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A close-up photo of a water droplet on a leaf is shown. The water droplet is not perfectly spherical.
The number of molecules in a single droplet of water is roughly 100 billion times greater than the number of people on earth. (credit: “tanakawho”/Wikimedia commons)

The relationships between formula mass, the mole, and Avogadro’s number can be applied to compute various quantities that describe the composition of substances and compounds. For example, if we know the mass and chemical composition of a substance, we can determine the number of moles and calculate number of atoms or molecules in the sample. Likewise, if we know the number of moles of a substance, we can derive the number of atoms or molecules and calculate the substance’s mass.

Deriving moles from grams for an element

According to nutritional guidelines from the US Department of Agriculture, the estimated average requirement for dietary potassium is 4.7 g. What is the estimated average requirement of potassium in moles?

Solution

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):

A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, “Mass of K atoms ( g )” while the one on the right contains the phrase, “Moles of K atoms ( mol ).” There is a phrase under the arrow that says, “Divide by molar mass (g / mol).”

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

4.7 g K ( mol K 39.10 g ) = 0.12 mol K

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

Check your learning

Beryllium is a light metal used to fabricate transparent X-ray windows for medical imaging instruments. How many moles of Be are in a thin-foil window weighing 3.24 g?

Answer:

0.360 mol

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Deriving grams from moles for an element

A liter of air contains 9.2 × 10 −4 mol argon. What is the mass of Ar in a liter of air?

Solution

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):

A diagram of two boxes connected by a right-facing arrow is shown. The box on the left contains the phrase, “Moles of A r atoms ( mol )” while the one on the right contains the phrase, “Mass of A r atoms ( g ).” There is a phrase under the arrow that says “Multiply by molar mass ( g / mol ).”

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

9.2 × 10 −4 mol Ar ( 39.95 g mol Ar ) = 0.037 g Ar

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

Check your learning

What is the mass of 2.561 mol of gold?

Answer:

504.4 g

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Deriving number of atoms from mass for an element

Copper is commonly used to fabricate electrical wire ( [link] ). How many copper atoms are in 5.00 g of copper wire?

A close-up photo of a spool of copper wire is shown.
Copper wire is composed of many, many atoms of Cu. (credit: Emilian Robert Vicol)

Solution

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:

A diagram of three boxes connected by a right-facing arrow in between each is shown. The box on the left contains the phrase, “Mass of C u atoms ( g ),” the middle box reads, “Moles of C u atoms ( mol ),” while the one on the right contains the phrase, “Number of C u atoms.” There is a phrase under the left arrow that says “Divide by molar mass (g / mol),” and under the right arrow it states, “Multiply by Avogadro’s number ( mol superscript negative one ).”

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 10 22 Cu atoms. Carrying out the two-step computation yields:

5.00 g Cu ( mol Cu 63.55 g ) ( 6.022 × 10 23 atoms mol ) = 4.74 × 10 22 atoms of copper

The factor-label method yields the desired cancellation of units, and the computed result is on the order of 10 22 as expected.

Check your learning

A prospector panning for gold in a river collects 15.00 g of pure gold. How many Au atoms are in this quantity of gold?

Answer:

4.586 × 10 22 Au atoms

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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