Question

The percent composition of bismuth oxide is \(89.7 \% \mathrm{Bi}\) and \(10.3 \%\) O. Calculate the empirical formula.

Short answer

The empirical formula of bismuth oxide is \(\mathrm{Bi_2O_3}\).

Step-by-step solution

Assume a Mass

For simplicity, assume you have 100 grams of bismuth oxide. This allows us to work directly with the given percentages as grams: \(89.7\) g of Bi and \(10.3\) g of O.

Convert Mass to Moles

Convert the mass of each element to moles by using their atomic masses. The atomic mass of Bi is \(208.98 \; \text{g/mol}\) and O is \(16.00 \; \text{g/mol}\).Calculate moles of Bi:\[ \frac{89.7 \; \text{g Bi}}{208.98 \; \text{g/mol}} = 0.429 \; \text{mol Bi} \]Calculate moles of O:\[ \frac{10.3 \; \text{g O}}{16.00 \; \text{g/mol}} = 0.644 \; \text{mol O} \]

Determine the Simplest Ratio

Divide the number of moles of each element by the smallest number of moles calculated.For Bi:\[ \frac{0.429 \; \text{mol Bi}}{0.429} = 1 \]For O:\[ \frac{0.644 \; \text{mol O}}{0.429} = 1.5 \]

Adjust to Whole Numbers

The ratio \(1:1.5\) is not whole, so multiply both parts of the ratio by 2 to get whole numbers:For Bi: \(1 \times 2 = 2\)For O: \(1.5 \times 2 = 3\)This gives the whole number ratio of \(2:3\).

Write the Empirical Formula

Using the whole number ratio obtained, the empirical formula of bismuth oxide is \(\mathrm{Bi_2O_3}\).

Key concepts

Percent Composition

Percent composition is a way to express the relative amount of each element in a compound by mass. It indicates what percentage of the total mass of the compound comes from each element. This concept helps chemists understand how much of each element is in a substance without needing to know the overall mass of a sample. To calculate percent composition:

• Determine the mass of each element in the compound.
• Divide the mass of each element by the total mass of the compound, then multiply by 100 to get a percentage.

In the case of bismuth oxide, the percent composition is given as 89.7% Bi and 10.3% O. This means that in any sample of bismuth oxide, 89.7% of the mass is due to bismuth and 10.3% due to oxygen. When solving problems involving percent composition, it is often useful to assume a total mass, such as 100 grams, to convert percentages directly into grams, facilitating calculations.

Mole Calculation

Mole calculations are a fundamental aspect of chemistry which involves converting between mass, moles, and number of particles using the concept of the mole. A mole represents Avogadro's number (approximately \(6.022 \times 10^{23}\)) of particles. For substances with a particular atomic or molecular mass, one mole of that substance has a mass in grams equal to the atomic or molecular mass.

• To convert from mass to moles, use the formula: \[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Atomic Mass (g/mol)}} \]
• This allows us to determine how many moles of an element are present in a given mass.

For instance, using the given atomic mass of bismuth (208.98 g/mol) and oxygen (16.00 g/mol), we can convert 89.7 grams of Bi to moles by performing the calculation: \( \frac{89.7 \text{ g}}{208.98 \text{ g/mol}}\). Similarly, for oxygen, calculate moles from mass using \( \frac{10.3 \text{ g}}{16.00 \text{ g/mol}} \). Such conversions are crucial in determining the simplest ratio of elements in a compound.

Atomic Mass

Atomic mass is a measure of the mass of an atom, typically expressed in atomic mass units (amu), which is roughly equivalent to \(1/12\) the mass of a carbon-12 atom. The atomic mass of an element is closely related to its isotopes and their abundance. When it comes to practical chemistry and calculations, atomic mass is used to convert between mass and moles, which is essential for determining empirical and molecular formulas of compounds.Key points about atomic mass:

• It provides a way to measure how heavy an average atom of an element is.
• Atomic mass shown on the periodic table reflects both the weighted average of all natural isotopes and the element's abundance.
• For calculations, the atomic mass is used as a conversion factor between grams and moles.

In solving for an empirical formula, like in the case of bismuth oxide, the atomic masses of Bi (208.98 g/mol) and O (16.00 g/mol) are used to convert the mass percentages to moles. Accurate application of atomic mass in these transformations allows us to maintain precision in deriving the simplest chemical formula of a compound.