<p class="MsoNormal"><b style="font-size: 1rem;">Medicines from plants</b><span lang="" style="font-size: 1rem;">: The first recorded use of plants for medicinal purposes was found on Egyptian
papyrus dated about 1500 BC. All over the world, from tribal medicine in Africa
to folk remedies in Britain, many thousands of plants have been used to cure
all sorts of ailments. Often, this knowledge has been handed down by word of
mouth, and there is now increasing concern the much of it will be lost for ever
as cultures change, tribes disperse and modern medicines of the industrialised
world dominate.</span><br></p><p class="MsoNormal"><span lang="">The bark of the cinchona tree was used for
centuries by the people of Peru to treat malaria, and we now know that it
contains the anti-malarial drug quinine. The Chinese use herbal medicines
extensively: one of their herbal cures for asthma has been shown to contain
ephedrine, which is used in modern medicine to enlarge the air passages of the
lungs. Even aspirin has its origins in willow bark.</span></p><p class="MsoNormal" style="text-align: center; "><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/7/76/Salix_alba_004.jpg/640px-Salix_alba_004.jpg" style="width: 527.528px;"><span lang=""><o:p><br></o:p></span></p><p class="MsoNormal" style="text-align: center; "><a href="https://commons.wikimedia.org/wiki/File:Salix_alba_004.jpg" target="_blank"><sup>The bark of willow trees contains salicylic acid, the active metabolite of aspirin, and has been used for millennia to relieve pain and reduce fever. Willow, CC BY-SA 2.5</sup></a><span lang=""><o:p><br></o:p></span></p><p class="MsoNormal"><span lang="">Interest in tribal and folk remedies has been
reawakened since the 1960s. Chemists investigate thousands of plants each year
to see if they contain biologically active chemicals that might be developed
into medicinal drugs. There is real hope that newly documented remedies will
lead to drugs that treat cancer, heart disease and HIV/AIDS.<o:p></o:p></span></p><h2><span lang="">INTRODUCTION<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">Many organic chemicals come from crude oil,
many come from plants and animals and others are made synthetically. Chemists
use separation techniques to isolate these substances, followed by analytical
techniques to find out their chemical structure. The methods chemists use to
identify the formulae of these organic compounds and determine their structures
are the subject of this post.<o:p></o:p></span></p><h2><span lang="">Pharmaceutical libraries and stage synthesis<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">Although you may think of a library as a
collection of books, to a pharmaceutical company it is a huge collection of
compounds that may be useful as drugs. Only a few years ago most of the
compounds in these libraries were naturally occurring compounds made by plants,
animals or microorganisms. Chemists would make these complex compounds one at a
time, and each would then be screened for any pharmaceutical activity. This was
a slow and expensive process. Typically, a medicinal chemist might synthesise a
hundred compounds in a year.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">In the past few years there has been a
revolution in building these libraries. Thousands of likely drug molecules can
be built up in stages in a matter of hours using computerized syringes. Basic
building blocks of chemicals are reacted at each stage; they combine and the
resulting molecules become larger and larger.<o:p></o:p></span></p><h3><span lang="">Building up molecules in stages<o:p></o:p></span></h3><p class="MsoNormal"><span lang="">Suppose the reactant molecules that a chemist
wishes to combine are called A, B and C. At stage 1, the A, B and C molecules
are placed in separate reaction vessels and shaken with polymer beads, often
polystyrene. These beads are tiny and one gram will contain approximately one
million. The A, B, or C molecules covalently bond to the polymer beads. This makes
it easier to wash and separate the growing molecules at the end of each stage. The
reactant molecules, attached to their polymer beads, are then split equally
into three reaction vessels, so that each one contains a third of each
reactant.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">At stage 2, more of A, B or C is added to each
of the three reaction vessels giving nine possible product molecules, all
attached to polymer beads:<o:p></o:p></span></p><h4 align="center" style="text-align:center"><span lang="">A-A, A-B, A-C, B-A,
B-B, B-C, C-A, C-B, C-C<o:p></o:p></span></h4><p class="MsoNormal"><span lang="">The polymer beads in each of the three
reaction vessels are washed to remove any unreacted reagent. They are again
split into three equal portions and mixed together in another three reaction
vessels.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">At stage 3, A, B or C is again added to give
27 different product molecules. Each is still attached to its polystyrene bead.
