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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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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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