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The molar gas volume in calculations, moles, gas volumes and Avogadro's Lawstudy examples carefully

9. The molar gas volume in calculations, moles, gas volumes and Avogadro's Law

• Avogadro's Law states that equal volumes of gases under the same conditions of temperature and pressure contain the same number of molecules.
  • So the volumes have equal moles of separate particles (molecules or individual atoms) in them.
  • Therefore one mole of any gas (formula mass in g), at the same temperature and pressure occupies the same volume .
  • This is 24dm³ (24 litres) or 24000 cm³, at room temperature of 25^oC/298K and normal pressure of 101.3 kPa/1 atmosphere (such conditions are often referred to as RTP).
  • The molar volume for s.t.p is 22.4 dm³ (22.4 litres) at 0^oC and 1atmosphere pressure.
  • Historically, s.t.p unfortunately stands for standard temperature and pressure, but these days 25^oC/298K is usually considered the standard temperature (RTP).
• Some handy relationships for substance Z below:

moles Z = mass of Z gas (g) / atomic or formula mass of gas Z (g/mol)

• mass of Z in g = moles of Z x atomic or formula mass of Z
• 
  
• atomic or formula mass of Z = mass of Z / moles of Z
• 1 mole = formula mass of Z  in g.

gas volume of Z = moles of Z x volume of 1 mole

• rearranging this equation gives ...
• moles of Z = gas volume of Z / volume of 1 mole

• moles = V(dm³) / 24   (at RTP)

The latter form of the equation can be used to calculate molecular mass from experimental data because

• moles = mass / molecular mass = gas volume / volume of 1 mole
• mass / molecular mass = gas volume / volume of 1 mole
• molecular mass = mass x volume of 1 mole/volume
• therefore at RTP: M_r = mass(g) x 24 / V(dm³)
• so, if you know the mass of a gas and its volume, you can work out moles of gas and then work out molecular mass.
• This has been done experimentally in the past, but these days, molecular mass is readily done very accurately in a mass spectrometer.

Note (i): In the following examples, assume you are dealing with room temperature and pressure i.e. 25^oC and 1 atmosphere pressure so the molar volume is 24dm³ or 24000cm³.

Note (ii):

• Apart from solving the problems using the mole concept (method (a) below, and reading any equations involved in a 'molar way' ...
• It is also possible to solve them without using the mole concept (method (b) below). You still use the molar volume itself, but you think of it as the volume occupied by the formula mass of the gas in g and never think about moles!

Methods of measuring how much gas is formed (volume can be compared with theoretical prediction!)

• (a) You can collect the gases in a calibrated gas syringe.
  

  • You must make sure too much gas isn't produced and too fast!
  • A gas syringe is more accurate than collecting the gas in an inverted measuring cylinder under water shown below, but its still only accurate to the nearest cm³.
  • You can collect any gas by this method.
• (b) The gas is collected in a measuring cylinder filled with water and inverted over a trough of water.Methods of measuring how much gas is formed

(volume can be compared with theoretical prediction!)

You can get a more accurate result by using an inverted burette instead of a measuring cylinder.
• However, this method is no good if the gas is soluble in water!
• Burettes are calibrated in 0.10 cm³ intervals. measuring cylinders to the nearest cm³ or worse!
• In both methods the reaction is carried out in conical flask fitted with a sealing rubber bung, but a tube enabling the gas evolved to be collected in some suitable container.
• (c) A third method is to measure the gas loss by carrying out the reaction in a flask set up on an accurate one-pan electronic balance.
• You need to put a cotton wool plug in the neck of the conical flask in case you lose any of the solution in a spray as the gas bubbles up - effervescence can produce an aerosol.
• This method can be used for any reaction that produces a gas, but the gas is released into the laboratory, ok if its harmless.
• It is potentially the most accurate method, BUT, the mass loss may be quite small especially hydrogen [M_r(H₂) = 2], better for the 'heavier' gas carbon dioxide [M_rCO₂) = 44]
• Molar

Introduction to MOLES

Introducing 'moles' and their the connection with mass and formula mass

(a) WHAT IS THE MOLE CONCEPT? and WHAT IS ONE MOLE OF A SUBSTANCE?

