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Chapter 2: Problem 48

Two substances have the same molecular and empirical formulas. Does this mean that they must be the same compound?

Short Answer

Expert verified
Yes, if two substances have the same molecular and empirical formulas, they must be the same compound. This is because the molecular formula shows the actual number of atoms of each element present in a molecule, and the empirical formula shows the simplest whole number ratio of the atoms of each element. If both formulas are the same for both substances, it indicates that they have the same atoms in the same proportion, making them the same compound.

Step by step solution

01

Understanding molecular and empirical formulas

Molecular formula shows the actual number of atoms of each element present in a molecule of a compound. Empirical formula, on the other hand, shows the simplest whole number ratio of the atoms of each element present in a molecule.
02

Comparing similarities between molecular and empirical formulas

When comparing molecular and empirical formulas, it's crucial to determine if their composition is similar. If two substances have the same molecular formula, it suggests that they consist of the same elements in the same proportion and the same number of atoms. Additionally, having the same empirical formula implies that both compounds have the same simplest whole number ratios of the elements present.
03

Analyzing if the same molecular and empirical formulas can indicate the same compound

Since two substances with the same molecular formula have the same atoms in the same proportion, this means that they are the same compound. Furthermore, having the same empirical formula supports this conclusion, as it indicates the simplest whole number ratio of atoms in both compounds is the same. Thus, if two substances have the same molecular and empirical formulas, they must be the same compound.

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Most popular questions from this chapter

The explosion of an atomic bomb releases many radioactive isotopes, including strontium- \(90 .\) Considering the location of strontium in the periodic table, suggest a reason for the fact that this isotope is particularly dangerous for human health.

The natural abundance of \(^{3} \mathrm{He}\) is 0.000137\(\%\) . (a) How many protons, neutrons, and electrons are in an atom of \(^{3} \mathrm{He?}\) (b) Based on the sum of the masses of their subatomic particles, which is expected to be more massive, an atom of \(^{3} \mathrm{He}\) or an atom of \(^{3} \mathrm{H}(\) which is also called tritium)? (c) Based on your answer to part (b), what would need to be the precision of a mass spectrometer that is able to differentiate between peaks that are due to \(^{3} \mathrm{He}^{+}\) and \(^{3} \mathrm{H}^{+}\) ?

Consider an atom of 10 \(\mathrm{B}\) . (a) How many protons, neutrons, and electrons does this atom contain? (b) What is the symbol of the atom obtained by adding one proton to \(^{10} \mathrm{B} ?\) (c) What is the symbol of the atom obtained by adding one neutron to 10 \(\mathrm{B}\) ? (c) What is the symbol of the atom obtained by adding one neutron to 10 \(\mathrm{B} ?(\mathbf{d})\) Are either of the atoms obtained in parts (b) and (c) isotopes of 10 \(\mathrm{B} ?\) If so which one?

The nucleus of 6 Li is a powerful absorber of neutrons. It exists in the naturally occurring metal to the extent of 7.5\(\%\) . In the era of nuclear deterrence, large quantities of lithium were processed to remove 6 Li for use in hydrogen bomb production. The lithium metal remaining after removal of 6 Li was sold on the market. (a) What are the compositions of the nuclei of 6 Li and 7 Li? (b) The atomic masses of 6 Li and 7 Li are 6.015122 and 7.016004 amu, respectively. A sample of lithium depleted in the lighter isotope was found on analysis to contain 1.442\(\%\) 6 Li. What is the average atomic weight of this sample of the metal?

Give the names and charges of the cation and anion in each of the following compounds: (a) \(\mathrm{CaO},(\mathbf{b}) \mathrm{Na}_{2} \mathrm{SO}_{4},(\mathbf{c}) \mathrm{KClO}_{4}\) (\mathbf{d} ) ~ \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2},(\mathbf{e}) \operatorname{Cr}(\mathrm{OH})_{3}\)
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