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):

\(\require{cancel}4.7 \mathrm{\cancel{g}K}\left ( \cfrac{\text{mol K}}{39.10 \mathrm{\cancel{g}}} \right ) \)\(= 0.12\text{ mol K}\)

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

## 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):

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

\(\require{cancel}9.2 × 10^{-4} \mathrm{\cancel{mol}}\text{ Ar} \left ( \cfrac{39.95\text{ g}}{\mathrm{\cancel{mol}}\text{ Ar}} \right ) \)\(= 0.037\text{ g Ar}\)

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

## Deriving Number of Atoms from Mass for an Element

Copper is commonly used to fabricate electrical wire (see image below). How many copper atoms are in 5.00 g of copper wire?

### 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:

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:

\(\require{cancel}5.00 \mathrm{\cancel{g}}\text{ Cu} \left ( \cfrac{\mathrm{\cancel{mol}}\text{ Cu}}{63.55 \mathrm{\cancel{g}}} \right ) \left (\cfrac{6.022 × 10^{23}\text{ atoms}}{\mathrm{\cancel{mol}}} \right ) \)\(= 4.74 × 10^{22}\text{ atoms of copper}\)

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