Question

calculate the empirical formula of a compound containing 87.5% by mass of nitrogen and 12.5% by mass of hydrogen( atomic weight of nitrogen is 14 and hydrogen is 1.)​

Answer 1

LET TOTAL MASS BE 100

N= 87.5

H=12.5

N:H

87.5/14. : 12.5/1

1:2

NH2 IS THE ANSWER

Answer 2

The empirical formula for the given compound is $NH_2$

Explanation:

We are given:

Percentage of N = 87.5 %

Percentage of H = 12.5 %

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

Mass of N = 87.5 g

Mass of H = 12.5 g

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

• Step 1: Converting the given masses into moles.

Moles of Nitrogen = $\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{87.5g}{14g/mole}=6.25moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{12.5g}{1g/mole}=12.5moles$

• 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 6.25 moles.

For Nitrogen = $\frac{6.25}{6.25}=1$

For Hydrogen = $\frac{12.5}{6.25}=2$

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

The ratio of N : H = 1 : 2

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