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These new mass ratio numbers are highly suggestive in thefollowing way. Notice that, in each case, we took the ratio of oxygen mass to a nitrogen mass of 1, and thatthe resultant ratios have a very simple relationship:

2.28 : 1.14 : 0.57 2 : 1 : 0.5 4 : 2 : 1
The masses of oxygen appearing in these compounds are in simple whole number ratios when we take a fixed amount of nitrogen. Theappearance of these simple whole numbers is very significant. These integers imply that the compounds contain amultiple of a fixed unit of mass of oxygen. The simplest explanation for this fixed unit of mass is that oxygen is particulate . We call the fixed unit of mass an atom . We now assume that the compounds have been formed from combinations of atoms with fixed masses, and thatdifferent compounds have differing numbers of atoms. The mass ratios make it clear that oxide B contains twice as manyoxygen atoms (per nitrogen atom) as does oxide C and half as many oxygen atoms (per nitrogen atom) as does oxide A. Thesimple mass ratios must be the result of the simple ratios in which atoms combine into molecules. If, for example, oxide C has the molecular formula N O , then oxide B has the formula N O 2 , and oxide A has the formula N O 4 . There are other possibilities: if oxide B has molecular formula N O , then oxide A has formula N O 2 , and oxide C has formula N 2 O . Or if oxide A has formula N O , then oxide B has formula N 2 O and oxide C has formula N 4 O . These three possibilities are listed in the following table .

Possible molecular formulae for nitrogen oxides
Assuming that: Oxide C is N O Oxide B is N O Oxide A is N O
Oxide A is N O 4 N O 2 N O
Oxide B is N O 2 N O N 2 O
Oxide C is N O N 2 O N 4 O

We don't have a way (from these data) to know which of these sets of molecular formulae are right. But wecan assert that either one of them or one analogous to them is right.

Similar data are found for any set of compounds formed from common elements. For example,there are two oxides of carbon, one with oxygen to carbon mass ratio 1.33:1 and the other with mass ratio 2.66:1. The secondoxide must have twice as many oxygen atoms, per carbon atom, as does the first. The general statement of this observation is the Law of Multiple Proportions .

Law of multiple proportions

When two elements combine to form more than one compound, the mass of element Awhich combines in the first compound with a given amount of element B has a simple whole number ratio with the mass ofelement A which combines in the second compound with the same given mass of element B.

This sounds confusing, but an example clarifies this statement. Consider the carbon oxides, and letcarbon be element B and oxygen be element A. Take a fixed given mass of carbon (element B), say 1 gram. The mass of oxygen whichcombines with 1 gram of carbon to form the first oxide is 1.33 grams. The mass of oxygen which combines with 1 gram of carbonto form the second oxide is 2.66. These masses are in ratio 2.66 : 1.33 2 : 1 , a simple whole number ratio.

In explaining our observations of the Law of Multiple Proportions for the carbon oxides and thenitrogen oxides, we have concluded that the simple mass ratio arises from the simple ratio of atoms contained in theindividual molecules. Thus, we have established the following postulates of the Atomic Molecular Theory .

Atomic molecular theory

  • the elements are comprised of identical atoms
  • all atoms of a single element have the same characteristic mass
  • these number and masses of these atoms do not change during a chemicaltransformation
  • compounds consist of identical molecules formed of atoms combined in simple whole number ratios

Review and discussion questions

Assume that matter does not consist of atoms. Show by example how this assumption leads tohypothetical predictions which contradict the Law of Multiple Proportions. Do these hypothetical examplescontradict the Law of Definite Proportions? Areboth observations required for confirmation of the atomic theory?

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Two compounds, A and B, are formed entirely from hydrogen and carbon. CompoundA is 80.0% carbon by mass, and 20.0% hydrogen, whereas Compound B is 83.3% carbon by mass and 16.7%hydrogen. Demonstrate that these two compounds obey the Law of Multiple Proportions. Explain whythese results strongly indicate that the elements carbon and hydrogen are composed ofatoms.

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In many chemical reactions, mass does not appear to be a conservedquantity. For example, when a tin can rusts, the resultant rusty tin can has a greater mass than before rusting. Whena candle burns, the remaining candle has invariably less mass than before it was burned. Provide an explanation ofthese observations, and describe an experiment which would demonstrate that mass is actually conserved in thesechemical reactions.

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The following question was posed on an exam:

An unknown non-metal element (Q) forms two gaseous fluorides of unknown molecular formula. A 3.2 g sampleof Q reacts with fluorine to form 10.8 g of the unknown fluoride A. A 6.4 g sample of Q reacts with fluorine toform 29.2 g of unknown fluoride B. Using these data only, demonstrate by calculation and explanation thatthese unknown compounds obey the Law of Multiple Proportions.
A student responded with the following answer:
The Law of Multiple Proportions states that when twoelements form two or more compounds, the ratios of the masses of the elements between the two compounds are ina simple whole number ratio. So, looking at the data above, we see that the ratio of the mass of element Q incompound A to the mass of element Q in compound B is 3.2 : 6.4 1 : 2 , which is a simple whole number ratio. This demonstrates that these compounds obey the Law of Multiple Proportions.
Assess the accuracy of the students answer. In your assessment, you must determine what information is correct or incorrect, provide the correct information where needed, explain whether the reasoning is logical or not, and provide logical reasoning where needed.

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Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
Idrissa Reply
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
Kevin Reply
a perfect square v²+2v+_
Dearan Reply
kkk nice
Abdirahman Reply
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
Kim Reply
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
Embra Reply
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
Nancy Reply
rolling four fair dice and getting an even number an all four dice
ramon Reply
Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
Emedobi Reply
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
Mary Reply
is it 3×y ?
Joan Reply
J, combine like terms 7x-4y
Bridget Reply
im not good at math so would this help me
Rachael Reply
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
linda Reply
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
Crystal Reply
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Chris Reply
what's the easiest and fastest way to the synthesize AgNP?
Damian Reply
China
Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
Sravani Reply
what is system testing?
AMJAD
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
AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
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
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
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
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, General chemistry i. OpenStax CNX. Jul 18, 2007 Download for free at http://cnx.org/content/col10263/1.3
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