Guide to Chemistry Practicals
Questions and Answers
Notes on Units and Formulas
A. Volume
The units of volume commonly used in this manual are cm3 and dm3. The conversion between cm3 and dm3 is 1 dm3 = 1000 dm 3. Note that a cm3 are the same as millilitres (ml) and dm3 are the same as litres (L). 1 cm3 = 1 ml and 1 dm3 = 1 L.
B. Concentration
Concentration is a measure of amount per volume. The amount can be grams, kilograms, number of moles, or any other unit of amount. Volume can be cm3, dm3, or any other volume. The two types of concentration used commonly are mol/dm3 and g/dm3.
- mol/dm3 is also called Molarity or molar concentration. The symbol for molarity is M. Molarity is used in the MAVA formula.
- g/dm3 is called gram concentration.
To convert from molar concentration to gram concentration, you multiply the molar concentration by the molar mass of the molecule.
To convert from gram concentration to molar concentration, you divide the gram concentration by the molar mass of the molecule.
C. The Mole Concept
A mole (mol) is a unit of amount. 1 mol of something is equal to 6.022x1023 of that thing. For example, 1 mol of H2 molecules is equal to 6.022x1023 H2 molecules. 1 mol of goats is 6.022x1023 goats.
There are two important formulas for determining the number of moles present of a compound. The first:
You see that when you cancel the units, you are left with moles.
The second formula:
where M = molarity (in mol/dm3) and V = volume (in dm3). Notice that when you cancel the units, you are left with moles.
D. Stoichiometry
Stoichiometry is the study of the relationships between amounts (number of moles or masses) of reactants and products. It is fundamental to all of chemistry, and it is essential that you understand the stroichiometry and are able to apply it to numbers of moles and masses in chemical reactions.
Quantitative reactions are reactions that continue until one of the reactants is finished. Reactions with a strong acid or strong base are quantitative, as are most of the reactions in redox titrations. If the reaction proceeds in only one direction (and is not reversible), it is a quantitative reaction. Usually the NECTA exam will state if a reaction is quantitative. If the question states that something "is in excess," this means that there is a very large amount of it, so much that it will not be finished in the reaction.
The mol ratio between two compounds is the ratio in which those two compounds react. For example, in the reaction:
NH3 + O2 → NO + H2O
One mol of NH3 reacts with one mol of O2 to form one mol of NO and one mol of H2O. If 8.5 g of NH3 react with 32 of O2, how many grams of NO will be formed? When comparing the relationships between two compounds in a reaction, you must look at the mole relation! The mass relation does not tell us very much. Converting to moles, we see that there are 0.5 moles of NH3 and 1 mole of O2.
Which of the reactants will be finished first, or will both be finished at the end of the reaction? NH3 will be finished first, since there are only 0.5 mols of it present and NH3 and O2 react in a 1:1 mole ratio.
How many moles of NO will be formed? NH3 and NO have a 1:1 mol ratio, so 0.5 mols of NH3 will produce 0.5 mols of NO. 0.5 moles of NO has a mass of 15 grams.
So you can see that in order to determine the relationship between reactants and products, we must first compare the number of moles! Once we have determined the mole relation, we can convert from number of moles to grams.
Another example: In the reaction: I2 + 2 S2O32- → S4O62- + 2 I-
1 mol of I2 reacts with 2 mol of S2O32- to produce 1 mol of S4O62 - and 2 mol of I-. This gives us no information about the actual number of moles of these compounds, only about the ratio in which they react. It means that 1 mol of I2 requires 2 mol of S2O32- to react completely. Or: 0.25 mol of I2
requires 0.5 mol of S2O32- to react completely. The ratio must me 1:2. But the actual number of moles can be 0.25 mol I2 and 0.5 mol S2O32-, or 2 mol I2 and 4 mol S 2O32-, or 10 mol I2 and 20 mol S 2O 32-.
To find the actual number of moles, you can use cross-multiplication. You could be asked: 0.0125 mol of S2O3 2- requires how many moles of I2 to react completely? You know that the mole ratio between S2O32- and I2 is 2:1. Thus:
We would find that x = 0.00625 mol I2 would be required to completely react with 0.0125 mol 2 S2O32-. Note that the relationship between the two compounds is between number of moles, not mass.
For example, it is true that 1 mol of I2 reacts with 2 mol of S2O32-, but this does not mean that 1 gram of I2 reacts with 2 grams of S 2O32-. You must think of these relationships in terms of numbers of moles!
E. The Dilution Formula
The dilution formula allows you to determine the molarity of a solution after it has been diluted. The formula is:
M1V1=M2V2
Where M1 and V1 are the molarity and volume before a dilution, and M2 and V2 are the molarity and volume after a dilution.
Consider this example from the 2006 NECTA: "Hydrochloric acid solution made by diluting 750 cm3 of 0.25 M HCl to 937.5 cm3 with distilled water." What is the molarity of this solution after dilution?
The volume before dilution (V1) was 730 cm3, and the molarity before dilution (M1) was 0.25 M. The volume after dilution (V2) was 937.5 cm3. To find the molarity after dilution, M 2, we simply substitute the known values into the dilution equation and solve for M2. We find that M2 = 0.2 M.
Whenever you are given both a molarity and volume before a change (M1 and V 1) and are asked to calculate a volume or molarity after a change (V2 or M2), the dilution formula is a good way to find the answer.
F. MAVA
The complete form of MAVA is shown below:
MA is the molarity of compound A, VA is the volume of compound A, MB is the molarity of compound B, and VB is the volume of compound B. nA and nB are the mole ratio between compounds A and B, found from the balanced chemical equation. This equation is extremely important in chemistry practicals. It is used in acid-base titrations, redox titrations, and is sometimes needed in kinetics and solvent extraction questions. You can use it to solve for any part of the formula: MA, VB, anything. If you can find values for five of the terms, you will be able to calculate the sixth.