Question

A binary compound of boron and hydrogen has the following percentage composition: \(78.14 \%\) boron, \(21.86 \%\) hydrogen. If the molar mass of the compound is determined by experiment to be between 27 and \(28 \mathrm{~g},\) what are the empirical and molecular formulas of the compound?

Short answer

The empirical formula of the compound is \(BH_3\) and the molecular formula is \(B_2H_6\).

Step-by-step solution

Calculate moles of boron and hydrogen

We will assume we have 100 grams of the compound since it makes calculations simpler using the percentage composition. Therefore, we have \(78.14g\) of boron and \(21.86g\) of hydrogen. Now, we need to find the moles of each element. To do this, we will divide the mass of each element by their respective molar masses. The molar mass of boron is \(10.81g/mol\), and that of hydrogen is \(1.01g/mol\). Moles of boron = \(\frac{78.14}{10.81} = 7.228 \space mol\) Moles of hydrogen = \(\frac{21.86}{1.01} = 21.683 \space mol\)

Find the molar ratio of boron to hydrogen

Next, we will determine the molar ratio of boron to hydrogen. To find the molar ratio, divide the moles of each element by the smallest number of moles: Molar ratio = \(\frac{7.228}{7.228} : \frac{21.683}{7.228}\) Molar ratio ≈ \(1 : 3\)

Determine the empirical formula

Using the molar ratio we found, the empirical formula of the compound is: Empirical formula = \(BH_3\)

Find the molecular formula

Using the empirical formula, we will find the molar mass of the empirical formula. The molar mass of the empirical formula \(BH_3\) is: Molar mass = \(10.81 + 3 \times 1.01 = 13.84 \space g/mol\) Since the given molar mass range of the compound is between \(27\) and \(28 g/mol\), we can find the integer multiple of the molar mass of the empirical formula that falls within this range: Multiple = \(\frac{27}{13.84} = 1.95\) Rounding to the nearest whole number, the multiple is \(2\). Now we can find the molecular formula by multiplying the empirical formula by this multiple: Molecular formula = \(2 \times BH_3 = B_2H_6\) The empirical formula of the compound is \(BH_3\), and the molecular formula is \(B_2H_6\).

Key concepts

Percentage Composition

Percentage composition is a way to express the relative amounts of each element in a compound based on mass. It's usually presented as a percentage. In our example, a compound made of boron and hydrogen has a percentage composition of 78.14% boron and 21.86% hydrogen.
To understand what these numbers mean, imagine you have 100 grams of the compound. Because percentages are parts per hundred, 78.14 grams would be boron and 21.86 grams would be hydrogen.
This information helps us figure out how much each element contributes to the total mass of the compound. It's a crucial first step in determining both empirical and molecular formulas.

Moles of Elements

After finding the percentage composition, the next step is to convert these masses into moles. Moles are a handy unit that chemists use to count particles, like atoms, in a sample. To find the moles, you divide the mass of each element by its molar mass (the mass of one mole of that element).
For boron, with a molar mass of 10.81 g/mol, we take 78.14 grams and divide by 10.81, resulting in approximately 7.228 moles of boron.
For hydrogen, with a molar mass of 1.01 g/mol, we calculate moles by dividing 21.86 grams by 1.01, giving us about 21.683 moles of hydrogen.
Understanding the mole concept is key to transforming mass data into molar data, which is more useful for calculations.

Molar Ratio

The molar ratio tells us the simplest whole number ratio of the moles of each element in a compound. We find this by dividing the calculated moles of each element by the smallest number of moles from our calculations.
In our case, we have approximately 7.228 moles of boron and 21.683 moles of hydrogen. By taking the smaller of the two numbers, which is 7.228, we divide the moles of both boron and hydrogen by this number:

• Boron: 7.228 / 7.228 = 1
• Hydrogen: 21.683 / 7.228 ≈ 3

This gives us a simplistic 1:3 molar ratio, indicating each boron is paired with approximately three hydrogens in the simplest form. This ratio is fundamental in deriving the empirical formula.

Empirical Formula Calculation

The empirical formula represents the simplest whole number ratio of elements in a compound. Using the molar ratio obtained, the empirical formula is constructed. From our case, we got a 1:3 ratio for boron to hydrogen. Thus, the empirical formula is denoted as \( BH_3 \).
To verify and further elaborate, we calculate the molar mass of the empirical formula \( BH_3 \), which adds up to 13.84 g/mol. If an experiment shows the compound's molar mass is between 27 and 28 g/mol, we can compare by using integer multiples of the empirical formula's molar mass.
The closest integer multiple that fits the observed molar mass is two, thus doubling the empirical formula gives us the molecular formula \( B_2H_6 \). This calculation shows that \( B_2H_6 \) is both the molecular and empirical basis of the compound.