Question

A compound being used in experimental medicine is 78.14% boron (B) and 21.86% hydrogen (H). The empirical formula of the compound will have a boron to hydrogen ratio of : .

Answer 1

Answer:The empirical formula of the compound will have a boron to hydrogen ratio of 1:3.

Explanation

Percentage of Boron = 78.14 %

Percentage of  Hydrogen = 21.86 %

Suppose  in 100 gram of medicine

Mass of boron in 100 grams of medicine = 78.18 g

Mass of the hydrogen in 100 grams of medicine = 21.86 g

Moles of boron:

=$\frac{given mass}{molar mass}=\frac{78.18 g}{10.81 g/mol}=7.23 mol$

Moles of hydrogen:

=$\frac{given mass}{molar mass}=\frac{21.86 g}{1.00g/mol}=21.83 mol$

For the ratio of boron and hydrogen divide the number moles of respective elements  with smallest number of moles calculated.

Here smallest numeric value moles  are of boron which is 7.23 moles

For boron = $\frac{7.23 moles}{7.23 moles}=1$

For hydrogen = $\frac{21.83 moles}{7.23 moles}=3.01\approx 3$

So ,the ratio of Boron to Hydrogen ; 1:3

The empirical formula of the compound will be :$B_1H_3$

The empirical formula of the compound will have a boron to hydrogen ratio of 1:3.