Question

A compound was analyzed and was found to contain the following percentages by mass: hydrogen, 3.09\%; phosphorus, 31.60\%; oxygen, 65.31\%. Determine the empirical formula of the compound.

Short answer

The empirical formula of the compound is \(H_3PO_4\).

Step-by-step solution

Convert percentages to grams

Since the percentages are given, we can assume we have 100 g of the compound. This makes it easy to convert the percentages to grams: - Hydrogen: \(3.09 \% \rightarrow 3.09 \, g\) - Phosphorus: \(31.60 \% \rightarrow 31.60 \, g\) - Oxygen: \(65.31 \% \rightarrow 65.31 \, g\)

Determine the number of moles of each element

Now, we will convert the grams for each element into moles, using their molar masses (H: \(1.01 \, g/mol\); P: \(30.97 \, g/mol\); O: \(16.00 \, g/mol\)): - Hydrogen: \(\frac{3.09 \, g}{1.01 \, g/mol} = 3.06 \, mol\) - Phosphorus: \(\frac{31.60 \, g}{30.97 \, g/mol} = 1.02 \, mol\) - Oxygen: \(\frac{65.31 \, g}{16.00 \, g/mol} = 4.08 \, mol\)

Determine the mole ratios

To find the mole ratios, we will divide each of the mole values by the smallest mole value. In this case, that is 1.02 moles (Phosphorus): - Hydrogen: \(\frac{3.06 \, mol}{1.02 \, mol} = 3\) - Phosphorus: \(\frac{1.02 \, mol}{1.02 \, mol} = 1\) - Oxygen: \(\frac{4.08 \, mol}{1.02 \, mol} = 4\)

Write the empirical formula

The empirical formula is written using the mole ratios obtained in Step 3 as subscripts for the elements. So, the empirical formula of the compound is: \(H_3PO_4\)

Key concepts

Mole Ratios

Mole ratios are a vital concept in chemistry, especially when determining empirical formulas. When we have a chemical compound, we seek to know the simplest integer ratio in which each element is combined. These ratios are not only theoretical but also reflect the actual number of atoms in a compound. To find these ratios, we first compute the moles of each element, often starting with a 100 g sample for simplicity. This initial conversion from percentages or masses into moles is crucial.

The next step involves normalizing these mole values by dividing each by the smallest mole value present. This scaling process helps eliminate the fractional parts, giving a clean-cut whole number ratio. For example, in our exercise, dividing the moles of hydrogen, phosphorus, and oxygen by the smallest mole value (1.02) gives the ratios 3:1:4.

• These ratios turn into subscript numbers in empirical formulas.
• This approach works for any compound, as long as it’s in typical conditions.

This simple but powerful calculation is at the heart of empirical formula determination.

Elemental Composition

Elemental composition is about understanding what percentage of a compound is made up of different elements. Knowing this helps in formulating chemical theories and balancing chemical equations. It's essentially the breakdown of a compound into its constituent masses. In practical terms, it allows us to infer the composition of unknown substances based on their mass contributions.

In our example, the elemental composition was provided as percentages (e.g., 3.09% hydrogen, 31.60% phosphorus, 65.31% oxygen). This can be used to determine how much of each element is present if we consider an arbitrary sample size — typically 100 grams. Converting these percentages into gram values offers a clear measurement for further calculations.

• These gram masses link to mole calculations.
• They enable chemists to back-calculate ratios and establish empirical formulas.

Mastering this concept is essential for anyone interested in uncovering the empirical and molecular formulas of chemical compounds.

Chemical Compound Analysis

Chemical compound analysis is the process we use to determine the structure and composition of a substance. It is a systematic method involving several steps such as identifying the elemental makeup and quantifying the present masses or percentages. The goal is to reach the empirical formula, which gives insight into the simplest whole-number ratio of atoms within a compound.

The process involves:

• Determining elemental composition of the compound.
• Calculating moles for each component using molar mass.
• Finding the simplest mole ratios to establish the empirical formula.

This type of analysis is fundamental in many scientific fields such as pharmaceuticals, materials science, and more, as it allows researchers to establish the basic framework of unknown compounds. Chemists can then use this data to synthesize new materials or identify the characteristics of existing ones. Understanding the nuts and bolts of chemical compound analysis allows anyone to dive deeper into the molecular world.