The beads are washed and the mix and split is repeated a fourth time.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">At stage 4, A, B or C is added. There are now
81 different product molecules.<o:p></o:p></span></p><h3><span lang="">Large-scale screening<o:p></o:p></span></h3><p class="MsoNormal"><span lang="">By the sixth stage there are 46 656 product
molecules. It would be no use producing this number of molecules if it took
years to screen each one. Drugs are described as fitting into the receptor
sites of molecules. These molecules are usually enzymes. To screen the
thousands of molecules produced from combinatorial chemistry rapidly for potential
drug activity, the molecules are split from their polymer beads and reacted
with enzymes. Only those that affect enzymes by fitting into the receptor sites
are developed further.</span></p><p class="MsoNormal" style="text-align: center; "><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Chemical_Genomics_Robot.jpg/640px-Chemical_Genomics_Robot.jpg" style="width: 527.528px;"><span lang=""><o:p><br></o:p></span></p><p class="MsoNormal" style="text-align: center; "><a href="https://commons.wikimedia.org/wiki/File:Chemical_Genomics_Robot.jpg" target="_blank"><sup>High-throughput screening robots used in the screening and manufacture of drugs. Maggie Bartlett, National Human Genome Research Institute, Public Domain</sup></a><span lang=""><o:p><br></o:p></span></p><h2><span lang="">WORKING OUT THE FORMULA OF A COMPOUND<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">I am going to show how we can work out the
formulae of completely unknown compounds, such as those that might be present in
a medicinal herb. First, we need to understand what percentage compositions and
empirical formulae mean.<o:p></o:p></span></p><h3><span lang="">PERCENTAGE COMPOSITIONS AND EMPIRICAL FORMULAE<o:p></o:p></span></h3><p class="MsoNormal"><span lang="">One very useful piece of information about any
compound is its percentage composition. This means the percentage by mass of
each element in the compound. If you have just discovered what you think is a
new compound, finding the percentage composition is one of the first
investigations you would carry out. You can either decompose a known mass of
the compound into elements, or burn it in oxygen and weigh the products formed,
such as carbon dioxide and water (called combustion analysis). The masses of
any other elements present in the compound are found by other means.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">In the combustion analysis for carbon and
hydrogen. A few milligrams of the compound being analysed are completely
combusted. The carbon and hydrogen in the compound form CO<sub>2</sub> and H<sub>2</sub>O,
respectively. The increased masses of the two absorbent materials are measured
to give the masses of CO<sub>2</sub> and H<sub>2</sub>O produced. Given that
the mass of the original sample is known, the percentage composition can be
calculated. Once the percentage composition is known, the empirical formula of
a compound can be calculated.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The empirical formula is the simplest
whole-number ratio of the number of atoms of each element in a compound. For
example, benzene has a molecular formula of C<sub>6</sub>H<sub>6</sub> but its
empirical formula is CH.<o:p></o:p></span></p><h3><span lang="">EXAMPLE 1<o:p></o:p></span></h3><p class="MsoNormal"><span lang="">Combustion analysis shows that the compound
that gives cinnamon its characteristic aroma has a percentage composition of 81.8
per cent carbon, 6.1 percent hydrogen and 12.1 per cent oxygen. Calculate its
empirical formula. (A<sub>r</sub>: C = 12.0, H = 1.0, O = 16.0)<o:p></o:p></span></p><h3><span lang="">SOLUTION<o:p></o:p></span></h3><p class="MsoNormal"><span lang="">The simplest way to deal with this problem is
to imagine that you have 100.0 g of the compound so that the masses of the
elements are easy to work out as shown below. Remember that the amount in moles
