Question

In a series of experiments, a chemist prepared three different compounds that contain only iodine and fluorine and determined the mass of each element in each compound: \(\begin{array}{ccc} \text { Compound } & \text { Mass of Iodine (g) } & \text { Mass of Fluorine (g) } \\ \hline 1 & 4.75 & 3.56 \\ 2 & 7.64 & 3.43 \\ 3 & 9.41 & 9.86 \end{array}\) (a) Calculate the mass of fluorine per gram of iodine in each compound. (b) How do the numbers in part (a) support the atomic theory?

Short answer

The mass ratios of fluorine to iodine in the three compounds are approximately 0.75, 0.45, and 1.05, which can be simplified to whole number ratios of \(\frac{3}{4}\), \(\frac{1}{2}\), and \(1\). These simple whole-number ratios support the atomic theory, as it states that elements combine in simple, whole-number ratios to form compounds.

Step-by-step solution

Determining the ratio of fluorine to iodine masses for each compound

To find the mass of fluorine per gram of iodine in each compound, divide the mass of fluorine by the mass of iodine in each respective compound. Compound 1: \(\frac{3.56}{4.75}\) Compound 2: \(\frac{3.43}{7.64}\) Compound 3: \(\frac{9.86}{9.41}\)

Calculating the mass ratios

Now we will calculate the mass ratios of fluorine to iodine in each compound. Compound 1: \(\frac{3.56}{4.75}= 0.749\) Compound 2: \(\frac{3.43}{7.64}= 0.449\) Compound 3: \(\frac{9.86}{9.41}= 1.047\) Notice that the ratios are approximately \(\frac{3}{4}\), \(\frac{1}{2}\), and \(1\). To make this clearer, we can round each mass ratio to the nearest hundredth: Compound 1: 0.75 Compound 2: 0.45 Compound 3: 1.05

Analyzing how the numbers support the atomic theory

The atomic theory states that elements combine in simple, whole-number ratios to form compounds. Observing the calculated mass ratios, each compound has a ratio of fluorine to iodine with a simple whole number relationship. Since the numbers are approximately \(\frac{3}{4}\), \(\frac{1}{2}\), and \(1\), this means that for every 3, 2, and 4 atoms of iodine, there are approximately 4, 1, and 4 atoms of fluorine in the respective compounds. These whole-number ratios support the atomic theory by indicating that the elements combine in simple, whole-number amounts to form compounds.

Key concepts

Mass Ratio Calculations

Understanding mass ratio calculations is essential for unraveling the composition of chemical compounds. It involves comparing the masses of different elements within a compound. To find the mass of one element per gram of another, like fluorine per gram of iodine, you simply divide the mass of fluorine by the mass of iodine.

For example, in Compound 1 with 4.75 grams of iodine and 3.56 grams of fluorine, the calculation would be \(\frac{3.56}{4.75}\), resulting in a ratio that can be rounded to 0.75. This ratio represents the consistent proportion of fluorine to iodine in that compound, providing insight into its fundamental chemical composition.

Educators stress the significance of mastering these calculations, as they help students to predict and understand how different quantities of elements relate within compounds—laying the foundation for more advanced chemistry concepts.

Chemical Compound Composition

The composition of chemical compounds is determined by the types of elements and the quantity of each element present. Each compound has a unique makeup that can often be expressed through empirical formulas, which show the simplest whole-number ratio of elements. Knowing the mass of each element, as shown in our exercise with the iodine and fluorine compounds, allows us to understand the compound's composition.

For instance, if a compound has exactly twice as much iodine by mass compared to fluorine, it might be represented with a formula where the subscript indicates twice as many iodine atoms as fluorine atoms. These ratios are pivotal in predicting product amounts in chemical reactions and aid in verifying theoretical models of atomic structure within a broader context of chemistry.

Stoichiometry

Stoichiometry deals with the quantitative aspect of chemical formulas and reactions. It is the calculation of the reactants and products in chemical reactions and relies heavily on balanced equations and mole concepts. Stoichiometry enables chemists to make predictions about the outcomes of reactions in terms of amounts of substances consumed or produced.

An understanding of stoichiometry is critical when interpreting our exercise's mass ratios—these ratios guide us in determining what amounts of one substance react with a given amount of another. For example, knowing the ratio of iodine to fluorine helps in predicting how much fluorine is needed to react completely with a certain mass of iodine. Stoichiometry is the heartbeat of practical chemistry as it connects the mass relationships between reactants and products to the theoretical world of atoms and molecules.