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A sample of a compound of \(\mathrm{Cl}\) and \(\mathrm{O}\) reacts with an excess of \(\mathrm{H}_{2}\) to give \(0.233 \mathrm{~g}\) of \(\mathrm{HCl}\) and \(0.403 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). Determine the empirical formula of the compound.

Short Answer

Expert verified
The empirical formula of the compound is \(ClO_7\).

Step by step solution

01

Calculate the moles of Cl

From the balanced reaction, we know that 1 mol of \(\mathrm{HCl}\) has 1 mol of \(\mathrm{Cl}\). To calculate the moles of \(\mathrm{Cl}\), we use the molar mass of \(\mathrm{HCl}\) which is \(36.461 \mathrm{~g/mol}\) from the periodic table. So, moles of \(\mathrm{Cl}\) = mass of \(\mathrm{HCl}\) / molar mass of \(\mathrm{HCl}\) = \(0.233 \mathrm{~g}\) / \(36.461 \mathrm{~g/mol}\) = \(0.00639 \mathrm{~mol}\) (rounded to 5 decimal places)
02

Calculate the moles of O

Following similar steps to Step 1, we calculate the moles of \(\mathrm{O}\) by dividing the mass of \(\mathrm{H2O}\) by the molar mass of \(\mathrm{H2O}\), \(18.015 \mathrm{~g/mol}\). So, moles of \(\mathrm{O}\) = mass of \(\mathrm{H2O}\) / molar mass of \(\mathrm{H2O}\) = \(0.403 \mathrm{~g}\) / \(18.015 \mathrm{~g/mol}\) = \(0.02238 \mathrm{~mol}\) (rounded to 5 decimal places)
03

Determine the simplest whole number ratio

The empirical formula is the simplest ratio of moles of each element. Thus, calculate the ratio of moles of Cl to O by dividing the moles of each by the smallest number of moles calculated. Ratio of Cl:O = moles of Cl / smallest moles = \(0.00639 / 0.00639\): moles of O / smallest moles = \(0.02238 / 0.00639\) = 3.5. This is a ratio of 1:3.5 but empirical formulas need to be in whole numbers. Multiplying by 2 gives the whole number ratio 1:7, so the empirical formula is \(ClO_7\).

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Key Concepts

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

Understanding the Mole Concept
The mole concept is fundamental in chemistry as it bridges the macroscopic world that we can see and measure, like grams or liters, with the atomic world of atoms and molecules.
Think of a mole like a dozen, but instead of 12, it represents a much larger number: Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles.
This standardized count is used to express the amount of chemical substances. For instance, one mole of
  • any element contains \(6.022 \times 10^{23}\) atoms of that element.
  • a compound, like \(\text{H}_2\text{O}\), contains \(6.022 \times 10^{23}\) molecules.
To use the mole concept in calculations, such as determining the empirical formula of a compound, we often convert masses into moles using the molar mass. This is the mass of one mole of a substance, expressed in grams per mole \( \text{g/mol}\). This conversion acts as a crucial step to reach conclusions about the chemical composition from empirical data.
Chemical Reactions and Mass Conservation
Chemical reactions involve the transformation of substances through the rearrangement of atoms. Starting materials, or reactants, are converted into products.
There's a principle in chemistry called the law of conservation of mass: matter is neither created nor destroyed in a chemical reaction.
In our exercise, the sample compound containing chlorine \(\text{Cl}\) and oxygen \(\text{O}\) reacts to form hydrochloric acid \(\text{HCl}\) and water \(\text{H}_2\text{O}\). This reaction involves a conservation of the number of atoms: the chlorine atoms present in the original compound appear in the
  • \(\text{HCl}\) produced.
  • The oxygen atoms appear in the \(\text{H}_2\text{O}\).
This is because the atoms are simply recombined into new molecules, not lost or created anew. Understanding this allows us to track the amount of reactants and products accurately by using moles and ratios.
Introduction to Stoichiometry
Stoichiometry is the field of chemistry that deals with the quantitative aspects of chemical reactions.
It involves calculations that derive from the law of conservation of mass, allowing chemists to predict the amounts of reactants and products involved in a given reaction.
In our problem, stoichiometry helped find the empirical formula by determining the
  • amount of chlorine and oxygen in the compound before the reaction,
  • then comparing it to the quantities of \(\text{HCl}\) and \(\text{H}_2\text{O}\) formed.
Firstly, it uses mole ratios from the balanced chemical equations to deduce information about reactants and products. Then, these mole ratios help us convert grams of substance to moles, which is crucial in finding the simplest whole-number ratio of elements in a compound, leading to the empirical formula. Stoichiometry thus enables chemists to quantify the reactants needed and products formed, making it an indispensable tool in chemical analysis.

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Most popular questions from this chapter

Carbohydrates are compounds containing carbon, hydrogen, and oxygen in which the hydrogen to oxygen ratio is 2: 1 . A certain carbohydrate contains 40.0 percent carbon by mass. Calculate the empirical and molecular formulas of the compound if the approximate molar mass is \(178 \mathrm{~g}\).

Define limiting reagent and excess reagent. What is the significance of the limiting reagent in predicting the amount of the product obtained in a reaction? Can there be a limiting reagent if only one reactant is present?

A sample of iron weighing \(15.0 \mathrm{~g}\) was heated with potassium chlorate \(\left(\mathrm{K} \mathrm{ClO}_{3}\right)\) in an evacuated container. The oxygen generated from the decomposition of \(\mathrm{KClO}_{3}\) converted some of the Fe to \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). If the combined mass of Fe and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) was \(17.9 \mathrm{~g},\) calculate the mass of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed and the mass of \(\mathrm{KClO}_{3}\) decomposed.

The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other sources is about 26 million tons. The equation for the reaction is $$ \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) $$ How much sulfur (in tons), present in the original materials, would result in that quantity of \(\mathrm{SO}_{2} ?\)

Consider the combustion of carbon monoxide (CO) in oxygen gas $$ 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) $$ Starting with 3.60 moles of \(\mathrm{CO},\) calculate the number of moles of \(\mathrm{CO}_{2}\) produced if there is enough oxygen gas to react with all of the CO.

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