The 2nd part of the heading is 'easy', the first part is a bit more 'abstract' to get your head round!

The mole concept is an invaluable way of solving many quantitative problems in chemistry!

Its a very important way of doing chemical calculations!

The theoretical basis is explained in section (b).

The mole is most simply expressed as the relative 'formula mass in g' or the 'molecular mass in g' of the defined chemical 'species', and that is how it is used in most chemical calculations. The mass of one mole of a substance is sometimes referred to as the molar mass.

The atomic/formula mass in grams = one mole of the defined substance.

If your are dealing with individual atoms, one mole of equals the relative atomic mass in grams.

This can be expressed as a simple formula ...

moles of species = (actual mass of species in g) / (atomic/formula mass of species)

therefore (using triangle on right if necessary)

 mass of species in g = moles species x atomic/formula mass of species

atomic/formula mass of species = mass of species in g / moles of species

Note these equations are for either an element or a compound,

but, whatever, you must clearly define the chemical species you mean for any mole calculation e.g.

Al metal element atom, H₂O covalent molecule, an element O₂ molecule, Na⁺Cl^- ionic compound or just any compound formula like CuSO₄ etc. etc.

This specificity cannot be overemphasised.

 

M_r is 'shorthand' for relative formula mass or molecular mass in amu (atomic mass units) and you must be able to work these out correctly from a given formula (Calculating relative formula/molecular mass of a compound or element molecule).

The term relative molecular mass (sum of the atomic masses of the atoms in a single molecule of the substance) is usually applied to definite molecular species.

Using the following atomic masses: H = 1, O = 16, N = 14, C = 12, Na = 23, Cl = 35.5, S = 32

and the three formulae above relating moles, mass and formula mass ...

molecular mass 18 for the water molecule H₂O, 17 for the ammonia molecule NH₃

so 1 mole of water is 18g, 0.333 mole = 0.333 x 18 = 6g

for ammonia 1mol = 17g, 34g = 34/17 = 2 mols ammonia

 

16 for the methane molecule CH₄ and 180 for the glucose sugar molecule C₆H₁₂O₆

so 0.5 mol methane = 0.5 x 16 8g, 72g = 72/16 = 4.5 mols methane

for glucose 18g = 18/180 = 0.10 moles, 0.05 mole = 0.05 x 180 = 9.0g glucose

the element nitrogen consists of N₂ molecules, molar mass = 28g, 0.25 moles = 0.25 x 28 = 7.0g

Relative atomic mass of iron is 56, 7g = 7/56 = 0.125 mol Fe (Relative atomic mass explained)

So, these calculations are quite simple, but they are often just one part of solving a more complex problem.

 

The term relative formula mass (sum of the atomic masses of the atoms in a specified formula) can be used for ANY specified formula of ANY chemical substance, though it is most often applied to ionic substances.

e.g. mass of 1 mole of ionic sodium chloride NaCl or Na⁺Cl^- is 58.5g (from 23 + 35.5)

mass of 1 mole of ammonium sulfate (ionic salt) (NH₄)₂SO₄ or (NH₄⁺)₂(SO₄^2-) =  130g

 

Empirical formula and formula mass from reacting masses (easy start, no moles!)

 Empirical formula and formula mass from reacting masses (easy start, no moles!)

The EMPIRICAL FORMULA of a compound can be worked out by knowing the exact masses of the elements that combine to form a given mass of a compound.
The empirical formula of a compound is the simplest whole number ratio of atoms present in a compound.


 Here the word 'empirical' means from experimental data.

Do not confuse with molecular formula which depicts the actual total numbers of each atom in a molecule.
The molecular formula and empirical formula can be different or the same.
They are the same if the molecular formula cannot be simplified on a whole number basis.