is the mass in grams divided by the mass of one mole in grams.<o:p></o:p></span></p><p class="MsoNormal"><span lang=""><br></span></p><table class="table table-bordered"><tbody><tr><td><br></td><td>Carbon<br></td><td>Hydrogen<br></td><td>Oxygen<br></td></tr><tr><td><span style="background-color: rgb(255, 255, 255);">Mass in grams:</span><br></td><td><span style="background-color: rgb(255, 255, 255);">81.8</span><br></td><td><span style="background-color: rgb(255, 255, 255);">6.1</span><br></td><td><span style="background-color: rgb(255, 255, 255);">12.1</span><br></td></tr><tr><td>Amount in moles:<br></td><td>81.8/12.0 = 6.82<br></td><td>6.1/1.0 = 6.1<br></td><td>12.1/16.0 = 0.756<br></td></tr><tr><td><span style="background-color: rgb(255, 255, 255);">Simplest ratio: (divide by the smallest number)</span><br></td><td><span style="background-color: rgb(255, 255, 255);">6.82/0.756 = 9.02</span><br></td><td><span style="background-color: rgb(255, 255, 255);">6.1/0.756 = 8.1</span><br></td><td><span style="background-color: rgb(255, 255, 255);">0.756/0.756 = 1.00</span><br></td></tr><tr><td>Simplest whole-number ratio:<br></td><td>9<br></td><td>8<br></td><td>1<br></td></tr></tbody></table><p class="MsoNormal"><span style="font-size: 1rem;">So its empirical formula is C</span><sub>9</sub><span style="font-size: 1rem;">H</span><sub>8</sub><span style="font-size: 1rem;">O</span><br></p><p class="MsoNormal"><span lang="">Another way of calculating the empirical
formula from combustion analysis data is to use the actual masses of the different
compounds produced. As shown in the next example below.<o:p></o:p></span></p><h3><span lang="">EXAMPLE 2<o:p></o:p></span></h3><p class="MsoNormal"><span lang="">0.100 g of a sugar, known to contain only
carbon, hydrogen and oxygen, is completely combusted to give 0.147 g CO<sub>2</sub>
and 0.0600 g H<sub>2</sub>O. Calculate its empirical formula.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">First, we need to calculate the masses of C, H
and O in the compound.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">There are 12.0 g of C in 44.0 g CO<sub>2</sub>
(Remember: 1 mol CO<sub>2</sub> = 12.0 + (16.0 </span><span lang="">×</span><span lang=""> 2) g)<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Therefore, the mass of C in 0.147 g = 0.147 </span><span lang="">×</span><span lang=""> 12.0/44.0 = 0.0401 g<o:p></o:p></span></p><p class="MsoNormal"><span lang="">There are 2.0 g H in 18.0 g H<sub>2</sub>O (1
mol H<sub>2</sub>O = (1.0 x 2) + 16.0 g)<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Therefore, the mass of H in 0.0600 g = 0.0600 </span><span lang="">× </span><span lang="">2.0/18.0 = 0.0067 g<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Since we now know the masses of C and H in
0.100 g of sugar, the rest of the mass must result from O:<o:p></o:p></span></p><p class="MsoNormal"><span lang="">0.0401 + 0.0067 = 0.0468 g<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Therefore, mass of O in 0.100 g = 0.100 - 0.0468
= 0.053 g<o:p></o:p></span></p><p class="MsoNormal"><span lang="">We can proceed now as we did in the previous example:</span></p><p class="MsoNormal"><span lang=""><o:p></o:p></span></p><table class="table table-bordered" style="width: 513.636px;"><tbody><tr><td><br></td><td>Carbon<br></td><td>Hydrogen<br></td><td>Oxygen<br></td></tr><tr><td><span style="background-color: rgb(255, 255, 255);">Mass in grams:</span><br></td><td><span style="background-color: rgb(255, 255, 255);">0.0401</span><br></td><td><span style="background-color: rgb(255, 255, 255);">0.0067</span><br></td><td><span style="background-color: rgb(255, 255, 255);">0.053</span><br></td></tr><tr><td>Amount in moles:<br></td><td>0.0401/12.0 = 0.00334</td><td>0.0067/1.0 = 0.0067</td><td>0.053/16.0 = 0.0033</td></tr><tr><td><span style="background-color: rgb(255, 255, 255);">Simplest ratio: (divide by the smallest number)</span><br></td><td><span style="background-color: rgb(255, 255, 255);">0.00334/0.0033 = 1.0</span></td><td><span style="background-color: rgb(255, 255, 255);">0.0067/0.0033 = 2.0</span></td><td><span style="background-color: rgb(255, 255, 255);">0.0033/0.0033 = 1.00</span></td></tr><tr><td>Simplest whole-number ratio:<br></td><td>1</td><td>2</td><td>1<br></td></tr></tbody></table><p class="MsoNormal"><span style="font-size: 1rem;">So its empirical formula is CH</span><sub>2</sub><span style="font-size: 1rem;">O</span><br></p><h2><span lang="">FINDING THE MOLECULAR FORMULA<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">The empirical formula tells you the simplest
