Question

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

Short answer

The empirical formulas for the given compounds are: (a) K3PO4 (b) Na2SiF6 (c) C6H6N2O3

Step-by-step solution

(a) Convert mass percentages to grams for compound (a)

For a 100g sample of the compound, the grams of each element are: - K: 55.3g - P: 14.6g - O: 30.1g

(a) Convert grams to moles for compound (a)

Using the molar mass of each element, convert grams to moles: - K: $\frac{55.3g}{39.10g/mol}=1.414$ moles - P: $\frac{14.6g}{30.97g/mol}=0.471$ moles - O: $\frac{30.1g}{16.00g/mol}=1.881$ moles

(a) Determine mole ratios and empirical formula for compound (a)

Now determine the mole ratios by dividing all mole values by the smallest one: - K: $\frac{1.414}{0.471}$ = 3 - P: $\frac{0.471}{0.471}$ = 1 - O: $\frac{1.881}{0.471}$ = 4 The empirical formula for compound (a) is K3P1O4 or K3PO4.

(b) Convert mass percentages to grams for compound (b)

For a 100g sample of the compound, the grams of each element are: - Na: 24.5g - Si: 14.9g - F: 60.6g

(b) Convert grams to moles for compound (b)

Using the molar mass of each element, convert grams to moles: - Na: $\frac{24.5g}{22.99g/mol}=1.065$ moles - Si: $\frac{14.9g}{28.09g/mol}=0.531$ moles - F: $\frac{60.6g}{19.00g/mol}=3.189$ moles

(b) Determine mole ratios and empirical formula for compound (b)

Now determine the mole ratios by dividing all mole values by the smallest one: - Na: $\frac{1.065}{0.531}$ = 2 - Si: $\frac{0.531}{0.531}$ = 1 - F: $\frac{3.189}{0.531}$ = 6 The empirical formula for compound (b) is Na2SiF6.

(c) Convert mass percentages to grams for compound (c)

For a 100g sample of the compound, the grams of each element are: - C: 62.1g - H: 5.21g - N: 12.1g - O: 20.59g (remainder)

(c) Convert grams to moles for compound (c)

Using the molar mass of each element, convert grams to moles: - C: $\frac{62.1g}{12.01g/mol}=5.162$ moles - H: $\frac{5.21g}{1.008g/mol}=5.169$ moles - N: $\frac{12.1g}{14.01g/mol}=0.863$ moles - O: $\frac{20.59g}{16.00g/mol}=1.287$ moles

(c) Determine mole ratios and empirical formula for compound (c)

Now determine the mole ratios by dividing all mole values by the smallest one: - C: $\frac{5.162}{0.863}$ = 6 - H: $\frac{5.169}{0.863}$ = 6 - N: $\frac{0.863}{0.863}$ = 1 - O: $\frac{1.287}{0.863}$ = 1.5 To correct the fractions in the mole ratio, multiply all mole ratios by 2: - C: 12 - H: 12 - N: 2 - O: 3 The empirical formula for compound (c) is C6H6N2O3.

Key concepts

Molecular Composition

Understanding the molecular composition of a compound involves analyzing its constituent elements and how they are present in terms of mass. In everyday terms, think of a compound as a cake. Just as a cake is made of precise amounts of flour, sugar, eggs, and butter, a chemical compound consists of specific quantities of different elements.
The molecular composition provides us with these proportions which are typically expressed as percentages. For example, in the given exercise, compound (a) has a molecular composition of 55.3% potassium (K), 14.6% phosphorus (P), and 30.1% oxygen (O). This means that if you had 100 grams of this compound, 55.3 grams would be potassium, 14.6 grams phosphorus, and 30.1 grams oxygen.
Understanding molecular composition is essential for converting these percentages into practical terms, such as grams, which in turn helps in finding the empirical formula through chemical analysis.

Mole Ratios

Mole ratios play a crucial role in understanding the stoichiometry of a compound, which allows us to understand the relationship between different elements within a compound.
To find mole ratios, you first convert the mass of each element into moles using its molar mass. The molar mass acts like a bridge to go from the world of grams to the world of chemical quantities. Once you've calculated the moles for each element, the next step is to determine the ratio of these moles by dividing each value by the smallest number of moles calculated.
For instance, for compound (a) in the exercise, moles of K, P, and O are calculated as 1.414, 0.471, and 1.881 respectively. Dividing each by 0.471 (the smallest number), we find that the mole ratios are approximately 3:1:4, leading to the empirical formula K3PO4.
These ratios are essential as they give us the simplest whole-number relationship among the atoms present in a compound.

Molar Mass

Molar mass is a key concept in understanding chemical compositions because it relates the mass of a substance to the amount of substance present, measured in moles. Each element has a defined molar mass, which can be found on the periodic table and is expressed in grams per mole (g/mol).
For example, in the exercise, the molar masses of potassium (39.10 g/mol), phosphorus (30.97 g/mol), and oxygen (16.00 g/mol) are used to convert the mass of these elements into moles. The conversion process is critical for finding both the mole ratios and the empirical formula of a compound.
Knowing the molar mass is fundamental because it allows chemists to proportion the ingredients of a chemical reaction accurately, much like knowing the exact amount of ingredients needed to bake a cake. Without molar mass, we wouldn't be able to accurately deduce how much of each element is reacting or being produced.

Chemical Analysis

Chemical analysis is the process by which we decipher the composition and structure of unknown chemical substances. This analysis can take various forms, from complex instrumentation to basic empirical calculations like those in the original exercise.
In the context of empirical formula determination, chemical analysis is often about using the percent composition by mass to find the simplest integer mole ratio of different atoms in a compound. This is done by taking a hypothetical 100 g sample of the compound, allowing easy conversion of percentages into grams.
For example, the problem listed provides for compound (c) a composition that includes 62.1% carbon, leading to calculations based on 62.1 grams of carbon. Such exercises enhance understanding by making abstract chemical equations tangible.
Therefore, chemical analysis serves as a powerful tool in chemistry, transforming theoretical data into practical, useful information about the compounds we encounter daily.