These equations only include the only ions that take part in the chemical reaction.
This equation will be broken down into ions.
After removing spectator ions from the equation, the ionic equation would come up to:
Different compounds may have the same empirical formula.
Benzene has carbon and hydrogen in the composition 92.3% carbon and 7.7% hydrogen. The relative molecular mass of benzene is 78.
By using this information, we can find both empirical and molecular formula of benzene.
Divide these two numbers by the smallest one. (7.7/7.7) for both.
The ratio for these two elements in the compound is 1:1. So the empirical formula is CH for benzene. The relative mass for CH is 13. Divide the relative molecular mass of benzene by the empirical mass. (78/13) à 6.
One mole of a substance has a mass equal to its relative atomic mass, relative molecular mass or relative formula mass as appropriate, expressed in grams. For example,
12 g of magnesium would have 0.5 mol as the relative atomic mass of magnesium is 24. (12g/24 à 0.5 mol)
2 g of H2O would have 0.11 mol as the relative atomic mass is 18. (2g/18 à 0.11 mol)
Reacting Mass And Volume
CH4(g) + 2O2(g) à CO2(g) + 2H2O(l)
1 molecule of methane and two molecules of oxygen react to form one molecule of carbon dioxide and 2 molecules of water. These molecules can also be referred as moles and hence in terms of grams as per the relative atomic mass. It would mean that 16g of methane react with 64g of oxygen to form 44g of carbon dioxide and 36g of water. This proportion would remain the same and would produce predictable results.
Finding The Emperical Formula Of A Compound
Magnesium oxide exists in different formulas such as MgO, Mg2O and MgO2
0.30g of magnesium burns in air to form 0.50g of magnesium oxide.
Ratio of the two is 1:1.67
One mole of magnesium is 24 and so by multiplying by the ratio (0.50/0.30 * 24) = 40g
24 g of magnesium in 40g Magnesium oxide, so remaining 16g for oxygen, therefore, one molecule of oxygen is there by relative atomic mass of oxygen (16)
16 + 24 = 40.
Predicting The Mass Of Products Of A Reaction
An equation for zinc nitrate is given that decomposes into zinc oxide, oxygen and nitrate on heating.
2Zn(NO3)2 (s) à 2ZnO(s) + O2 (g) + 4NO2 (g)
5g of zinc nitrate is used in the reaction.
As per the equation 2 moles of zinc nitrate will decompose to form 2 moles of zinc oxide.
The molar mass of these two as per the molecular mass in grams is 379g for zinc nitrate and 163g for zinc oxide.
163/379 * 5g = 2.15g of zinc oxide will form.
Reactions Involving Gases
• One Mole Of Any Gas At Room Temperature Which Is 250c Will Have A Volume Of 24 dm3 (Avagadro’s Law)
Zinc reacts with excess of dilute sulphuric acid to form zinc sulphate and 100cm3 hydrogen gas.
Zn + H2SO4 à ZnSO4 + H2
1 mole of zinc produces 1 mole of hydrogen gas.
Zinc mass as per Mr is 65.5g à 1 mol
Hydrogen volume 24000cm3 à 1 mol
For hydrogen, 24000/24000 * 100 = 100cm3
For zinc, 65.5/24000 * 100 = 0.27g
0.27g of zinc will produce 100cm3 of hydrogen gas.
Reacting Gases Only
For these reactions we will take into consideration the equation of oxidization of carbon monoxide to form carbon dioxide, 250 cm3 of carbon monoxide is oxidized at r.t.p.
2CO(g) + O2(g) à 2CO2(g)
2 moles of carbon monoxide will oxidize with 1 mol of oxygen to form 2 moles of carbon dioxide. 2 moles is 250cm3. By ration of 2:1:2 with 2 representing 250cm3.
125cm3 of oxygen will be required to form 250cm3 carbon dioxide.
Reactions In Solution
A one molar concentration of a substance consists of one mole of the substance dissolved in 1dm3 of solution.
Amount of substance (moles) = Volume of solution (dm3) x concentration (mol/dm3)
The concentration of potassium hydroxide solution is 0.10g/cm3. What is the concentration expressed in mol/dm3?
The molar mass of potassium hydroxide is 56g. 56g à 1dm3.
100g = 100/56 = 1.79 mole.
The concentration is 1.79mol/dm3.
What mass of potassium nitrate would be required to produce potassium nitrate of volume 50cm3 and concentration 0.5 mol/dm3.
Volume of solution 0.050 dm3.
Number of moles of potassium nitrate = 0.050 x 0.5 = 0.025 mol
Mr of potassium nitrate = 101
Mass of potassium nitrate = 101 x 0.025 = 2.53g
Reactions Involving Solids And Solutions
Solid iron fillings and dilute hydrochloric acid will react to form iron (II)chloride. How much of dilute hydrochloric acid would be required of concentration 0.5mol/dm3 to react completely with 1g of iron fillings.
Fe(s) + 2HCl(aq) à FeCl2(aq) + H2 (g)
1g of iron fillings would be 0.018 moles. As molar mass is 56g and 56g is 1 mole.
The equation shows the ratio in which HCl and Fe react which is 2:1 and so 0.036moles of HCl would be required to react. The concentration of HCL is 0.5mol/dm3. Which means in 1dm3 there will be 0.5 mol. So for 0.036 à (1000/0.5 * 0.036) = 72cm3.
Reactions Involving Only Solutions
Potassium iodide solution reacts with aqueous lead(II) nitrate solution forming an insoluble precipitate of lead(II) nitrate and a solution of potassium nitrate.
Pb(NO3)2 (aq) + 2KI(aq) à PbI2(s) + 2KNO3(aq)
What volume of potassium iodide solution of concentration 0.5mol/dm3 is needed to react exactly with 10cm3 of lead(II) nitrate solution of concentration 0.1mol/dm3, what mass of lead(II) iodide is formed?
1dm3 or 1000cm3 of lead(II) nitrate à 1 mol, 10cm3 à 0.01 mol.
1 mol of lead(II) nitrate reacts with 2 mol of potassium iodide and so 0.02 mol of potassium iodide is required for this reaction.
1000cm3 of potassium iodide contains 0.5 mol and so a solution of 0.02 mol will be of 40 cm3.
1 mole of lead(II) nitrate produces 1 mole lead(II) iodide as per the ratio and 0.01 mol of lead(II) nitrate must be produced.
The Mr of lead(II) iodide is 461g and so the precipitate would be 4.61g (461 x 0.01)
Percentage Yield And Percentage Purity
Percentage yield is the ratio of actual and theoretical yields in percentage. The formula for it is as follows:
Actual yield x 100
The ionic equation for lead(II) nitrate with excess of potassium iodide to form lead(II) iodide.
3.31g of lead(II) nitrate was dissolved to obtained 3.89g of lead(II) iodide.
1 mol of lead(II) nitrate produces 1 mol of lead(II) iodide.
Pb2+(aq) + 2I-(aq) à PbI2(s)
1mol/dm3 solution of lead nitrate has a concentration of 2.0 mol/dm3 of nitrate ions.
Lead(II) nitrate Mr in mass = 331g à produces 461g of Lead(II) iodide
3.31x331/331 = 3.31g of lead(II) nitrate produces 3.31x461/331 = 4.61g of lead(II) nitrate.
Theoretical yield is 4.61g
Actual yield is 3.89g
Percentage yield = 3.89 x 100/ 4.61 = 84.4%
A sample of 5g of impure sodium chloride has 4.85g of pure sodium chloride. The purity of this 5g would be
4.85/5 x 100 = 97%