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Chapter 8: Problem 75

How does the molecular formula of a compound differ from the empirical formula? Can a compound's empirical and molecular formulas be the same? Explain.

Short Answer

Expert verified
The molecular formula of a compound represents the actual number of atoms of each element in the compound, whereas the empirical formula represents the simplest whole-number ratio of atoms for each element. A compound's empirical and molecular formulas can be the same if the molecular formula is already in its simplest whole-number ratio, as seen in the case of water (H2O).

Step by step solution

01

Define molecular formula and empirical formula

The molecular formula of a compound represents the actual number of atoms of each element present in the compound. It is a representation of the compound's molecular structure. The empirical formula, on the other hand, represents the simplest whole-number ratio of atoms for each element in a compound. It is the reduced form of the molecular formula.
02

Provide examples to illustrate the difference

For example, consider the molecular formula of glucose, which is C6H12O6. In this case, there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms in the glucose molecule. The empirical formula for glucose is obtained by dividing each subscript in the molecular formula by the greatest common divisor (in this case, 6). The resulting empirical formula for glucose is CH2O. Another example is water (H2O). Both its empirical and molecular formulas are the same: H2O. This indicates that its molecular structure is already expressed in the simplest whole-number ratio.
03

Explain if a compound's empirical and molecular formulas can be the same

Yes, a compound's empirical and molecular formulas can be the same. As explained in the previous step, this occurs when the molecular formula is already expressed in the simplest whole-number ratio.
04

Summarize the answer

In conclusion, the molecular formula of a compound represents the actual number of atoms of each element in the compound, while the empirical formula represents the simplest whole-number ratio of atoms for each element in the compound. A compound's empirical and molecular formulas can be the same if the molecular formula is already in its simplest whole-number ratio.

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Most popular questions from this chapter

When \(1.00 \mathrm{g}\) of metallic chromium is heated with elemental chlorine gas, \(3.045 \mathrm{g}\) of a chromium chloride salt results. Calculate the empirical formula of the compound.

Give the name and calculate the molar mass for each of the following substances. a. \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) b. \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(\mathrm{P}_{4} \mathrm{O}_{6}\) d. \(\mathrm{Bi}_{2} \mathrm{O}_{3}\) e. \(C S_{2}\) f. \(\mathrm{H}_{2} \mathrm{SO}_{3}\)

Calculate the mass in grams of each of the following samples. a. 0.251 mol of ethyl alcohol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\) b. 1.26 mol of carbon dioxide c. \(9.31 \times 10^{-4}\) mol of gold(III) chloride d. 7.74 mol of sodium nitrate e. 0.000357 mol of iron

If an average sodium atom weighs 22.99 amu, how many sodium atoms are contained in \(1.98 \times 10^{13}\) amu of sodium? What will \(3.01 \times 10^{23}\) sodium atoms weigh?

When \(3.269 \mathrm{g}\) of zinc is heated in pure oxygen, the sample gains \(0.800 \mathrm{g}\) of oxygen in forming the oxide. Calculate the empirical formula of zinc oxide.
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