The mass of K is provided, and the corresponding amount of K in moles is requested. Referring to the periodic table, the atomic mass of K is 39.10 amu, and so its molar mass is 39.10 g/mol. The given mass of K (4.7 g) is a bit more than one-tenth the molar mass (39.10 g), so a reasonable “ballpark” estimate of the number of moles would be slightly greater than 0.1 mol.

The molar amount of a substance may be calculated by dividing its mass (g) by its molar mass (g/mol):

The factor-label method supports this mathematical approach since the unit “g” cancels and the answer has units of “mol:”

[latex]4.7\cancel{\text{ g K}}\left(\frac{\text{1 mol K}}{\text{39.10 }\cancel{\text{g K}}}\right)=0.12\text{ mol K}[/latex]

The calculated magnitude (0.12 mol K) is consistent with our ballpark expectation, since it is a bit greater than 0.1 mol.

The molar amount of Ar is provided and must be used to derive the corresponding mass in grams. Since the amount of Ar is less than 1 mole, the mass will be less than the mass of 1 mole of Ar, approximately 40 g. The molar amount in question is approximately one-one thousandth (~10^–3) of a mole, and so the corresponding mass should be roughly one-one thousandth of the molar mass (~0.04 g):

In this case, logic dictates (and the factor-label method supports) multiplying the provided amount (mol) by the molar mass (g/mol):

[latex]9.2\times {10}^{-4}\cancel{\text{mol Ar}}\left(\frac{\text{39.95 }\text{g Ar}}{\text{1 }\cancel{\text{mol Ar}}}\right)=\text{0.037 }\text{g Ar}[/latex]

The result is in agreement with our expectations as noted above, around 0.04 g Ar.

The number of Cu atoms in the wire may be conveniently derived from its mass by a two-step computation: first calculating the molar amount of Cu, and then using Avogadro’s number (N_A) to convert this molar amount to number of Cu atoms:

Considering that the provided sample mass (5.00 g) is a little less than one-tenth the mass of 1 mole of Cu (~64 g), a reasonable estimate for the number of atoms in the sample would be on the order of one-tenth N_A, or approximately 10²² Cu atoms. Carrying out the two-step computation yields:

[latex]\text{5.00 }\cancel{\text{g Cu}}\left(\frac{\text{1 }\cancel{\text{mol Cu}}}{\text{63.55 }\cancel{\text{g Cu}}}\right)\left(\frac{\text{6.022}\times{10}^{23}\text{ }\text{atoms Cu}}{\text{1 }\cancel{\text{mol Cu}}}\right)=4.74\times {10}^{22}\text{ atoms Cu}[/latex]

The factor-label method yields the desired cancellation of units, and the computed result is on the order of 10²² as expected.

We can derive the number of moles of a compound from its mass following the same procedure we used for an element in Example 3:

The molar mass of glycine is required for this calculation, and it is computed in the same fashion as its molecular mass. One mole of glycine, C₂H₅O₂N, contains 2 moles of carbon, 5 moles of hydrogen, 2 moles of oxygen, and 1 mole of nitrogen:

The provided mass of glycine (~28 g) is a bit more than one-third the molar mass (~75 g/mol), so we would expect the computed result to be a bit greater than one-third of a mole (~0.33 mol). Dividing the compound’s mass by its molar mass yields:

[latex]\text{28.35 }\cancel{\text{g }\text{C}_{2}\text{H}_{5}\text{O}_{2}\text{N}}\left(\frac{\text{1 }\text{mol }\text{C}_{2}\text{H}_{5}\text{O}_{2}\text{N}}{\text{75.07 }\cancel{\text{g }\text{C}_{2}\text{H}_{5}\text{O}_{2}\text{N}}}\right)=0.378\text{ mol }\text{C}_{2}\text{H}_{5}\text{O}_{2}\text{N}[/latex]

This result is consistent with our rough estimate.

As for elements, the mass of a compound can be derived from its molar amount as shown:

The molar mass for this compound is computed to be 176.124 g/mol. The given number of moles is a very small fraction of a mole (~10^-4 or one-ten thousandth); therefore, we would expect the corresponding mass to be about one-ten thousandth of the molar mass (~0.02 g). Performing the calculation, we get:

[latex]1.42\times {10}^{-4}\cancel{\text{mol }\text{C}_{6}\text{H}_{8}\text{O}_{6}}\left(\frac{\text{176.124 }\text{g }\text{C}_{6}\text{H}_{8}\text{O}_{6}}{\text{1 }\cancel{\text{mol }\text{C}_{6}\text{H}_{8}\text{O}_{6}}}\right)=\text{0.0250 }\text{g }\text{C}_{6}\text{H}_{8}\text{O}_{6}[/latex]

This is consistent with the anticipated result.

