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Therefore, the atomic masses of non-gaseous elements can be determined by mass and volume measurements ongaseous compounds containing these elements. This procedure is fairly general, and most atomic masses can be determined in thisway.

Moles, molecular formulae and stoichiometric calculations

We began with a circular dilemma: we could determine molecular formulae provided that we knew atomic masses,but that we could only determine atomic masses from a knowledge of molecular formulae. Since we now have a method for determining allatomic masses, we have resolved this dilemma and we can determine the molecular formula for any compound for which we have percentcomposition by mass.

As a simple example, we consider a compound which is found to be 40.0% carbon, 53.3% oxygen, and 6.7% hydrogenby mass. Recall from the Law of Definite Proportions that these mass ratios areindependent of the sample, so we can take any convenient sample to do our analysis. Assuming that we have 100.0g of the compound, wemust have 40.0g of carbon, 53.3g of oxygen, and 6.7g of hydrogen. If we could count or otherwise determine the number of atoms ofeach element represented by these masses, we would have the molecular formula. However, this would not only be extremelydifficult to do but also unnecessary.

From our determination of atomic masses, we can note that 1 atom of carbon has a mass which is 12.0 times themass of a hydrogen atom. Therefore, the mass of N atoms of carbon is also 12.0 times the mass of N atoms of hydrogen atoms, no matter what N is. If we consider this carefully, we discover that 12.0g of carbon contains exactly the same number of atoms as does 1.0g ofhydrogen. Similarly, we note that 1 atom of oxygen has a mass which is 16.0 12.0 times the mass of a carbon atom. Therefore, the mass of N atoms of oxygen is 16.0 12.0 times the mass of N atoms of carbon. Again, we can conclude that 16.0g of oxygen containsexactly the same number of atoms as 12.0g of carbon, which in turn is the same number of atoms as 1.0g of hydrogen. Without knowing (or necessarily even caring) what the number is, we can say that itis the same number for all three elements.

For convenience, then, we define the number of atoms in 12.0g of carbon to be 1 mole of atoms. Note that 1 mole is a specific number ofparticles, just like 1 dozen is a specific number, independent of what objects we are counting. The advantage to defining themole in this way is that it is easy to determine the number of moles of a substance we have, and knowing the number ofmoles is equivalent to counting the number of atoms (or molecules) in a sample. For example, 24.0g of carbon contains 2.0 moles ofatoms, 30.0g of carbon contains 2.5 moles of atoms, and in general, x grams of carbon contains x 12.0 moles of atoms. Also, we recall that 16.0g of oxygen contains exactly as many atoms as does 12.0gof carbon, and therefore 16.0g of oxygen contains exactly 1.0 mole of oxygen atoms. Thus, 32.0g of oxygen contains 2.0 moles of oxygenatoms, 40.0g of oxygen contains 2.5 moles, and x grams of oxygen contains x 16.0 moles of oxygen atoms. Even more generally, then, if we have m grams of an element whose atomic mass is M , the number of moles of atoms, n , is

n m M

Now we can determine the relative numbers of atoms of carbon, oxygen, and hydrogen in our unknown compoundabove. In a 100.0g sample, we have 40.0g of carbon, 53.3g of oxygen, and 6.7g of hydrogen. The number of moles of atoms in eachelement is thus

n C 40.0 g 12.0 g mol 3.33 moles
n O 53.3 g 16.0 g mol 3.33 moles
n H 6.7 g 1.0 g mol 6.67 moles

We note that the numbers of moles of atoms of the elements are in the simple ratio n C : n O : n H = 1 : 1 : 2 . Since the number of particles in 1 mole is the same for all elements, then itmust also be true that the number of atoms of the elements are in the simple ratio 1 : 1 : 2. Therefore, the molecular formula of thecompound must be C O H 2 .

Or is it? On further reflection, we must realize that the simple ratio 1 : 1 : 2 need not represent the exactnumbers of atoms of each type in a molecule of the compound, since it is indeed only a ratio. Thus the molecular formula could just aseasily be C 2 O 2 H 4 or C 3 O 3 H 6 . Since the formula C O H 2 is based on empirical mass ratio data, we refer to this as the empirical formula of the compound. To determine the molecular formula , we need to determine the relative mass of a molecule of the compound,i.e. the molecular mass. One way to do so is based on the Law of Combining Volumes,Avogadro's Hypothesis, and the Ideal Gas Law . To illustrate, however, if we were to find that the relative mass of one molecule of the compound is60.0, we could conclude that the molecular formula is C 2 O 2 H 4 .

Review and discussion questions

State the Law of Combining Volumes and provide an example of yourown construction which demonstrates this law.

Explain how the Law of Combining Volumes, combined with theAtomic-Molecular Theory, leads directly to Avogadro's Hypothesis that equal volumes of gas atequal temperatures and pressure contain equal numbers of particles.

Use Avogadro's Hypothesis to demonstrate that oxygen gasmolecules cannot be monatomic.

The density of water vapor at room temperature and atmospheric pressure is 0.737 g L . Compound A is 80.0% carbon by mass, and 20.0% hydrogen. Compound B is 83.3%carbon by mass and 16.7% hydrogen. The density of gaseous Compound A is 1.227 g L , and the density of Compound B is 2.948 g L . Show how these data can be used to determine the molar masses of Compounds A and B, assuming that water hasmolecular mass 18.

From the results above , determine the mass of carbon in a molecule of Compound A and in a molecule ofCompound B. Explain how these results indicate that a carbon atom has atomic mass 12.

Explain the utility of calculating the number of moles in a sample of asubstance.

Explain how we can conclude that 28g of nitrogen gas ( N 2 ) contains exactly as many molecules as 32g of oxygen gas ( O 2 ), even though we cannot possibly count this number.

Questions & Answers

how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Do somebody tell me a best nano engineering book for beginners?
s. Reply
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
is Bucky paper clear?
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Do you know which machine is used to that process?
how to fabricate graphene ink ?
for screen printed electrodes ?
What is lattice structure?
s. Reply
of graphene you mean?
or in general
in general
Graphene has a hexagonal structure
On having this app for quite a bit time, Haven't realised there's a chat room in it.
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
many many of nanotubes
what is the k.e before it land
what is the function of carbon nanotubes?
I'm interested in nanotube
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
good afternoon madam
what is system testing
what is the application of nanotechnology?
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
anybody can imagine what will be happen after 100 years from now in nano tech world
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
silver nanoparticles could handle the job?
not now but maybe in future only AgNP maybe any other nanomaterials
I'm interested in Nanotube
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
can nanotechnology change the direction of the face of the world
Prasenjit Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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