Problem 50 - Chemistry The Central Science

Question

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{HCO}_{2},\) molar mass \(=90.0 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O},\) molar mass \(=88 \mathrm{~g} / \mathrm{mol}\).

Step-by-Step Solution

Answer

The molecular formulas for (a) and (b) are \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) and \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\), respectively.

Step 1: (a) Molar mass of HCO₂

Find the molar mass of the given empirical formula \(\mathrm{HCO}_{2}\) by adding up the molar masses of hydrogen (H), carbon (C), and oxygen (O) multiplied by 2. We have \(1.008 \;\text{g/mol}\) for H, \(12.011 \;\text{g/mol}\) for C, and \(15.999 \;\text{g/mol}\) for O. \(M_{\text{empirical}} = (1.008) + (12.011) + (2\times15.999) = 45.0 \;\text{g/mol}\) In the same way, find the molar mass of the empirical formula \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\). We have \(2 \times 12.01 \;\text{g/mol}\) for C, \(4 \times 1.008 \;\text{g/mol}\) for H, and \(15.999 \;\text{g/mol}\) for O. \(M_{\text{empirical}} = (2\times12.011) + (4\times1.008) + (15.999) = 44.0 \;\text{g/mol}\) ##Step 2: Calculating the scaling factor##

Step 2: (a) Scaling factor for HCO₂

Divide the given molar mass by the molar mass of the empirical formula to find the scale factor: Scale factor \(= \frac{M_{\text{molecular}}}{M_{\text{empirical}}}=\frac{90.0}{45.0}=2\)

Step 3: (b) Scaling factor for C₂H₄O

Divide the given molar mass by the molar mass of the empirical formula to find the scale factor: Scale factor \(= \frac{M_{\text{molecular}}}{M_{\text{empirical}}}=\frac{88.0}{44.0}=2\) ##Step 3: Finding the molecular formula##

Step 4: (a) Molecular formula for HCO₂

Multiply the empirical formula by the scale factor (2) to find the molecular formula: Molecular formula: \(\mathrm{(HCO}_{2}) \times 2 = \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\)

Step 5: (b) Molecular formula for C₂H₄O

Multiply the empirical formula by the scale factor (2) to find the molecular formula: Molecular formula: \(\mathrm{(C}_{2} \mathrm{H}_{4} \mathrm{O}) \times 2 = \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\) The molecular formulas for (a) and (b) are \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) and \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{2}\), respectively.

Key Concepts

Empirical Formula

Starting with the basics, the empirical formula of a compound represents the simplest whole-number ratio of the elements within it. Think of it as the building block or 'recipe' that shows the proportional amounts of each type of atom. For instance, the empirical formula \(\mathrm{HCO}_{2}\) means for every hydrogen atom, there is one carbon atom and two oxygen atoms.

This concept is vital as it serves as a stepping stone to determining the actual molecular formula. The empirical formula does not necessarily tell us how many atoms are in a molecule, but rather the ratio between the different types of atoms. Understanding that empirical and molecular formulas might differ significantly, but could also be identical, is key. For instance, the empirical formula of water, \(\mathrm{H}_2\mathrm{O}\), is also its molecular formula; however, glucose with a molecular formula of \(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\) has the empirical formula \(\mathrm{CH}_2\mathrm{O}\).

Molar Mass

Molar mass is a crucial topic in chemistry, representing the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It's akin to a 'chemical weight ticket' that tells us how much a given number of atoms weigh. Molar mass can be calculated by summing the atomic masses of all atoms in a compound's empirical formula.

For example, to calculate the molar mass of \(\mathrm{HCO}_{2}\), one would add the atomic masses of one hydrogen atom, one carbon atom, and two oxygen atoms. Understanding how to determine molar mass is not only important in deriving molecular formulas but also in various other chemical calculations, such as finding out how many molecules are in a given sample or converting between mass and moles in stoichiometric calculations.

Scaling Factor

The scaling factor bridges the gap between the empirical formula and the molecular formula. It's essentially the number by which we multiply the subscripts of the atoms in the empirical formula to arrive at the correct molecular formula. To find the scaling factor, we divide the molar mass of the molecular formula (provided or experimentally determined) by the molar mass of the empirical formula.

Let's put this into practice; with a given molar mass of 90 g/mol for a compound with an empirical formula of \(\mathrm{HCO}_{2}\), and knowing that the molar mass of the empirical formula is 45 g/mol, the scaling factor is calculated to be 2. This means that the actual molecule has twice the number of each atom as indicated by the empirical formula. Identifying and applying the correct scaling factor is a crucial skill in accurately determining the chemical makeup of molecules.

Chemical Composition

Chemical composition pertains to the identity and quantity of the elements that make up a chemical compound. It's a more detailed account compared to the empirical formula, as it provides the actual number of atoms of each element in a molecule. The determination of a compound's chemical composition involves not only its empirical formula but also requires knowledge of the molecular formula.

The molecular formula offers a clear picture of the number of each type of atom in a molecule, which is essential when predicting properties and behaviors of the substance. For example, \(\mathrm{H}_{2}\mathrm{C}_{2}\mathrm{O}_{4}\) tells us precisely that there are two hydrogen atoms, two carbon atoms, and four oxygen atoms in a molecule of that substance. Without a clear understanding of chemical composition, it would be challenging for scientists and students alike to discuss, analyze, or predict the characteristics of chemical compounds.