Question

A new compound contains nitrogen, hydrogen, boron, and fluorine. The assay values are: nitrogen, 13.360%; hydrogen, 3.8455%; boron, 10.312%. Determine its empirical formula.

Answer 1

The compound contains:

Nitrogen = 13.360%

Hydrogen = 3.8455%

Boron = 10.312%

This implies that for every 100 g of the substance:

Mass of nitrogen = 13.360 g

Mass of hydrogen = 3.8455 g

Mass of Boron = 10.312 g

Atomic mass of N = 14 g/mol

Atomic mass of H = 1 g/mol

Atomic mass of B = 11 g/mol

# moles of N = 13.360/14 = 0.9543

# moles of H = 3.8455/1 = 3.8455

# moles of B = 10.312/11 = 0.9375

Divide by the smallest # moles:

N = 0.9543/0.9375 = 1.02

H = 3.8455/0.9375 = 4.10

B = 0.9375/0.9375 = 1

Empirical formula = NH4B

Answer 2

Answer: The empirical formula for the given compound is $NH_4BF_4$

Explanation:

We are given:

Percentage of N = 13.360 %

Percentage of H = 3.8455 %

Percentage of B = 10.312 %

Percentage of F = [100 - 13.360 - 3.8455 - 10.312] = 72.4825 %

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

Mass of N = 13.360 g

Mass of H = 3.8455 g

Mass of B = 10.312 g

Mass of H = 72.4825 g

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

• Step 1: Converting the given masses into moles.

Moles of Nitrogen =$\frac{\text{Given mass of Nitrogen}}{\text{Molar mass of Nitrogen}}=\frac{13.360g}{14g/mole}=0.954moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{3.8455g}{1g/mole}=3.8455moles$

Moles of Boron = $\frac{\text{Given mass of Boron}}{\text{Molar mass of Boron}}=\frac{10.312g}{11g/mole}=0.937moles$

Moles of Fluorine = $\frac{\text{Given mass of Fluorine}}{\text{Molar mass of Fluorine}}=\frac{72.4825g}{19g/mole}=3.815moles$

• 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 0.937 moles.

For Nitrogen = $\frac{0.954}{0.937}=1.02\approx 1$

For Hydrogen = $\frac{3.8455}{0.937}=4.10\approx 4$

For Boron = $\frac{0.937}{0.937}=1$

For Fluorine = $\frac{3.815}{0.937}=4.07\approx 4$

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

The ratio of N : H : B : F = 1 : 4 : 1  : 4

Hence, the empirical formula for the given compound is $NH_4BF_4$