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Chapter 3: Problem 30

What is the difference between the molar mass and the empirical formula mass of a compound? When are these masses the same, and when are they different? When different, how is the molar mass related to the empirical formula mass?

Short Answer

Expert verified
The molar mass of a compound is the mass of one mole of the substance, while the empirical formula mass is the mass of the simplest formula representing the compound's lowest whole number element ratio. These masses are the same when the compound has the simplest possible ratio of elements, meaning its molecular formula is the same as its empirical formula. They are different when the molecular formula is not in the simplest ratio, and in this case, the molar mass is an integer multiple of the empirical formula mass, expressed as M = n × m, where n is the n-factor, and can be found by dividing the molar mass by the empirical formula mass.

Step by step solution

01

Define Molar Mass and Empirical Formula Mass

Molar mass is the mass of one mole of a substance, usually represented in grams per mole (g/mol). It is calculated by summing up the atomic masses of all the atoms in the compound. Empirical formula mass is the mass of the most simplified formula of a compound, which represents the lowest whole number ratio of elements present in the compound. It is calculated by summing up the atomic masses of all the atoms in the empirical formula.
02

Describe when molar mass and empirical formula mass are the same

Molar mass and empirical formula mass are the same when a compound has the simplest possible ratio of elements. In other words, the compound's molecular formula is the same as its empirical formula. For example, the molecular formula of water (H₂O) is the same as its empirical formula, so the molar mass and empirical formula mass are the same.
03

Describe when molar mass and empirical formula mass are different

Molar mass and empirical formula mass are different when a compound has a molecular formula that is not the simplest possible ratio of elements. In other words, the compound's molecular formula is a multiple of its empirical formula. For example, the molecular formula of benzene (C₆H₆) is different from its empirical formula (CH), so the molar mass and empirical formula mass are different.
04

Explain the relation between molar mass and empirical formula mass when they are different

When the molar mass and empirical formula mass are different, the molar mass is an integer multiple of the empirical formula mass. The integer multiple is known as the "n-factor". The relationship between the molar mass (M) and empirical formula mass (m) can be expressed as: M = n × m Where "n" is the n-factor, an integer greater than 1. To find the n-factor, divide the molar mass by the empirical formula mass: n = M / m

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Most popular questions from this chapter

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react 52.9 g of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\) with excess red phosphorus, what mass of tetraphosphorus decaoxide $\left(\mathrm{P}_{4} \mathrm{O}_{10}\right)$ could be produced? $\mathrm{KClO}_{3}(s)+\mathrm{P}_{4}(s) \longrightarrow \mathrm{P}_{4} \mathrm{O}_{10}(s)+\mathrm{KCl}(s) \quad$ (unbalanced)

An element X forms both a dichloride \(\left(\mathrm{XCl}_{2}\right)\) and a tetrachloride \(\left(\mathrm{XCl}_{4}\right) .\) Treatment of 10.00 $\mathrm{g} \mathrm{XCl}_{2}\( with excess chlorine forms 12.55 \)\mathrm{g} \mathrm{XCl}_{4}\( . Calculate the atomic mass of \)\mathrm{X},$ and identify \(\mathrm{X}\) .

The compound cisplatin, $\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2},$ has been studied as an antitumor agent. Cisplatin is synthesized as follows: $$ \mathrm{K}_{2} \mathrm{PtCl}_{4}(a q)+2 \mathrm{NH}_{3}(a q) \rightarrow \operatorname{Pt}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}(s)+2 \mathrm{KCl}(a q) $$ What mass of cisplatin can be produced from \(100 .\) g of $\mathrm{K}_{2} \mathrm{PtCl}_{4}\( and sufficient \)\mathrm{NH}_{3} ?$

The aspirin substitute, acetaminophen $\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{N}\right),$ is produced by the following three-step synthesis: $$ \mathrm{I} . \quad \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{N}(s)+3 \mathrm{H}_{2}(g)+\mathrm{HCl}(a q) \longrightarrow $$ $$ \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ $$ \mathrm{II}\quad \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+\mathrm{NaOH}(a q) \longrightarrow $$ $$ \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCl}(a q) $$ $$ \mathrm{III.} \quad \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}(l) \longrightarrow $$ $$ \mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{N}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(l) $$ The first two reactions have percent yields of 87\(\%\) and 98\(\%\) by mass, respectively. The overall reaction yields 3 moles of acetaminophen product for every 4 moles of \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{N}\) reacted. a. What is the percent yield by mass for the overall process? b. What is the percent yield by mass of Step III?

One of the components that make up common table sugar is fructose, a compound that contains only carbon, hydrogen, and oxygen. Complete combustion of 1.50 \(\mathrm{g}\) of fructose produced 2.20 \(\mathrm{g}\) of carbon dioxide and 0.900 \(\mathrm{g}\) of water. What is the empirical formula of fructose?
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