Examples where molecular formula = empirical formula

e.g. for sodium sulfate Na₂SO₄  and  propane C₃H₈
You cannot simplify the atomic ratios 2 : 1 : 4 or 3 : 8 to smaller whole number (integer) ratios

Examples of where molecular formula and empirical formula are different e.g.

butane molecular formula C₄H₁₀, empirical formula C₂H₅

numerically, the empirical formula of butane is 'half' of its molecular formula

4 : 10 ==> 2 : 5

glucose molecular formula C₆H₁₂O₆, empirical formula CH₂O

numerically, the empirical formula of glucose is '¹/₆th' of the full molecular formula

6 : 12 : 6 ==> 1 : 2 : 1

The following examples illustrate the ideas using numbers more easily appreciated than in real experiments.

In real laboratory experiments only a fraction of a gram or a few grams of elements would be used, and a more 'tricky' mole calculation method is required than shown here

However the examples below show in principal how formulae are worked out from experiments.

Any calculation method must take into account the different relative atomic masses of the elements in order to get to the actual ratio of the atoms in the formula.

For example, just because 10g of X combines with 20g of Y, it does not mean that the formula of the compound is XY₂ !

If you divide the mass of each element by its atomic mass, you actually get the atomic ratio.

• Empirical formula calculation Example 5.1 The compound formed between lead and sulfur
• It is found that 207g of lead combined with 32g of sulphur to form 239g of lead sulphide.
• From the data work out the formula of lead sulphide. (Relative atomic masses: Pb = 207 and S = 32)
• In this case it easy to see that by the atomic mass ratio, 239 splits on a 1 to 1 basis of 1 atom of lead to 1 atom of sulphur (1 x 207 to 1 x 32 by mass)
• so the formula is simply PbS
• You can set out the calculation in a simple table format, in this case the numbers are very easy to deal with!
 

RATIOS ...	lead (Ar = 207)	sulphur S (Ar = 32)	Comments and tips
Reacting mass	207g	32g	not the real atom ratio
atom ratio from mass / atomic mass values	207/207 = 1	32/32 = 1	work out the simplest whole number ratio
simplest whole number atom ratio by trial & error	1	1
therefore the integer simplest ratio of 1 : 1 gives the empirical formula for lead sulphide as PbS

• -
• Empirical formula calculation Example 5.2 The empirical formula of a lead oxide
• It is found that 207g of lead combined with oxygen to form 239g of a lead oxide.
• From the data work out the formula of the lead oxide. (Relative atomic masses: Pb = 207 and O = 16)
• In this case, you first have to work out the amount of oxygen combined with the lead.
• By simple logic from the law of conservation of mass, this is 239 - 207 = 32g
• In atomic ratio terms, the 207 is equivalent to 1 atom of lead and the 32 is equivalent to 2 atoms of oxygen (1 x 207 to 2 x 16),
• so the formula is simply PbO₂
• Note: The mass of oxygen combined with the lead is deduced by subtracting the original mass of lead from final total mass of lead oxide.

 

RATIOS ...	lead (Ar = 207)	oxygen O (Ar = 16)	Comments and tips
Reacting mass	207g	239-207 = 32g	not the real atom ratio
atom ratio from mass / atomic mass values	207/207 = 1	32/16 = 2	work out the simplest whole number ratio
simplest whole number atom ratio by trial & error	1	2
therefore the simplest whole number ratio of 1 : 2 gives the empirical formula for this lead oxide as PbO2 Its actually called lead(IV) oxide

Credit to Docbrown

 

The Law of Conservation of Mass

 

What is the Law of Conservation of Mass?

 

When elements and compounds react to form new products, mass cannot be lost or gained.

"The Law of Conservation of Mass" definition states that mass cannot be created or destroyed, but changed into different forms.

So, in a chemical change, the total mass of reactants must equal the total mass of products.

