Stoichiometry: Chemical Formulas and Equations

Empirical formulas from analyses


Empirical Formulas from Analyses

An empirical formula tells us the relative ratios of different atoms in a compound. The ratios hold true on the molar level as well. Thus, H2O is composed of two atoms of hydrogen and 1 atom of oxygen. Likewise, 1.0 mole of H2O is composed of 2.0 moles of hydrogen and 1.0 mole of oxygen.

We can also work backwards from molar ratios:

if we know the molar amounts of each element in a compound we can determine the empirical formula.


Mercury forms a compound with chlorine that is 73.9% mercury and 26.1% chlorine by mass. What is the empirical formula?

Let's say we had a 100 gram sample of this compound. The sample would therefore contain 73.9 grams of mercury and 26.1 grams of chlorine. How many moles of each atom do the individual masses represent?

For Mercury:

(73.9 g)*(1 mol/200.59 g) = 0.368 moles

For Chlorine:

(26.1 g)*(1 mol/35.45 g) = 0.736 mol

What is the molar ratio between the two elements?

( 0.736 mol Cl/0.368 mol Hg) = 2.0

Thus, we have twice as many moles (i.e. atoms) of Cl as Hg. The empirical formula would thus be (remember to list cation first, anion last):

HgCl2


Molecular formula from empirical formula

The chemical formula for a compound obtained by composition analysis is always the empirical formula. We can obtain the chemical formula from the empirical formula if we know the molecular weight of the compound.

The chemical formula will always be some integer multiple of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula).


Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What is the empirical and chemical formula for ascorbic acid?

In 100 grams of ascorbic acid we would have:

40.92 grams C

4.58 grams H

54.50 grams O

This would give us how many moles of each element?

Determine the simplest whole number ratio by dividing by the smallest molar amount (3.406 moles in this case - see Oxygen):

The relative molar amounts of carbon and oxygen appear to be equal, but the relative molar amount of hydrogen is higher. Since we cannot have "fractional" atoms in a compound, we need to normalize the relative amount of hydrogen to be equal to an integer. 1.333 would appear to be 1 and 1/3, so if we multiply the relative amounts of each atom by '3', we should be able to get integer values for each atom.

C = (1.0)*3 = 3

H = (1.333)*3 = 4

O = (1.0)*3 = 3

or, C3H4O3

This is our empirical formula for ascorbic acid. What about the chemical formula? We are told that the experimentally determined molecular mass is 176 amu. What is the molecular mass of our empirical formula?

(3*12.011) + (4*1.008) + (3*15.999) = 88.062 amu

The molecular mass from our empirical formula is signficantly lower than the experimentally determined value. What is the ratio between the two values?

(176 amu/88.062 amu) = 2.0

Thus, it would appear that our empirical formula is essentially one half the mass of the actual molecular mass. If we multiplied our empirical formula by '2', then the molecular mass would be correct. Thus, the actual molecular formula is:

2* C3H4O3 = C6H8O6


The general flow chart for solving empirical formulas from known mass percentages is:


Combustion analysis

When a compound containing carbon and hydrogen is subject to combustion with oxygen in a special combustion apparatus all the carbon is converted to CO2 and the hydrogen to H2O.

The amount of carbon produced can be determined by measuring the amount of CO2 produced. This is trapped by the sodium hydroxide, and thus we can monitor the mass of CO2 produced by determining the increase in mass of the CO2 trap. Likewise, we can determine the amount of H produced by the amount of H2O trapped by the magnesium perchlorate.


Consider the combustion of isopropyl alcohol. The sample is known to contain only C, H and O. Combustion of 0.255 grams of isopropyl alcohol produces 0.561 grams of CO2 and 0.306 grams of H2O. From this information we can quantitate the amount of C and H in the sample:

Since one mole of CO2 is made up of one mole of C and two moles of O, if we have 0.0128 moles of CO2 in our sample, then we know we have 0.0128 moles of C in the sample. How many grams of C is this?

How about the hydrogen?

Since one mole of H2O is made up of one mole of oxygen and two moles of hydrogen, if we have 0.017 moles of H2O, then we have 2*(0.017) = 0.034 moles of hydrogen. Since hydrogen is about 1 gram/mole, we must have 0.034 grams of hydrogen in our original sample.

When we add our carbon and hydrogen together we get:

0.154 grams (C) + 0.034 grams (H) = 0.188 grams

But we know we combusted 0.255 grams of isopropyl alcohol. The 'missing' mass must be from the oxygen atoms in the isopropyl alcohol:

0.255 grams - 0.188 grams = 0.067 grams oxygen

This much oxygen is how many moles?

Overall therfore, we have:

0.0128 moles Carbon

0.0340 moles Hydrogen

0.0042 moles Oxygen

Divide by the smallest molar amount to normalize:

C = 3.05 atoms

H = 8.1 atoms

O = 1 atom

Within experimental error, the most likely empirical formula for propanol would be:

C3H8O


1996 Michael Blaber