Question

(1). Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% oxygen by mass (Atomic Mass : Fe = 55.85 amu, O = 16.00 amu).

(2). Calculate the amount of carbon dioxide that could be produced when

(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 mole of carbon is burnt in 16 g of dioxygen.

The balanced chemical equation for combustion of carbon in dioxygen/air is

$\sf{C + O_2 \longrightarrow CO_2}$

C = 1 mole
O2 = 32 g
CO2 = 44 g​

Answer 1

Answer:

Empirical formula of a compound may be defined as the formula which gives the simplest whole number ratio of the atoms of various elements present in one molecule of the compound.

 

Form the available data Percentage of iron = 69.9

 

Percentage of oxygen= 30.1 Total percentage of iron & oxygen= 69.9+30.1= 100%

 

Step 1 calculation of simplest whole number ratios of the elements

Element

Percentage

Atomic mass

Atomic ratio

Simplest ratio

Simplest whole no ratio

Fe

69.9

55.84

69.9/55.84=1.25

1.25= 1

2

O

30.1

16

30.1/16 = 1.88

1.88=1.5

3

Step 2 writing the empirical formula of the compound

 

The empirical formula of the compound = Fe2 O3

Explanation:

Answer 2

Explanation:

Aloha !

$\text { This is Sweety Adihya }$❤️

Fe=Given percentage /molecular weight =69.9/55.83=1.25

O=Given percentage /molecular weight

=30.1/16=1.88

Round off the

Fe=1×2=2

O=1.5×2=3

Fe²O³ is empirical formula.

Hope it will help you

@ Sweety Adihya

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