A mole is the Ar or Mr expressed in grams e.g. 1 mole of Carbon-12 is equal to 12 grams. It is equal to 6.02 × 1023 atoms, this number is called Avogrado’s constant. 1 mole of a gas at RTP (25°C and 1atm or room temperature and pressure) occupies a volume of 24dm3. The symbol for a mole is “mol” e.g. 1mol. Concentration is measured in mol/dm3 or M for short.
Moles calculations:
(NOTE: when you read this say the “/” sign as “per” e.g. 12g/1mol is “12 grams per mole”) Grams to moles: number of grams × 1 / Mr
Moles of one substance to another:
Moles back to grams: Number of moles × (number of grams per mole e.g. 20g/mole) Moles of gas to volume (at RTP): Number of moles × 24dm3
Gas volume to moles: volume / 24dm3
Moles to atoms: Number of moles × 6.02 × 1023
Volume to moles: volume × concentration (concentration=moles of solute/total volume)
Percent purity = amount of wanted substance / total amount of substance × 100%
Percent yield = actual yield / theoretical yield × 100%
Emperical formula: simplest ratio of atoms in the chemical formula.
Molecular formula: the formula using the actual number of atoms in a molecule. To find out the emperical formula you:
-make the percent ratio into the simplest whole number ratio (NOTE: if it is percent mass, then you have to divide each percentage by its Ar to convert the mass ratio to an atom ratio (because atoms have different weights) then make that ratio into the smallest whole number ratio by dividing the coefficients of each element symbol by the lowest coefficient, for example:
- A substance is 40% carbon, 6666% hydrogen and 53.3334% oxygen by mass. First you must convert the mass ration (40:6.6666:53.3334) to an atom ratio:
- 40/12 = 3.3333 → C3.3333
- 6.6666/1 = 6.6666 → H6.6666
- 53.3334/16 = 3.3333 → O3.3333
- The ratio is C3.3333 H6.6666 O3.3333
- Just by looking we can see there are twice as many hydrogen atoms in the formula as either oxygen or carbon, but let us suppose that the ratio wasn’t so easy to simplify, you would then have to:
- Convert moles of carbon to the simplest whole number: 3333 / 3.3333 = 1
- Convert moles of hydrogen to the simplest whole number: 6666 / 3.3333 = 2
- Convert moles of oxygen to the simplest whole number: 3333 / 3.3333 = 1
- The ratio is therefore 1:2:1 so the empirical formula is CH20
-simplify the molecular formula
for example take C6H1206 and divide each coefficient by the smallest number (6) → CH20
-if you are given the masses of each element in the compound you divide the masses by the Ar, and convert into the simplest whole number ratio.
To calculate the molecular mass:
-if you have the emperical formula you know the emperical mass as well. Then if you are given the molecular mass you just do molecular mass/emperical mass to give you “n”. Then you multiply the emperical ratio by “n”.
Now let’s suppose that they tell you the molecular mass of the compound that has the empirical formula CH2O is 180, and you are asked to find the molecular formula
- Find out the empirical mass: 12 + (2 × 1) + 16 = 30
- Find the number you must multiply the empirical formula by, let’s call it ‘n’: 180 / 30 = n = 6
- Multiply the empirical ratio by 6 → 1:2:1 × 6 is 6:12:6
- The molecular formula is therefore C6H1206
The limiting reagent is the reactant that is in shorter supply than the other reactant (taking into account the number of moles of each reactant in the balanced equation) which will stop the chemical reaction. To find out which is the limiting reagent you convert the amounts of reactant if it is in grams, volume etc. into moles. Then you find out how many moles of product will form with that many moles of reactant by using the balanced equation. Whichever reactant will produce the least number of moles of product is the limiting reagent.
For example take this equation:
2KOH (aq) + H2SO4 (aq) → 2H2O + K2SO4
Let’s say you have 30dm3 of sulphuric acid with a concentration of 1mol/dm3, and 2.8 kilograms of KOH 2.8kg × 1000g/1kg × 1mol/(39+1+16)g = 50mol of KOH
50mol × 1molK2SO4 / 2molKOH = 25mol of K2SO4 produced 30dm3 × 1mol/dm3 = 30mol of sulphuric acid
30mol × 1molK2SO4 / 1mol H2SO4 = 30mol of potassium sulphate produced Potassium hydroxide is the limiting reagent.