Question

Two elements A (At. mass 75) and B (At.
mass 16) combine to give a compound
having 75.8% of A. The formula of the
compound is
a.AB
b.A2B
c.AB2
D.A2B3​

Answer 1

Answer:

D. A2B3

ELEMENT % MASS RATIO ATOMIC RATIO

A 76 76/75 = 1.01

1.01/1.01 = 1× 2 = 2

B 24 24/16 = 1.5

1.5/1.01 = 1.48

1.5 × 2 = 3

SO EMPIRICAL FORMULA OF COMPOUND = A2B3

Answer 2

The formula for the given compound is $A_2B_3$

Explanation:

We are given:

Percentage of A = 75.8 %

Percentage of B = (100 - 75.8) = 24.2 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of A = 75.8 g

Mass of B = 24.2 g

To formulate the empirical formula, we need to follow some steps:

• Step 1: Converting the given masses into moles.

Moles of A =$\frac{\text{Given mass of A}}{\text{Molar mass of A}}=\frac{75.8g}{75g/mole}=1.01moles$

Moles of B = $\frac{\text{Given mass of B}}{\text{Molar mass of B}}=\frac{24.2g}{16g/mole}=1.51moles$

• Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.01 moles.

For A = $\frac{1.01}{1.01}=1$

For B  = $\frac{1.51}{1.01}=1.51$

To make the mole ratio a whole number, we multiply each of the mole ratio by '2'

Mole ratio of A = (1 × 2) = 2

Mole ratio of B = (1.5 × 2) = 3

• Step 3: Taking the mole ratio as their subscripts.

The ratio of A : B = 2 : 3

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