Quantitative chemistry

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C7.20 Avogadro’s Law and use it to calculate volumes of gases in reactions
Avogadro's law states that equal volumes of gases measured at the same temperature and pressure contain equal numbers of molecules.
The numbers of molecules shown in a chemical equation give the ratio of volumes of reacting gases. e.g. if steam is made the equation is:
2H₂(g) +O₂(g) ---> 2H₂O(g)
This means that 2 molecules of hydrogen react with 1 molecule of oxygen to form 2 molecules of steam.
So 2dm³ of hydrogen reacts with 1 dm³ of oxygen and form 2dm³ of steam.
Or 4dm³ of hydrogen reacts with 2dm³ of oxygen and form 4dm³ of steam etc.
volume of H₂(g)/ volume of O₂(g)= molecules of H₂(g)/ molecules of O₂(g)
Write two other equations linking volumes and molecules in the above equation.
Calculate the volume of steam formed if 10cm3 of hydrogen is burned in oxygen.
volume of steam/volume of hydrogen = molecules of steam/molecules of hydrogen
volume of steam= volume of hydrogen * molecules of steam/molecules of hydrogen
volume of steam = 10cm³ * 2/2 = 10cm³
Task C7.20
Calculate the volume of steam formed if 10cm³of oxygen is used to burn hydrogen.
Calculate the volume of HCl gas formed if 2dm³ of hydrogen is burned in chlorine.
What volume of hydrogen and nitrogen is needed to make 30dm³ of ammonia by the equation
N₂ (g) + 3H₂ (g)= 2NH₃(g)
When ammonia is oxidised by oxygen what volume of NO and steam is formed by 25cm³of ammonia? 4NH₃(g) + 5O₂(g) = 4NO(g) + 6H₂O(g)

C7.21 Converting amounts into masses and vice versa

quantity	unit	symbol
mass	gram	g
amount	mole	mol
molar mass	gram per mole	g/mol

The chemical amount represents a number of particles. One mole is a special very large number called the Avogadro number. It is 6*10²³. We cannot count particles but can weigh substances. The molar mass is the mass of one mole of particles e.g. atoms, of a substance. The molar mass of an element made of atoms is the relative atomic mass of the element in grams. The molar mass of a compound of element made of molecules is the relative molecular mass or relative formula mass of that substance in grams. These 3 quantities are connected:
amount = mass/molar mass
e.g What amount of methane molecule CH₄ is there is 4g of methane?
C=12 H=1 so molar mass of methane = 12 + (4*1) = 16g/mol
Amount = mass/molar mass = 4g/16g/mol = 0.25mol
What are the molar masses of the following; He, Na, Cl, Cl₂, O₂, N, FeS, MgO, KF, HCl, H₂O, NH₃, NaOH, HNO₃, H₂SO₄, Ca(OH)₂.
What amounts are the following masses; 2g of H, 2g of He, 7g of Cl2, 11.2g of FeS, 73g of HCl, 8g of NaOH, 25g of CaCO₃.
What is the mass of: 1 mol of Li, 2 mol of C, 3 mol of S, 1 mol of O₂, 1 mol of O₃, 1 mol of NaCl, 0.1 mol of NH₃, 0.5 mol of H₂O, 0.2 mol of CaCO₃.

C7.22 Calculating the volume of a given mass of gas and vice versa
The volume of one mole of any gas is a constant known as the molar volume.
At room temperature and pressure it is 24dm³/mol.
amount of gas molecules = volume of gas/molar volume
e.g. What is the mass of 6dm³ of hydrogen H₂at room temperature and pressure if the molar volume under these conditions is 24dm³/mol?
H=1 so Molar mass of H₂ = 1*2 = 2g/mol
amount of H₂ = volume of H₂/molar volume =6dm³/24dm3/mol =0.25mol
mass of H₂ =amount of H₂ * molar mass of H₂ = 0.25mol*2g/mol = 0.5g
What is the mass of: 24dm³ of He, 12dm³ of N₂, 6dm³ of CO₂, 4dm³ of O₂, 3dm³ of F₂, 48dm³ of SO₂, 2dm³ of H₂, 1dm³ of H₂S.
What is the volume of 1g of He, 2.8g of N₂, 22g of CO₂, 64g of O₂, 19g of F₂, 64g SO₂, 0.5g H₂, 0.34g of H₂S. (assume that the molar volume is 24dm³ at room temp)

