/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 Can the results of a combustion ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 65

Can the results of a combustion analysis ever give the true molecular formula of a compound?

Short Answer

Expert verified
In conclusion, one can obtain the true molecular formula of a compound from the results of a combustion analysis by first determining the empirical formula, which represents the simplest whole number ratio of the elements in the compound. To convert this empirical formula into the true molecular formula, additional information such as the molecular weight or molar mass of the compound is required. Comparing the molecular weight with the empirical formula mass and applying the appropriate multiple will yield the true molecular formula of the compound.

Step by step solution

01

Understand the distinction between empirical and molecular formulas

An empirical formula is the simplest whole number ratio of atoms of each element in a compound. A molecular formula, on the other hand, shows the actual number of atoms of each element in a molecule of the compound. The molecular formula may be a multiple of the empirical formula. For example, glucose has the molecular formula C6H12O6, while its empirical formula is CH2O. The molecular formula is a multiple of the empirical formula (C6H12O6 = 6 * (CH2O)).
02

Perform combustion analysis to determine the empirical formula

Combustion analysis involves burning a known mass of the compound in excess oxygen, resulting in the formation of carbon dioxide and water. By measuring the masses of carbon dioxide and water produced, one can determine the amounts of carbon and hydrogen present in the sample. The ratio of these amounts gives the simplest whole number ratio of carbon and hydrogen atoms in the compound. If elements other than carbon, hydrogen, and oxygen are present in the compound, additional analytical techniques may be required to determine their amounts and include them in the empirical formula.
03

Consider the possibility of obtaining the true molecular formula from the empirical formula

The empirical formula obtained from combustion analysis is the simplest whole number ratio representation of the compound's elemental composition. To determine the true molecular formula, one needs additional information such as the molecular weight or molar mass of the compound. By comparing the empirical formula mass (the sum of the atomic masses of the elements in the empirical formula) with the molecular weight (or molar mass), the ratio between the two can be found. This ratio represents the multiple of the empirical formula required to obtain the true molecular formula. For example, if the empirical formula is CH2O and the molecular weight is found to be 180 g/mol, the ratio of molecular weight to empirical formula mass is (180 g/mol) / (30 g/mol) = 6. Therefore, the molecular formula for this compound is 6 * (CH2O) = C6H12O6.
04

Conclusion

The results of a combustion analysis can be used to determine the empirical formula of a compound, which represents the simplest whole number ratio of the elements present. To obtain the true molecular formula, additional information like the molecular weight or molar mass of the compound is required. Combining the empirical formula and this additional information allows the determination of the true molecular formula of the compound.

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Most popular questions from this chapter

There are two ways to write the equation for the combustion of ethane: $$\begin{aligned}\mathrm{C}_{2} \mathrm{H}_{6}(g)+\frac{7}{2} \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g) \\\2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \rightarrow 6 \mathrm{H}_{2} \mathrm{O}(g)+4 \mathrm{CO}_{2}(g)\end{aligned}$$ Do the different ways of writing the equation affect the calculation of how much \(\mathrm{CO}_{2}\) is produced from a known quantity of \(\mathrm{C}_{2} \mathrm{H}_{6} ?\)

Mining for Gold Unlike most metals, gold occurs in nature as the pure element. Miners in California in 1849 searched for gold nuggets and gold dust in streambeds, where the denser gold could be easily separated from sand and gravel. However, larger deposits of gold are found in veins of rock and can be separated chemically in the following two-step process: (1) \(4 \mathrm{Au}(s)+8 \mathrm{NaCN}(a q)+\mathrm{O}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$4 \mathrm{NaAu}(\mathrm{CN})_{2}(a q)+4 \mathrm{NaOH}(a q)$$ (2) \(2 \mathrm{NaAu}(\mathrm{CN})_{2}(a q)+\mathrm{Zn}(s) \rightarrow\) $$2 \mathrm{Au}(s)+\mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{CN})_{4}\right](a q)$$ If 23 kilograms of ore is \(0.19 \%\) gold by mass, how much Zn is needed to react with the gold in the ore? Assume that reactions 1 and 2 are \(100 \%\) efficient.

You are given a \(0.6240 \mathrm{g}\) sample of a substance with the generic formula \(\mathrm{MCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O} .\) After completely drying the sample (which means removing the 2 mol of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{MCl}_{2}\), , the sample has a mass of \(0.5471 \mathrm{g}\). What is the identity of element M?

GRAS List for Food Additives The alcohol geraniol is on the U.S. Food and Drug Administration's GRAS (generally recognized as safe) list and can be used in foods and personal care products. By itself, geraniol has a roselike odor, but it is frequently blended with other scents to produce the fruity fragrances of some personal care products. Complete combustion of 175 mg of geraniol produces \(499 \mathrm{mg} \mathrm{CO}_{2}\) and \(184 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula of geraniol?

A reaction vessel contains equal masses of magnesium metal and oxygen gas. The mixture is ignited, forming MgO. After the reaction has gone to completion, the mass of the \(\mathrm{MgO}\) is less than the mass of the reactants. Is this result a violation of the law of conservation of mass? Explain your answer.
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