Question

One compound of mercury with a molar mass of 519 contains \(77.26 \% \mathrm{Hg}, 9.25 \% \mathrm{C},\) and \(1.17 \% \mathrm{H}\) (with the balance being \(\mathrm{O}\) ). Calculate the empirical formula.

Short answer

The empirical formula of the mercury compound is HgC2H2O.

Step-by-step solution

Calculate mass of elements

Determine the mass of each element in 100 g of the compound. For mercury (Hg), mass of Hg = 77.26 g; for carbon (C), mass of C = 9.25 g; for hydrogen (H), mass of H = 1.17 g. The balance consists of oxygen (O), so mass of O = 100 g - (mass of Hg + mass of C + mass of H) = 100 - (77.26 + 9.25 + 1.17) = 12.32 g.

Calculate moles of elements

Divide the mass of each element by its atomic mass to calculate the moles. Moles of Hg = 77.26 g / 200.59 g/mol; moles of C = 9.25 g / 12.01 g/mol; moles of H = 1.17 g / 1.008 g/mol; moles of O = 12.32 g / 16.00 g/mol.

Determining the mole ratio

Find the mole ratio of the elements by dividing each element's moles by the smallest number of moles among the elements calculated in Step 2.

Find the lowest whole number ratio

Divide all of the mole ratios by the smallest ratio to get a whole number ratio. In cases where values are close to .5 or a whole number, multiply all ratios by a common factor (usually 2, 3, etc.) to get whole numbers.

Write the empirical formula

Using the whole number mole ratios as subscripts, write the empirical formula of the compound.

Key concepts

Mole Ratio

Understanding the mole ratio is essential when determining an empirical formula. In simple terms, the mole ratio compares the relative amount of moles of different elements in a compound. This ratio is central to the study of stoichiometry, which is the calculation of reactants and products in chemical reactions.

For example, if we have a compound where the moles of hydrogen are twice the moles of oxygen, the mole ratio of hydrogen to oxygen is 2:1. When calculating the empirical formula, we use the mole ratio to determine the simplest whole number ratio of atoms in a compound. It's like a recipe that tells you how many cups of flour to sugar you need, but instead of cups, we talk about moles, and instead of flour and sugar, we discuss elements.

Molar Mass

Molar mass is like the 'weight' of a mole. To put it simply, it is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). The molar mass is calculated by summing the atomic masses of all the atoms in a single molecule of the substance. Think of it as if you're adding up the weights of each ingredient to find out how heavy your cake will be once it's all mixed up.

The molar mass lets us convert between grams of a substance and the number of moles. This conversion is crucial when working out the empirical formula, as we first need to know how many moles of each element we have, which we get by dividing the mass of each element by its molar mass.

Atomic Mass

Atomic mass is the 'weight' of a single atom, usually measured in atomic mass units (u or amu). For an individual element, the atomic mass roughly equals the mass number, which is the sum of protons and neutrons in the nucleus. However, because most elements exist as a mixture of isotopes (atoms of the same element with different numbers of neutrons), the atomic mass is an average value accounting for the isotopes and their abundances.

For example, the atomic mass of carbon is roughly 12.01 amu, which is used when calculating molar mass or converting between grams and moles in empirical formula calculations. Understanding atomic mass helps us to accurately determine the molar mass of compounds, which is a stepping stone in figuring out the empirical formula.

Stoichiometry

The concept of stoichiometry lies at the heart of chemical reactions. It's all about the quantitative relationships between the amounts of reactants used and products formed in a chemical reaction. Stoichiometry is like a mathematical recipe using balanced equations, ensuring you have the right amounts of ingredients (reactants) to get the product you need.

To apply stoichiometry to empirical formula calculations, you use the mole ratios to ensure each element is in proportion. In the context of the exercise, we use stoichiometry to convert between the mass of elements to the moles, and then to a simple whole number ratio, which gives us the empirical formula that represents the simplest proportion of elements in the compound.