By using this law, together with atomic and formula masses, you can calculate the quantities of reactants and products involved in a reaction and the simplest formula of a compound

... before tackling the first calculations based on the Law of Conservation of mass, its worth describing a simple experiment to demonstrate the validity of the law. The experiment is illustrated in the diagram above. You prepare solutions of copper sulfate (blue) and sodium hydroxide (colourless, light grey in diagram!). The most impressive way to demonstrate this is to use a sealed system on an accurate electronic one pan balance. You can use 50 cm³ of 1 molar copper sulfate solution and pour into conical flask. The concentrated sodium hydroxide solution is suspended by a string in a suitable container - small test tube or weighing/sample bottle. The whole lot is weighed (fictitiously 67.25g) with the rubber bung on sealing the 'system'. Then, releasing the bung and string, the sodium hydroxide container is lowered into the copper sulfate solution and shaken gently to thoroughly mix the reactants. The reaction is immediate and a dark blue precipitate of copper hydroxide is formed and the solution eventually turns colourless because only colourless sodium sulfate is left in solution. The recorded mass will still be 67.25g showing that no mass was created or destroyed in the chemical reaction, though to observe the law in action, you must do the experiment in a sealed system where nothing can get in or get out i.e. no atoms have been gained or lost.

The equation for this reaction is ...

copper sulfate + sodium hydroxide ==> copper hydroxide + sodium sulfate

CuSO₄ +  2NaOH  ===>  Cu(OH)₂  +  Na₂SO₄

Teacher note

50 cm³ of 1 molar copper sulfate = 1.0 x 50 / 1000 = 0.05 mol CuSO₄, M_r(NaOH) = 40, you need 2x 0.05 = 0.10 mol NaOH,

which equals 0.10 x 40 = 4.0g NaOH pellets dissolved in the minimum volume of water, 4.1g should complete the precipitation.

 

NOTE that in calculations ...

(1) the symbol equation must be correctly balanced to get the right answer!

(2) You convert all the formula in the equations into their formula masses AND take into account any balancing numbers to get the true theoretical reactiing masses.

(2) There are good reasons why, when doing a real chemical preparation-reaction to make a substance you will not get 100% of what you theoretically calculate.

• Law of conservation of mass calculation Example 3.1
• Magnesium + Oxygen ==> Magnesium oxide
• 2Mg + O₂ ==> 2MgO (atomic masses required: Mg=24 and O=16)
• think of the ==> as an = sign, so the mass changes in the reaction are:
• (2 x 24) + (2 x 16) = 2 x (24 + 16)
• 48 + 32 = 2 x 40 and so 80 mass units of reactants = produces 80 mass units of products.
• You can work with any mass units such as g, kg or tonne (1 tonne = 1000 kg), as long as you use the same units for all the masses involved.
• Law of conservation of mass calculation Example 3.2
• iron + sulphur ==> iron sulphide (see the diagram at the top of the page!)
• Fe + S ==> FeS (atomic masses: Fe = 56, S = 32)
• If 59g of iron is heated with 32g of sulphur to form iron sulphide, how much iron is left unreacted? (assuming all the sulphur reacted)
• From the atomic masses, 56g of Fe combines with 32g of S to give 88g FeS.
• This means 59 - 56 = 3g Fe unreacted.
• Law of conservation of mass calculation Example 3.3
• When limestone (calcium carbonate) is strongly heated, it undergoes thermal decomposition to form lime (calcium oxide) and carbon dioxide gas.
• CaCO₃ ==> CaO + CO₂ (relative atomic masses: Ca = 40, C = 12 and O = 16)
• Calculate the mass of calcium oxide and the mass of carbon dioxide formed by decomposing 50 tonnes of calcium carbonate.
• (40 + 12 + 3x16) ==> (40 + 16) + (12 + 2x16)
• 100 ==> 56 + 44
• scaling down by a factor of two
• gives
• 50 ==> 28 + 22
• so decomposing 50 tonnes of limestone produces 28 tonnes of lime and 22 tonnes of carbon dioxide gas.
• Example 3.4:
• For more complicated examples and more practice of calculations based on reacting masses in accordance with the Law of Conservation of Mass ...

 

credit to : doc brown chemistry as level