ratio of different atoms in a compound. It does not tell you the actual number
of atoms of each element in a molecule. To find the molecular formula, you need
to know the relative molecular mass, M, of a compound, which is:<o:p></o:p></span></p><p class="MsoNormal"><span lang="">M<sub>r</sub> = mass of 1 molecule of the
compound/one-twelfth mass of one atom of carbon-12<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The molecular formula may be the same as the
empirical formula, or it may be a multiple of it. In the example above, the empirical
formula of the substance found in the vinegar is CH<sub>2</sub>O. So we have:<o:p></o:p></span></p><p class="MsoNormal"><span lang="">Relative mass of CH<sub>2</sub>O = 12 + (2 </span><span lang="">×</span><span lang=""> 1) + 16 = 30<o:p></o:p></span></p><p class="MsoNormal"><span lang="">However, M<sub>r</sub> (CH<sub>2</sub>O) = 60,
so because 30 </span><span lang="">×</span><span lang=""> 2 = 60, the empirical
formula must be multiplied by 2 to find the molecular formula:<o:p></o:p></span></p><p class="MsoNormal"><span lang="">(CH<sub>2</sub>O) </span><span lang="">×</span><span lang=""> 2 = C<sub>2</sub>H<sub>4</sub>O<sub>2</sub><o:p></o:p></span></p><p class="MsoNormal"><span lang="">If the empirical formula of sugar is CH<sub>2</sub>O,
and in this case, M<sub>r</sub> = 180. This is 30 </span><span lang="">×</span><span lang=""> 6, so the molecular formula is:<o:p></o:p></span></p><p class="MsoNormal"><span lang="">(CH<sub>2</sub>O) </span><span lang="">×</span><span lang=""> 6 = C<sub>6</sub>H<sub>12</sub>O<sub>6</sub><o:p></o:p></span></p><p class="MsoNormal"><span lang="">So, when you analyse a compound, you need to
know its relative molecular mass. This is where the mass spectrometer comes in.<o:p></o:p></span></p><h2><span lang="">MASS SPECTROMETRY<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">Mass spectrometry is the most accurate method
of determining relative atomic and molecular masses, but it has many other
applications. The mass spectrometer is used by geologists to date rocks, by
anaesthetists to analyse compounds in a patient’s breath, by pharmaceutical
companies to determine the structure of novel compounds and by the oil industry
to work out where samples of crude oil originated. A mass spectrometer has even
been taken to Mars to analyse rocks and dust on the planet’s surface and gases
in its atmosphere. Clearly, the potential of mass spectrometry is enormous.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The mass spectrometer was developed in 1919 by
the eminent English physicist Francis Aston, from apparatus used by J.J.
Thomson (the discoverer of the electron). In 1922, Aston received a Nobel prize
for his work in developing the mass spectrometer.</span></p><p class="MsoNormal" style="text-align: center; "><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/9/9a/Early_Mass_Spectrometer_%28replica%29.jpg/640px-Early_Mass_Spectrometer_%28replica%29.jpg" style="width: 527.528px;"><span lang=""><o:p><br></o:p></span></p><p class="MsoNormal" style="text-align: center; "><a href="https://commons.wikimedia.org/wiki/File:Early_Mass_Spectrometer_(replica).jpg" target="_blank"><sup>Replica of J.J. Thomson's third mass spectrometer. Jeff Dahl, CC BY-SA 3.0</sup></a><span lang=""><o:p><br></o:p></span></p><p class="MsoNormal"><span lang="">Aston used his mass spectrometer to show that
