Question

A 0.5998 -g sample of a new compound has been analyzed and found to contain the following masses of elements: carbon, \(0.2322 \mathrm{~g}\); hydrogen, \(0.05848 \mathrm{~g}\); oxygen, \(0.3091 \mathrm{~g}\). Calculate the empirical formula of the compound.

Short answer

The empirical formula of the compound is \(CH_3O\).

Step-by-step solution

Calculate the moles of each element

To find the moles of each element, we will divide the mass given by the molar mass of the respective element Moles of carbon: Molar mass of carbon (C) = 12.01 g/mol Moles of C = mass of C (g) / molar mass of C (g/mol) = \(0.2322 / 12.01\) Moles of Hydrogen: Molar mass of hydrogen (H) = 1.008 g/mol Moles of H = mass of H (g) / molar mass of H (g/mol) = \(0.05848 / 1.008\) Moles of Oxygen: Molar mass of oxygen (O) = 16.00 g/mol Moles of O = mass of O (g) / molar mass of O (g/mol) = \(0.3091 / 16.00\)

Determine the mole ratio of the elements

To determine the mole ratio of the elements, divide each of the moles obtained in step 1 by the smallest mole value found Moles of C = \(0.2322 / 12.01 = 0.01933\) Moles of H = \(0.05848 / 1.008 = 0.05799\) Moles of O = \(0.3091 / 16.00 = 0.01932\) Smallest mole value = 0.01932 Mole ratio of C:H:O = \(\frac{0.01933}{0.01932} : \frac{0.05799}{0.01932} : \frac{0.01932}{0.01932}\) Mole ratio of C:H:O = \(1.000 : 2.999 : 1.000\) Since we can't have a fraction in the empirical formula, we'll round the mole ratio to whole numbers. In this case, approximately 1:3:1.

Write the empirical formula based on the mole ratio

Now that we have the mole ratio of 1:3:1 for C:H:O, we can write the empirical formula of the compound Empirical Formula: \(CH_3O\) So the empirical formula of the compound is \(CH_3O\).

Key concepts

Molar Mass

Molar mass is a fundamental concept in chemistry that refers to the mass of one mole of a substance, measured in grams per mole (g/mol). It is essential to determine the molar mass of each element within a compound to calculate the empirical formula accurately. The molar mass is numerically equivalent to the atomic or molecular weight of the substance but expressed in units that allow chemists to work with a convenient quantity of matter, the mole, which corresponds to Avogadro's number, approximately 6.022 x 10²³ particles.

For example, in the provided exercise, the molar mass of carbon (C) is 12.01 g/mol, hydrogen (H) is 1.008 g/mol, and oxygen (O) is 16.00 g/mol. These values are crucial to finding out how many moles of each element are present in a given sample, which is the first step towards determining the empirical formula of the compound.

Mole Ratio

The mole ratio in a chemical compound reflects the proportion of elements present and is determined by comparing the moles of each element in the compound. By finding the number of moles for each element and then dividing these values by the smallest number of moles calculated, we get the mole ratio. It's important to keep in mind that the mole ratio should be expressed in the simplest whole numbers, as it is used to determine the empirical formula of the compound.

In this case, the mole ratio calculation leads to a ratio of approximately 1:3:1 for carbon, hydrogen, and oxygen, respectively. It's essential to convert these ratios to the nearest whole numbers to accurately represent the atoms in a compound. This ratio indicates the simplest whole number ratio between the elements that make up the compound.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. This area of chemistry is grounded in the conservation of mass and the concept of the mole, allowing chemists to predict the amounts of substances consumed and produced in reactions. For empirical formula calculations, stoichiometry is employed to understand the exact proportions of each element within a compound.

The process involves converting mass measurements to moles, establishing the mole ratio, and then using this ratio to deduce the empirical formula. For example, with a mole ratio of 1:3:1 for C:H:O, as determined in the exercise, stoichiometry guides us to the empirical formula of the compound, which represents the smallest whole-number ratio of the elements. Thus, the empirical formula is written as CH₃O, reflecting the found stoichiometric relationship.