The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in Figure 5, and then multiplying by Avogadro’s number:

Using the provided mass and molar mass for saccharin yields:

[latex]\text{0.0400 }\cancel{\text{g }{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}}\left(\frac{\text{1 }\cancel{\text{mol }}{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}}{\text{183.18 }\cancel{\text{g }{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}}}\right)\left(\frac{6.022\times {10}^{23}\text{ molecules}{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S }}{\text{1 }\cancel{\text{mol }{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}}}\right)=1.31\times {10}^{20}\text{molecules }{\text{C}}_{7}{\text{H}}_{5}{\text{NO}}_{3}\text{S}[/latex]

[latex]9.545\times {10}^{22}\text{molecules }{\text{C}}_{4}{\text{H}}_{10}[/latex]

Since we are being asked to determine the number of moles of each element (C, H, and O) within a mole sample of C₁₂H₁₄O₃, we can use a mole to mole ratio from the chemical formula.

Moles of carbon: For one mole of C₁₂H₁₄O₃, there will be twelve moles of C.

[latex]\text{2.50 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}\left(\frac{\text{12 }\text{mol C}}{\text{1 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}}\right)=30 \text{ mol C} [/latex]

Moles of hydrogen: For one mole of C₁₂H₁₄O₃, there will be fourteen moles of H.

[latex]\text{2.50 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}\left(\frac{\text{14 }\text{mol H}}{\text{1 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}}\right)=35 \text{ mol H} [/latex]

Moles of oxygen: For one mole of C₁₂H₁₄O₃, there will be three moles of O.

[latex]\text{2.50 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}\left(\frac{\text{3 }\text{mol O}}{\text{1 }\cancel{\text{mol C}_{12}\text{H}_{14}\text{O}_{3}}}\right)=7.5 \text{ mol O} [/latex]

[latex]\text{10.0 }\cancel{\text{mol N}_{2}\text{O}_{4}}\left(\frac{\text{2 }\text{mol N}}{\text{1 }\cancel{\text{mol N}_{2}\text{O}_{4}}}\right)=20.0 \text{ mol N} [/latex]

Since we are given grams of isoamyl acetate, we need the molar mass of isoamyl acetate.  Using the periodic table we can calculate the molar mass of isoamyl acetate to be 130.18 g/mol.  Since we are looking for grams of carbon, we also need the molar mass of carbon (12.01 g/mol).  We also need to utilize a mole to mole ratio from a chemical formula since we are looking for “units” of carbon within “units” of the larger isoamyl acetate sample.  Using the provided formula, we can derive the mole to mole:  [latex]\left(\frac{\text{7 }\text{mol C}}{\text{1 }\text{mol C}_{7}\text{H}_{14}\text{O}_{2}}\right) [/latex]

[latex]\text{0.500 }\cancel{\text{g C}_{7}\text{H}_{14}\text{O}_{2}}\left(\frac{\text{1 }\cancel{\text{mol C}_{7}\text{H}_{14}\text{O}_{2}}}{\text{130.18 }\cancel{\text{g C}_{7}\text{H}_{14}\text{O}_{2}}}\right)\left(\frac{\text{7 }\cancel{\text{mol C}}}{\text{1 }\cancel{\text{mol C}_{7}\text{H}_{14}\text{O}_{2}}}\right)\left(\frac{\text{12.01 }\text{g C}}{\text{1 }\cancel{\text{mol C}}}\right)=0.323 \text{ g C} [/latex]

[latex]\text{5.0 }\cancel{\text{g N}_{2}\text{O}_{4}}\left(\frac{\text{1 }\cancel{\text{mol N}_{2}\text{O}_{4}}}{\text{92.02 }\cancel{\text{g N}_{2}\text{O}_{4}}}\right)\left(\frac{\text{4 }\cancel{\text{mol O}}}{\text{1 }\cancel{\text{mol N}_{2}\text{O}_{4}}}\right)\left(\frac{\text{16.00 }\cancel{\text{g O}}}{\text{1 }\cancel{\text{mol O}}}\right)=3.5 \text{ g O} [/latex]

Since we have grams of methane given, we will need the molar mass of methane (16.04 g/mol).  We are looking for atoms of H, meaning we will need Avogadro’s number, and since we are going from “units” of methane to “units” of carbon, we will need a mole to mole ratio [latex]\left(\frac{\text{4 }\text{mol H}}{\text{1 }\text{mol C}\text{H}_{4}}\right) [/latex]

[latex]\text{260.0 }\cancel{\text{g C}\text{H}_{4}}\left(\frac{\text{1 }\cancel{\text{mol C}\text{H}_{4}}}{\text{16.04 }\cancel{\text{g C}\text{H}_{4}}}\right)\left(\frac{\text{4 }\cancel{\text{mol H}}}{\text{1 }\cancel{\text{mol C}\text{H}_{4}}}\right)\left(\frac{6.022\times {10}^{23}\text{atoms H}}{\text{1 }\cancel{\text{mol H}}}\right)=3.905\times {10}^{25} \text{ atoms H} [/latex]

[latex]\text{260.0 }\cancel{\text{g C}\text{H}_{4}}\left(\frac{\text{1 }\cancel{\text{mol C}\text{H}_{4}}}{\text{16.04 }\cancel{\text{g C}\text{H}_{4}}}\right)\left(\frac{\text{1 }\cancel{\text{mol C}}}{\text{1 }\cancel{\text{mol C}\text{H}_{4}}}\right)\left(\frac{6.022\times {10}^{23}\text{atoms C}}{\text{1 }\cancel{\text{mol C}}}\right)=9.761\times {10}^{24} \text{ atoms C} [/latex]