neon gas was composed of isotopes. Neon atoms were ionised, separated, and then
made to hit a photographic plate. Two lines were produced, one much darker than
the other, corresponding to relative isotopic masses of 20 and 22. By measuring
the relative darkness of the two lines, Aston found that neon-20 atoms were ten
times more abundant than neon-22.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">From this information, he worked out the
average atomic mass of neon to be 20.2. As mass spectrometers became more
accurate, a third isotope of neon, neon-21, was discovered. It did not show up
in Aston’s instrument because only 0.26 per cent of naturally occurring neon is
<sup>21</sup>Ne (only 26 atoms in 10 000).<o:p></o:p></span></p><p class="MsoNormal"><span lang="">As mass spectrometers have become smaller they
have been used to analyse the frozen hydrocarbon surface of Titan, the largest
moon of the planet Saturn, and to search for evidence of life on Mars. Mass
spectrometers are used to determine the ratio of carbon-12 to carbon-14 in dead
organic matter. This is the basis of carbon-14 dating.<o:p></o:p></span></p><h2><span lang="">HOW A MASS SPECTROMETER WORKS<o:p></o:p></span></h2><p class="MsoNormal"><span lang="">All mass spectrometers have three basic
operations:<o:p></o:p></span></p><ul><li><span lang="">producing gaseous ions from a sample;</span></li><li>separating the ions according to their mass
(and charge):</li><li>detecting the ions.</li></ul><p class="MsoNormal"><span lang="">The type of mass spectrometer that dominated
much of the 20th century was based on Aston’s apparatus and it is still in use
today. A vaporised sample is injected into the instrument. The exceptional
sensitivity of the instrument means that only nanograms (10<sup>-9</sup> g) of
a sample are required. This is then ionised to form positive ions by bombarding
the atoms or molecules with high-energy electrons from an electron gun. The bombarding
electrons knock electrons from the atoms or molecules in the sample, creating
positively charged ions.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">The ions are separated according to their mass
and charge, first by accelerating them in an electric field of several thousand
volts between two negatively charged plates. Next, they enter a strong magnetic
field, which deflects them into a series of separate circular paths according to
their mass/charge ratio. Positive ions with higher mass/charge ratios are
deflected less than those with lower ratios. So by varying precisely the
strength of the magnetic field, ions of a particular mass/charge ratio can be
focused on the detector in sequential order to build up a spectrum.<o:p></o:p></span></p><p class="MsoNormal"><span lang="">On the mass spectra, the x-axis is labelled
mass/charge ratio (m/e). Although 2+ ions do occur (when two electrons are
removed). 1+ ions are far more abundant. For these, the charge e = +1, which
makes the mass/charge ratios of the peaks equal to the relative masses of the
ions. This is why you will sometimes see the r-axis labelled simply: mass.</span></p><p class="MsoNormal" style="text-align: center; "><img src="https://upload.wikimedia.org/wikipedia/commons/7/7b/ObwiedniaPeptydu.gif" style="width: 400px;"><span lang=""><o:p><br></o:p></span></p><p class="MsoNormal" style="text-align: center; "><a href="https://commons.wikimedia.org/wiki/File:ObwiedniaPeptydu.gif" target="_blank"><sup>Mass spectrum of a peptide showing the isotopic distribution. Mkotl, GNU General Public License</sup></a><span lang=""><o:p><br></o:p></span></p><p>























































































































































