, Topic 1: Quantitative aspects of chemical change
● Concentration: The concentration of a solution is the
quantity of dissolved solute mol per dm3 of solution
● 1dm3 = 1000cm3
Eg. 2 mol of salt is dissolved in 250cm3 of water
c = n/v
c = 2 mol/ 0.25dm3
C= 8 mol dm-3
● If the mol value is not given, we use a different
equation which includes the mass given, & molar mass
(atomic number) of the element
Eg. 12g KCl dissolved in 650cm3 of water F
c= m/MV
c= 12/ 75.4x 0.65
c= 0.25 mol dm-3
Percentage Composition
the ratio of the amount of each element to the total amount of individual elements present in
the compound multiplied by 100
● Eg. Determine the percentage composition by mass of NaHCO3
● 23+1+12+3(16) = 84g mol-1
○ Na=23/84 x 100= 27.58%
○ H= 1.19%
○ C= 14.28%
○ O= 57.14%
Empirical Formula
Simplest ratio of the atoms in a compound
Steps:
1. Divide each mass by their atomic masses to calculate the no. of mols
2. Bond in a ratio
3. Divide by the smallest no.
4. Obtain whole numbers by either rounding off or multiplying
(all of the molar masses can be found in the periodic table, below the element symbol)
Eg. 40% of C; 6.6% of H; 53.3% of O
n[C] = 40/12 = 3.33 mol (12 is the molar mass)
n[H] = 6.6/1 = 6.66 mol ( 1 is the molar mass)
n[O] = 53.3/16 = 3.33 mol (16 is the molar mass)
Divide all by the lowest no. which is 3.33
C; H ;O is 1 ; 2 ; 1
=CH2O
Empirical & Molecular Formula
- The empirical formula is the simplest ratio of atoms in a compound
- & determine the molecular formula, you will need the Empirical formula & the molar
mass of the compound (usually given in tests)
- The molecular formula gives us the total number of atoms in the molecule. (It tells
you the amount of the compound that is found in a certain g/mol)
Eg. A sample of 100g, consists of 63% carbon ; 5% hydrogen & 32% oxygen. Its molar mass is 291
g/mol. Determine both the empirical & molecular formula
● Concentration: The concentration of a solution is the
quantity of dissolved solute mol per dm3 of solution
● 1dm3 = 1000cm3
Eg. 2 mol of salt is dissolved in 250cm3 of water
c = n/v
c = 2 mol/ 0.25dm3
C= 8 mol dm-3
● If the mol value is not given, we use a different
equation which includes the mass given, & molar mass
(atomic number) of the element
Eg. 12g KCl dissolved in 650cm3 of water F
c= m/MV
c= 12/ 75.4x 0.65
c= 0.25 mol dm-3
Percentage Composition
the ratio of the amount of each element to the total amount of individual elements present in
the compound multiplied by 100
● Eg. Determine the percentage composition by mass of NaHCO3
● 23+1+12+3(16) = 84g mol-1
○ Na=23/84 x 100= 27.58%
○ H= 1.19%
○ C= 14.28%
○ O= 57.14%
Empirical Formula
Simplest ratio of the atoms in a compound
Steps:
1. Divide each mass by their atomic masses to calculate the no. of mols
2. Bond in a ratio
3. Divide by the smallest no.
4. Obtain whole numbers by either rounding off or multiplying
(all of the molar masses can be found in the periodic table, below the element symbol)
Eg. 40% of C; 6.6% of H; 53.3% of O
n[C] = 40/12 = 3.33 mol (12 is the molar mass)
n[H] = 6.6/1 = 6.66 mol ( 1 is the molar mass)
n[O] = 53.3/16 = 3.33 mol (16 is the molar mass)
Divide all by the lowest no. which is 3.33
C; H ;O is 1 ; 2 ; 1
=CH2O
Empirical & Molecular Formula
- The empirical formula is the simplest ratio of atoms in a compound
- & determine the molecular formula, you will need the Empirical formula & the molar
mass of the compound (usually given in tests)
- The molecular formula gives us the total number of atoms in the molecule. (It tells
you the amount of the compound that is found in a certain g/mol)
Eg. A sample of 100g, consists of 63% carbon ; 5% hydrogen & 32% oxygen. Its molar mass is 291
g/mol. Determine both the empirical & molecular formula
