Question

A compound was analyzed and was found to contain the following percentages of the elements by mass: boron, \(78.14 \% ;\) hydrogen, \(21.86 \%\). Determine the empirical formula of the compound.

Short answer

The empirical formula of the compound is B_2H_6.

Step-by-step solution

Convert given percentages to grams

Since we are dealing with percentages, let's assume we have a 100-gram sample of the compound. In this case, we will have 78.14 grams of boron and 21.86 grams of hydrogen.

Calculate moles of each element

Now, we need to convert these grams into moles. To do this, we will use the molar mass of the elements. For boron (B), the molar mass is 10.81 g/mol, and for hydrogen (H), the molar mass is 1.01 g/mol. The formulas for calculating the moles are: Moles of B = (Mass of B) / (Molar mass of B) Moles of H = (Mass of H) / (Molar mass of H) Moles of B = \( \frac{78.14g}{10.81g/mol} \) Moles of H = \( \frac{21.86g}{1.01g/mol} \)

Calculate the mole ratio

Next, we divide the moles of each element by the smallest number of moles to find the mole ratio: Mole ratio of B = \(\frac{ moles\:of\:B }{ smallest\:number\:of\:moles }\) Mole ratio of H = \(\frac{ moles\:of\:H }{ smallest\:number\:of\:moles }\) Calculate the moles and mole ratio for both elements.

Determine the empirical formula

From the mole ratio in Step 3, we can determine the empirical formula of the compound. Round off the mole ratio to the nearest whole number if needed; consider that sometimes you might need to multiply all the ratios by a small whole number to obtain whole numbers ratios that are suitable for the formula. Combine the rounded mole ratios of both elements to represent the empirical formula of the compound.

Key concepts

Molar Mass Calculations

Understanding molar mass is a fundamental step in chemistry for converting between grams and moles of a substance. The molar mass of an element is the mass of one mole of that element, typically expressed in grams per mole (g/mol). Each element has a unique molar mass, which can usually be found on the periodic table.
For example, in the given problem, we have boron and hydrogen with molar masses of 10.81 g/mol and 1.01 g/mol, respectively. Calculating moles from mass involves using the formula:

• Moles = \( \frac{\text{Mass in grams}}{\text{Molar mass}} \)

This means for 78.14 grams of boron you calculate \( \frac{78.14}{10.81} \), and for hydrogen, \( \frac{21.86}{1.01} \) to find the moles. These calculations are crucial as they allow us to relate the mass of an element to the amount of substance present, facilitating further calculations like determining the mole ratio.

Percent Composition

Percent composition involves finding the percentage by mass of each element in a compound. This is what helps chemists understand the makeup of a compound in simpler terms. In any analysis, percent composition is calculated using the basic formula:

• Percent Composition = \( \frac{\text{Mass of element}}{\text{Total mass of compound}} \times 100\)

In our exercise, it was given that the compound contains 78.14% boron and 21.86% hydrogen. These percentages allow us to easily convert to mass if we assume a 100-gram sample, making our initial calculation straightforward. With this assumption, percent composition serves as a direct path to understand how much of each element is present in one mole of the substance.

Mole Ratio

Calculating the mole ratio is a key step in determining the empirical formula. The mole ratio gives the simplest whole-number ratio of moles of elements in a compound. Once you have calculated the moles for each component, the next step is to find the simplest ratio by dividing each by the smallest number of moles.
In the example, once you've obtained the moles of boron and hydrogen, divide each by the smaller number of the two. The formula used is:

• Mole Ratio = \( \frac{\text{Moles of element}}{\text{Smallest number of moles}} \)

For instance, if boron has fewer moles, then you divide both elements' moles by the moles of boron to find the ratio. The goal is to find ratios close to whole numbers, which can then be directly used in the empirical formula. Round the values to the nearest whole numbers or scale up if necessary to obtain whole numbers, achieving the empirical formula which represents the simplest form of the compound.