Thursday, November 27, 2008

Quantitative chemistry( chemical calculation)-IGCSE/GCSE

Avogadro’s Law and use it to

calculate volumes of gases in reactionsAvogadro'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 hydrogenvolume of steam= volume of hydrogen * molecules of steam/molecules of hydrogenvolume of steam = 10cm3 * 2/2 = 10cm3

Calculate the volume of steam formed if 10cm3 of 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 equationN2 (g) + 3H2 (g)= 2NH3(g)

When ammonia is oxidised by oxygen what volume of NO and steam is formed by 25cm3 of ammonia? 4NH3(g) + 5O2(g) = 4NO(g) + 6H2O(g)

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 masse.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/molAmount = 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 versaThe 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 volumee.g.

What is the mass of 6dm3 of hydrogen H2 at room temperature and pressure if the molar volume under these conditions is 24dm3/mol?

H=1 so Molar mass of H2 = 1*2 = 2g/molamount of H2 = volume of H2/molar volume =6dm3/24dm3/mol =0.25molmass 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)

Calculating reacting masses of substances or volumes of gasesA 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 =108O = 16 so relative formula mass of 2Al2O3 = 2(27*2 + 16*3) = 204so 108g of aluminium forms 204g of aluminium oxideso 1g of aluminium forms 204/108g of aluminium oxideso 216g of aluminium forms 216*204/108g of aluminium oxideso 216g of aluminium forms 408g of aluminium oxide

Method 2

4Al(s) + 3O2(g) ---> 2Al2O3(s)mass of Al = 216gamount of Al = mass/molar mass = 216g/27g/mol = 8molfrom equation: amount Al2O3/amount Al =2/4amount Al2O3 =amount of Al * 2/4= 8 mol *2/4 =4molmass Al2O3 = amount*molar mass =4mol*204gmol =408gFrom equation amount of O2/amount of Al =3/4amount of O2 =amount of Al*3/4 = 8mol *3/4mol = 6molmolar mass of O2 = 32g/molmass of O2 = amount of O2 * molar mass of O2mass of O2 = 6mol *32g/mol = 192gFor the reaction N2(g) +3H2(g) ---> 2NH3(g)

Calculate the masses and volumes of nitrogen and hydrogen needed to make 17g of ammonia NH3.

Converting mass-concentration into mol dm-3 and vice versaconcentration = 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 NaOHamount of NaOH in 1 dm3 = 4g/40g/mol = 0.1mol so concentration of NaOH is 0.1mol/dm3 = 0.1MNB a concentration of 1M = 1mol/dm3.

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?

Simple calculations from the results of titrationse.g.

What is the concentration of a solution of sodium hydroxide NaOH if 10cm3 of 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 H2SO4amount of H2SO4 = 0.5mol/dm3 *20/1000dm3 = 0.01molH2SO4 + 2NaOH ---> Na2SO4(aq) + 2H2O(l)so amount of NaOH/amount of H2SO4=2/1so amount of NaOH = amount of H2SO4*2/1 = 0.01mol *2/1 = 0.02molconcentration of NaOH = amount of NaOH/volume of NaOH concentration of NaOH = 0.02mol/20/1000dm3 = 1mol/dm3NB 1cm3 = 1/1000dm3

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 25.0cm3 of a solution of sodium carbonate of concentration 0.2moldm-3? (ans = 0.2M)

No comments:

chemistry notes / igcse-gcse- Olevel