Page 72 - Chemistry_Form_2

Chemistry
for Secondary Schools

Step 2: Divide the mass of each element in the sample by its R.A.M., i.e.
Mass in sample

R AM.. .

15.8 g
FOR ONLINE READING ONLY
C ! !1.32
12 g
84.2 g
S ! ! 2.63
32 g

Step 3: Divide throughout by the smallest value, in this case 1.32:

1.32 : 2.63 = 1:1.99
1.32 1.32

Step 4: Obtain whole number ratios:
C : S
1 : 2
The empirical formula is therefore CS .
2
The empirical formula together with the relative molecular mass can then be used
to establish the molecular formula of the compound.
Molecular formula = n empirical formula, where n is a whole number.

Hint: R.M.M. = n sum of R.A.M.

Therefore, the molecular formula = n empirical formula = n(CS )
2
Now, R.M.M. = n sum of R.A.M.

76 = n [12 + (2 × 32)]

76 = n (12 + 64)

76 = 76 n
n = 1

Therefore, the molecular formula is CS .
2

Student’s Book Form Two 65

01/08/2025 11:20:56
CHEMISTRY FORM TWO NEW 2025 DUMMY.indd 65 01/08/2025 11:20:56
CHEMISTRY FORM TWO NEW 2025 DUMMY.indd 65