C7.23 Calculating reacting masses of substances or volumes of gases
A balanced chemical equation shows the amounts which react so masses or volumes of gases can be worked out from an equation.
E.g. What mass of aluminium oxide can be made from 216g of aluminium and what volume of oxygen is needed?
Method 1
4Al(s) + 3O₂(g) ---> 2Al₂O₃(s)
Al = 27, so relative formula mass of 4Al = 4*27 =108
O = 16 so relative formula mass of 2Al₂O₃ = 2(27*2 + 16*3) = 204
so 108g of aluminium forms 204g of aluminium oxide
so 1g of aluminium forms 204/108g of aluminium oxide
so 216g of aluminium forms 216*204/108g of aluminium oxide
so 216g of aluminium forms 408g of aluminium oxide
Method 2
4Al(s) + 3O₂(g) ---> 2Al₂O₃(s)
mass of Al = 216g
amount of Al = mass/molar mass = 216g/27g/mol = 8mol

from equation: amount Al₂O₃/amount Al =2/4
amount Al₂O₃ =amount of Al * 2/4= 8 mol *2/4 =4mol
mass Al₂O₃ = amount*molar mass =4mol*204gmol =408g
From equation amount of O₂/amount of Al =3/4
amount of O₂ =amount of Al*3/4 = 8mol *3/4mol = 6mol
molar mass of O₂ = 32g/mol
mass of O₂= amount of O2 * molar mass of O2
mass of O₂ = 6mol *32g/mol = 192g
For the reaction N₂(g) +3H₂(g) ---> 2NH₃(g)
Calculate the masses and volumes of nitrogen and hydrogen needed to make 17g of ammonia NH₃.
C7.24 Converting mass-concentration into mol dm-3 and vice versa
concentration = mass/volume
OR concentration = amount/volume
To convert them just change masses into amounts or vice versa.
e.g. What is the concentration in mol/dm³ of a solution of sodium hydroxide NaOH of concentration 4g/dm³?
amount of NaOH in 1 dm³= mass of NaOH/molar mass of NaOH
amount of NaOH in 1 dm³= 4g/40g/mol = 0.1mol
so concentration of NaOH is 0.1mol/dm³= 0.1M
NB a concentration of 1M = 1mol/dm³.
Task C7.24
1. Some dilute sulphuric acid, H₂SO₄, had a concentration of 4.90gdm^-3. What is its concentration in mol dm^-3?
2. What is the concentration in gdm^-3 of some potassium hydroxide, KOH, solution with a concentration of 0.200 mol dm^-3?
3. What mass of sodium carbonate, Na₂CO₃, would be dissolved in 100cm³ of solution in order to get a concentration of 0.100 mol dm^-3?

C7.25 Simple calculations from the results of titrations
e.g. What is the concentration of a solution of sodium hydroxide NaOH if 10cm³of the NaOH require 20cm³ of a 0.5M solution of sulphuric acid H₂SO₄ for neutralisation in a titration.
amount of H₂SO₄ = concentration of H₂SO₄ * volume of H₂SO₄
amount of H₂SO₄ = 0.5mol/dm³ *20/1000dm³ = 0.01mol
H₂SO₄+ 2NaOH ---> Na₂SO₄(aq) + 2H₂O(l)
so amount of NaOH/amount of H₂SO₄=2/1
so amount of NaOH = amount of H₂SO₄*2/1 = 0.01mol *2/1 = 0.02mol
concentration of NaOH = amount of NaOH/volume of NaOH
concentration of NaOH = 0.02mol/20/1000dm³= 1mol/dm³NB 1cm³ = 1/1000dm³Task C7.25
1. What is the concentration of hydrochloric acid, 25.0cm³ of which neutralise 20.0cm³ of sodium hydroxide solution of concentration 0.15moldm^-3. (ans = 0.15M)
2. What is the concentration of sulphuric acid, 20.0cm³ of which neutralise 30.0cm³ of a potassium hydroxide solution of concentration 0.1moldm^-3 (ans=0.1M)
3. What is the concentration of sodium hydroxide, 10.0cm³ of which neutralise 15.0cm³ of hydrochloric acid solution of concentration 2.5moldm^-3? (ans=2.5M)
4. What is the concentration of nitric acid, 10.0cm³ of which react with 25.0cm³of a solution of sodium carbonate of concentration 0.2moldm^-3? (ans = 0.2M)