# Analysis of Compounds

Submitted by ChemPRIME Staff on Wed, 12/08/2010 - 23:48

Up to this point we have obtained all stoichiometric ratios from the coefficients of balanced chemical equations. Chemical formulas also indicate relative amounts of substanceA material that is either an element or that has a fixed ratio of elements in its chemical formula., however, and stoichiometric ratios may be derived from them, too. For example, the formula HgBr2 tells us that no matterAnything that occupies space and has mass; contrasted with energy. how large a sample of mercuric bromide we have, there will always be 2 mol of bromine atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume. for each moleThat chemical amount of a substance containing the same number of units as 12 g of carbon-12. of mercury atoms. That is, from the formula HgBr2 we have the stoichiometric ratio

$\text{S}\left( \frac{\text{Br}}{\text{Hg}} \right)=\frac{\text{2 mol Br}}{\text{1 mol Hg}}$

We could also determine that for HgBr2

$S\left( \frac{\text{Hg}}{\text{HgBr}_{\text{2}}} \right)=\frac{\text{1 mol Hg}}{\text{1 mol HgBr}_{\text{2}}}$

$\text{ }S\left( \frac{\text{Br}}{\text{HgBr}_{\text{2}}} \right)=\text{ }\frac{\text{2 mol Br}}{\text{1 mol HgBr}_{\text{2}}}$

(The reciprocals of these stoichiometric ratios are also valid for HgBr2.)

A combustion train. H2O and CO2, produced by combination of O2 with H and C in the sample, are selectively absorbed by tubes containing Dehydrite [Mg(ClO4)2•3H2O] and ascarite (NaOH on asbestos).

Stoichiometric ratios derived from formulas instead of equations are involved in the most common procedure for determining the empirical formulas of compounds which contain only C, H, and O. A weighed quantity of the substance to be analyzed is placed in a combustion train and heated in a stream of dry O2. All the H in the compound is converted to H2O(g) which is trapped selectively in a previously weighed absorptionPermeation of a solid by a gas or liquid, or permeation of a liquid by a gas. Absorption differs from adsorption in that the substance absorbed is found throughout the absorbent. tube. All the C is converted to CO2(g) and this is absorbed selectively in a second tube. The increase of mass of each tube tells, respectively, how much H2O and CO2 were produced by combustion of the sample.

EXAMPLE 1 A 6.49-mg sample of ascorbic acidIn Arrhenius theory, a substance that produces hydrogen ions (hydronium ions) in aqueous solution. In Bronsted-Lowry theory, a hydrogen-ion (proton) donor. In Lewis theory, a species that accepts a pair of electrons to form a covalent bond. (vitamin C) was burned in a combustion train. 9.74 mg CO2 and 2.64 mg H2O were formed. Determine the empirical formula of ascorbic acid.

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. We need to know the amount of C, the amount of H, and the amount of O in the sample. The ratio of these gives the subscripts in the formula. The first two may be obtained from the masses of CO2 and H2O using the molar masses and the stoichiometric ratios

$\text{S}\left( \frac{\text{C}}{\text{CO}_{\text{2}}} \right)=\frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}}$

$\text{S}\left( \frac{\text{H}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}}$

Thus

$n_{\text{C}}=\text{9}\text{.74}\times \text{10}^{\text{-3}}\text{g CO}_{\text{2}}\times \frac{\text{1 mol CO}_{\text{2}}}{\text{44}\text{.01 g CO}_{\text{2}}}\times \frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}}=\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}$

$n_{\text{H}}=\text{2}\text{.64}\times \text{10}^{\text{-3}}\text{g H}_{\text{2}}\text{O}\times \frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18}\text{.02 g H}_{\text{2}}\text{O}}\times \frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}}=\text{2}\text{.93}\times \text{10}^{\text{-4}}\text{ mol H}$

The compound may also have contained oxygen. To see if it does, calculate the masses of C and H and subtract from the total mass of sample

\begin{align} & m_{\text{C}}=\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}\times \frac{\text{12}\text{.01 g C}}{\text{1 mol C}}=\text{2}\text{.65}\times \text{10}^{\text{-3}}\text{g C}=\text{2}\text{.65 mg C} \ & \ & m_{\text{H}}=\text{2}\text{.93}\times \text{10}^{\text{-4}}\text{ mol H}\times \frac{\text{1}\text{.008 g H}}{\text{1 mol H}}=\text{2}\text{.95}\times \text{10}^{\text{-4}}\text{g H}=\text{0}\text{.295 mg H} \ \end{align}

Thus we have

6.49 mg sample – 2.65 mg C – 0.295 mg H = 3.54 mg O

and

$n_{\text{O}}=\text{3}\text{.54}\times \text{10}^{\text{-3}}\text{ g O}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g O}}=\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol O}$

The ratios of the amounts of the elements in ascorbic acid are therefore

$\frac{n_{\text{H}}}{n_{\text{C}}}=\frac{\text{2}\text{.93}\times \text{10}^{\text{-4}}\text{ mol H}}{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}}=\frac{\text{1}\text{.33 mol H}}{\text{1 mol C}}=\frac{\text{1}\tfrac{1}{3}\text{mol H}}{\text{1 mol C}}=\frac{\text{4 mol H}}{\text{3 mol C}}$

$\frac{n_{\text{O}}}{n_{\text{C}}}=\frac{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol O}}{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}}=\frac{\text{1 mol O}}{\text{1 mol C}}=\frac{\text{3 mol O}}{\text{3 mol C}}$

Since nC:nH:nO is 3 mol C:4 mol H:3 mol O, the empirical formula is C3H4O3.

A drawing of a moleculeA set of atoms joined by covalent bonds and having no net charge. of ascorbic acid is shown here. You can determine by counting the atoms that the molecular formulaThe chemical formula of a substance written using the subscripts that reflect the number of each kind of atom present in a molecule of the substance. For example, the simplest formula for ethane is CH3, but the molecular formula is C2H6 because there are 2 C atoms and 6 H atoms in a molecule of ethane. is C6H8O6—exactly double the empirical formula. It is also evident that there is more to know about a molecule than just how many atoms of each kind are present. In ascorbic acid, as in other molecules, the way the atoms are connected together and their arrangement in three-dimensional space are quite important. A picture showing which atoms are connected to which is called a structural formulaA representation of the structure of a species that indicates the arrangement of the atoms in space and the bonds between atoms.. Empirical formulas may be obtained from percent composition or combustion-train experiments, and, if the molecular weightThe mass of one mole of molecules of a substance; the molar mass of a molecular substance. is known, molecular formulas may be determined from the same data. More complicated experiments are required to find structural formulas. In Example 2 we obtained the mass of O by subtracting the masses of C and H from the total mass of sample. This assumed that only C, H, and O were present. Sometimes such an assumption may be incorrect. When penicillin was first isolated and analyzed, the fact that it contained sulfur was missed. This mistake was not discovered for some time because the atomic weightThe average mass of the naturally occurring isotopes of an element, taking into account the different natural abundances of the isotopes. Expressed relative to the value of exactly 12 for carbon-12; also called atomic mass. of sulfur is almost exactly twice that of oxygen. Two atoms of oxygen were substituted in place of one sulfur atom in the formula.

A 3D representation of L-Ascorbic Acid