## Deriving Moles from Grams for a Compound

Our bodies synthesize protein from amino acids. One of these amino acids is glycine, which has the molecular formula C_{2}H_{5}O_{2}N. How many moles of glycine molecules are contained in 28.35 g of glycine?

### Solution

We can derive the number of moles of a compound from its mass following the same procedure we used for an element in the example on deriving moles from grams for an element in the previous lesson.

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, C_{2}H_{5}O_{2}N, 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:

\(\require{cancel}28.35 \mathrm{\cancel{g}}\text{ glycine} \left ( \cfrac{\text{mol glycine}}{75.07 \mathrm{\cancel{g}}} \right ) \)\(= 0.378\text{ mol glycine}\)

This result is consistent with our rough estimate.

## Deriving Grams from Moles for a Compound

Vitamin C is a covalent compound with the molecular formula C_{6}H_{8}O_{6}. The recommended daily dietary allowance of vitamin C for children aged 4–8 years is 1.42 × 10^{−4} mol. What is the mass of this allowance in grams?

### Solution

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:

\(\require{cancel}1.42 × 10^{−4} \mathrm{\cancel{mol}}\text{ vitamin C} \left ( \cfrac{176.124\text{ g}}{\mathrm{\cancel{mol}}\text{ vitamin C}} \right ) \)\(= 0.0250\text{ g vitamin C}\)

This is consistent with the anticipated result.

## Deriving the Number of Atoms and Molecules from the Mass of a Compound

A packet of an artificial sweetener contains 40.0 mg of saccharin (C_{7}H_{5}NO_{3}S), which has the structural formula:

Given that saccharin has a molar mass of 183.18 g/mol, how many saccharin molecules are in a 40.0-mg (0.0400-g) sample of saccharin? How many carbon atoms are in the same sample?

### Solution

The number of molecules in a given mass of compound is computed by first deriving the number of moles, as demonstrated in the example above on deriving moles from grams for a compound, and then multiplying by Avogadro’s number:

Using the provided mass and molar mass for saccharin yields:

\(\require{cancel}0.0400 \mathrm{\cancel{g}} \mathrm{C_7H_5NO_3S} \)\(\left ( \cfrac{\mathrm{\cancel{mol}} \mathrm{C_7H_5NO_3S}}{183.18 \mathrm{\cancel{g}} \mathrm{C_7H_5NO_3S}} \right ) \)\(\left ( \cfrac{6.022 × 10^{23} \mathrm{C_7H_5NO_3S}\text{ molecules}}{1 \mathrm{\cancel{mol}} \mathrm{C_7H_5NO_3S}} \right )\)

\(= 1.31 × 10^{20} \mathrm{C_7H_5NO_3S} \text{ molecules}\)

The compound’s formula shows that each molecule contains seven carbon atoms, and so the number of C atoms in the provided sample is:

\(1.31 × 10^{20} \mathrm{C_7H_5NO_3S}\text{ molecules} \)\(\left ( \cfrac{7\text{ C atoms}}{1 \mathrm{C_7H_5NO_3S}\text{ molecule}} \right ) \)\(= 9.20 × 10^{21}\text{ C atoms}\)