</p><h2><span lang="">REFERENCES</span></h2><p><a href="https://www.ru.nl/systemschemistry/equipment/analysis-separation/about-analysis/" target="_blank">https://www.ru.nl/systemschemistry/equipment/analysis-separation/about-analysis/</a><span lang=""><br></span></p><p><a href="http://unaab.edu.ng/funaab-ocw/attachments/474_CHM%20703.pdf" target="_blank">http://unaab.edu.ng/funaab-ocw/attachments/474_CHM%20703.pdf</a><br></p><p><a href="https://www.ncbi.nlm.nih.gov/books/NBK207662/" target="_blank">https://www.ncbi.nlm.nih.gov/books/NBK207662/</a><br></p><p><a href="https://www.bbc.co.uk/bitesize/guides/zqqtrwx/revision/1" target="_blank">https://www.bbc.co.uk/bitesize/guides/zqqtrwx/revision/1</a><br></p><p><a href="https://www.bbc.co.uk/bitesize/guides/zp2wrwx/revision/3" target="_blank">https://www.bbc.co.uk/bitesize/guides/zp2wrwx/revision/3</a><br></p><p><a href="https://en.wikipedia.org/wiki/Medicinal_plants" target="_blank">https://en.wikipedia.org/wiki/Medicinal_plants</a><br></p><p><a href="https://www.thoughtco.com/drugs-and-medicine-made-from-plants-608413" target="_blank">https://www.thoughtco.com/drugs-and-medicine-made-from-plants-608413</a><br></p><p><a href="https://thesunlightexperiment.com/blog/2018/6/7/9-famous-examples-of-drugs-that-came-from-plants" target="_blank">https://thesunlightexperiment.com/blog/2018/6/7/9-famous-examples-of-drugs-that-came-from-plants</a><br></p><p><a href="https://syrris.com/applications/drug-discovery-and-development/" target="_blank">https://syrris.com/applications/drug-discovery-and-development/</a><br></p><p><a href="https://en.wikipedia.org/wiki/Combinatorial_chemistry" target="_blank">https://en.wikipedia.org/wiki/Combinatorial_chemistry</a><br></p><p><a href="https://syrris.com/applications/discovery-medicinal-chemistry-applications/what-is-library-synthesis-in-chemistry/" target="_blank">https://syrris.com/applications/discovery-medicinal-chemistry-applications/what-is-library-synthesis-in-chemistry/</a><br></p><p><a href="https://www.pharmatutor.org/articles/combinatorial-chemistry-modern-synthesis-approach" target="_blank">https://www.pharmatutor.org/articles/combinatorial-chemistry-modern-synthesis-approach</a><br></p><p><a href="https://www.khanacademy.org/science/chemistry/atomic-structure-and-properties/names-and-formulas-of-ionic-compounds/e/find-the-formula-for-ionic-compounds" target="_blank">https://www.khanacademy.org/science/chemistry/atomic-structure-and-properties/names-and-formulas-of-ionic-compounds/e/find-the-formula-for-ionic-compounds</a><br></p><p><a href="https://chem.libretexts.org/Courses/University_of_British_Columbia/CHEM_100%3A_Foundations_of_Chemistry/06%3A_Chemical_Composition/6.9%3A_Calculating_Molecular_Formulas_for_Compounds" target="_blank">Calculating_Molecular_Formulas_for_Compounds</a><br></p><p><a href="https://chem.libretexts.org/Courses/Los_Angeles_Trade_Technical_College/DMA_Chem_51_Su_19/2%3A_Beginning_Chemistry_(Ball)/05%3A_Stoichiometry_and_the_Mole/5.4_Percent_Composition%2C_Empirical_and_Molecular_Formulas" target="_blank">Stoichiometry_and_the_Mole/5.4_Percent_Composition%2C_Empirical_and_Molecular_Formulas</a><br></p><p><a href="https://www.chem.tamu.edu/class/fyp/stone/tutorialnotefiles/fundamentals/empirical.htm">https://www.chem.tamu.edu/class/fyp/stone/tutorialnotefiles/fundamentals/empirical.htm</a><a href="https://www.chem.tamu.edu/class/fyp/stone/tutorialnotefiles/fundamentals/empirical.htm" target="_blank"></a></p><p><a href="https://study.com/academy/lesson/calculating-percent-composition-and-determining-empirical-formulas.html#:~:text=Determining%20the%20Empirical%20Formula&amp;text=To%20do%20this%2C%20you%20need,number%20mole%20ratio%20of%20atoms.&amp;text=Use%20the%20composition%20in%20moles,whole%20number%20ratio%20of%20atoms." target="_blank">calculating-percent-composition-and-determining-empirical-formulas</a><br></p><p><a href="https://socratic.org/questions/how-do-you-find-molecular-formula-of-a-compound" target="_blank">https://socratic.org/questions/how-do-you-find-molecular-formula-of-a-compound</a><br></p><p><a href="https://chemed.chem.purdue.edu/genchem/probsolv/stoichiometry/molecular2/mf2.4.html" target="_blank">https://chemed.chem.purdue.edu/genchem/probsolv/stoichiometry/molecular2/mf2.4.html</a><br></p><p><br><br></p><h2><span lang=""><o:p></o:p></span></h2>