Everything Science Grade 11

CHAPTER 6. QUANTITATIVE ASPECTS OF CHEMICAL CHANGE 6.5

SOLUTION

Step 1 : Calculate the mass of each element in 100 g ofacetic acid.

In 100 g of acetic acid,there is 39.9 g C, 6.7 g Hand

53.4 g O

Step 2 : Calculate the number of moles of each element in 100 g of acetic acid.

n =mM

nC =

39. 9

12

= 3. 33 mol

nH =

6. 7

1

= 6. 7 mol

nO =

53. 4

16

= 3. 34 mol

Step 3 : Divide the number ofmoles of each elementby the lowest number to get the

simplest mole ratio of the elements (i.e. the empirical formula) in acetic acid.

Empirical formula is CH 2 O

Step 4 : Calculate the molecular formula, using the molar mass of acetic acid.

The molar mass of acetic acid using the empirical formula is 30 g· mol−^1.

Therefore the actual number of moles of each element must be double what it is

in the empirical formula.

The molecular formula is therefore C 2 H 4 O 2 or CH 3 COOH

Exercise 6 - 5

1. Calcium chloride is produced as the productof a chemical reaction.

(a) What is the formulaof calcium chloride?

(b) What percentage does each of the elements contribute to the mass ofa molecule of calcium

chloride?

(c) If the sample contains 5 g of calcium chloride, what is the mass of calcium in the sample?

(d) How many moles ofcalcium chloride are inthe sample?

2. 13g of zinc combines with 6.4g of sulphur. What is the empirical formula of zinc sulphide?

(a) What mass of zinc sulphide will be produced?

(b) What percentage does each of the elementsin zinc sulphide contribute to its mass?

(c) Determine the formula of zinc sulphide.