Question

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(55.3 \% \mathrm{~K}, 14.6 \% \mathrm{P}\), and \(30.1 \% \mathrm{O}\) (b) \(24.5 \% \mathrm{Na}, 14.9 \% \mathrm{Si},\) and \(60.6 \% \mathrm{~F}\) (c) \(62.1 \% \mathrm{C}, 5.21 \% \mathrm{H}, 12.1 \% \mathrm{~N},\) and the remainder \(\mathrm{O}\)

Short answer

The empirical formulas of the given compounds are: (a) \(\mathrm{K}_3\mathrm{P}\mathrm{O}_4\) (b) \(\mathrm{Na}_2\mathrm{Si}\mathrm{F}_6\) (c) \(\mathrm{C}_{12}\mathrm{H}_{12}\mathrm{N}_2\mathrm{O}_3\)

Step-by-step solution

Convert mass percentages into grams

Assuming we have 100 grams of the compound, the mass of each element will be the same as its mass percentage. In this case, we have 55.3 grams of \(\mathrm{K}\), 14.6 grams of \(\mathrm{P}\), and 30.1 grams of \(\mathrm{O}\).

Convert grams to moles

Using the molar masses of the elements, we can convert the grams to moles: - Moles of \(\mathrm{K} = \frac{55.3~\text{g}}{39.10~\text{g/mol}} = 1.414~\text{mol}\) - Moles of \(\mathrm{P} = \frac{14.6~\text{g}}{30.97~\text{g/mol}} = 0.4713~\text{mol}\) - Moles of \(\mathrm{O} = \frac{30.1~\text{g}}{16.00~\text{g/mol}} = 1.881~\text{mol}\)

Find the simplest whole-number ratio

Divide all the mole values by the smallest value, which is 0.4713. After rounding to the nearest whole number, we get the ratios: - \(\mathrm{K}: \frac{1.414}{0.4713} \approx 3\) - \(\mathrm{P}: \frac{0.4713}{0.4713} = 1\) - \(\mathrm{O}: \frac{1.881}{0.4713} \approx 4\) Thus, the empirical formula for compound (a) is: \(\mathrm{K}_3\mathrm{P}\mathrm{O}_4\). #(b) Compound # We'll follow the same steps as above.

Convert mass percentages into grams

In this case, we have 24.5 grams of \(\mathrm{Na}\), 14.9 grams of \(\mathrm{Si}\), and 60.6 grams of \(\mathrm{F}\).

Convert grams to moles

Using the molar masses of the elements, we can convert the grams to moles: - Moles of \(\mathrm{Na} = \frac{24.5~\text{g}}{22.99~\text{g/mol}} = 1.065~\text{mol}\) - Moles of \(\mathrm{Si} = \frac{14.9~\text{g}}{28.09~\text{g/mol}} = 0.5306~\text{mol}\) - Moles of \(\mathrm{F} = \frac{60.6~\text{g}}{19.00~\text{g/mol}} = 3.189~\text{mol}\)

Find the simplest whole-number ratio

Divide all the mole values by the smallest value, which is 0.5306. After rounding to the nearest whole number, we get the ratios: - \(\mathrm{Na}: \frac{1.065}{0.5306} \approx 2\) - \(\mathrm{Si}: \frac{0.5306}{0.5306} = 1\) - \(\mathrm{F}: \frac{3.189}{0.5306} \approx 6\) The empirical formula for compound (b) is: \(\mathrm{Na}_2\mathrm{Si}\mathrm{F}_6\). #(c) Compound # Once more, we'll follow the same steps as before.

Convert mass percentages into grams

Given the mass percentages and the remaining mass is occupied by \(\mathrm{O}\), we have 62.1 grams of \(\mathrm{C}\), 5.21 grams of \(\mathrm{H}\), 12.1 grams of \(\mathrm{N}\), and the remainder is \(\mathrm{O}\). The percent of \(\mathrm{O}\) by mass is \(100 - (62.1 + 5.21 + 12.1) = 20.59\%\). Therefore, we have 20.59 grams of \(\mathrm{O}\).

Convert grams to moles

Using the molar masses of the elements, we can convert the grams to moles: - Moles of \(\mathrm{C} = \frac{62.1~\text{g}}{12.01~\text{g/mol}} = 5.169~\text{mol}\) - Moles of \(\mathrm{H} = \frac{5.21~\text{g}}{1.008~\text{g/mol}} = 5.168~\text{mol}\) - Moles of \(\mathrm{N} = \frac{12.1~\text{g}}{14.01~\text{g/mol}} = 0.8640~\text{mol}\) - Moles of \(\mathrm{O} = \frac{20.59~\text{g}}{16.00~\text{g/mol}} = 1.287~\text{mol}\)

Find the simplest whole-number ratio

Divide all the mole values by the smallest value, which is 0.8640. After rounding to the nearest whole number, we get the ratios: - \(\mathrm{C}: \frac{5.169}{0.8640} \approx 6\) - \(\mathrm{H}: \frac{5.168}{0.8640} \approx 6\) - \(\mathrm{N}: \frac{0.8640}{0.8640} = 1\) - \(\mathrm{O}: \frac{1.287}{0.8640} \approx 1.5\) To obtain whole numbers, we multiply all ratios by 2, giving us the empirical formula for compound (c) as: \(\mathrm{C}_{12}\mathrm{H}_{12}\mathrm{N}_2\mathrm{O}_3\).

Key concepts

Mass Percentage Composition

Understanding the mass percentage composition of a compound is crucial for determining its empirical formula. It represents the proportion of each element's mass to the total mass of the compound and is expressed as a percentage.

The mass percentage composition is helpful because it allows us to calculate the amount of each element in a given mass of the compound - typically, we assume 100 grams for simplicity. This then sets the stage for converting these masses into moles, which is a critical step for finding the mole-to-mole ratio needed to determine the empirical formula.

In our example exercises, the first step was to interpret the percentage values as masses. Using these mass values, the next steps involve converting them to moles using the molar mass of each element, and from here, the empirical formula can be determined.

Mole-to-Mole Ratio

The mole-to-mole ratio is a comparison of the moles of one element to the moles of another within a compound. When we have converted the mass of each element to moles, we then compare these values to get a ratio.

This ratio is vital in determining the empirical formula of the compound. An empirical formula is essentially the simplest, whole-number ratio of atoms in the compound. To find this ratio, divide the moles of each element by whichever of the elements has the smallest mole value. If the resulting numbers aren't whole numbers, we may have to multiply them by a common factor to achieve a set of whole numbers, as was necessary in the case of compound (c).

It's important not to round off prematurely in the case of close decimals and sometimes, multipliers are necessary to resolve fractional parts in ratios - like the multiplication by 2 in the case of compound (c) to get the whole-number empirical formula.

Molar Mass

The molar mass is the weight in grams of one mole of any element or compound. It serves as a bridge between the mass of a substance and the number of moles present. Every element has a unique molar mass that can be found on the periodic table, often listed in the unit of grams per mole (g/mol).

In the provided exercise, knowing the molar mass was crucial for converting the mass percentages to moles. For each element, the given mass (assuming 100 grams of substance) was divided by the element's molar mass to find out how many moles of each element were present. It is always important to use precise values for the molar masses to ensure that the subsequent calculations, especially the mole-to-mole ratios, are accurate. Consistency in this is key to obtaining the correct empirical formula.