Quantifying Compounds in Bulk

PREVIOUS: Concept of Mole: From Atomic Scale to Large Scale

In the previous post, we devised a technique to estimate number of atoms of an element in a given mass. In brief, if an element \(^{A}_{Z}X\) has mass number \(A\), then \(1 \, mole\) of atoms of that element will have mass \(A \, gm\). The question now is: How do we extend this concept to find number of molecules in compounds? After all, much of the world we see around us and materials we use in daily life are compounds that consist of multiple elements. Concepts of \(mole\) and \(amu\) would be of limited use if we cannot apply them to analyze compounds. So, let's invest some time in generalizing concepts developed in previous posts to include compounds.

Say you are given \(50 \, gm\) of common salt (\(NaCl\)) and are asked to find the number of formula units (or molecules) in it. How will you solve this problem?

To begin with, note that one formula unit (or molecule) of salt contains one \(Na\) atom and one \(Cl\) atom. One sodium atom, \(^{23}_{11}Na\), has mass of \(23 \, amu\); and one chlorine atom, \(^{35.5}_{17}Cl\), has mass of \(35.5 \, amu\). Therefore, one \(NaCl\) molecule has a total mass of \((23+35.5) \, amu = 58.5 \, amu\). This means that one \(NaCl\) molecule has \(58.5\) nucleons and \(58.5\) times heavier than one hydrogen atom which has mass \(1 \, amu\). So, if we take \(1 \, mole\) of \(NaCl\), it must have mass of:
\[6.023 \times 10^{23} \times (58.8 \times 1.66 \times 10^{-24} \, gm) = 58.8 \, gm\] Therefore, \(50 \, gm\) of common salt (\(NaCl\)) must have : \[\frac{50 \, gm}{58.8 \, gm \, (\mbox{mass of 1 mole of} \, NaCl)} = 0.85 \, moles\].

Consider the case of water (\(H_{2}O\)). It consists of one oxygen atom (\(^{16}_{8}O\)) and two hydrogen atoms (\(^{1}_{1}H\)). Therefore, one \(H_{2}O\) molecule contains \(16+1+1=18\) nucleons, meaning it has a mass of \(18 \, amu\). One mole of \(H_{2}O\) has mass of: \[6.023 \times 10^{23} \times (18 \times 1.66 \times 10^{-24} \, gm) = 18 \, gm\] If we are given \(5 \, gm\) water, there would be: \[\frac{5 \, gm}{18 \, gm \, (\mbox{mass of 1 mole of} \, H_{2}O)} = 0.2778 \, moles\].

Thus, the process used to quantify compounds is not all that different from that we used to quantify elements in the previous post. All we have to do is:

First, find the total number of nucleons in one molecule of the given compound. Say this number is \(N\).
Second, realize that the mass of \(1 \, mole\) of this compound is \(N \, gm\).
Third, the number of moles of this compound in \(P \, gm\) is \(\frac{P \, gm}{N \, gm \, (i.e., \, mass \, of \,1 \, mole)} \).

Consider sugar (\(C){6}H_{12}O_{6}\)) as a final example. Suppose you have prepared lemonade for four people in which you have added \(100 \, gm\) of sugar. And now you are curious about how many sugar molecules will each person consume. How are you going to figure out this number? Here's what you do:

In one sugar molecule, there are six \(^{12}_{6}C\) atoms, twelve \(^{1}_{1}H\) atoms and six \(^{16}_{8}O\) atoms. The total number of nucleons adds up to 180. This means there are 180 nucleons in one sugar molecules, and one mole of sugar would have mass of \(180 \, gm\). Therefore, in \(100 \, gm\) of sugar, there are \(100 \, gm/180 \, gm = 0.556 \, moles\) of sugar molecules in the entire lemonade. Each person will consume \(0.556 \, moles / 4 = 0.139 \, moles\) of sugar. In \(0.139 \, moles\), there are \(0.139 \times 6.023 \times 10^{23} = 8.372 \times 10^{22} \) sugar molecules.

SUMMARY:
\(1 \, amu = 1.66 \times 10^{-24} \, gm = \mbox{mass of one proton (or neutron)} = \mbox{mass of one hydrogen atom}\)

\(1 \, mole = \frac{\mbox{1 gm}}{\mbox{mass of one nucleon (or mass of one hydrogen atom) in grams}} =\frac{1 \, gm}{1.66 \times 10^{-24} \, gm}= 6.023 \times 10^{23}\)

\(1 \, mole \, \mbox{of element} \, ^{A}_{Z}X \, \mbox{has mass of} \, A \, gm\).