Question

An Oxide of nitrogen gave the following percentage composition of N = 25. 94 O =74.06.calculate the empirical formula

Answer 1

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Answer 2

The empirical formula of the given compound is $N_2O_5$

Explanation:

We are given:

Percentage of N = 25.94 %

Percentage of O = 74.06 %

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

Mass of N = 25.94 g

Mass of O = 74.06 g

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

• Step 1: Converting the given masses into moles.

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{25.94g}{14g/mole}=1.85moles$

Moles of Oxygen = $\frac{\text{Given mass of Oxygen}}{\text{Molar mass of Oxygen}}=\frac{74.06g}{16g/mole}=4.63moles$

• 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 1.85 moles.

For Nitrogen = $\frac{1.85}{1.85}=1$

For Oxygen = $\frac{4.63}{1.85}=2.5$

To convert the mole ratios into whole numbers, we multiply individual mole ratios by 2

Mole ratio of Nitrogen = (2 × 1) = 2

Mole ratio of Oxygen = (2 × 2.5) = 5

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

The ratio of N : O = 2 : 5

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