Question

Give the empirical formula of each of the following com$\mathrm{H}$, and $0.0065\mathrm{mol}\mathrm{O}$; (b) $11.66\mathrm{g}$ iron and $5.01\mathrm{g}$ oxygen; (c) $40.0\mathrm{\%}\mathrm{C},6.7\mathrm{\%}\mathrm{H}$, and $53.3\mathrm{\%}\mathrm{O}$ by mass.

Short answer

The empirical formulas for the given compounds are: (a) C₂H₅O, (b) FeO₂, and (c) C₂H₄O.

Step-by-step solution

Case (a): 0.010 mol C, 0.030 mol H, and 0.0065 mol O.

To find the empirical formula, we first need to find the mole ratio of these elements. 1. Divide each mole amount by the smallest value among them: C: $\frac{0.010}{0.0065}\approx 1.54$ moles H: $\frac{0.030}{0.0065}\approx 4.62$ moles O: $\frac{0.0065}{0.0065}=1$ mole 2. Next, round each number to the nearest whole number. This will give us the smallest whole number ratio for each element: C = 2 H = 5 O = 1 3. Finally, write the empirical formula using these ratios: C₄H₁₀O

Case (b): 11.66 g iron and 5.01 g oxygen.

To find the empirical formula, we first need to find the mole ratio of iron and oxygen. 1. Convert each mass to moles using molar mass(Iron=55.85 g/mol, Oxygen=16.00 g/mol): Iron: $\frac{11.66}{55.85}\approx 0.2086$ moles Oxygen: $\frac{5.01}{16.00}\approx 0.3131$ moles 2. Divide each mole amount by the smallest value among them: Iron: $\frac{0.2086}{0.2086}=1$ mole Oxygen: $\frac{0.3131}{0.2086}\approx 1.5$ moles 3. Next, round each number to the nearest whole number. This will give us the smallest whole number ratio for each element: Iron = 1 Oxygen = 2 4. Finally, write the empirical formula using these ratios: FeO₂

Case (c): 40.0 % C, 6.7 % H, and 53.3 % O by mass.

To find the empirical formula, we first need to find the mole ratio of Carbon, Hydrogen, and Oxygen. 1. Assume a 100 g sample, and convert each mass percentage to grams and then to moles using molar mass(Carbon=12.01 g/mol, Hydrogen=1.01 g/mol, Oxygen=16.00 g/mol): Carbon: $\frac{40}{12.01}=3.33$ moles Hydrogen: $\frac{6.7}{1.01}\approx 6.63$ moles Oxygen: $\frac{53.3}{16.00}\approx 3.33$ moles 2. Divide each mole amount by the smallest value among them: Carbon: $\frac{3.330}{3.330}=1$ mole Hydrogen: $\frac{6.63}{3.330}\approx 1.99$ moles Oxygen: $\frac{3.33}{3.330}=1$ mole 3. Next, round each number to the nearest whole number. This will give us the smallest whole number ratio for each element: Carbon = 1 Hydrogen = 2 Oxygen = 1 4. Finally, write the empirical formula using these ratios: C₂H₄O

Key concepts

Mole Ratio Calculation

Understanding mole ratio calculation is vital when trying to determine the empirical formula of a compound. An empirical formula represents the simplest whole number ratio of elements in a compound. To calculate mole ratios, you must first convert the given quantities of elements into moles, as this exercise demonstrates.

To find the mole ratio, divide the number of moles of each element by the number of moles of the element that is present in the smallest amount. This establishes a baseline and allows us to see how many times more of one element there is compared to another. If the ratios aren't whole numbers, we then multiply the ratios by a common factor to obtain the smallest whole number ratio. For instance, in the exercise, we adjusted the ratios by rounding them to the nearest whole numbers. Students should note that sometimes it may be necessary to multiply by two or more to achieve whole numbers, especially when dealing with decimals close to .5.

Practical Tip:

Always double-check your calculations before finalizing the mole ratios, and be mindful of rounding. Large ratios may require multiplying by a common factor to simplify to the smallest whole numbers.

Chemical Composition

Chemical composition is the identification of the different elements that make up a chemical compound as well as their relative amounts. This concept is key in chemistry because it defines the real nature of the substance. In our exercise, we calculated the chemical composition in terms of percentage by mass and then converted these percentages to moles. For instance, considering a 100 g sample simplifies the conversion of mass percentages to grams, and subsequently, to moles because the percentage can be directly equated to grams in a 100 g sample.

It is also important to express the chemical composition in terms of moles since it reflects the number of particles, and atoms combine in fixed ratios. Understanding the actual number of atoms that combine helps us to construct the empirical formula, which is a central part of chemical identity.

Exercise Improvement Advice:

When improving exercises on chemical composition, emphasize consistent units and be explicit about assumptions, such as using a 100 g sample to simplify calculations.

Molecular Mass

The molecular mass (or molecular weight) of a substance is the mass of one molecule of that substance, usually expressed in atomic mass units (u). It is the sum of the atomic masses of all the atoms in the molecule. In practice, especially for calculating empirical formulas, we use the molar mass, which is the mass of one mole of a substance and is expressed in grams per mole (g/mol).

In the given exercise, molar mass is used to convert the mass of an element into moles, a crucial step in the empirical formula determination. For example, the molar mass of iron (Fe) is 55.85 g/mol, which means each mole of iron weighs 55.85 grams. By knowing the molar mass, one can relate the mass of an element in grams to the number of moles, setting the stage for comparing the proportions of different elements within a compound.

Important Note:

Keep in mind that the molar mass is different for every element and is determined by the combined number of protons and neutrons in the nucleus of the atom. In exercises, it is important to strategically use the molar mass to convert mass to moles accurately for further calculations.