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Hydrogen peroxide has the empirical formula HO and an empirical formula weight of \(17.0\) amu. If the molecular mass is \(34.0\) amu, what is the molecular formula?

Short Answer

Expert verified
The molecular formula is Hâ‚‚Oâ‚‚.

Step by step solution

01

Understand Empirical and Molecular Formulas

The empirical formula of a compound represents the simplest whole-number ratio of elements in that compound. The molecular formula is an actual representation of the number of each type of atom in a molecule. The problem gives us the empirical formula as HO and the empirical formula weight as \(17.0\, \text{amu}\). We are also provided with the molecular mass, which is \(34.0\, \text{amu}\).
02

Calculate the Number of Empirical Units in the Molecular Formula

To find the molecular formula, we need to determine how many empirical units fit into the molecular formula's weight. We do this by dividing the molecular mass by the empirical formula weight: \[ \frac{34.0\, \text{amu}}{17.0\, \text{amu}} = 2. \] This means the molecular formula contains 2 empirical units.
03

Determine the Molecular Formula

Since the molecular formula consists of 2 empirical units, we multiply the subscripts in the empirical formula by this factor of 2. The empirical formula HO becomes \((\text{H}_1\text{O}_1) \times 2 = \text{H}_2\text{O}_2\), which is the molecular formula for hydrogen peroxide.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Empirical Formula
The empirical formula of a compound provides the simplest, basic overview of its composition. It shows the simplest whole-number ratio of atoms in the compound. For instance, in our example of hydrogen peroxide, the empirical formula is HO. This indicates that for every hydrogen (H) atom, there is one oxygen (O) atom. It’s essential to understand that the empirical formula doesn’t necessarily reflect the actual number of atoms in the compound. Rather, it gives insight into the relative proportions of each element. For example, while HO is the empirical formula of hydrogen peroxide, it doesn’t tell us there are precisely two hydrogen atoms and two oxygen atoms in the actual molecule. Calculating the empirical formula involves simplifying the ratio of the number of moles of each element present in a sample. Once you have the relative ratios, you express them in the smallest possible whole numbers. Understanding empirical formulas greatly helps in predicting some properties of compounds, though the complete structure is revealed by the molecular formula.
Molecular Mass
Molecular mass, often referred to as molecular weight, is the sum of the atomic masses of atoms in a molecule. It's measured in atomic mass units (amu) and provides insight into the actual physical mass of a molecule. In our hydrogen peroxide example, its molecular mass is given as 34.0 amu. This measurement is critical because it helps us distinguish between different compounds that might share the same empirical formula but differ in actual molecular size or structure. One key use of molecular mass is to determine the molecular formula from the empirical formula. By knowing the molecular mass, you can see how many "units" of the empirical formula combine to form the actual molecule. Calculating molecular mass involves multiplying the number of each type of atom by their respective atomic masses from the periodic table and summing these values. Understanding this concept is crucial for tasks involving stoichiometry, which deals with the calculation of reactants and products in chemical reactions.
Molecular Formula
The molecular formula of a compound indicates the actual number of each type of atom present in a molecule. Unlike the empirical formula, which only shows the simplest ratio, the molecular formula provides a true depiction of the compound's structure. In the case of hydrogen peroxide, the empirical formula is HO. However, based on its molecular mass of 34.0 amu and the fact that the empirical formula mass is 17.0 amu, we find that the molecular formula consists of two empirical formula units. This leads to the molecular formula being Hâ‚‚Oâ‚‚. The process to find the molecular formula from the empirical formula involves:
  • Calculating the empirical formula weight.
  • Dividing the molecular mass by the empirical formula weight to find the number of empirical units in the molecular formula.
  • Multiplying the subscripts in the empirical formula by the result to get the molecular formula.
Knowing the molecular formula gives you insight into the precise arrangement and number of atoms in a compound, which is essential for understanding its chemical behavior and properties.

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Most popular questions from this chapter

Caffeine, the stimulant in coffee and tea, has the molecular formula \(\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} .\) Calculate the mass percentage of each element in the substance. Give the answers to three significant figures.

Ammonium nitrate, \(\mathrm{NH}_{4} \mathrm{NO}_{3}\), is used as a nitrogen fertilizer and in explosives. What is the molar mass of \(\mathrm{NH}_{4} \mathrm{NO}_{3} ?\)

While cleaning out your closet, you find a jar labeled "2.21 moles lead nitrite." Since Stock convention was not used, you do not know the oxidation number of the lead. You weigh the contents, and find a mass of \(6.61 \times 10^{5} \mathrm{mg} .\) What is the percentage composition of nitrite?

A titanium ore contains rutile \(\left(\mathrm{TiO}_{2}\right)\) plus some iron oxide and silica. When it is heated with carbon in the presence of chlorine, titanium tetrachloride, \(\mathrm{TiCl}_{4}\), is formed. $$ \mathrm{TiO}_{2}(s)+\mathrm{C}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{TiCl}_{4}(g)+\mathrm{CO}_{2}(g) $$ Titanium tetrachloride, a liquid, can be distilled from the mixture. If \(35.4 \mathrm{~g}\) of titanium tetrachloride is recovered from \(17.4 \mathrm{~g}\) of crude ore, what is the mass percentage of \(\mathrm{TiO}_{2}\) in the ore (assuming all \(\mathrm{TiO}_{2}\) reacts)?

White phosphorus, \(\mathrm{P}_{4}\), is prepared by fusing calcium phosphate, \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\), with carbon, \(\mathrm{C}\), and sand, \(\mathrm{SiO}_{2}\), in an electric furnace. $$ \begin{aligned} 2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+10 \mathrm{C}(s) & \longrightarrow \\ \mathrm{P}_{4}(g) &+6 \mathrm{CaSiO}_{3}(l)+10 \mathrm{CO}(g) \end{aligned} $$ How many grams of calcium phosphate are required to give \(15.0 \mathrm{~g}\) of phosphorus?

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