas in the lighter we must subtract the vapour pressure of water from the atmospheric pressure (answer is 98.083 Pa)

R=8.314 J K^-1 mol^-1

We then insert those values into the equation

M= 0.230 x 8.314 x 295 / 98083 x 1.000 x10^-4

M= 57.5 g mol^-1

70
Q

When experimentally calculating the molar mass of the gas- why is the measured value LOWER than the actual value?

A

The lighter was not dried completely

The pressure of gas was not equalized with the pressure of the room

The lighter may contain a mixture of propane and butane

71
Q

When experimentally calculating the molar mass of the gas- why is the measured value HIGHER than the actual value?

A

Some bibles of gas escaped from the cylinder

72
Q

Calculating concentration of solutions

A

The concentration of a solution can be measured in g dm^-3, mol dm^-3, and ppm

c(g dm^-3) = mass of solute (g) / volume of solution (dm^3)

c(mol dm^-3) = amount of solute (mol) / volume of solution (dm^3)

ppm = ( mass of solute (g) / mass of solution (g) ) x 10^6

73
Q

Determining an unknown concentration by titration

A

Titration is a useful technique for finding the concentration of a solution. It involved reacting the solution with a stoic biometric amount of a standard solution.

A known accurate volume of one of the solutions is placed in a conical flask using a pipette. A burette is the used to add the other solution dropwise until the reaction is complete. This can be seen when one drop causes the solution to change colour.

For acid-base titrations it is usually to add an indicator using acidified potassium permanganate, as the reactant itself causes the colour change

74
Q

How to use a burette

A

Use a beaker and funnel to fill the burette.

Leave an air gap when filling.

The burette reading is taken from the bottom of the meniscus.

Use the left hand to control the flow rate. Swirl the flask with your right hand whilst the drops are being added.

75
Q

Determining concentration by titration

e. g
25. 00cm^3 of a solution of sodium hydroxide of an unknown concentration required 23.65cm^3 of 0.100mol dm^-3 hydrochloric acid solution for complete neutralization. Calculate the concentration of the NaOH solution

A

First calculate the amount of HCl present in 23.65cm^3

23.65cm^3 = (23.65/1000) x 0.100 = 2.365x10^-3 mol

Since one mol of NaOH reacts with one mol of HCl, the amount of NaOH present in 25.00cm^3 = 2.365x10^-3 mol

Concentration of NaOH = 2.365x10^-3 mol x (1000/25.00) = 0.0946 mol dm^-3

SL Chem (11 decks)

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