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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 is the mass of one mole of a substance, usually in grams per mole (g/mol), while the empirical formula mass is the mass of one mole of the simplest formula representing a compound's composition. These masses are the same when the molecular formula is the same as the empirical formula, often in ionic compounds. They are different when the molecular formula is not the same as the empirical formula, typically in covalent compounds. When different, the molar mass is an integer multiple of the empirical formula mass, found by dividing the molar mass by the empirical formula mass.

Step by step solution

01

Defining Molar Mass and Empirical Formula Mass

Molar mass is the mass of one mole of a substance, which is the mass of 6.022 x 10^23 units (atoms, molecules, or formula units) of that substance. It is usually expressed in grams per mole (g/mol). Empirical formula mass is the mass of one mole of the simplest formula that represents the compound's composition. The empirical formula represents the simplest whole-number ratio of atoms in the compound.
02

When Molar Mass and Empirical Formula Mass Are the Same

The molar mass and empirical formula mass will be the same when the molecular formula of a compound is the same as its empirical formula, meaning that the compound already exists in its simplest whole-number ratio. This often occurs for ionic compounds. For example, consider table salt, which has the molecular formula NaCl. Since its empirical formula is also NaCl, its molar mass and empirical formula mass are the same.
03

When Molar Mass and Empirical Formula Mass Are Different

The molar mass and empirical formula mass will be different when the molecular formula of a compound is not the same as its empirical formula, meaning that the compound does not exist in its simplest whole-number ratio. This frequently occurs for covalent compounds. For example, consider glucose, which has the molecular formula C6H12O6. Its empirical formula is CH2O because the compound's molecular formula can be reduced to a simpler whole-number ratio of atoms. In this case, the molar mass is not equal to the empirical formula mass.
04

Relating Molar Mass and Empirical Formula Mass

When the molar mass and empirical formula mass are different, they are related by an integer multiple. To find the multiple, divide the molar mass by the empirical formula mass, which should give a whole number result (considering reasonable experimental error). Using the glucose example from Step 3, the empirical formula mass of CH2O is 12.01 g/mol (C) + 2.02 g/mol (H) + 16.00 g/mol (O) = 30.03 g/mol. The molar mass of glucose is approximately 180.16 g/mol. Dividing the molar mass by the empirical formula mass: \[ n = \frac{180.16 \ \text{g/mol}}{30.03 \ \text{g/mol}} \approx 6 \] The whole-number multiple, n, is approximately 6. This means that the molecular formula of glucose (C6H12O6) is approximately six times the empirical formula (CH2O).

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Key Concepts

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

Empirical formula
An empirical formula represents the simplest whole-number ratio of atoms within a compound. It provides a minimalistic overview of the elements present in a compound without detailing how the atoms are connected or the quantity of each atom type in a molecule.
This formula is derived directly from the composition by reducing the subscript numbers of the molecular formula. For example, for glucose with the molecular formula C\(_6\)H\(_{12}\)O\(_6\), the empirical formula would be CH\(_2\)O.
It is important to remember:
  • Empirical formulas focus on the relative ratios, not the actual atom count in a specific molecule.
  • While empirical formulas are helpful, they can't convey the molecule's true structure or specific atom arrangement.
These simplified formulas are especially useful for identifying compounds in initial analysis.
Molecular formula
The molecular formula gives the exact number of each type of atom in a molecule. Unlike the empirical formula, it does not simplify the ratio but indicates the precise makeup of the compound.
Consider glucose: its molecular formula C\(_6\)H\(_{12}\)O\(_6\) reveals that each molecule consists of 6 carbon, 12 hydrogen, and 6 oxygen atoms.
Some key points to understand about molecular formulas include:
  • The molecular formula is always an integer multiple of the empirical formula. For glucose, this multiple is 6.
  • It provides detailed structural insights, useful in determining reactivity and interaction with other substances.
  • Molecular formulas are crucial when calculating the molar mass of the compound.
Knowing both the empirical and molecular formulas enriches our understanding of a compound's chemical structure.
Covalent compounds
Covalent compounds form when non-metals share electrons to stabilize their outer electron shells. They can have both empirical and molecular formulas.
  • Covalent compounds frequently have molecular formulas that differ from their empirical counterparts. For instance, glucose has a molecular formula of C\(_6\)H\(_{12}\)O\(_6\) and an empirical formula of CH\(_2\)O.
  • This is because covalent compounds often have multiples of the basic atom ratio, showing real atom count and bonds.
  • Molecular and empirical variations in covalent compounds offer insights into both their elemental makeup and potential syntheses.
Understanding covalent compounds helps students realize why detailed formulas and precise molar measurements are essential for chemistry.

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Most popular questions from this chapter

With the advent of techniques such as scanning tunneling microscopy, it is now possible to "write" with individual atoms by manipulating and arranging atoms on an atomic surface. a. If an image is prepared by manipulating iron atoms and their total mass is \(1.05 \times 10^{-20} \mathrm{~g}\), what number of iron atoms were used? b. If the image is prepared on a platinum surface that is exactly 20 platinum atoms high and 14 platinum atoms wide, what is the mass (grams) of the atomic surface? c. If the atomic surface were changed to ruthenium atoms and the same surface mass as determined in part \(\mathrm{b}\) is used, what number of ruthenium atoms is needed to construct the surface?

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(\mathrm{g})\) or \(\mathrm{NO}_{2}(\mathrm{~g})\) according to these unbalanced equations: $$ \begin{array}{l} \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ In a certain experiment \(2.00 \mathrm{~mol} \mathrm{NH}_{3}(g)\) and \(10.00 \mathrm{~mol}\) \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, \(6.75 \mathrm{~mol} \mathrm{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations, because you cannot assume that the two reactions will occur with equal probability.)

Consider the following unbalanced equation: \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q)\) What masses of calcium sulfate and phosphoric acid can be produced from the reaction of \(1.0 \mathrm{~kg}\) calcium phosphate with \(1.0\) \(\mathrm{kg}\) concentrated sulfuric acid \(\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4}\right.\) by \(\left.\mathrm{mass}\right)\) ?

A sample of urea contains \(1.121 \mathrm{~g} \mathrm{~N}, 0.161 \mathrm{~g} \mathrm{H}, 0.480 \mathrm{~g} \mathrm{C}\), and \(0.640 \mathrm{~g} \mathrm{O} .\) What is the empirical formula of urea?

Consider the reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ Identify the limiting reagent in each of the reaction mixtures given below: a. 50 molecules of \(\mathrm{H}_{2}\) and 25 molecules of \(\mathrm{O}_{2}\) b. 100 molecules of \(\mathrm{H}_{2}\) and 40 molecules of \(\mathrm{O}_{2}\) c. 100 molecules of \(\mathrm{H}_{2}\) and 100 molecules of \(\mathrm{O}_{2}\) d. \(0.50 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}\). e. \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}_{2}\) f. \(1.0 \mathrm{~g} \mathrm{H}_{2}\) and \(0.25 \mathrm{~mol} \mathrm{O}_{2}\) g. \(5.00 \mathrm{~g} \mathrm{H}_{2}\) and \(56.00 \mathrm{~g} \mathrm{O}_{2}\)

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